Excerpt
Power Converter
Machines & Drives
Amit Sachan
(Assistant Professor)
Electrical Engineering
Acknowledgment
"I Am Indebted To My Parent & Brother in
Law for the Efflorescence of My Knowledge"
First, I give my sincere thanks to the Almighty God for his blessing and showing
me the right direction. This work would not have been possible without the support
and generous help of My Parent, Brother in Law, sister for providing me
supportive and excellent research facilities in and around the region, exposure to
the scientific world and a platform to rise.
Foremost, I would like to express my sincere thanks to my wife for his advice
during work. His consistent support and intellectual guidance encouraged me to
remain focused on achieving my goal. His observations and comments helped me
to establish the overall direction of the research and to move forward with
investigation in depth.
Amit Sachan
Objectives
Power has two distinct effects in our lives: On one hand, we have to produce it, transmit
it and receive it in a way that is affordable, reliable, and with minimal negative effect on
the environment. On the other hand, we have to control its use, so that we can make the
best out of it, i.e. use it safely and efficiently, and harness it to achieve qualities like
speed, accuracy and efficiency. Electrical Engineers should understand the path that
energy takes from coal, water, or other fuel to robot, heater or computer, and be able
based on this understanding to specify or design equipment and systems. In their work
related to this material, they will have to use concepts and methods that they learn in
courses as diverse as computers, mathematics, electronics and control theory. This course
serves only as an introduction to both electrical machines and power electronics. It
focuses on the most common devices and systems that an electrical engineer will
encounter: AC machines, transformers, rectifiers and inverters, as well as Electrical
Drives and Uninterruptible Power Supplies:
Commutation of DC machines,
Dynamic models of Electrical Machines,
Design of Electrical Machines,
Analytical discussion of PWM inverters and matrix converters
Control of Electrical Drives
Some of this material will be cover in the lectures and the laboratory, and in graduate
courses. Students are expected to know exactly what is discuss in class and handed out
homework, notes, assignments,
Topic
1  Introduction  Threephase systems
2  Magnetic circuits and
Materials
3 Transformers
4
Concepts of Electrical Machines: DC Motors
5 Threephase Windings
6 Induction Machines
7 Synchronous Machines and Drives
8 Introduction of Power Electronics
Chapter 1: Introduction and Three Phase Power
1.1
Review of Basic Circuit Analysis
Definitions:
Node  Electrical junction between two or more devices.
Loop  Closed path formed by tracing through an ordered sequence of nodes without passing through
any node more than once.
Element Constraints:
Ohm's Law
iR
v
Capacitor Equation
dt
dv
C
i
Inductor Equation
dt
di
L
v
Connection Constraints:
Kirchhoff's Current Law  The algebraic sum of currents entering a node is zero at every instant in
time.
0
¦
Node
k
i
(1.1)
Kirchhoff's Voltage Law  The algebraic Sum of all voltages around a loop is zero at every instant in
time.
0
¦
Loop
k
v
(1.2)
Passive Sign Convention:
Whenever the reference direction of current into a two terminal device is in the direction of the
reference voltage drop across the device, then the power absorbed (or dissipated) is positive.
+
_
v
i
Figure 1. Circuit element and passive sign convention.
t
v
t
i
t
p
(1.3)
1 1
When the above convention is used,
p(t) 0 for absorbed power, and p(t) 0 for delivered power.
Time Varying Signals
Although a number of exceptions can be found throughout the world, the predominance of
electric power follows 60Hz or 50Hz frequencies. North America, part of Japan, and ships at sea use
60Hz while most of the rest of the world uses 50Hz. The historical reasons for these two frequencies
stem from the differences in lighting (filaments in vacuum or filaments in a gas atmosphere). The
lower frequencies caused an annoying flicker for lights having filaments in a gas atmosphere, and
thus a higher frequency was adopted in part of the world that initially used such lighting.
While not strictly adhered to within your textbook and these notes, an attempt to use the
following conventions has been made.
Scalar time varying signals (examples):
t
v
v
Z
sin
max
(V)
t
v
t
v
Z
cos
max
(V)
t
v
377
sin
185
(kV)
t
i
t
i
S
100
cos
max
(A)
Spatial vectors (bold or arrow overhead or line overhead):
F or F
G
or
B
If necessary, unit vectors will be used, (for example):
z
B
y
B
x
B
z
y
x
^
^
^
B
These unit vectors should not be confused with phasors below
Phasors (phasor representation of a time varying signal):
A phasor can be represented as a complex number with real and imaginary components
such that a phasor of magnitude (or length) R that is at a phase angle of
T
with respect to
the xaxis (the Real axis) can be written as
T
T
T
sin
cos
^
jR
R
e
R
jy
x
R
j
where Euler's formula
T
T
T
sin
cos
j
e
j
was used for the last representation. Note
that R is the magnitude and
T
the phase of the phasor, and
2
2
y
x
R
and
¸
¹
·
¨
©
§
x
y
arctan
T
. The phasor can be shown graphically in Figure 2.
1 2
T
Zt
Real
Imaginary
R
x
y
Figure 2. Phasor of magnitude R and phase
T
.
For a sinusoidal function
T
Z
t
V
t
v
cos
, the phasor representation is
V
Ve
V
j
T
^
/
T
If the phasor is rotating counterclockwise about the origin at a rate of
Z radians per
second, then we multiply the phasor by
such that
t
j
e
Z
T
Z
Z
T
Z
t
j
t
j
j
t
j
Ae
e
Ve
e
V^
and
using Euler's formula
T
Z
T
Z
T
Z
Z
t
jV
t
V
Ve
e
V
t
j
t
j
sin
cos
^
. Then we see that
T
Z
t
V
t
v
cos
is the real part of
, or
t
j
e
V
Z
^
^ `
T
Z
Z
t
V
e
V
t
v
t
j
cos
^
Re
.
In the above description of the phasor, the peak value is used, however we will use the
RMS value of V as described in (1.6) instead of the peak value. This simplifies many of
the calculations, particularly those associated with power as shown below.
Phasors will be represented with a hat (or caret) above the variable. Impedance is understood to be a
complex quantity in general, and the hat (or caret ^) is left off the impedance variable
where R is the resistance, and X is the reactance component respectively.
jX
R
Z
For a circuit element such as the one shown in Figure 1 representing a load in the circuit with i(t)
as the instantaneous value of current through the load and v(t) is the instantaneous value of the
voltage across the load.
