Notes

for an Introductory Course

On Electrical Machines
and Drives

E.G.Strangas

MSU Electrical Machines and Drives Laboratory

 Contents

Preface vii

1 Three Phase Circuits and Power 1

1.1 Electric Power with steady state sinusoidal quantities 1
1.2 Solving 1-phase problems 5
1.3 Three-phase Balanced Systems 6
1.4 Calculations in three-phase systems 9

2 Magnetics 15
2.1 Introduction 15
2.2 The Governing Equations 15
2.3 Saturation and Hysteresis 19
2.4 Permanent Magnets 21
2.5 Faraday’s Law 22
2.6 Eddy Currents and Eddy Current Losses 25
2.7 Torque and Force 27

3 Transformers 29
3.1 Description 29
3.2 The Ideal Transformer 30
3.3 Equivalent Circuit 32
3.4 Losses and Ratings 36
3.5 Per-unit System 37
v
vi CONTENTS

3.6 Transformer tests 40

3.6.1 Open Circuit Test 41
3.6.2 Short Circuit Test 41
3.7 Three-phase Transformers 43
3.8 Autotransformers 44

4 Concepts of Electrical Machines; DC motors 47

4.1 Geometry, Fields, Voltages, and Currents 47

5 Three-phase Windings 53
5.1 Current Space Vectors 53
5.2 Stator Windings and Resulting Flux Density 55
5.2.1 Balanced, Symmetric Three-phase Currents 58
5.3 Phasors and space vectors 59
5.4 Magnetizing current, Flux and Voltage 60
 Preface

The purpose of these notes is be used to introduce Electrical Engineering students to Electrical Ma-
chines, Power Electronics and Electrical Drives. They are primarily to serve our students at MSU,
who come to the course on Energy Conversion and Power Electronics with a solid background in
Electric Circuits and Electromagnetics, and want to acquire a basic working knowledge of the mate-
rial, but plan a career in a different area (venturing as far as computer or mechanical engineering), and
to students interested in continuing in the study of electrical machines and drives, power electronics
or power systems.
Staring from basic concepts, the student is led to the understanding of how force, torque, induced
voltages and currents are developed in an electrical machine. Then models of the machines are
developed, in terms of both simplified equations and of equivalent circuits, leading to the basic
understanding of modern machines and drives. Power Electronics are introduced, at the device and
systems level, and electrical drives are discussed.
Equations are kept to a minimum, and in the examples only the basic equations are used to solve
simple problems.
These notes do not aim to cover completely the subjects of Energy Conversion and Power
Electronics, nor to be used as a reference book, not even to be useful for an advanced course. They
are meant only to be an aid for the instructor who is working with intelligent and interested students,
who are taking their first (and perhaps their last) course on the subject. How successful this endeavor
has been will be tested in the class and in practice.
In the present form this text is to be used solely for the purposes of teaching the introductory
course on Energy Conversion and Power Electronics at MSU.

E.G.STRANGAS
E. Lansing, Michigan and Pyrgos, Tinos

vii
 A Note on Symbols

Throughout this text an attempt has been made to use symbols in a consistent way. Hence a script
letter, say v denotes a scalar time varying quantity, in this case a voltage. Hence one can see
v = 5 sin ωt or v = v̂ sin ωt
The same letter but capitalized denotes the rms value of the variable, assuming it is periodic.
Hence: √
v = 2V sinωt
The capital letter, but now bold, denotes a phasor:
V = V ejθ
Finally, the script letter, bold, denotes a space vector, i.e. a time dependent vector resulting from
three time dependent scalars:
v = v1 + v2 ejγ + v3 ej2γ
In addition to voltages, currents, and other obvious symbols we have:
B Magnetic flux Density (T)
H Magnetic filed intensity (A/m)
Φ Flux (Wb) (with the problem that a capital letter is used to show a time
dependent scalar)
λ, Λ, λ flux linkages (of a coil, rms, space vector)
ωs synchronous speed (in electrical degrees for machines with more than
two-poles)
ωo rotor speed (in electrical degrees for machines with more than two-poles)
ωm rotor speed (mechanical speed no matter how many poles)
ωr angular frequency of the rotor currents and voltages (in electrical de-
grees)
T Torque (Nm)
<(·), =(·) Real and Imaginary part of ·
viii
 1
Three Phase Circuits and Power

Chapter Objectives
In this chapter you will learn the following:
• The concepts of power, (real reactive and apparent) and power factor
• The operation of three-phase systems and the characteristics of balanced loads in Y and in ∆
• How to solve problems for three-phase systems

1.1 ELECTRIC POWER WITH STEADY STATE SINUSOIDAL QUANTITIES

We start from the basic equation for the instantaneous electric power supplied to a load as shown in
figure 1.1

i(t)

 
+
v(t)


 
Fig. 1.1 A simple load

p(t) = i(t) · v(t) (1.1)

1
2 THREE PHASE CIRCUITS AND POWER

where i(t) is the instantaneous value of current through the load and v(t) is the instantaneous value
of the voltage across it.
In quasi-steady state conditions, the current and voltage are both sinusoidal, with corresponding
amplitudes î and v̂, and initial phases, φi and φv , and the same frequency, ω = 2π/T − 2πf :

v(t) = v̂ sin(ωt + φv ) (1.2)

i(t) = î sin(ωt + φi ) (1.3)

In this case the rms values of the voltage and current are:
s
Z
1 T 2 v̂
V = v̂ [sin(ωt + φv )] dt = √ (1.4)
T 0 2
s
Z
1 T 2 î
I = î [sin(ωt + φi )] dt = √ (1.5)
T 0 2
6 φv 6 φi
and these two quantities can be described by phasors, V = V and I = I .
Instantaneous power becomes in this case:

p(t) = 2V I [sin(ωt + φv ) sin(ωt + φi )]

1
= 2V I [cos(φv − φi ) + cos(2ωt + φv + φi )] (1.6)
2
The first part in the right hand side of equation 1.6 is independent of time, while the second part
varies sinusoidally with twice the power frequency. The average power supplied to the load over
an integer time of periods is the first part, since the second one averages to zero. We define as real
power the first part:
P = V I cos(φv − φi ) (1.7)
If we spend a moment looking at this, we see that this power is not only proportional to the rms
voltage and current, but also to cos(φv − φi ). The cosine of this angle we define as displacement
factor, DF. At the same time, and in general terms (i.e. for periodic but not necessarily sinusoidal
currents) we define as power factor the ratio:
P
pf = (1.8)
VI
and that becomes in our case (i.e. sinusoidal current and voltage):

pf = cos(φv − φi ) (1.9)

Note that this is not generally the case for non-sinusoidal quantities. Figures 1.2 - 1.5 show the cases
of power at different angles between voltage and current.
We call the power factor leading or lagging, depending on whether the current of the load leads
or lags the voltage across it. It is clear then that for an inductive/resistive load the power factor is
lagging, while for a capacitive/resistive load the power factor is leading. Also for a purely inductive
or capacitive load the power factor is 0, while for a resistive load it is 1.
We define the product of the rms values of voltage and current at a load as apparent power, S:

S =VI (1.10)
 ELECTRIC POWER WITH STEADY STATE SINUSOIDAL QUANTITIES 3

0.5

i(t)
0

−0.5

−1
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

1
u(t)

−1

−2
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

1.5

1
p(t)

0.5

−0.5
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Fig. 1.2 Power at pf angle of 0o . The dashed line shows average power, in this case maximum

0.5
i(t)

−0.5

−1
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

1
u(t)

−1

−2
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

1.5

1
p(t)

0.5

−0.5
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Fig. 1.3 Power at pf angle of 30o . The dashed line shows average power

and as reactive power, Q

Q = V I sin(φv − φi ) (1.11)

Reactive power carries more significance than just a mathematical expression. It represents the
energy oscillating in and out of an inductor or a capacitor and a source for this energy must exist.
Since the energy oscillation in an inductor is 1800 out of phase of the energy oscillating in a capacitor,
4 THREE PHASE CIRCUITS AND POWER

0.5

i(t)
0

−0.5

−1
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

1
u(t)

−1

−2
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0.5
p(t)

−0.5

−1
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Fig. 1.4 Power at pf angle of 90o . The dashed line shows average power, in this case zero

0.5
i(t)

−0.5

−1
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

1
u(t)

−1

−2
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0.5

0
p(t)

−0.5

−1

−1.5
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Fig. 1.5 Power at pf angle of 180o . The dashed line shows average power, in this case negative, the opposite
of that in figure 1.2

the reactive power of the two have opposite signs by convention positive for an inductor, negative for
a capacitor.
The units for real power are, of course, W , for the apparent power V A and for the reactive power
V Ar.
 SOLVING 1-PHASE PROBLEMS 5

Using phasors for the current and voltage allows us to define complex power S as:

S = VI∗ (1.12)
6 φv 6 −φi
= V I (1.13)

and finally
S = P + jQ (1.14)

For example, when

p π
v(t) = (2 · 120 · sin(377t + )V (1.15)
6
p π
it = (2 · 5 · sin(377t + )A (1.16)
4

then S = V I = 120 · 5 = 600W , while pf = cos(π/6 − π/4) = 0.966 leading. Also:

6 π/6 6 −π/4
S = VI∗ = 120 5 = 579.6W − j155.3V Ar (1.17)

Figure 1.6 shows the phasors for lagging and leading power factors and the corresponding complex
power S.