In quasisteady state conditions, the current and voltage are both sinusoidal, with corresponding
amplitudes of V
max
, I
max
, and initial phases,
I
v
and
I
i
, and the same frequency
T
f
S
S
Z
2
2
v
t
V
t
v
I
Z
cos
max
(1.4)
i
t
I
t
i
I
Z
cos
max
(1.5)
The rootmeansquared (RMS) value of the voltage and current are then:
@
2
cos
1
max
2
max
V
dt
t
V
T
V
T
o
v
³
I
Z
(1.6)
1 3
@
2
cos
1
max
2
max
I
dt
t
I
T
I
T
o
i
³
I
Z
(1.7)
Phasor representations for the above signals use the RMS values as
V
V
^
/
I
v
and
I
I
^
/
I
i
.
The instantaneous power is the product of voltage times current or:
i
v
i
v
t
t
I
V
t
I
t
V
t
i
t
v
t
p
I
Z
I
Z
I
Z
I
Z
cos
cos
cos
cos
max
max
max
max
i
v
t
t
VI
I
Z
I
Z
cos
cos
2
i
v
i
v
t
VI
I
I
Z
I
I
2
cos
cos
2
1
2
@
(1.8)
The average value is found by integrating over a period of time and then dividing the result by
that same time interval. The first term in (1.8) is independent of time, while the second term varies
from 1 to +1 symmetrically about zero. Because it is symmetric about zero, integration over an
integer number of cycles (or periods) gives a value of zero for the second term in (1.8), and the
average power which we define as the real power with units of watts (W) is:
i
v
VI
P
I
I
cos
(1.9)
This shows the power is not only proportional to the RMS values of voltage and current, but also
proportional to
i
v
I
I
cos
. The cosine of this angle is defined as the displacement factor, DF. In
more general terms for periodic, but not necessarily sinusoidal signals, the power factor is defined as:
VI
P
pf
{
(1.10)
For sinusoidal signals, the power factor equals the displacement factor, or
i
v
pf
I
I
cos
(1.11)
For comparison, the voltage, current, and power for various angles between voltage and current
are shown below:
For leading power factors:
10
5
0
5
10
0
0.5
1
1.5
2
2.5
3
V
o
lt
ag
e (V
),
C
u
rr
e
n
t (A)
,
P
o
we
r (W
)
# of Periods
t
t
v
Z
cos
5
t
t
i
Z
cos
2
5
cos
2
2
cos
max
max
i
v
i
v
I
V
VI
P
I
I
I
I
(W)
Real
Imaginary
V
I
1 4
1 5
10
5
0
5
10
0
0.5
1
1.5
2
2.5
3
V
o
lt
ag
e (V
),
C
u
rr
e
n
t (A)
,
Po
w
er
(W
)
# of Periods
t
t
v
Z
cos
5
º
30
cos
2
t
t
i
Z
33
.
4
cos
2
2
cos
max
max
i
v
i
v
I
V
VI
P
I
I
I
I
(W)
Real
Imaginary
V
I
10
5
0
5
10
0
0.5
1
1.5
2
2.5
3
V
o
lt
ag
e (V
),
C
u
rr
e
n
t (A)
,
Po
w
er
(W
)
# of Periods
t
t
v
Z
cos
5
º
60
cos
2
t
t
i
Z
5
.
2
cos
2
2
cos
max
max
i
v
i
v
I
V
VI
P
I
I
I
I
(W)
Real
Imaginary
V
I
10
5
0
5
10
0
0.5
1
1.5
2
2.5
3
Vol
tag
e (
V
),
Cur
re
n
t (A
)
,
Po
w
er
(W
)
# of Periods
t
t
v
Z
cos
5
º
90
cos
2
t
t
i
Z
0
cos
2
2
cos
max
max
i
v
i
v
I
V
VI
P
I
I
I
I
(W)
Real
Imaginary
V
I
10
5
0
5
10
0
0.5
1
1.5
2
2.5
3
Vol
tag
e (
V
),
Cur
re
n
t (A
)
,
Po
w
er
(W
)
# of Periods
t
t
v
Z
cos
5
º
120
cos
2
t
t
i
Z
5
.
2
cos
2
2
cos
max
max
i
v
i
v
I
V
VI
P
I
I
I
I
(W)
Real
Imaginary
V
I
10
5
0
5
10
0
0.5
1
1.5
2
2.5
3
V
o
lt
ag
e (V
),
C
u
rr
e
n
t (A)
,
Po
w
er
(W
)
# of Periods
t
t
v
Z
cos
5
º
150
cos
2
t
t
i
Z
33
.
4
cos
2
2
cos
max
max
i
v
i
v
I
V
VI
P
I
I
I
I
(W)
Real
Imaginary
V
I
10
5
0
5
10
0
0.5
1
1.5
2
2.5
3
V
o
lt
ag
e (V
),
C
u
rr
e
n
t (A)
,
Po
w
er
(W
)
# of Periods
t
t
v
Z
cos
5
º
180
cos
2
t
t
i
Z
5
cos
2
2
cos
max
max
i
v
i
v
I
V
VI
P
I
I
I
I
(W)
Real
Imaginary
V
I
For lagging power factors:
10
5
0
5
10
0
0.5
1
1.5
2
2.5
3
V
o
lt
ag
e (V
),
C
u
rr
e
n
t (A)
,
P
o
we
r (W
)
# of Periods
t
t
v
Z
cos
5
t
t
i
Z
cos
2
5
cos
2
2
cos
max
max
i
v
i
v
I
V
VI
P
I
I
I
I
(W)
Real
Imaginary
V
I
1 6
1 7
10
5
0
5
10
0
0.5
1
1.5
2
2.5
3
V
o
lt
ag
e (V
),
C
u
rr
e
n
t (A)
,
Po
w
er
(W
)
# of Periods
t
t
v
Z
cos
5
º
30
cos
2
t
t
i
Z
33
.
4
cos
2
2
cos
max
max
i
v
i
v
I
V
VI
P
I
I
I
I
(W)
Real
Imaginary
V
I
10
5
0
5
10
0
0.5
1
1.5
2
2.5
3
V
o
lt
ag
e (V
),
C
u
rr
e
n
t (A)
,
Po
w
er
(W
)
# of Periods
t
t
v
Z
cos
5
º
60
cos
2
t
t
i
Z
5
.
2
cos
2
2
cos
max
max
i
v
i
v
I
V
VI
P
I
I
I
I
(W)
Real
Imaginary
V
I
10
5
0
5
10
0
0.5
1
1.5
2
2.5
3
V
o
lt
ag
e (V
),
C
u
rr
e
n
t (A)
,
Po
w
er
(W
)
# of Periods
t
t
v
Z
cos
5
º
90
cos
2
t
t
i
Z
0
cos
2
2
cos
max
max
i
v
i
v
I
V
VI
P
I
I
I
I
(W)
Real
Imaginary
V
I
10
5
0
5
10
0
0.5
1
1.5
2
2.5
3
V
o
lt
ag
e (V
),
C
u
rr
e
n
t (A)
,
Po
w
er
(W
)
# of Periods
t
t
v
Z
cos
5
º
120
cos
2
t
t
i
Z
5
.
2
cos
2
2
cos
max
max
i
v
i
v
I
V
VI
P
I
I
I
I
(W)
Real
Imaginary
V
I
10
5
0
5
10
0
0.5
1
1.5
2
2.5
3
V
o
lt
ag
e (V
),
C
u
rr
e
n
t (A)
,
Po
w
er
(W
)
# of Periods
t
t
v
Z
cos
5
º
150
cos
2
t
t
i
Z
33
.
4
cos
2
2
cos
max
max
i
v
i
v
I
V
VI
P
I
I
I
I
(W)
Real
Imaginary
V
I
10
5
0
5
10
0
0.5
1
1.5
2
2.5
3
V
o
lt
ag
e (V
),
C
u
rr
e
n
t (A)
,
Po
w
er
(W
)
# of Periods
t
t
v
Z
cos
5
º
180
cos
2
t
t
i
Z
5
cos
2
2
cos
max
max
i
v
i
v
I
V
VI
P
I
I
I
I
(W)
Real
Imaginary
V
I
We call the power factor leading or lagging, depending on whether the current of the load leads or
lags the voltage across it. We know that current does not change instantly through an inductor, so it
is clear that for a resistive  inductive (RL) load, the power factor is lagging. Likewise the voltage
can not change instantly across a capacitor, therefore for a resistive capacitive (RC) load, the power
factor is leading (the current changes before the voltage can). Also, for a purely inductive or
capacitive load the power factor is 0, while for a purely resistive load it is 1.
The product of the RMS values of voltage and current at a load is the apparent power, S having
units of voltamperes (VA):
VI
S
(1.12)
The reactive power is
Q with units of voltamperes reactive (VA reactive, or VAr):
i
v
VI
Q
I
I
sin
(1.13)
The reactive power represents the energy oscillating in and out of an inductor or capacitor. Since
the energy oscillation in an inductor is 180º out of phase with the energy oscillating in an capacitor,
the reactive power of these two have opposite signs with the convention that it is positive for the
inductor and negative for the a capacitor.
Using phasors, the complex apparent power, is:
S^
V
*
^
^
^
I
V
S
/
I
v
I
/
I
i
(1.14)
or
1 8
(1.15)
jQ
P
S
^
As an example, consider the following voltage and current for a given load:
¸
¹
·
¨
©
§
6
377
sin
2
120
S
t
t
v
(V) (1.16)
¸
¹
·
¨
©
§
4
377
sin
2
5
S
t
t
i
(A) (1.17)
then
(W), while the power factor is
600
5
120
VI
S
966
.
0
4
6
cos
¸
¹
·
¨
©
§
S
S
pf
leading. Also,
the complex apparent power is:
120
*
^
^
^
I
V
S
/
S/6 · 5/S/4 = 600/ S/12 = 579.6 (W) j155.3 (VAr)
(1.18)
It should also be noted that when the angles are represented in radian, care must be taken to
assure your calculator is in radians mode. Often we represent the argument of the sine or cosine term
in mixed units. For example we might write cos(377t + 30º). The first term (377t) has units of
radians, while the second (30º) has units of degrees, and one of these must be converted before
making calculations with your calculator or computer either in radians mode or in degrees mode.
As a phasor the complex apparent power can be shown on a complex plane for the cases of
leading and lagging power factors as shown in Figure 3.
Real
Imaginary
V
I
Real
Imaginary
P
Q
S
Real
Imaginary
V
I
Real
Imaginary
P
S
Q
(a) Leading power factor
(b) Lagging power factor
Figure 3. Phasor of magnitude R and phase
T
.
1 9
Recall that the leading power factor corresponds to a resistivecapacitive load. We see from
Figure 3(a), this also corresponds to a negative value for Q. Similarly, lagging power factor has the
current lagging the voltage corresponding to an inductiveresistive load, and Figure 3(b) shows this
corresponds to a positive value for Q.
To summarize, we can use the following tables:
Equation
Units
Real Power
pf
S
pf
VI
VI
P
i
v
I
I
cos
Watts
Reactive Power
leading)
(for
1
lagging)
(for
1
sin
2
2
pf
S
pf
S
VI
Q
i
v
I
I
VoltAmperesReactive
Apparent Power
VI
S
*
^
^
^
I
V
S
V
/
I
v
I
/
I
i
jQ
P
S
^
2
2
2
Q
P
S
VoltAmperes
Type of Load Reactive Power Power Factor
Inductive
Q 0
lagging
Capacitive
Q 0
leading
Resistive
Q = 0
1
From the interdependence of the four quantities, S, P, Q, pf, if we know any two of these
quantities, the other two can be determined. For example, if S = 100 (kVA), and pf = 0.8 leading,
then:
@
i
v
i
v
S
Q
pf
S
Q
pf
S
P
I
I
I
I
sin
then
8
.
0
arccos
sin
sin
or
(kVAR),
60
1
(kW)
800
2
It is important to notice that Q 0, such that
i
v
I
I
sin
is a negative quantity. This can be seen
when it is understood that there are two possible answers for the arccos(0.8), that is cos(36.87º) = 0.8,
and cos(36.87º) = 0.8 so to obtain a Q 0 we use (
I
v
I
i
) =
36.87º.
Generally, in systems that contain more than one load (or source), the real and reactive power can
be found by adding individual contributions, but this is not the case with the apparent power. That is
¦
i
i
total
P
P
¦
(1.19)
i
i
total
Q
Q
¦
z
i
i
total
S
S
1 10
For the above example, if the load voltage is V
L
= 2000 (V), then the load current would be
I
L
= S/V
L
= [100
u10
3
(VA)]/[2
u10
3
(V)] = 50 (A). If we use the load voltage as the reference, then:
2000
^
V
/0º (V)
50
^
I
/
I
i
= 50
/+36.87º (A)
*
^
^
^
I
V
S
[2000
/0º][50 /
36.87º] = P + jQ = 80u10
3
(W) j60
u10
3
(VAr)
1.2
Three Phase Balanced Systems
Threephase systems offer significant advantages over single phase systems: for the same power
and voltage there is less copper in the windings, and the total power absorbed remains constant rather
than oscillating about an average value.