P
S jQ
jQ
S

P I
V

V
I
Fig. 1.6 (a) lagging and (b) leading power factor

1.2 SOLVING 1-PHASE PROBLEMS

Based on the discussion earlier we can construct the table below:

Type of load Reactive power Power factor

Reactive Q>0 lagging
Capacitive Q<0 leading
Resistive Q=0 1
6 THREE PHASE CIRCUITS AND POWER

We also notice that if for a load we know any two of the four quantities, S, P , Q, pf , we can
calculate the other two, e.g. if S = 100kV A, pf = 0.8 leading, then:

P = S · pf = 80kW
q
Q = −S 1 − pf 2 = −60kV Ar , or
sin(φv − φi ) = sin [arccos 0.8]
Q = S sin(φv − φi )

Notice that here Q < 0, since the pf is leading, i.e. the load is capacitive.
Generally in a system with morePthan one loads P (or sources) real and P
reactive power balance, but
not apparent power, i.e. Ptotal = i Pi , Qtotal = i Qi , but Stotal 6= i Si .
In the same case, if the load voltage were VL = 2000V , the load current would be IL = S/V
= 100 · 103 /2 · 103 = 50A. If we use this voltage as reference, then:
6
V = 2000 0 V
6 6 o
I = 50 φi = 50 36.9 A
6 6 −36.9o
S = V I∗ = 2000 0 · 50 = P + jQ = 80 · 103 W − j60 · 103 V Ar

1.3 THREE-PHASE BALANCED SYSTEMS

Compared to single phase systems, three-phase systems offer definite advantages: for the same power
and voltage there is less copper in the windings, and the total power absorbed remains constant rather
than oscillate around its average value.
Let us take now three sinusoidal-current sources that have the same amplitude and frequency, but
their phase angles differ by 1200 . They are:
√
i1 (t) = 2I sin(ωt + φ)
√ 2π
i2 (t) = 2I sin(ωt + φ − ) (1.18)
3
√ 2π
i3 (t) = 2I sin(ωt + φ + )
3
If these three current sources are connected as shown in figure 1.7, the current returning though node
n is zero, since:
2π 2π
sin(ωt + φ) + sin(ωt − φ + ) + sin(ωt + φ + )≡0 (1.19)
3 3
Let us also take three voltage sources:
√
va (t) = 2V sin(ωt + φ)
√ 2π
vb (t) = 2V sin(ωt + φ − ) (1.20)
3
√ 2π
vc (t) = 2V sin(ωt + φ + )
3
connected as shown in figure 1.8. If the three impedances at the load are equal, then it is easy
to prove that the current in the branch n − n0 is zero as well. Here we have a first reason why
 THREE-PHASE BALANCED SYSTEMS 7

i1

i2 n

i3

Fig. 1.7 Zero neutral current in a Y -connected balanced system

+
v1

’
n’
n

+ v2 v3 +

Fig. 1.8 Zero neutral current in a voltage-fed, Y -connected, balanced system.

three-phase systems are convenient to use. The three sources together supply three times the power
that one source supplies, but they use three wires, while the one source alone uses two. The wires
of the three-phase system and the one-phase source carry the same current, hence with a three-phase
system the transmitted power can be tripled, while the amount of wires is only increased by 50%.
The loads of the system as shown in figure 1.9 are said to be in Y or star. If the loads are connected
as shown in figure 1.11, then they are said to be connected in Delta, ∆, or triangle. For somebody
who cannot see beyond the terminals of a Y or a ∆ load, but can only measure currents and voltages
there, it is impossible to discern the type of connection of the load. We can therefore consider the
two systems equivalent, and we can easily transform one to the other without any effect outside the
load. Then the impedances of a Y and its equivalent ∆ symmetric loads are related by:

1
ZY = Z∆ (1.21)
3

Let us take now a balanced system connected in Y , as shown in figure 1.9. The voltages
6 6 2π
between the neutral and each of the three phase terminals are V1n = V φ , V2n = V φ− 3 , and
6 2π
V3n = V φ+ 3 . Then the voltage between phases 1 and 2 can be shown either through trigonometry
or vector geometry to be:
8 THREE PHASE CIRCUITS AND POWER
I1
1
3 I1
1
13 I 12
+ I 31
+ +
1 V1n V I2 I 12
+ - 3n
V + - + 3
12 - 2
V1n V I2
-
V
2n - 3n I 23
V -
12 - 2
2 + I3
V I 23
- 2n
-V1n
Fig. 1.9 Y Connected Loads: Voltages and Currents

2 V31 + I3
-I
V2n V3n -V 23
1n
I2 I
-V3n V31 V12 31
- I12 I3
V I23
V 1n
23
V2n V3n
-V
2n I2
I12 I
-V3n V12 I1
31
- I12
V I
V 1n - I23
31
23
-V2n
I12
Fig. 1.10 Y Connected Loads: Voltage phasors I1
- I3

√ 6 φ+ π
V12 = V1 − V2 = 3V 3 (1.22)
This is shown in the phasor
√ diagrams of figure 1.10, and it says that the rms value of the line-to-line
voltage at a Y load, Vll , is 3 times that of the line-to-neutral or phase voltage, Vln . It is obvious
that the phase current is equal to the line current in the Y connection. The power supplied to the
system is three times the power supplied to each phase, since the voltage and current amplitudes and
the phase differences between them are the same in all three phases. If the power factor in one phase
is pf = cos(φv − φi ), then the total power to the system is:

S3φ = P3φ + jQ3φ

= 3V1 I∗1
√ √
= 3Vll Il cos(φv − φi ) + j 3Vll Il sin(φv − φi ) (1.23)

Similarly, for a connection in ∆, the phase voltage is equal to the line voltage. On the other hand,
6 6 2π 6 2π
if the phase currents phasors are I12 = I φ , I23 = I φ− 3 and I31 = I φ+ 3 , then the current of
 12 I 31
+
+ +
V1n V I2
- 3n
V - 3
12 - 2
V I 23
- 2n

2 + I3

-V1n CALCULATIONS IN THREE-PHASE SYSTEMS 9

V31
I1
1
-I
V2n V3n 23
I2 I
I 12 31
-V3n V12 I 31 - I I3
+ V 12 I23
V 1n
V 23 I2
1n - 3n
- -V2n
- 2 3 I12
V I 23 I1
2n
- I 31
+ I3

-V1n Fig. 1.11 ∆ Connected Loads: Voltages and Currents

31
line 1, as shown in figure 1.11 is:
-I √
V3n 23
I1 = I12 − I31 = 3I
6 φ− π
3 (1.24)
I2 I
V12 31
To calculate the power in the three-phase, Y connected load,
- I12 I3
V I23 = P3φ + jQ3φ
S3φ
1n
= 3V1 I∗1
-V2n √ √
= I 3Vll Il cos(φv − φi ) + j 3Vll Il sin(φv − φi ) (1.25)
12
I1
1.4 -I
CALCULATIONS IN THREE-PHASE SYSTEMS
31
It is often the case that calculations have to be made of quantities like currents, voltages, and power,
in a three-phase system. We can simplify these calculations if we follow the procedure below:
1. transform the ∆ circuits to Y ,
2. connect a neutral conductor,
3. solve one of the three 1-phase systems,
4. convert the results back to the ∆ systems.
1.4.1 Example
For the 3-phase system in figure 1.12 calculate the line-line voltage, real power and power factor at
the load.
First deal with only one phase as in the figure 1.13:
120 6 −40.6o
I = = 13.02 A
j1 + 7 + j5
6 o 6 −5o
Vln = I Zl = 13.02 −40.6 (7 + j5) = 111.97 V
SL,1φ = VL I∗ = 1.186 · 103 + j0.847 · 103
PL1φ = 1.186kW, QL1φ = 0.847kV Ar
pf = cos(−5 − (−40.6o )) = 0.814 lagging
o
 10 THREE PHASE CIRCUITS AND POWER

j1Ω

+
120V 7+5j Ω

’
n’
n

+ +

Fig. 1.12 A problem with Y connected load.

j1 Ω I L

+
7+j5 Ω
120V

Fig. 1.13 One phase of the same load

For the three-phase system the load voltage (line-to-line), and real and reactive power are:
√
12

VL,l−l = 3 · 111.97 = 193.94V

0V

PL,3φ = 3.56kW, QL,3φ = 2.541kV Ar

1.4.2 Example
For the system in figure 1.14, calculate the power factor and real power at the load, as well as the
phase voltage and current. The source voltage is 400V line-line.

j1Ω

18+j6 Ω
’
n’

+ +

Fig. 1.14 ∆-connected load

 CALCULATIONS IN THREE-PHASE SYSTEMS 11

√ the load to Y and work with one phase. The line to neutral voltage of the source
First we convert
is Vln = 400/ 3 = 231V .

j1 Ω

+
6+j2 Ω
231V

’
n’
n

+ +

Fig. 1.15 The same load converted to Y

j1 Ω I L

+
6+j2 Ω
231V

Fig. 1.16 One phase of the Y load

231 6 −26.6o
IL = = 34.44 A
j1 + 6 + j2
6 −8.1o
VL = IL (6 + j2) = 217.8 V

The power factor at the load is:

pf = cos(φv − φi ) = cos(−8.1o + 26.6o ) = 0.948lag

Converting back to ∆:
√ √
Iφ = IL / 3 = 34.44/ 3 = 19.88A
√
Vll = 217.8 · 3 · 377.22V

At the load √ √
P3φ = 3Vll IL pf = 3 · 377.22 · 34.44 · 0.948 = 21.34kW

1.4.3 Example
Two loads are connected as shown in figure 1.17. Load 1 draws from the system PL1 = 500kW at
0.8 pf lagging, while the total load is ST = 1000kV A at 0.95 pf lagging. What is the pf of load 2?
12 THREE PHASE CIRCUITS AND POWER

Power System Load 1

load 2

Fig. 1.17 Two loads fed from the same source

Note first that for the total load we can add real and reactive power for each of the two loads:

PT = PL1 + PL2
QT = QL1 + QL2
ST 6 = SL1 + SL2

From the information we have for the total load

PT = ST pfT = 950kW
QT = ST sin(cos−1 0.95) = 312.25kV Ar

Note positive QT since pf is lagging

For the load L1, PL1 = 500kW , pf1 = 0.8 lag,

500 · 103
SL1 = = 625kV A
q 0.8
QL1 = 2 − P 2 = 375kV Ar
SL1 L1

QL1 is again positive, since pf is lagging.

Hence,

PL2 = PT − PL1 = 450kW (1.26)

QL2 = QT − QL1 = −62.75kV Ar

and
PL2 450
pfL2 = =√ = 0.989 leading.
SL2 4202 + 62.752
 CALCULATIONS IN THREE-PHASE SYSTEMS 13

Notes

• A sinusoidal signal can be described uniquely by:

1. as e.g. v(t) = 5 sin(2πf t + φv )
2. by its graph,
3. as a phasor and the associated frequency.
one of these descriptions is enough to produce the other two.
• The phase of a sinusoidal quantity does not really matter. We need it to solve circuit problems,
after we take one quantity (a voltage or a current) as reference, i.e. we assign to it an arbitrary
value, often 0. There is no point in giving the phase of currents and voltages as answers, and,
especially for line-line voltages or currents in ∆ circuits, these numbers are often wrong and
anyway meaningless.
• In both 3-phase and 1-phase systems the sum of the real and the sum of the reactive powers of
individual loads are equal respectively to the real and reactive power of the total load. This is
not the case for apparent power and of course not for power factor.