For a three phase system consisting of three current sources having the same amplitude and
frequency, but with phases differing by 120º as:
¸
¹
·
¨
©
§
¸
¹
·
¨
©
§
3
2
sin
2
3
2
sin
2
sin
2
3
2
1
S
I
Z
S
I
Z
I
Z
t
I
t
i
t
I
t
i
t
I
t
i
(1.20)
If these are connected as shown in Figure 4, then at node n or n', the current adds to zero, and the
neutral line nn' (dashed) is not needed.
0
3
2
sin
3
2
sin
sin
2
3
2
1
¿
¾
½
¯
®
¸
¹
·
¨
©
§
¸
¹
·
¨
©
§
S
I
Z
S
I
Z
I
Z
t
t
t
I
t
i
t
i
t
i
i
1
n
n'
i
2
i
3
Figure 4. Balanced three phase Yconnected system with zero neutral current.
If instead we had three voltage sources Yconnected as in Figure 5 with the following values
1 11
¸
¹
·
¨
©
§
¸
¹
·
¨
©
§
3
2
sin
2
3
2
sin
2
sin
2
3
2
1
S
I
Z
S
I
Z
I
Z
t
V
t
v
t
V
t
v
t
V
t
v
(1.21)
then, the current through each of the three loads (assuming the loads are equal), would have equal
magnitudes, but each current would have a phase that is shifted by an equal amount with respect to
the voltages, v
1
(t), v
2
(t), v
3
(t).
v
1
n
n'
v
2
v
3
+
+
+
i
a
i
b
i
c
Figure 5. Balanced voltage fed three phase Yconnected system with zero neutral current.
With equal impedances for the loads, then
¸
¹
·
¨
©
§
¸
¹
·
¨
©
§
T
S
I
Z
T
S
I
Z
T
I
Z
3
2
sin
2
3
2
sin
2
sin
2
t
Z
V
t
i
t
Z
V
t
i
t
Z
V
t
i
c
b
a
(1.22)
and again the currents sum to zero at nodes n or n', and for this balanced three phase system, the
neutral wire (dashed) is not required.
In comparison to a single phase system, where two wires are required per phase, the three phase
system delivers three times the power, and requires only three transmission wires total. This is a
significant advantage considering the hundreds of miles of wire needed for power transmission.
Y and
' Connections
The loads in the previous two figures, as well as in Figure 6 are connected in a Y or star
configuration. If the load of Figure 6 is for a balanced Y system, then the voltages between each
phase and the neutral are:
V
V
n
1
^
/
I
,
V
V
n
2
^
/
I
S , and
V
V
n
3
^
/
I
S .
Kirchhoff's voltage law (KVL) states that the sum of voltages around a closed loop equals zero.
This is also the case here however the voltages are complex numbers or phasors, and as such must be
1 12
added as vectors. The phase
I
can be any value, but the relative position of the phase to neutral
phasors must be 120º with respect to each other as shown in Figure 7.
n
V
12
1
3
2
+
+
+
+
V
1n
V
3n
V
2n
Figure 6. Yconnected loads with voltages relative to neutral identified.
By KVL:
, or
as shown in
0
^
^
^
n
2
n
1
12
V
V
V
n
2
n
1
12
^
^
^
V
V
V
Figure 7.
V
3n
V
1n
V
2n
2n
V
V
3n
1n
V
V
12
V
31
V
23
Figure 7. Voltage phasors of the Yconnected loads shown in Figure 6.
We could also use the phasor representation
V
V
n
1
^
/
I
,
V
V
n
2
^
/
I
S ,
and
V
V
n
3
^
/
I
S to determine the linetoline voltages as
V
V
V
V
n
n
3
^
^
^
2
1
12
/
I
+
S/6 (1.23)
This shows the RMS value of the linetoline voltage, V
ll
, at a Y load is 3 times the lineto
neutral or phase voltage, V
ln
. In the Y connection, the phase current is equal to the line current, and
the power supplied to the system is three times the power supplied to each phase, since the voltage
and current amplitudes and phase differences between them are the same in all three phases. If the
power factor in one phase is
i
v
pf
I
I
cos
, then the total power to the system is:
1 13
i
v
l
l
l
i
v
l
l
l
n
n
I
V
j
I
V
I
V
jQ
P
S
I
I
I
I
I
I
I
sin
3
cos
3
^
^
3
^
*
1
1
3
3
3
(1.24)
Similarly, for a connection of the loads in the
' configuration (as in Figure 8), the phase voltage
is equal to the line voltage however; the phase currents are not equal to the line currents for the
'
configuration. If the phase currents are
I
I
12
^
/
I
,
I
I
23
^
/
I
S , and
I
I
31
^
/
I
S then
using Kirchhoff's current law (KCL) the current of line 1, as shown in Figure 8 is:
I
I
I
I
3
^
^
^
31
12
1
/
I

S/6
(1.25)
Thus for the
' configuration, the line current is 3 times the I
'
current.
I
1
2
1
3
I
2
I
3
I
12
I
31
I
23
I
12
I
31
I
23
I
23
I
1
I
3
I
2
I
12
I
31
Figure 8.
' connected load, line, and phase currents.
To calculate the power in the threephase
' connected load:
i
v
l
l
l
i
v
l
l
l
I
V
j
I
V
I
V
jQ
P
S
I
I
I
I
I
I
I
sin
3
cos
3
^
^
3
^
*
12
12
3
3
3
(1.26)
which is the same value as for the Y connected load.
For a balanced system, the loads of the three phases are equal. Also, a
' configured load, can be
replaced with a Y configured load (and visa versa) if:
'
Z
Z
Y
3
1
(1.27)
1 14
Under these conditions, the two loads are indistinguishable by the power transmission lines.
You might recall the
' to Y transformation for resistor circuits can be remembered by overlaying
the
' and Y configurations such as in Figure 9.
R
C
B
A
B
R
A
R
C
R
1
R
3
R
2
Figure 9.
' to Y or Y to ' resistor network transformation.