• Of the four quantities, real power, reactive power, apparent power and power factor, two
describe a load adequately. The other two can be calculated from them.
 2
Magnetics

Chapter Objectives
In this chapter you will learn the following:
• How Maxwell’s equations can be simplified to solve simple practical magnetic problems
• The concepts of saturation and hysteresis of magnetic materials
• The characteristics of permanent magnets and how they can be used to solve simple problems
• How Faraday’s law can be used in simple windings and magnetic circuits
• Power loss mechanisms in magnetic materials
• How force and torque is developed in magnetic fields

2.1 INTRODUCTION

Since a good part of electromechanical energy conversion uses magnetic fields it is important early
on to learn (or review) how to solve for the magnetic field quantities in simple geometries and under
certain assumptions. One such assumption is that the frequency of all the variables is low enough
to neglect all displacement currents. Another is that the media (usually air, aluminum, copper, steel
etc.) are homogeneous and isotropic. We’ll list a few more assumptions as we move along.

2.2 THE GOVERNING EQUATIONS

We start with Maxwell’s equations, describing the characteristics of the magnetic field at low fre-
quencies. First we use:
∇·B=0 (2.1)
15
16 MAGNETICS

the integral form of which is:

Z
B · dA ≡ 0 (2.2)

for any path. This means that there is no source of flux, and that all of its lines are closed.
Secondly we use I Z
H · dl = J · dA (2.3)
A

where the closed loop is defining the boundary of the surface A. Finally, we use the relationship
between H, the strength of the magnetic field, and B, the induction or flux density.

B = µr µ0 H (2.4)

where µ0 is the permeability of free space, 4π10−7 T m/A, and µr is the relative permeability of the
material, 1 for air or vacuum, and a few hundred thousand for magnetic steel.
There is a variety of ways to solve a magnetic circuit problem. The equations given above, along
with the conditions on the boundary of our geometry define a boundary value problem. Analytical
methods are available for relatively simple geometries, and numerical methods, like Finite Elements
Analysis, for more complex geometries.
Here we’ll limit ourselves to very simple geometries. We’ll use the equations above, but we’ll add
boundary conditions and some more simplifications. These stem from the assumption of existence
of an average flux path defined within the geometry. Let’s tackle a problem to illustrate it all. In

r
g

airgap

Fig. 2.1 A simple magnetic circuit

figure 2.1 we see an iron ring with cross section Ac , average diameter r, that has a gap of length g
and a coil around it of N turns, carrying a current i. The additional assumptions we’ll make in order
to calculate the magnetic field density everywhere are:

• The magnetic flux remains within the iron and a tube of air, the airgap, defined by the cross
section of the iron and the length of the gap. This tube is shown in dashed lines in the figure.

• The flux flows parallel to a line, the average flux path, shown in dash-dot.
 THE GOVERNING EQUATIONS 17

• Flux density is uniform at any cross-section and perpendicular to it.

Following one flux line, the one coinciding with the average path, we write:
I Z
H · dl = J · dA (2.5)

where the second integral extends over any surface (a bubble) terminating on the path of integration.
But equation 2.2, together
R with the first assumption assures us that for any cross section of the
geometry the flux, Φ = Ac B · dA = Bavg Ac , is constant. Since both the cross section and the flux
are the same in the iron and the air gap, then

Biron = Bair
µiron Hiron = µair Hair
(2.6)

and finally

Hiron (2πr − g) + Hgap · g = N i

· ¸
µair
(2πr − g) + g Hgap = N i
µiron

Ac
Ay
H y1 H y2

Hr
g

Hl l
c

y y

Fig. 2.2 A slightly complex magnetic circuit

Let us address one more problem: calculate the magnetic field in the airgap of figure 2.2,
representing an iron core of depth d . Here we have to use two loops like the one above, and we have
18 MAGNETICS

a choice of possible three. Taking the one that includes the legs of the left and in the center, and the
outer one, we can write:

Hl · l + Hy1 · y + Hc · c + Hg · g + Hc · c + Hy1 · y = Ni (2.7)

Hl · h + 2Hy1 · y + Hr · l + 2Hy2 · y = Ni

Applying equation 2.3 to the closed surface shown shaded we also obtain:

Bl Ay − Bc Ac − Br Ay = 0

and of course

Bl = µ Hl , Bc = µ Hc , Br = µ Hr Bg = µ0 Hg

The student can complete the problem. We notice though something interesting: a similarity
between Kirchoff’s equations and the equations above. If we decide to use:

Φ = BA (2.8)
l
R = (2.9)
Aµ
F = Ni (2.10)

then we notice that we can replace the circuits above with the one in figure 2.3, with the following
correspondence:

R y1 R y2

R c1
Rl
+ R c2
Rr
-
R c3

Fig. 2.3 Equivalent electric circuit for the magnetic circuit in figure 2.2
.

Magnetic Electrical
F, magnetomotive force V , voltage, or electromotive force
Φ, flux I, current
R, reluctance R, resistance
 SATURATION AND HYSTERESIS 19

This is of course a great simplification for students who have spent a lot of effort on electrical
circuits, but there are some differences. One is the nonlinearity of the media in which the magnetic
field lives, particularly ferrous materials. This nonlinearity makes the solution of direct problems a
little more complex (problems of the type: for given flux find the necessary current) and the inverse
problems more complex and sometimes impossible to solve without iterations (problems of the type:
for given currents find the flux).

2.3 SATURATION AND HYSTERESIS

Although for free space a equation 2.3 is linear, in most ferrous materials this relationship is nonlinear.
Neglecting for the moment hysteresis, the relationship between H and B can be described by a curve
of the form shown in figure 2.4. From this curve, for a given value of B or H we can find the other
one and calculate the permeability µ = B/H.

Fig. 2.4 Saturation in ferrous materials

In addition to the phenomenon of saturation we have also to study the phenomenon of hysteresis
in ferrous materials. The defining difference is that if saturation existed alone, the flux would be a
unique function of the field intensity. When hysteresis is present, flux density for a give value of
field intensity, H depends also on the history of magnetic flux density, B in it. We can describe the
relationship between field intensity, H and flux density B in homogeneous, isotropic steel with the
curves of 2.5. These curves show that the flux density depends on the history of the magnetization of
the material. This dependence on history is called hysteresis. If we replace the curve with that of the
locus of the extrema, we obtain the saturation curve of the iron, which in itself can be quite useful.
Going back to one of the curves in 2.5, we see that when the current changes sinusoidally between
the two values, î and −î, then the point corresponding to (H, B) travels around the curve. During
this time, power is transferred to the iron, referred to as hysteresis losses, Physt . The energy of these
losses for one cycle is proportional to the area inside the curve. Hence the power of the losses is
proportional to this surface, the frequency, and the volume of iron; it increases with the maximum
value of B:
Physt = kf B̂ x 1 < x < 2 (2.11)
20 MAGNETICS

Fig. 2.5 Hysteresis loops and saturation

If the value of H, when increasing towards Ĥ, does so not monotonously, but at one point, H1 ,
decreases to H2 and then increases again to its maximum value, Ĥ, a minor hysteresis loop is created,
as shown in figure 2.6. The energy lost in one cycle includes these additional minor loop surfaces.

Fig. 2.6 Minor loops on a hysteresis curve

 PERMANENT MAGNETS 21

Br

Hc H

Fig. 2.7 Hysteresis curve in magnetic steel

2.4 PERMANENT MAGNETS

If we take a ring of iron with uniform cross section and a magnetic characteristic of the material
that in figure 2.7, and one winding around it, and look only at the second quadrant of the curve, we
notice that for H = 0, i.e. no current in an winding there will be some nonzero flux density, Br . In
addition, it will take current in the winding pushing flux in the opposite direction (negative current)
in order to make the flux zero. The iron in the ring has became a permanent magnet. The value
of the field intensity at this point is −Hc . In practice a permanent magnet is operating not at the
second quadrant of the hysteresis loop, but rather on a minor loop, as shown on figure 2.6 that can
be approximated with a straight line. Figure 2.8 shows the characteristics of a variety of permanent
magnets. The curve of a permanent magnet can be described by a straight line in the region of
interest, 2.9, corresponding to the equation:

Hm + Hc
Bm = Br (2.12)
Hc

2.4.1 Example
In the magnetic circuit of figure 2.10 the length of the magnet is lm = 1cm, the length of the air gap
is g = 1mm and the length of the iron is li = 20cm. For the magnet Br = 1.1T , Hc = 750kA/m.
What is the flux density in the air gap if the iron has infinite permeability and the cross section is
uniform?
22 MAGNETICS

1.4

1.2

1.0

0.8

B
(T)

0.6

0.4

0.2

-H (kA/m)

Fig. 2.8 Minor loops on a hysteresis curve

Since the cross section is uniform, B is the same everywhere, and there is no current:

Hi · 0.2 + Hg · g + Hm · li = 0

for infinite iron permeability Hi = 0, hence,

µ ¶
1 Hc
Bair g + (Bm − 1.1) li = 0
µo Br
⇒ B · 795.77 + (b − 1.1) · 6818 = 0
B = 0.985T

2.5 FARADAY’S LAW

We’ll see now how voltage is generated in a coil and the effects it may have on a magnetic material.
This theory, along with the previous chapter, is essential in calculating the transfer of energy through
a magnetic field.
First let’s start with the governing equation again. When flux through a coil changes for whatever
reason (e.g. change of the field or relative movement), a voltage is induced in this coil. Figure 2.11
 FARADAY’S LAW 23

Br

B
m

Hc

H H
m

Fig. 2.9 Finding the flux density in a permanent magnet

g
l
m

Fig. 2.10 Magnetic circuit for Example 2.4.1

shows such a typical case. Faraday’s law relates the electric and magnetic fields. In its integral form:
I Z
d
E · dl = − B · dA (2.13)
C dt A
 24 MAGNETICS

and in the cases we study it becomes:

dΦ(t)
v(t) = (2.14)
dt

V Φ

Fig. 2.11 Flux through a coil

If a coil has more than one turns in series, we define as flux linkages of the coil, λ, the sum of the
flux through each turn,
X
λ= Φi (2.15)
i

and then:
dλ(t)
v(t) = (2.16)
dt

2.5.1 Example
For the magnetic circuit shown below µiron = µo · 105 , the air gap is 1mm and the length of the iron
core at the average path is 1m. The cross section of the iron core is 0.04m2 . The winding labelled
‘primary’ has 500 turns. A sinusoidal voltage of 60Hz is applied to it. What should be the rms
value of it if the flux density in the iron (rms) is 0.8T ? What is the current in the coil? The voltage
induced in the coil will be

But if B(t) = B̂ sin(2πf t) ⇒ Φ(t) = AB̂ sin(2πf t)

√
⇒ Φ(t) = 0.04( 2 · 0.8) sin(377t)W b
dΦ
e1 (t) =
h dt
√ π i
⇒ e1 (t) = 500 0.04 2 · 0.8 · 377 sin(377t + ) V
2
eˆ1
⇒ E1 = √ = 500 · 0.04 · 0.8 · 377 = 6032V
2
 EDDY CURRENTS AND EDDY CURRENT LOSSES 25

primary secondary

Fig. 2.12 Magnetic circuit for Example 2.5.1

To calculate the current we integrate around the loop of the average path:

Hiron l + Hair g = N i
√
√ 2 · 0.8
biron = Bair = 2 · 0.8 sin(377t) ⇒ Hair = sin(377t)A/m
µo
√
2 · 0.8
⇒ Hiron = sin(377t)A/m
µo · 105

Finally
√ µ ¶
2 · 0.8 sin(377t) 1 1 · 10−3
500 · i = 5
+
µo 10 1
î
⇒ i = 1.819 sin(377t)A ⇒ I = √ = 1.286A
2

2.6 EDDY CURRENTS AND EDDY CURRENT LOSSES

When the flux through a solid ferrous material varies with time, currents are induced in it. Figure
2.13 gives a simple explanation: Let’s consider a ring of iron defined within the material shown in
black and a flux Φ through it, shown in grey. As the flux changes, a voltage e = dΦ/dt is induced
in the ring. Since the ring is shorted, and has a resistance R, a current flows in it, and Joule losses,
Peddy = e2 /R, result. We can consider a multitude of such rings in the material, resulting into Joule
losses, but the method discussed above is not the appropriate one to calculate these losses. We can,
though, estimate that for sinusoidal flux, the flux, voltage, and losses are:
26 MAGNETICS

 
   





Fig. 2.13 Eddy currents in solid iron

Φ = Φ̂ sin(ω t) = AB̂ sin(ω t) (2.17)

e = ω Φ̂ cos(ω t) = 2πAf B̂ cos(ω t) (2.18)
Peddy = k f 2 B̂ 2 (2.19)

which tells us that the losses are proportional to the square of both the flux density and frequency.
A typical way to decrease losses is to laminate the material, as shown in figure 2.14, decreasing the
paths of the currents and the total flux through them.

Iron insulation

Fig. 2.14 Laminated steel

 TORQUE AND FORCE 27

2.7 TORQUE AND FORCE

Calculating these is quite more complex, since Maxwell’s equations do not refer directly to them.
One can use the Maxwell stress tensor to find forces or torques on enclosed volumes, calculate forces
using the Lorenz force equation, here F = liB, or use directly the balance of energy. Here we’ll use
only this last method, e.g. balance the mechanical and electrical energies.
In a mechanical system with a force F acting on a body and moving it at velocity v in its direction,
the power Pmech is
Pmech = F · v (2.20)
This eq. 2.20, becomes for a rotating system with torque T , rotating a body with angular velocity
ωmech :
Pmech = T · wmech (2.21)
On the other hand, an electrical source e, supplying current i to a load provides electrical power
Pelec
Pelec = e · i (2.22)
Since power has to balance,

Pelec = Pmech ⇒ T · wmech = e · i (2.23)

 3
Transformers

Although transformers have no moving parts, they are essential to electromechanical energy conver-
sion. They make it possible to increase or decrease the voltage so that power can be transmitted at
a voltage level that will lead to decreased costs, and can be distributed and used at a safe voltage
level. In addition, they can provide matching of impedances, and regulate the flow of power (real or
reactive) in a network.
In this chapter we’ll start from basic concepts and build the equations and circuits corresponding
first to an ideal transformer and then to typical transformers in use. We’ll introduce and work with
the per unit system and will cover three-phase transformers as well.
After working on this chapter, you’ll be able to:

• Choose the correct rating and characteristics of a transformer for a specific application

• Calculate the losses, efficiency, and voltage regulation of a transformer under specific operating
conditions

• Experimentally determine the transformer parameters given its ratings

3.1 DESCRIPTION

When we see a transformer on a utility pole all we see is a cylinder with a few wires sticking out.
These wires enter the transformer through bushings. Inside the tank there is an iron core linking
coils, made most probably with copper and insulated. The system of insulation is associated with
that of cooling the core/coil assembly. In most cases the insulation is paper, and the whole assembly
is often immersed in insulating oil, used to both increase the dielectric characteristics of the paper
and to transfer heat from the assembly to the outer walls of the tank to the air. Figure 3.1 shows the
cutout of a typical distribution transformer
29
30 TRANSFORMERS

HV bushing

Surge suppressor

LV bushing

Insulating
oil

Core
Coil

Fig. 3.1 Cutaway view of a single phase distribution transformer. Notice only one HV bushing and lightning
arrester

3.2 THE IDEAL TRANSFORMER

Few ideal versions of human constructions exist, and the transformer offers no exception. An ideal
transformer is based on very simple concepts, and a large number of assumptions. This is the
transformer one learns about in high school.
Let us take an iron core with infinite permeability and two coils (with zero resistance), one with
N1 and the other with N2 turns wound around it, as shown in figure 3.2. All the flux is to remain in
the iron. We assign dots at one terminal of each coil in the following fashion: if the flux in the core

Φm
i1 i2

+ +
e1 e2

Fig. 3.2 Magnetic Circuit of an ideal transformer

changes, inducing a voltage in the coils, and the dotted terminal of one coil is positive with respect
its other terminal, so is the dotted terminal of the other coil. Or, the corollary to this, current into
dotted terminals produces flux in the same direction.
 THE IDEAL TRANSFORMER 31

Assume that somehow a time varying flux, Φ(t), is established in the iron. Then the flux linkages
in each coil will be λ1 = N1 Φ(t) and λ2 = N2 Φ(t). Voltages will be induced in these two coils:

dλ1 dΦ
e1 (t) = = N1 (3.1)
dt dt
dλ2 dΦ
e2 (t) = = N2 (3.2)
dt dt
and dividing:
e1 (t) N1
= (3.3)
e2 (t) N2
On the other hand, currents flowing in the coils are related to the field intensity H. If currents
flowing in the direction shown, i1 into the dotted terminal of coil 1, and i2 out of the dotted terminal
of coil 2, then:
N1 · i1 (t) − N2 i2 (t) = H · l (3.4)
but B = µiron H, and since B is finite and µiron is infinite, then H = 0. We recognize that this is
impossible, but so is the existence of an ideal transformer.
Finally:
i1 N2
= (3.5)
i2 N1
Equations 3.3 and 3.5 describe this ideal transformer, a two port network. The symbol of a
network that is defined by these two equations is in the figure 3.3. An ideal transformer has an

N1 N2

Fig. 3.3 Symbol for an ideal transformer

interesting characteristic. A two-port network that contains it and impedances can be replaced by an
equivalent other, as discussed below. Consider the circuit in figure 3.4a. Seen as a two port network

I1 Z I2 I1 Z I2

+ + + + + + + +
V1 E1 E2 V2 V E1 E2 V2
1

N1 N2 N1 N2

I1 I2 I1 I2
Z’ Z’
(a) a (b) b
+ + + + + + + +

V1 E1 E2 V2 V E1 E2 V2
1
Fig. 3.4 Transferring an impedance from one side to the other of an ideal transformer
N1 N2 N1 N2
32 TRANSFORMERS

with variables v1 , i1 , v2 , i2 , we can write:

e1 = u1 − i1 Z (3.6)
N2 N2 N2
e2 = e1 = u1 − i1 Z (3.7)
N1 N1 N1
µ ¶2
N2 N2 N2
v2 = e2 = e1 = u1 − i2 Z (3.8)
N1 N1 N1

which could describe the circuit in figure 3.4b. Generally a circuit on a side 1 can be transferred to
side 2 by multiplying its component impedances by (N2 /N1 )2 , the voltage sources by (N2 /N1 ) and
the current sources by (N1 /N2 ), while keeping the topology the same.

3.3 EQUIVALENT CIRCUIT

To develop the equivalent circuit for a transformer we’ll gradually relax the assumptions that we had
first imposed. First we’ll relax the assumption that the permeability of the iron is infinite. In that
case equation 3.4 does not revert to 3.5, but rather it becomes:

N1 i1 − N2 i2 = RΦm (3.9)

where R is the reluctance of the path around the core of the transformer and Φm the flux on this path.
To preserve the ideal transformer equations as part of our new transformer, we can split i1 to two
components: one i01 , will satisfy the ideal transformer equation, and the other, i1,ex will just balance
the right hand side. Figure 3.5 shows this.