1
1
3
3
2
2
1
1
R
R
R
R
R
R
R
R
R
R
R
R
R
R
A
C
B
A
C
B
(1.28)
Equation (1.28) gives a similar result to that of (1.27) when R
A
= R
B
= R
B
C
and R
1
= R
2
= R
3
.
1.3
Calculations in ThreePhase Systems
Calculations of quantities like currents, voltages, and power in threephase systems can be
simplified by the following procedure:
1. transform the
' circuits to Y,
2. connect a neutral conductor,
3. solve one of the three 1phase systems
4. convert the results back to the
' systems
1.3.1 Example
For the 3phase system in Figure 10 calculate the linetoline voltage, real power, and power
factor at the load.
To solve this by the procedure outlined above, first consider only one phase as shown in Figure
11.
1 15
v
1
n
n'
v
2
v
3
+
+
+
= 120 (V)
j1 (
:)
7 + j5 (
:)
Figure 10. Three phase system with Y connected load, and line impedance.
n
v
1
n'
+
= 120 (V)
j1 (
:)
7 + j5 (
:)
I
Figure 11. One phase of the three phase system shown in Figure 10.
For the onephase in Figure 11,
02
.
13
5
7
1
120
^
j
j
I
/40.6º (A)
02
.
13
^
^
1
L
n
Z
I
V
/40.6º (7+j5) = 112/5.1º (V)
*
1
,
^
^ I
V
S
L
L
I
(112
/5.1º)(13.02/40.6º) = 1458.3/35.5º = 1.187
u10
3
+ j0.848
u10
3
I
1
,
L
P
1.187 (kW),
0.848 (kVAr)
I
1
,
L
Q
pf
= cos(5.1º  (40.6º)) = 0.814 lagging
For the threephase system of Figure 10 the load voltage (linetoline), the real, and reactive
power are:
194
112
3
,
l
l
L
V
(V)
I
3
,
L
P
3.56 (kW)
I
3
,
L
Q
2.544 (kVAr)
1 16
1.3.2 Example
For the Y to
' threephase system in Figure 12, calculate the power factor and the real power at
the load, as well as the phase voltage and current. The source voltage is 400 (V) linetoline.
v
1
n'
v
2
v
3
+
+
+
j1 (
:)
18 + j6 (
:)
Figure 12.
' connected load.
First convert the load to an equivalent Y connected load, then work with one phase of the system.
The line to neutral voltage of the source is
231
3
400
ln
V
(V).
231 (V)
n
n'
+
+
+
j1 (
:)
6 + j2 (
:)
Figure 13. Equivalent Y connected load.
n
v
1
n'
+
= 231 (V)
j1 (
:)
6 + j2 (
:)
I
L
Figure 14. One phase of the Y connected load.
1 17
44
.
34
2
6
1
231
^
j
j
I
L
/26.6º (A)
8
.
217
2
6
^
^
j
I
V
L
L
/8.1º (V)
The power factor at the load is:
948
.
0
º
6
.
26
º
1
.
8
cos
cos
i
v
pf
I
I
lagging
Converting back to a
' connected load gives:
88
.
19
3
44
.
34
3
L
I
I
I
(A)
22
.
377
3
8
.
217
l
l
V
(V)
At the load the power is:
,3
3
3 377.22 34.44 0.948
21.34
L
l l L
P
V I pf
I
(kW)
1.3.3 Example
Two threephase loads are connected as shown in . Load 1 draws from the system P
L1
= 500
(kW) at 0.8 pf lagging, while the total load is S
T
= 1000 (kVA) at 0.95 pf lagging. What is the power
factor of load 2?
Power System
Load 2
Load 1
Figure 15. Two threephase loads connected to the same power source.
1 18
For the total load we can add the real and reactive power for each of the two loads (we can not
add the apparent power).
2
1
2
1
2
1
L
L
T
L
L
T
L
L
T
S
S
S
Q
Q
Q
P
P
P
z
From the information we have for the total load we can write the following:
T
T
T
pf
S
P
950 (kW)
@
25
.
312
95
.
0
cos
sin
1
T
T
S
Q
(kVAr)
The reactive power, Q
T
, is positive since the power factor is lagging.
For the load L1, P
L1
= 500 (kW), pf
1
= 0.8 lagging, thus:
u
8
.
0
10
500
3
1
L
S
625 (kVA)
2
1
2
1
1
L
L
L
P
S
Q
375 (kVA)
Again, Q
L1
is positive since the power factor is lagging. This leads to:
1
2
L
T
L
P
P
P
450 (kW)
1
2
L
T
L
Q
Q
Q
62.75 (kVAr)
and
99
.
0
75
.
62
450
450
2
2
2
2
2
L
L
L
S
P
pf
leading
.
Chapter Notes
x
A sinusoidal signal can be described uniquely by:
1.
Time dependent form as for example:
v
ft
t
v
I
S
2
sin
5
(V)
2.
by a time dependent graph of the signal
3.
as a phasor along with the associated frequency of the phasor
one of these descriptions is enough to produce the other two.
x
It is the phase difference that is important in power calculations, not the phase. The phase is
arbitrary depending on the defined time (t = 0). We need the phase to solve circuit problems
after we take one quantity (some voltage or current) as a reference. For that reference quantity
we assign an arbitrary phase (often zero).
x
In both threephase and onephase systems the total real power is the sum of the real power
from the individual loads. Likewise the total reactive power is the sum of the reactive power of
the individual loads. This is not the case for the apparent power or the power factor.
1 19
1 20
x
Of the four quantities: real power, reactive power, apparent power, and power factor, any two
describe a load adequately. The other two quantities can be calculated from the two given.
x
To calculate real, reactive, and apparent power when using equations (1.9), (1.12), and (1.13)
we must use absolute values, not complex values for the currents and voltages. To calculate the
complex power using equation (1.14) we do use complex currents and voltages and find
directly both the real and reactive power (as the real and imaginary components respectively).
x
When solving a circuit to calculate currents and voltages, use complex impedances, currents
and voltages.
HOME WORK # 1
ECE 320
1. The circuit of figure below is excited by a 240V, 60Hz sinusoidal source. If
F
C
P
10
,
determine:
(a)
The magnitude of current
I
(b)
The average value of power dissipated by resistor
R
:
10
R
I
1
I
2
I
C
mH
L
25
V
o
p
2.
A set of balanced threephase voltages is impressed on a balanced, threephase, delta
connected load. If
q
0
7200
ab
V
V and
q
0
12
a
I
A. Determine:
(a)
The phase current
ab
I
(b)
The total average power supplied to the load.