’
i1 i1 i2

i +
+ 1, ex

? e2
e1

N1 N2
- -

ideal transformer

Fig. 3.5 First step to include magnetizing current

i1 = i01 + i1,ex (3.10)

N1 i1,ex = RΦm (3.11)
N1 · i01 (t) − N2 i2 (t) = H ·l (3.12)
 EQUIVALENT CIRCUIT 33

We can replace the current source, i1,ex , with something simpler if we remember that the rate of
change of flux Φm is related to the induced voltage e1 :

dΦm
e1 = N1 (3.13)
dt
d (N1 i1,ex /R)
= N1 (3.14)
µ 2 ¶ dt
N1 di1,ex
= (3.15)
R dt

Since the current i1,ex flows through something, where the voltage across it is proportional to its
derivative, we can consider that this something could be an inductance. This idea gives rise to the
N2
equivalent circuit in figure 3.6, where Lm = R1 Let us now relax the assumption that all the flux has
’
i1 i1 i2

i +
+ 1, ex

e2
e1

N1 N2
- -

ideal transformer

Fig. 3.6 Ideal transformer plus magnetizing branch

to remain in the iron as shown in figure 3.7. Let us call the flux in the iron Φm , magnetizing flux, the
flux that leaks out of the core and links only coil 1, Φl1 , leakage flux 1, and for coil 2, Φl2 , leakage
flux 2. Since Φl1 links only coil 1, then it should be related only to the current there, and the same
should be true for the second leakage flux.

Fig. 3.7 If the currents in the two windings were to have cancelling values of N · i then the only flux left
would be the leakage fluxes. This is the case shown here, designed to point out these fluxes.
 34 TRANSFORMERS

Φl1 = N1 i1 /Rl1 (3.16)

Φl2 = N2 i2 /Rl2 (3.17)

where Rl1 and Rl2 correspond to paths that are partially in the iron and partially in the air. As these
currents change, so do the leakage fluxes, and a voltage is induced in each coil:
µ ¶ µ 2¶
dλ1 dΦm dΦl1 N1 di1
e1 = = N1 + N1 = e1 + (3.18)
dt dt dt Rl1 dt
µ ¶ µ 2¶
dλ2 dΦm dΦl2 N2 di2
e2 = = N2 + N2 = e2 + (3.19)
dt dt dt Rl2 dt

. N1 2 . N12
If we define Ll1 = Rl1 , Ll2 = Rl2 , then we can arrive to the equivalent circuit in figure 3.8. To this

’
i1 i1 i2
+ +
+ i 1, ex

L l2 Ll2
v1 e2 v2
e
1

N1 N2
- - -
Ideal transformer

Fig. 3.8 Equivalent circuit of a transformer plus magnetizing and leakage inductances

circuit we have to add:

1. The winding (ohmic) resistance in each coil, R1,wdg , R2,wdg , with losses P1,wdg = i21 R1,wdg ,
P22,wdg = i22 R2,wdg , and

2. some resistance to represent iron losses. These losses (at least the eddy-current ones) are
proportional to the square of the flux. But the flux is proportional to the square of the induced
voltage e1 , hence Piron = ke21 . Since this resembles the losses of a resistance supplied by
voltage e1 , we can develop the equivalent circuit 3.9.

3.3.1 Example
Let us now use this equivalent circuit to solve a problem. Assume that the transformer has a
turns ratio of 4000/120, with R1,wdg = 1.6Ω, R2,wdg = 1.44mΩ, Ll1 = 21mH, Ll2 = 19µH,
Rc = 160kΩ, Lm = 450H. assume that the voltage at the low voltage side is 60Hz, V2 = 120V ,
and the power there is P2 = 20kW , at pf = 0.85 lagging. Calculate the voltage at the high voltage
side and the efficiency of the transformer.

Xm = Lm ∗ 2π60 = 169.7kΩ
X1 = 7.92Ω
X2 = 7.16mΩ
 EQUIVALENT CIRCUIT 35

i1 i1 ’

i2
+ i1, ex + + +
R wdg,1 R wdg, 2
v1 L l2 e1 e2 Ll2 v2
Rc
- N1 N2 - -

Ideal transformer

i1 i2

+ R L l1 + + R L l2 +
wdg,1 wdg,2
v1 Rc Lm e1 e2 v2

N1 : N2

(a)

Fig. 3.9 Equivalent circuit for a real transformer

From the power the load:

6 −31.80 6 −31.80
I2 = PL /(VL pf ) = 196.1336 A
E2 = V2 + I2 (Rwdg,2 + jXl2 ) = 120.98 + j1.045V
µ ¶
N1
E1 = E2 = 4032.7 + j34.83V
N2
µ ¶
N2
I01 = I2 = 5.001 − j3.1017A
N1
µ ¶
1 1
I1,ex = E1 + = 0.0254 − j0.0236A
Rc jXm
I1 = I1,ex + I01 = 5.0255 − j3.125A
6 0.90
V1 = E1 + I1 (Rwdg,1 + jXl,1 ) = 4065.5 + j69.2V = 4066 V

The power losses are concentrated in the windings and core:

Pwdg,2 = I22 Rwdg,2 = 196.132 · 1.44 · 10−3 = 55.39W

Pwdg,1 = I12 Rwdg,1 = 5.9182 · 1.6 = 65.37W
Pcore = E12 /Rc = 4032.82 /(160 · 103 ) = 101.64W
Ploss = Pwdg,1 + Pwdg,2 + Pcore = 213.08W
Pout Pout 20kW
η= = = = 0.9895
Pin (Pout + Ploss ) 20kW + 213.08W
 36 TRANSFORMERS

3.4 LOSSES AND RATINGS

Again for a given frequency, the power losses in the core (iron losses) increase with the voltage
e1 (or e2 ). These losses cannot be allowed to exceed a limit, beyond which the temperature of the
hottest spot in the transformer will rise above the point that will decrease dramatically the life of the
insulation. Limits therefore are put to E1 and E2 (with a ratio of N1 /N2 ), and these limits are the
voltage limits of the transformer.
Similarly, winding Joule losses have to be limited, resulting in limits to the currents I1 and I2 .
Typically a transformer is described by its rated voltages, E1N and E2N , that give both the limits
and turns ratio. The ratio of the rated currents, I1N /I2N , is the inverse of the ratio of the voltages
if we neglect the magnetizing current. Instead of the transformer rated currents, a transformer is
described by its rated apparent power:

SN = E1N I1N = E2N I2N (3.20)

Under rated conditions, i.e. maximum current and voltage, in typical transformers the magnetizing
current I1,ex , does not exceed 1% of the current in the transformer. Its effect therefore on the voltage
drop on the leakage inductance and winding resistance is negligible.
Under maximum (rated) current, total voltage drops on the winding resistances and leakage
inductances do not exceed in typical transformers 6% of the rated voltage. The effect therefore of
the winding current on the voltages E1 and E2 is small, and their effect on the magnetizing current
can be neglected.
These considerations allow us to modify the equivalent circuit in figure 3.9, to obtain the slightly
inaccurate but much more useful equivalent circuits in figures 3.10a, b, and c.

3.4.1 Example
Let us now use these new equivalent circuits to solve the previous problem 3.3.1. We’ll use the circuit
in 3.10b. Firs let’s calculate the combined impedances:
µ
¶2
N1
Rwdg = Rwdg,1 + Rwdg,2 = 3.2Ω
N2
µ ¶2
N1
Xl = Xl,1 + Xl,2 = 15.8759Ω
N2

then, we solve the circuit:

6 −31.80 6 −31.80
I2 = PL /(VL · pf ) = 196.1336 A
E2 = V2
µ ¶
N2
I01 = I2 · = 5 + j3.102A
N1
µ ¶
N1
E1 = E2 · = 4000V
N2
µ ¶
1 1
I1,ex = E1 + = 0.0258 − j0.0235A
Rc jXm
I1 = I1,ex + I01 = 5.0259 − j3.125A
6 10
V1 = E1 + I01 (Rwdg + jXl ) = 4065 + j69.45V = 4065 V
 PER-UNIT SYSTEM 37

i1 i2

+ R wdg,1 L l1 + + R L l2 +
wdg,2
v1 Rc Lm e2 v2
e1

N1 : N2
’ ’
i1 R wdg,1 L l1 R wdg,2 L l2 i2

+ + ’ ’’
+ + +
v1 Rc Lm e1 e2 v2

N1 : N2
i1 R ’wdg,1 L’l1 R wdg,2 L l2 i2

+ ’
+ + +
v1 R ’c Lm’ v2
e1 e2

N1 : N2

Fig. 3.10 Simplified equivalent circuits of a transformer

The power losses are concentrated in the windings and core:

Pwdg = I10 Rwdg = 110.79W

Pcore = V12 /Rc = 103.32W
Ploss = Pwdg + Pcore = 214.11W
Pout Pout 20kW
η= = = = 0.9894
Pin (Pout + Ploss ) 20kW + 221.411W

3.5 PER-UNIT SYSTEM

The idea behind the per unit system is quite simple. We define a base system of quantities, express
everything as a percentage (actually per unit) of these quantities, and use all the power and circuit
equations with these per unit quantities. In the process the ideal transformer in 3.10 disappears.
Working in p.u. has a some other advantages, e.g. the range of values of parameters is almost the
same for small and big transformers.
Working in the per unit system adds steps to the solution process, so one hopes that it simplifies
the solution more than it complicates it. At first attempt, the per unit system makes no sense. Let us
look at an example:
 38 TRANSFORMERS

3.5.1 Example
A load has impedance 10 + j5Ω and is fed by a voltage of 100V . Calculate the current and power
at the load.
Solution 1 the current will be
VL 100 6 −26.570
IL = = = 8.94 A
ZL 10 + j5
and the power will be
PL = VL IL · pf = 100 · 8.94 · cos(26.57) = 800W
Solution 2 Let’s use the per unit system.
1. define a consistent system of values for base. Let us choose Vb = 50V , Ib = 10A. This means
that Zb = Vb /Ib = 5Ω, and Pb = Vb · Ib = 500W , Qb = 500V Ar, Sb = 500V A.
2. Convert everything to pu. VL,pu = VL /Bb = 2pu, ZL,pu = (10 + j5)/5 = 2 + j1 pu.
3. solve in the pu system.
VL,pu 2 6 −26.570
IL,pu = = = 0.894 pu
ZL,pu 2 + j1
PL,pu = VLp u IL,pu · pf = 2 · 0.894 · cos(26.570 ) = 1.6 pu
4. Convert back to the SI system
IL = IL,pu · Ib = 0.894 · 10 = 8.94A
Pl = PL,pu · Pb = 1.6 · 500 = 800W

The second solution was a bit longer and appears to not be worth the effort. Let us now apply this
method to a transformer, but be shrewder in choosing our bases. Here we’ll need a base system for
each side of the ideal transformer, but in order for them to be consistent, the ratio of the voltage and
current bases should satisfy:
V1b N1
= (3.21)
V2b N2
I1b N2
= (3.22)
I2b N1
⇒ S1b = V1b I1b = V2b I2b = S2b (3.23)
i.e. the two base apparent powers are the same, as are the two base real and reactive powers.
We often choose as bases the rated quantities of the transformer on each side, This is convenient,
since the transformer most of the time operates at rated voltage (making the pu voltage unity), and
the currents and power are seldom above rated, above 1pu.
Notice that the base impedances on the two sides are related:
V1,b
Z1,b = (3.24)
I1,b
µ ¶2
V2,b N2 V1,b
Z2,b = = (3.25)
I2,b N1 I1,b
µ ¶2
N2
Z2,b = Z1,b (3.26)
N1
 PER-UNIT SYSTEM 39

We notice that as we move impedances from the one side of the transformer to the other, they get
³ ´2
multiplied or divided by the square of the turns ratio, N 2
N1 , but so does the base impedance, hence
the pu value of an impedance stays the same, regardless on which side it is.
Also we notice, that since the ratio of the voltages of the ideal transformer is E1 /E2 = N1 /N2 ,
is equal to the ratio of the current bases on the two sides on the ideal transformer, then

E1,pu = E2,pu

and similarly,
I1,pu = I2,pu
This observation leads to an ideal transformer where the voltages and currents on one side are
identical to the voltages and currents on the other side, i.e. the elimination of the ideal transformer,
and the equivalent circuits of fig. 3.11 a, b. Let us solve again the same problem as before, with

i1 i2 i1 i2

+ ’ ’
R wdg,1 L l1 R wdg,2 L l2 + + R wdg,1 L l1 R wdg,2 L l2 +
v Rc Lm v2 Rc Lm
1 v v2
1

(a) (b)

Fig. 3.11 Equivalent circuits of a transformer in pu

some added information:

3.5.2 Example
A transformer is rated 30kV A, 4000V /120V , with Rwdg,1 = 1.6Ω, Rwdg,2 = 1.44mΩ, Ll1 =
21mH, Ll2 = 19µH, Rc = 160kΩ, Lm = 450H. The voltage at the low voltage side is 60Hz,
V2 = 120V , and the power there is P2 = 20kW , at pf = 0.85 lagging. Calculate the voltage at the
high voltage side and the efficiency of the transformer.

1. First calculate the impedances of the equivalent circuit:

V1b = 4000V
S1b = 30kV A
30 · 103
I1b = = 7.5A
4 · 103
2
V1b
Z1b = = 533Ω
S1b
V2b = 120V
S2b = S1b = 30kV A
S1b
I2b = = 250A
V2b
V1b
Z2b = = 0.48Ω
I1b
40 TRANSFORMERS

2. Convert everything to per unit: First the parameters:

Rwdg,1,pu = Rwdg,1 /Z1b = 0.003 pu
Rwdg,2,pu = Rwdg,2 /Z2b = 0.003 pu
2π60Ll1
Xl1,pu = = 0.017 pu
Z1b
2π60Ll2
Xl2,pu = = 0.01149 pu
Z2b
Rc
Rc,pu = = 300 pu
Z1b
2π60Llm
Xm,pu = = 318pu
Z1b
Then the load:
V2
V2,pu = = 1pu
V2b
P2
P2,pu = = 0.6667pu
S2b

3. Solve in the pu system. We’ll drop the pu symbol from the parameters and variables:
µ ¶6 arccos(pf )
P2
I2 = = 0.666 − j0.413pu
V2 · pf
V1 = V2 + I [Rwdg,1 + Rwdg,2 + j(Xl1 + Xl2 )] = 1.0172 + j0.0188pu
V1 V1
Im = + = 0.0034 − j0.0031pu
Rc jXm
I1 = Im + I2 = 0.06701 − j0.416 pu
Pwdg = I22 (Rwdg,1 + Rewg,2 ) = 0.0037 pu
V12
Pcore = = 0.0034pu
Rc
P2
η = = 0.9894
Pwdg + Pcore + P2

4. Convert back to SI. The efficiency, η, is dimensionless, hence it stays the same. The voltage,
V2 is
6 0
V1 = V1,pu V1b = 4069 1 V

3.6 TRANSFORMER TESTS

We are usually given a transformer, with its frequency, power and voltage ratings, but without the
values of its impedances. It is often important to know these impedances, in order to calculate voltage
regulation, efficiency etc., in order to evaluate the transformer (e.g. if we have to choose from many)
or to design a system. Here we’ll work on finding the equivalent circuit of a transformer, through
two tests. We’ll use the results of these test in the per-unit system.
First we notice that if the relative values are as described in section 3.4, we cannot separate the
values of the primary and secondary resistances and reactances. We will lump R1,wdg and R2,wdg
 TRANSFORMER TESTS 41

together, as well as Xl1 and Xl2 . This will leave four quantities to be determined, Rwdg , Xl , Rc and
Xm .

3.6.1 Open Circuit Test

We leave one side of the transformer open circuited, while to the other we apply rated voltage (i.e.
Voc = 1pu) and measure current and power. On the open circuited side of the transformer rated
voltage appears, but we just have to be careful not to close the circuit ourselves. The current that
flows is primarily determined by the impedances Xm and Rc , and it is much lower than rated. It is
reasonable to apply this voltage to the low voltage side, since (with the ratings of the transformer in
our example) is it easier to apply 120V , rather than 4000V . We will use these two measurements to
calculate the values of Rc and Xm .
Dropping the subscript pu, using the equivalent circuit of figure 3.11b and neglecting the voltage
drop on the horizontal part of the circuit, we calculate:

Voc 2 1
Poc = = (3.27)
Rc Rc
Voc Voc
Ioc = +
Rc jXm
s
1 1
Ioc = 1 2 + (3.28)
Rc Xm 2

Equations 3.27 and 3.28, allow us to use the results of the short circuit test to calculate the vertical
(core) branch of the transformer equivalent circuit.

3.6.2 Short Circuit Test

To calculate the remaining part of the equivalent circuit, i.e the values of Rwdg and Xl , we short
circuit one side of the transformer and apply rated current to the other. We measure the voltage of
that side and the power drawn. On the other side, (the short-circuited one) the voltage is of course
zero, but the current is rated. We often apply voltage to the high voltage side, since a) the applied
voltage need not be high and b) the rated current on this side is low.
Using the equivalent circuit of figure 3.11a, we notice that:

2
Psc = Isc Rwdg = 1 · Rwdg (3.29)
Vsc = Isc (Rwdg + jXl )
q
Vsc = 2
1 · Rwdg + Xl2 (3.30)

Equations 3.29 and 3.30 can give us the values of the parameters in the horizontal part of the
equivalent circuit of a transformer.

3.6.1 Example
A 60Hz transformer is rated 30kV A, 4000V /120V . The open circuit test, performed with the high
voltage side open, gives Poc = 100W , Ioc = 1.1455A. The short circuit test, performed with the
low voltage side shorted, gives Psc = 180W , Vsc = 129.79V . Calculate the equivalent circuit of
the transformer in per unit.
 42 TRANSFORMERS

First define bases:

V1b = 4000V
S1b = 30kV A
30 · 103
I1b = = 7.5A
4 · 103
2
V1b
Z1b = = 533Ω
S1b
V2b = 120V
S2b = S1b = 30kV A
S1b
I2b = = 250A
V2b
V1b
Z2b = = 0.48Ω
I1b

Convert now everything to per unit:

180
Psc,pu = = 0.006ppu
30 · 103
129.79
Vsc,pu = = 0.0324pu
4000
100
Poc,pu = = 0.003333pu
30 · 103
1.1455
Ioc,pu = = 0.0046pu
250

Let’s calculate now, dropping the pu subscript:

2 2
Psc = Isc Rwdg ⇒ Rwdg = Psc Isc Psc = 1 · Psc = 0.06pu
q q
|Vsc | = |Isc | · |Rwdg + jXl | = 1 · Rwdg2 + Xl2 ⇒ Xl = Vsc 2 − R2
wdg = 0.017pu
2
Voc 12
Poc = ⇒ Rc = = 300pu
Rc Poc
¯ ¯ s
¯ Voc Voc ¯¯ 1 1 1
¯
|Ioc | = ¯ Rc + jXm ¯ = R2 + X 2 ⇒ Xm = q 2 1
= 318pu
c m Ioc − Rc2

A more typical problem is of the type:

3.6.2 Example
A 60Hz transformer is rated 30kV A, 4000V /120V . Its short circuit impedance is 0.0324pu and the
open circuit current is 0.0046pu. The rated iron losses are 100W and the rated winding losses are
180W . Calculate the efficiency and the necessary primary voltage when the load at the secondary
is at rated voltage, 20kW at 0.8pf .
 THREE-PHASE TRANSFORMERS 43

Working in pu:
Zsc = 0.0324pu
180
Psc = Rwdg = 3
= 6 · 10−3 pu
q 30 · 10
⇒ Xl = 2 − R2
Zsc wdg = 0.017pu
1 1 1
Poc = ⇒ Rc = = = 300pu
Rc Poc 100/30 · 103
s ,s
1 1 2 −
1
Ioc = 2
+ 2 ⇒ Xm = 1 Ioc = 318pu
Rc Xm Rc2
Having finished with the transformer data, let us work with the load and circuit. The load power
is 20kW , hence:
20 · 103
P2 = = 0.6667pu
30 · 103
but the power at the load is:
P2 = V2 I2 pf ⇒ 0.6667 = 1 · I2 · 0.8 ⇒ I2 = 0.8333pu
Then to solve the circuit, we work with phasors. We use the voltage V2 as reference:
V2 = V2 = 1pu
6 cos−1 0.8
I2 = 0.8333 = 0.6667 − j0.5pu
V1 = V2 + I2 (Rwdg + jXl ) = 1.0125 + j0.0083pu ⇒ V1 = 1.0125pu
Pwdg = I22 · Rwdg = 0.0042pu
Pc = V12 /Rc = 0.034pu
P2
η = = 0.989
P2 + Pwdg + Pc
Finally, we convert the voltage to SI
V1 = V1,pu · Vb1 = 1.0125 · 4000 = 4050V