3. Capacitors
C are added to the 60Hz, threephase network given below to correct the
input power factor to unity. Determine:
(a)
The value of
C
(b)
If
V, find the total kVAR rating of the capacitor bank.
480
V
AB
C
C
C
A
B
C
:
4
j
:
3
HOME WORK # 1 SOLUTIONS
ECE 320
1. The circuit of figure below is excited by a 240V, 60Hz sinusoidal source. If
F
C
P
10
,
determine:
(a)
The magnitude of current
I
(b)
The average value of power dissipated by resistor
R
:
10
R
I
1
I
2
I
C
mH
L
25
V
o
p
Solution:
(a) It is convenient to assume
V
on the reference. The two branch impedances are found as
1
6
1
1
1
90
90
265.258
90
120
10 10
j
C
C
Z
Z
S
q
§
·
q
q :
¨
¸
©
¹
u
Z
2
10
120
0.025
10
9.425 13.742 43.3
R
j L
j
j
Z
S
q :
Z
The branch current can now be determined.
1
1
240 0
0.905 90 A
265.258
90
V
I
q
q
q
Z
2
2
240 0
17.465
43.3 A
13.742 43.3
V
I
q
q
q
Z
Applying KCL,
1
2
0.905 90
17.465
43.3
I
I
I
q
q
12.710
11.073 16.857
41.06 A
j
q
Hence,
16.857 A
I
(b) The power dissipated by R is
2
2
2
17.465
10
3050.26 W
R
P
I R
2.
A set of balanced threephase voltages is impressed on a balanced, threephase, delta
connected load. If
q
0
7200
ab
V
V and
q
0
12
a
I
A. Determine:
(a)
The phase current
ab
I
(b)
The total average power supplied to the load.
Solution:
Assume
phase sequence.
a b c
(a) The phase current
ab
I
12 0
3 30
3 30
a
ab
I
I
q
q
q
6.928 30 A
ab
I
q
(b) The total average power supplied to the load.
30
ab
ab
V
I
T
q
,
3
cos
3 7200 12 cos
30
129.60 kW
T
L L
P
V I
T
q
3. Capacitors
C are added to the 60Hz, threephase network given below to correct the
input power factor to unity. Determine:
(a)
The value of
C
(b)
If
V, find the total kVAR rating of the capacitor bank.
480
V
AB
C
C
C
A
B
C
:
4
j
:
3
Solution:
(a) The RL branches of the delta network can be converted to an equivalent wye network by
1
1
3
3
3
4
Y
j
'
Z
Z
:
In order for a unity PF to exist, the reactive power supplied by the capacitor of each phase
must equal the reactive power supplied to the inductor of each phase.
2
2
Y
Y
V
CV
L
Z
I
I
Z
Z
§
·
¨
¸
©
¹
or
2
2
2
2
2
1
3
4 /120
1273.2 F
3
4
3
3
3
3
Y
Y
Y
L
L
C
Z
R
L
S
P
Z
§ ·
§
·
§ ·
§ ·
¨ ¸
¨
¸
¨ ¸
¨ ¸
© ¹
©
¹
© ¹
© ¹
(b)
2
2
6
480
3
3 120
1273.2 10
110.59 kVARs
3
T
Q
CV
I
Z
S
§
·
u
¨
¸
©
¹
Chapter 2: Magnetic Circuits and Materials
Chapter Objectives
In this chapter you will learn the following:
x
How Maxwell's equations can be simplified to solve simple practical magnetic problems
x
The concepts of saturation and hysteresis of magnetic materials
x
The characteristics of permanent magnets and how they can be used to solve simple problems
x
How Faraday's law can be used in simple windings and magnetic circuits
x
Power loss mechanisms in magnetic materials
x
How force and torque is developed in magnetic fields
2.1
Ampere's Law and Magnetic Quantities
Ampere's experiment is illustrated in Figure 1 where there is a force on a small current element
I
2
l when it is placed a distance, r, from a very long conductor carrying current I
1
and that force is
quantified as:
l
I
r
I
F
2
1
2
S
P
(N)
(2.1)
I
1
r
I
2
F
Conductor
1
Current Element
of Length l
Figure 1. Ampere's experiment of forces between current carrying wires.
The magnetic flux density, B, is defined as the first portion of equation (2.1) such that:
l
BI
F
2
(N)
(2.2)
1 1
From (2.1) and (2.2) we see the magnetic flux density around conductor 1 is proportional to the
current through conductor 1, I
1
, and inversely proportional to the distance from conductor 1. Looking
at the units and constants handout given in class, or from (2.2) the units of, B, are seen as
¸
¹
·
¨
©
§
m
A
N
,
thus µ, called permeability, has units of
¸
¹
·
¨
©
§
2
A
N
. More commonly, the relative permeability of a given
material is given where
0
P
P
P
r
and
9
0
2
N
400
10
A
P
S
§
u
¨© ¹
·
¸ . Since a Newtonmeter is a Joule, and
a Joule is a Wattsecond:
¸
¹
·
¨
©
§
¸
¹
·
¨
©
§
¸
¹
·
¨
©
§
¸
¹
·
¨
©
§
¸
¹
·
¨
©
§
2
2
2
2
m
s
V
m
A
s
A
V
m
A
J
m
A
m
N
m
A
N
. This shows B is a
per meter squared quantity, and the (V·s) units represents the magnetic flux and is given units of
Webers (Wb). This flux can be found by integrating the normal component of B over the area of a
given surface:
³
S
ds
n
B ^
I
(2.3)
The magnetic field intensity is related to the magnetic flux density by the permeability of the
media in which the magnetic flux exists.
P
B
H
{ (2.4)
For the system in Figure 1,
r
I
B
H
S
P
2
1
and have units of
¸
¹
·
¨
©
§
m
A
. If there were multiple
conductors in place of conductor 1, for example in a coil, then the units would be ampereturns per
meter. A line integration of H over a closed circular path gives the current enclosed by that path, or
for the system in Figure 1:
1
2
0
1
2
I
dl
r
I
dl
H
H
r
C
³
³
S
S
(2.5)
again, if the system contained multiple conductors within the enclosed path, the result would give
ampereturns. Equation (2.5) is Ampere's circuital law.
An alternative approach as described in your textbook is to begin with Maxwell's equations
which include Ampere's circuital law in a more general form as shown in Table I below:
Table I. Maxwell's equations.
Name
Point Form
Integral Form
Faraday's Law
t
B
w
w
u
E
³
³
C
S
ds
n
B
dt
d
dl
^
E
Ampere's Law Modified by Maxwell
t
D
J
H
w
w
u
³
³
»¼
º
«¬
ª
w
w
S
C
ds
n
t
D
J
dl
H
^
Gauss's Law
U
D
³
³
V
S
^
dv
ds
n
D
U
Gauss's Law for Magnetism
0
B
0
^
S
³
ds
n
B
1 2
where
³
C
represents the integral over a closed path,
³
S
represents the integral over the surface of a
closed volume of space, is a surface normal vector,
n^
E is the spatial vector of electric field, B is the
spatial vector of magnetic flux density,
J
is the spatial vector of the electric current density, D is
the displacement charge vector,
U
is the electric charge density,
³
V
is a volume integral
u
is the
curl and
the divergence of the vector being acted upon.
Some assumptions commonly used in electromechanical energy conversion include a low enough
frequency, that the displacement current,
t
D
w
w
, can be neglected, and the assumption of homogeneous
and isotropic media used in the magnetic circuit. Under these assumptions, Ampere's circuital law is
modified to remove the displacement current component such that
³
³
S
C
ds
n
J
dl
H
^
(2.6)
which for the system of Figure 1 reduces to equation (2.5).
2.2 Magnetic
Circuits
From Ampere's circuital law,
³
³
S
C
ds
n
J
dl
H
^
, we see the magnetic field intensity around a
closed contour is a result of the total electric current density passing through any surface linking that
contour. Gauss's Law for magnetism,
0
^
S
³
ds
n
B
states that there are no magnetic monopoles
that is to say there for a closed surface there is as much magnetic field density leaving that closed
surface as there is entering the closed surface. If the integration is for an area, but not a closed
surface area, then we obtain the flux or
³
S
ds
n
B ^
I
.
The permeability of free space is 1, while the permeability of magnetic steel is a few hundred
thousand. Magnetic flux can be confined to the structures or paths formed by high permeability
materials. In this way, magnetic circuits can be formed such as the one shown in Figure 2.
Figure 2. Simple magnetic circuit [1].
1 3
The driving force for the magnetic field is the magnetomotive force (mmf),
F, which equals the
ampereturn product
i
N
F
(2.7)
The analysis of a magnetic circuit is similar to the analysis of an electric circuit, and an analogy
can be made for the individual variables as shown in Table II.
Table II. Comparison of electric and magnetic circuits.
Electric Circuit
Magnetic Circuit
V
r
I
I
a
b
c
Electrically Conductive
Material
V
r
I
I
a
b
c
High Permeability
Material
Driving Force
applied battery voltage = V
applied ampereturns =
F
Response
resistance
electric
force
driving
current
or
R
V
I
reluctance
magnetic
force
driving
flux
or
R
F
I
Impedance
Impedance is used to indicate the impediment to the driving force in establishing a response.
A
l
R
V
resistance
where l = 2
Sr, V = electrical conductivity, A =
crosssectional area
A
l
P
R
reluctance
where l = 2
Sr,
P
= permeability, A = cross
sectional area
Equivalent Circuit
V
I
R
V = IR
F
I
R
F =
I
R
1 4
1 5
Electric Circuit
Magnetic Circuit
Fields
Electric Field Intensity
r
V
l
V
S
2
{
E
(V/m)
or
³
V
dl
E
Magnetic Field Intensity
r
l
H
S
2
F
F
{
(At/m)
or
³
F
dl
H
Potential
Electric Potential Difference
ab
ab
ab
b
a
b
a
ab
IR
l
A
l
l
I
l
l
IR
dl
l
V
dl
V
³
³
V
E
Magnetic Potential Difference
ab
ab
ab
ab
b
a
ab
l
A
l
l
l
l
l
l
dl
H
R
R
F
F
I
P
I
I
³
Flow Densities
Current Density
E
E
V
V
{
A
l
A
l
AR
V
A
I
J
Flux Density
H
A
l
A
Hl
A
A
B
P
P
I
¸¹
·
¨©
§
{
R
F
An example magnetic circuit is shown in Figure 3, below.
Figure 3. Simple magnetic circuit with an air gap [1].
For this circuit, we will assume:
x
the magnetic flux density is uniform throughout the magnetic core's crosssectional area and is
perpendicular to the crosssectional area
x
the magnetic flux remains within the core and the air gap defined by the crosssectional area of the
core and the length of the gap (no leaking of field, no fringe fields at the gap).
Then
g
g
c
c
A
A
B
A
B
A
d
B
c
³
I
(2.8)
with A
c
= A
g
H
H
B
B
g
c
g
c
P
P
(2.9)
Since,
F = Hl
g
g
g
c
c
c
g
c
c
l
B
l
B
g
H
l
H
P
P
F
(2.10)
or
g
c
g
g
c
c
c
A
g
A
l
R
R
F
¸
¸
¹
·
¨
¨
©
§
I
P
P
I
(2.11)
Thus the magnetic circuit shown in Figure 3 can be represented as
F
I
R
R
c
g
g
c
g
c
i
N
R
R
R
R
F
I
Figure 4. Equivalent circuit for the magnetic circuit in Figure 3.
This concept is helpful for more complex configurations such as shown in Figure 5.
1 6
Area
A
1
Area
A
1
Area
A
1
Area A
2
Area
A
3
g
2
I
Area A
2
g
1
I
2
1
Area
A
1
Figure 5. Magnetic circuit with various crosssectional areas and two coils.
Using the lengths defined in Figure 6, and paying attention to the direction of the magnetomotive
force from each of the coils by use of the right hand rule, the equivalent circuit can be drawn as seen
in Figure 7.
Area A
Area A
2
Area
A
3
g
2
I
g
1
I
2
1
1
Area A
1
Area A
1
l
1
·l
2
2
1
·l
3
2
1
Figure 6. Lengths for each section of the magnetic circuit of Figure 5.
Then,
1
2
1
1
1
1
1
1
I
I
I
g
l
I
N
R
R
F
(2.12)
and
2
2
3
1
2
1
2
2
2
l
g
l
g
I
N
R
R
R
R
F
I
I
(2.13)
For a system with the dimensions, number of turns, current, and permeability known, then
equations (2.12) and (2.13) give two equations with two unknowns such that
I
1
and
I
2
can be found.