3.7 THREE-PHASE TRANSFORMERS

We’ll study now three-phase transformers, considering as consisting of three identical one-phase
transformers. This method is accurate as far as equivalent circuits and two-port models are our
interest, but it does not give us insight into the magnetic circuit of the three-phase transformer. The
primaries and the secondaries of the one-phase transformers can be connected either in ∆ or in Y .
In either case, the rated power of the three-phase transformer is three times that of the one-phase
transformers. For ∆ connection,
Vll = V1φ (3.31)
√
Il = 3I1φ (3.32)
For Y connection
√
Vll = 3V1φ (3.33)
Il = I1φ (3.34)
 44 TRANSFORMERS

i1 i2 i1 i2

V1 V1 V2
V2

i1 i2 i1 i2

V1 V2 V1 V2

i1 i2 i1 i2

V1 V2 V1 V2

(a) (b)

Fig. 3.12 Y − Y and Y − ∆ Connections of three-phase Transformers

i1 i2 i1 i2

V1 V1 V2
V2

i1 i2 i1 i2

V1 V1 V2
V2

i1 i2 i1 i2

V1 V2 V1 V2

(a) (b)

Fig. 3.13 ∆ − Y and ∆ − ∆ Connections of three-phase Transformers

3.8 AUTOTRANSFORMERS

An autotransformer is a transformer where the two windings (of turns N1 and N2 ) are not isolated
from each other, but rather connected as shown in figure 3.14. It is clear form this figure that the
 AUTOTRANSFORMERS 45

voltage ratio in an autotransformer is

V1 N1 + N2
= (3.35)
V2 N2
while the current ratio is
I2 N1 + N2
= (3.36)
I1 N2
The interesting part is that the coil of turns of N1 carries current I1 , while the coil of turns N2 carries
the (vectorial) sum of the two currents, I1 − I2 . So if the voltage ratio where 1, no current would flow
through that coil. This characteristic leads to a significant reduction in size of an autotransformer
compared to a similarly rated transformer, especially if the primary and secondary voltages are of
the same order of magnitude. These savings come at a serious disadvantage, the loss of isolation
between the two sides.

I

V
N

I -I

Fig. 3.14 An Autotransformer

 4
Concepts of Electrical
Machines; DC motors

We’ll study DC machines here, at a conceptual level, for two reasons:

1. DC machines although complex in construction, can be useful in establishing the concepts of

emf and torque development, and are described by simple equations.

2. The magnetic fields in them, along with the voltage and torque equations can be used easily to
develop the ideas of field orientation.

4.1 GEOMETRY, FIELDS, VOLTAGES, AND CURRENTS

Let us start with the geometry shown in figure 4.1

+ 1

2
+

3
+

4
Fig. 4.1 Geometry of an elementary DC motor

47

+
5
48 CONCEPTS OF ELECTRICAL MACHINES; DC MOTORS

This geometry describes an outer iron window (stator), through which (i.e. its center part) a
uniform magnetic flux is established, say Φ̂. How this is done (a current in a coil, or a permanent
magnet) is not important here.
In the center part of the window there is an iron cylinder (called rotor), free to rotate around its
axis. A coil of one turn is wound diametrically around the cylinder, parallel to its axis. As the
cylinder and its coil rotate, the flux through the coil changes. Figure 4.2 shows consecutive locations
of the rotor and we can see that the flux through the coil changes both in value and direction. The

+
+

+ +

1 2 3 4 5

Fig. 4.2 Flux through a coil of a rotating DC machine

1.5
1
2
1

0.5
3
coil

0
λ

4
−0.5
5

−1

−1.5

0 100 200 300 400 500 600 700

1.5

0.5
1 5
coil

0
v

2
4
−0.5 3

−1

−1.5

0 100 200 300 400 500 600 700

Fig. 4.3 Flux and voltage in a coil of the DC machine in 4.2. Points 1 – 5 represent the coil positions.

top graph of figure 4.3 shows how the flux linkages of the coil through the coil would change, if the
 GEOMETRY, FIELDS, VOLTAGES, AND CURRENTS 49

rotor were to rotate at a constant angular velocity, ω.

λ = Φ̂ cos [ωt] (4.1)

Since the flux linking the coil changes with time, then a voltage will be induced in this coil, vcoil ,

dλ
vcoil = = −Φ̂ω sin (ωt) (4.2)
dt

shown in the second graph of figure 4.3. The points marked there correspond to the position of the
rotor in figure4.2.
This alternating voltage has to somehow be rectified, since this is a DC machine. Although this
can be done electronically, a very old mechanical method exists. The coil is connected not to the DC
source or load, but to two ring segments, solidly attached to it and the rotor, and hence rotating with
it. Two ‘brushes’, i.e. conducting pieces of material (often carbon/copper) are stationary and sliding
on these ring segments as shown in figure 4.4

Copper Half Ring

Stationary rotating with the
Brush coil

Coil

Fig. 4.4 A coil of a DC motor and a commutator with brushes

The structure of the ring segments is called a commutator. As it rotates, the brushes make contact
with the opposite segments just as the induced voltage goes through zero and switches sign.
Figure 4.5 shows the induced voltage and the terminal voltage seen at the voltmeter of figure 4.4.
If a number of coils are placed on the rotor, as shown in figure 4.6, each connected to a commutator
segment, the total induced voltage to the coils, E will be:

E = k Φ̂ω (4.3)

where k is proportional to the number of coils.

50 CONCEPTS OF ELECTRICAL MACHINES; DC MOTORS

1.5

0.5

vcoil
0

−0.5

−1

−1.5

0 100 200 300 400 500 600 700

1.5

0.5
vterm

−0.5

−1

−1.5

0 100 200 300 400 500 600 700

Fig. 4.5 Induced voltage in a coil and terminal voltage in an elementary DC machine
+
+

+
+

+ +
+ +

- -

- -
-
-

-
-

Fig. 4.6 Coils on the rotor of DC machine

Going back to equation 2.23,

E · i = Tω (4.4)
k Φ̂ω = T ω (4.5)
T = k Φ̂ (4.6)

If the electrical machine is connected to a load or a source as in figure4.7, the induced voltage and
terminal voltage will be related by:
 GEOMETRY, FIELDS, VOLTAGES, AND CURRENTS 51

i
m

i
g

Load
R
wdg
or
E

Source

Fig. 4.7 Circuit with a DC machine

Vterm = E − ig Rwdg f or a generator (4.7)

Vterm = E + im Rwdg f or a motor (4.8)

4.1.1 Example
A DC motor, when connected to a 100V source and to no load runs at 1200rpm. Its stator resistance
is 2Ω. What should be the torque and current if it is fed from a 220V supply and its speed is
1500rpm? Assume that the field is constant.
The first piece of information gives us the constant k. Since at no load the torque is zero and
T = ki, then the current is zero as well. This means that for this operation:

V = E = kω

but ω is 1200 rpm, or in SI units:

2π
ω = 1200 = 125.66rad/s
60
And
100V = k · 125.66 ⇒ k = 0.796V s
At the operating point of interest:
2π
ωo = 1500rpm = 1500 = 157.08rad/s ⇒ E = kω = 125V
60
For a motor:

V = E + IR
⇒ 220 = 125 + I · 2Ω
⇒I = 47.5A
⇒T = kI = 37.81N m
 5
Three-phase Windings

Understanding the geometry and operation of windings in AC machines is essential in understanding

how these machines operate. We introduce here the concept of Space Vectors, (or Space Phasors) in
its general form, and we’ll see how they are applied to three-phase windings.

5.1 CURRENT SPACE VECTORS

Let us assume that in a uniformly permeable space we have placed three identical windings as shown
in figure 5.1. Each carries a time dependent current, i1 , i2 and i3 . We require that:

i1 (t) + i2 (t) + i3 (t) ≡ 0 (5.1)

Each current produces a flux in the direction of the coil axis, and if we assume the magnetic
medium to be linear, we can find the total flux by adding the individual fluxes. This means that we
could produce the same flux by having only one coil, identical to the three, but placed in the direction
of the total flux, carrying an appropriate current. Figure 5.2 shows such a set of coils carrying for
i1 = 5A, i2 = −8A and i3 − = 3A and the resultant coil.
To calculate the direction of the resultant one coil and the current it should carry, all we have to
do is create three vectors, each in the direction of one coil, and of amplitude equal to the current of
each coil. The sum of these vectors will give the direction of the total flux and hence of the one coil
that will replace the three. The amplitude of the vectors will be that of the current of each coil.
Let us assume that the coils are placed at angles 00 , 1200 and 2400 . Then their vectorial sum will
be:
6 φ 0 0
i=i = i1 + i2 ej120 + i3 ej240 (5.2)

We call i, defined thus, a space vector, and we notice that if the currents i1 , i2 and i3 are functions
of time, so will be the amplitude and the angle of i. By projecting the three constituting currents on
the horizontal and vertical axis, we can find the real (id = <[i]) and imaginary (iq = =[i]) parts of
53
 54 THREE-PHASE WINDINGS

Φ 2

Ι I1
2

Φ 1

Ι3

Φ
3

Fig. 5.1 Three phase concentrated windings

5A
3A

-8A 5A
__
I1

3A
__
-8A I2
__
I __
I3

(a) (b)

Fig. 5.2 Resultant space vector and corresponding winding

it. Also, from the definition of the current space vector we can reconstruct the constituent currents:
2
i1 (t) =
<[i(t)]
3
2
i2 (t) = <[i(t)e−jγ ] (5.3)
3
2
i3 (t) = <[i(t)e−j2γ ]
3
2π
γ = 1200 = rad (5.4)
3
 STATOR WINDINGS AND RESULTING FLUX DENSITY 55

5.2 STATOR WINDINGS AND RESULTING FLUX DENSITY

Fig. 5.3 A Stator Lamination

Stator Winding

Stator

Airgap

Rotor
Fig. 5.4 A sinusoidal winding on the stator
56 THREE-PHASE WINDINGS

Assume now that these windings are placed in a fixed structure, the stator, which is surrounds a
rotor. Figure 5.3 shows a typical stator cross-section, but for the present we’ll consider the stator
as a steel tube. Figure 5.5 shows the windings in such a case. Instead of being concentrated, they
are sinusoidally distributed as shown in figure 5.4. Sinusoidal distribution means that the number of
turns dNs covering an angle dθ at a position θ and divided by dθ is a sinusoidal function of the angle
θ. This turns density, ns1 (θ), is then:
dns
= ns1 (θ) = n̂s sin θ
dθ
and for a total number of turns Ns in the winding:
Z π
Ns
Ns = ns1 (θ)dθ ⇒ ns1 (θ) = sin θ
0 2
We now assign to the winding we are discussing a current i1 . To find the flux density in the airgap

Current in

positive direction

Airgap
θ

Stator

Rotor
Current in
negative direction Stator winding

Fig. 5.5 Integration path to calculate flux density in the airgap

between rotor and stator we choose an integration path as shown in figure 5.5. This path is defined by
the angle θ and we can notice that because of symmetry the flux density at the two airgap segments
in the path is the same. If we assume the permeability of iron to be infinite, then Hiron = 0 and:
Z θ+π
2Hg1 (θ)g = i1 ns1 (φ)dφ
θ
2Bg1 (θ)
g = i1 Ns cos θ
µ0
Ns µ0
Bg1 (θ) = i1 cos θ (5.5)
2g
This means that for a given current i1 in the coil, the flux density in the air gap varies sinusoidally with
angle, but as shown in figure 5.6 it reaches a maximum at angle θ = 0. For the same machine and
 STATOR WINDINGS AND RESULTING FLUX DENSITY 57

Fig. 5.6 Sketch of the flux in the airgap

1
ns(θ)

−1
0 90 180 270 360

1
Bg(θ)

−1
0 90 180 270 360
θ

Fig. 5.7 Turns density on the stator and air gap flux density vs. θ

conditions as in 5.6, 5.7 shows the plots of turns density, ns (θ) and flux density, Bg (θ) in cartesian
coordinates with θ in the horizontal axis.
If the current i1 were to vary sinusoidally in time, the flux density would also change in time,
maintaining its space profile but changing only in amplitude. This would be considered a wave, as it
changes in time and space. The nodes of the wave, where the flux density is zero, will remain at 900
and 2700 , while the extrema of the flux will remain at 00 and 1800 .
58 THREE-PHASE WINDINGS

Consider now an additional winding, identical to the first, but rotated with respect to it by 1200 .
For a current in this winding we’ll get a similar airgap flux density as before, but with nodes at
900 + 1200 and at 2700 + 1200 . If a current i2 is flowing in this winding, then the airgap flux density
due to it will follow a form similar to equation 5.5 but rotated 1200 = 2π3 .

Ns µ0 2π
Bg2 (θ) = i2 cos(θ − ) (5.6)
2g 3

Similarly, a third winding, rotated yet another 1200 and carrying current i3 , will produce airgap
flux density:
Ns µ0 4π
Bg3 (θ) = i3 cos(θ − ) (5.7)
2g 3
Superimposing these three flux densities, we get yet another sinusoidally distributed airgap flux
density, that could equivalently come from a winding placed at an angle φ and carrying current i:

Ns µ0
Bg (θ) = Bg1 (θ) + Bg2 (θ) + Bg3 (θ) = i cos(θ + φ) (5.8)
2g
This means that as the currents change, the flux could be due instead to only one sinusoidally
distributed winding with the same number of turns. The location, φ(t), and current, i(t), of this
winding can be determined from the current space vector:

6 φ(t) 0 0
i(t) = i(t) = i1 (t) + i2 (t)ej120 + i3 (t)ej240

5.2.1 Balanced, Symmetric Three-phase Currents

If the currents i1 , i2 , i3 form a balanced three-phase system of frequency fs = ωs /2π, then we can
write:
√
√ 2 £ jωs t ¤
i1 = 2I cos(ωs t + φ1 ) = Is e + Is e−jωs t
2 √ h
√ 2π 2 i
i2 = 2I cos(ωs t − φ1 + )= Is ej(ωs t−2π/3) + Is e−j(ωs t−2π/3) (5.9)
3 √2 h
√ 4π 2 i
i3 = 2I cos(ωs t − φ1 + )= Is ej(ωs t−4π/3) + Is e−j(ωs t−4π/3)
3 2

where I is the phasor corresponding to the current in phase 1. The resultant space vector is
√ √
3 2 jωs t 3 2 j(ωs t+φ1 ) π
is (t) = Ie = Ie I = Iej(φ1 + 2 ) (5.10)
2 2 2 2
The resulting flux density wave is then:

3 √ Ns µ0
B(θ, t) = 2I cos(ωs t + φ1 − θ) (5.11)
2 2g
√
which shows a travelling wave, with a maximum value B̂ = 3
2 2I N s
µ0 . This wave travels around the
stator at a constant speed ωs , as shown in figure 5.8
 PHASORS AND SPACE VECTORS 59

t1 t2 t3
Bgθ)

−1
0 90 180 270 360
θ

Fig. 5.8 Airgap flux density profile, t3 > t2 > t1

5.3 PHASORS AND SPACE VECTORS

current phasor, I = Iejφ0 , describes

It is easy at this point to confuse space vectors and phasors. A √
one sinusoidally varying current, of frequency ω, amplitude 2I and initial phase φ0 . We can
reconstruct the sinusoid from the phasor:
√
2 £ jωt ¤ √ ¡ ¢
i(t) = Ie + I∗ e−jωt = 2I cos(ωt + φ0 ) = < Iejωt (5.12)
2

Although rotation is implicit in the definition of the phasor, no rotation is described by it.
On the other hand, the definition of a current space vector requires three currents that sum to
zero. These currents are implicitly in windings symmetrically placed, but the currents themselves
are not necessarily sinusoidal. Generally the amplitude and angle of the space vector changes with
time, but no specific pattern is a priori defined. We can reconstruct the three currents that constitute
the space vector from equation 5.3.
When these constituent currents form a balanced, symmetric system, of frequency ωs , then the
resultant space vector is of constant amplitude, rotating at constant speed. In that case, the relationship
between the phasor of one current and the space vector is shown in equation 5.10.

5.3.1 Example √
Let us take three balanced sinusoidal currents with amplitude 1, i.e. rms value of 1/ 2A. Choose
an initial phase angle such that:

i1 (t) = 1 cos(ωt)A
i2 (t) = 1 cos(ωt − 2π/3)A
i2 (t) = 1 cos(ωt − 4π/3)A
60 THREE-PHASE WINDINGS

When ωt = 0, as shown in figure 5.9a,

i1 = 1A
i2 = −0.5A
i3 = −0.5A
0 0 6 0
i = i1 + i2 ej120 + i3 ej240 = 1.5 A
0
and later, when ωt = 20 = π/9 rad, as shown in figure 5.9b,
i1 = 0.939A
i2 = −0.766A
i3 = −0.174A
0 0 6 200
i = i1 + i2 ej120 + i3 ej240 = 1.5 A

_
i

i = -0.5A
3 i =-0.1737A
3

_ i = 0.9397A
i = 1A i 1
1

i = -0.5A
2 i = -0.766A
2

(a) (b)

Fig. 5.9 Space vector movement for sinusoidal, symmetric three-phase currents

5.4 MAGNETIZING CURRENT, FLUX AND VOLTAGE

Let us now see what results this rotating flux has on the windings, using Faraday’s law. From this
point on we’ll use sinusoidal symmetric three-phase quantities.
We look again at our three real stationary windings linked by a rotating flux. For example, when
the current is maximum in phase 1, the flux is as shown in figure 5.10a, linking all of the turns in
phase 1. Later, the flux has rotated as shown in figure 5.10b, resulting in lower flux linkages with
the phase 1 windings. When the flux has rotated 900 , as in 5.10c the flux linkages with the phase 1
winding are zero.
To calculate the flux linkages λ we have to take a turn of the winding, placed at angle θ, as shown
in figure 5.11. The flux through this coil is:
Z θ Z θ+
Φ(t, θ) = Bg (t, φ)dA = lr Bg (t, φ)dφ (5.13)
θ−π θ−π
 MAGNETIZING CURRENT, FLUX AND VOLTAGE 61

(a) (b) (c)

Fig. 5.10 Rotating flux and flux linkages with phase 1

Fig. 5.11 Flux linkages of one turn

But the number of turns linked by this flux is dns (θ) = ns (θ)dθ, so the flux linkages for these turns
are:
dλ = ns (θ)dθ · Φ(θ)
To find the flux linkages λ1 for all of the coil, we have to integrate the flux linkages over all turns of
coil 1: Z π
λ1 = λ(θ)dθ
0

giving us at the end:

Ns2 lr √ √
λ1 (t) = 3πµ0 I cos(ωt + φ1 ) = LM 2I cos(ωt + φ1 ) (5.14)
8g

which means that the flux linkages in coil 1 are in phase with the current in this coil and proportional
to it. The flux linkages of the other two coils, 2 and 3, are identical to that of coil 1, and lagging in
62 THREE-PHASE WINDINGS

time by 1200 and 2400 . With these three quantities we can create a flux-linkage space vector, λ .
0 0
λ ≡ λ1 + λ2 ej120 + λ3 ej240 = LM i (5.15)

Since the flux linkages of each coil vary, and in our case sinusoidally, a voltage is induced in each
of these coils. The induced voltage in each coil is 900 ahead of the current in it, bringing to mind
the relationship of current and voltage of an inductor. Notice though, that it is not just the current in
the winding that causes the flux linkages and the induced voltages, but rather the current in all three
windings. Still, we call the constant LM magnetizing inductance.

dλ1 √ π
e1 (t) = = ω 2I cos(ωt + φ1 + )
dt 2
dλ2 √ π 2π
e2 (t) = = ω 2I cos(ωt + φ1 + − ) (5.16)
dt 2 3
dλ3 √ π 4π
e3 (t) = = ω 2I cos(ωt + φ1 + − )
dt 2 2
and of course we can define voltage space vectors e:
0 0
e = e1 + e2 ej120 + e3 ej240 = jωLM i (5.17)

Note that the flux linkage space vector λ is aligned with the current space vector, while the voltage
space vector e is ahead of both by 900 . This agrees with the fact that the individual phase voltages
lead the currents by 900 , as shown in figure 5.12.

e
λ
i

ωt

Fig. 5.12 Magnetizing current, flux-linkage and induced voltage space vectors