1 7
F
R
g
R
g
F
2
1
1
2
R
l
1
R
l2
R
l3
I
1
I
2
Figure 7. Equivalent circuit for the configuration of Figure 5.
Assuming the gaps are air gaps, the value of each reluctance can be found as:
1
1
1
A
l
l
P
R
1
0
1
1
A
g
g
P
R
2
2
2
A
l
l
P
R
3
3
3
A
l
l
P
R
2
0
2
2
A
g
g
P
R
Then we can find the magnetic flux density for each gap as:
2
2
1
1
2
1
A
B
A
B
g
g
I
I
(2.14)
Note that the permeability of the core material can be much larger than the permeability of the gaps
such that the total reluctance is dominated by the reluctances of the gaps.
2.3 Inductance
From circuit theory we recall that the voltage across an inductor is proportional to the time rate of
change of the current through the inductor.
dt
t
di
L
t
v
L
L
(2.15)
and while the power of an inductor can be positive or negative, the energy is always positive as
2
2
1
L
L
i
L
t
w
(2.16)
In Table I, Faraday's law is
³
³
C
S
ds
n
B
dt
d
dl
^
E
or the electric field intensity around a
closed contour C is equal to the time rate of change of the magnetic flux linking that contour.
Integrating over the closed contour of the coil itself gives us the negative of the voltage at the
terminals of the coil. On the right side of Faraday's law we then must integrate over the surface of
the full coil, thus including the N turns of the coil. Then Faraday's law gives
1 8
dt
t
d
N
t
v
L
I
(2.17)
Comparing equation (2.15) and (2.17) gives
di
d
N
L
I
(2.18)
For linear inductors, the flux
I
is directly proportional to current, i, for all values such that
i
N
L
I
(2.19)
with units of (Weberturns per ampere), or Henry's (H). The flux linkage,
O
, is defined as
I
O
N
and combining this with the relationship between flux and total circuit reluctance
tot
tot
Ni
R
R
F
I
along with (2.19) gives
tot
N
L
R
2
(2.20)
Thus inductance can be increased by increasing the number of turns, using a metal core with a higher
permeability, reducing the length of the metal core, and by increasing the crosssectional area of the
metal core. This is information that is not readily seen by either the circuit laws for inductors, or
through Faraday's equation alone.
Mutual Inductance
For a magnetic circuit containing two coils and an air gap such as in Figure 8, with each coil
wound such that the flux is additive, then the total magnetomotive force is given by the sum of
contributions from the two coils as
2
2
1
1
i
N
i
N
F
(2.21)
i
i
2
1
Area A
c
l
c
I
g
O
1
O
2
Figure 8. Mutual inductance magnetic circuit.
The equivalent circuit is shown in Figure 9.
1 9
F
2
R
g
R
lc
F
1
I
Figure 9. Equivalent circuit for Figure 8.
If the permeability of the core is large such that
g
l
c
R
R
and the crosssectional area of the gap
is assumed equal to the crosssectional area of the core (A
g
= A
c
), then
g
A
i
N
i
N
c
0
2
2
1
1
P
I

(2.22)
The flux linkage,
O
1
, in Figure 8 is
I
O
1
1
N
, or
2
12
1
11
2
0
2
1
1
0
2
1
1
1
i
L
i
L
i
g
A
N
N
i
g
A
N
N
c
c
¸¸¹
·
¨¨©
§
¸¸¹
·
¨¨©
§
P
P
I
O
(2.23)
where
¸¸¹
·
¨¨©
§
g
A
N
L
c
0
2
1
11
P
(2.24)
is the selfinductance of coil 1 and L
11
i
1
is the flux linkage of coil 1 due to its own current i
1
. The
mutual inductance between coils 1 and 2 is
¸¸¹
·
¨¨©
§
g
A
N
N
L
c
0
2
1
12
P
(2.25)
and L
12
i
2
is the flux linkage of coil 1 due to current i
2
in the other coil. Similarly, for coil 2
2
22
1
21
2
0
2
2
1
0
2
1
2
2
i
L
i
L
i
g
A
N
i
g
A
N
N
N
c
c
¸¸¹
·
¨¨©
§
¸¸¹
·
¨¨©
§
P
P
I
O
(2.26)
where L
21
= L
12
is the mutual inductance and L
22
is the selfinductance of coil 2.
¸¸¹
·
¨¨©
§
g
A
N
L
c
0
2
2
22
P
(2.27)
1 10
2.4
Magnetic Material Properties
A simplification we have used is that the permeability of a given material is constant for different
applied magnetic fields. This is true for air, but not for magnetic materials. Materials that have a
relatively large permeability are ferromagnetic materials in which the magnetic moments of the atoms
can align in the same direction within domains of the material when and external field is applied. As
more of these domains align, saturation is reached when there is no further increase in flux density of
that of free space for further increases in the magnetizing force. This leads to a changing
permeability of the material and a nonlinear B vs. H relationship as shown in Figure 10.
H
B
Figure 10. Nonlinear B vs H normal magnetization curve.
When the field intensity is increased to some value and is then decreased, it does not follow the
curve shown in Figure 10, but exhibits hysteresis as shown by the abcdea loop in Figure 11. The
deviation from the normal magnetization curve is caused by some of the domains remaining oriented
in the direction of the originally applied field. The value of B that remains after the field intensity H
is removed is called residual flux density. Its value varies with the extent to which the material is
magnetized. The maximum possible value of the residual flux density is called retentivity and results
whenever values of H are used that cause complete saturation. When the applied magnetic field is
cyclically applied so as to form the hysteresis loop such as abcdea in Figure 11, the field intensity
required to reduce the residual flux density to zero is called the coercive force. The maximum value
of the coercive force is called the coercivity.
The delayed reorientation of the domains leads to the hysteresis loops. The units of BH is
3
2
m
J
m
N
meter

ampere
newtons
meter
amperes
of
units
u
HB
(2.28)
1 11
or an energy density.
H (A·t/m)
B (Wb/m
2
)
O
a
b
c
d
e
f
Residual
flux
Retentivity
Coercive force
Coercivity
Figure 11. Hysteresis loops. The normal magnetization curve is in bold.
H
B
O
a
b
c
d
e
f
Figure 12. Energy relationship for hysteresis loop per halfcycle.
1 12
Excerpt out of 207 pages
 Quote paper
 Assist. Professor Amit Sachan (Author), 2013, Power Converter: Machines and Drives, Munich, GRIN Verlag, https://www.grin.com/document/213396
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