ELEC 3105 BASIC EM AND
POWER ENGINEERING
Losses in Transformers
Ideal / Real Transformer
Some common applications
Some common issues with
transformers
TRANSFORMER LOSSES
There are two dominant loss mechanisms in transformers
Eddy currents
Hysteresis
EDDY CURRENTS
Lenz’s Law

 
emf   v  B   d 

motional emf
The induced emf always
opposes the change in
flux
• Move loop towards magnet
• B increases in loop
• Flux increases in loop
• Current induced
• Current produces magnetic field in loop
• This magnetic field in opposite direction to magnetic field of magnet

B
Move loop
m a g n et

B

v
I
N
S
lo o p
The changing magnetic field on the
conducting sheet will induce current in the
sheet. These currents are known as Eddy
currents. Energy loss occurs through
Joule’s heating.
The changing flux induces an
emf about the loop and as a
result a current flow in the loop.
Provided the loop resistance is
not infinite “or zero” power loss
will be observed through Joule’s
heating.
2
I R
EDDY CURRENTS
B increasing on sheet
I direction such as to create B in opposite
direction to magnets B as expected from
Lenz’s law.
Magnet slides along sheet. In the direction
of motion the electrons of the sheet are
first introduced to an increasing magnetic
field. I induced opposes this increase.
Once the magnet has passed over the
electrons, they will then be subjected to a
decreasing magnetic field and as such will
induce a current which attempts to
reinforce the dropping external magnet
field.
EDDY CURRENTS
The Eddy Current Story
• Suppose a current i1 in the primary is increased
• Magnetic field increases
• Flux in the core increases
• This induces an emf in the core
• This sets up a counter current which tries to keep B constant
• These currents are Eddy Currents.
The Eddy Currents heat the core. This heat energy must be
extracted from the electric energy supplied by the primary.
EDDY CURRENTS
Eddy currents are reduced
by lamination of the core.
Lamination breaks the
Eddy Current paths
EDDY CURRENTS
Eddy current loss
Eddy current losses occur whenever the core material is electrically conductive.
Most ferromagnetic materials contain iron: a metal that has fairly low resistivity
(roughly 10-7 Ω m). The problem is intuitively obvious if you consider that the
magnetic field is contained within a 'circuit' or loop formed by the periphery of the
core in the same way as it is contained within a turn on the windings. Around that
periphery a current will be induced in the same way as it is in an ordinary turn
which is shorted at its ends. What is needed, then, is some method of increasing the
resistance of the core to current flow without inhibiting the flow of magnetic flux. In
mains transformers this is achieved by alloying the iron with about 3% of silicon.
This lifts the resistivity to 4.5×10-7 Ω m. Depending
upon the amount of silicon this material is called
'transformer iron', 'electrical iron' or 'armature iron'.
The alternative name 'silicon steel' is a misnomer
because steel is iron alloyed with carbon; and carbon
does no good in a transformer core. The silicon does,
though, increase the mechanical hardness of iron in
the same way as carbon - try sawing up a transformer
core and you'll discover this quickly.
EDDY CURRENTS
Eddy current loss
In any resistive circuit the power is proportional to the square of the applied
voltage. The induced voltage is itself proportional to f × B and so the Eddy losses
are proportional to f2B2. The flux is also related to the size of the loop. Figure PLM
shows how the idea of lamination is used to reduce the power losses caused by
eddy currents in mains transformers. The same principle applies to motors and
generators too. Using a solid iron core (as in cross-section B) results in a large
circulating current. So, instead, the core is made up of a stack of thin (~0.5
millimeter) sheets (cross section C). Here I have shown only four laminations but
there will normally be many more. The lines of magnetic flux can still run around
the core within the plane of the laminations. The
situation for the Eddy currents is different. The
surface of each sheet carries an insulating
oxide layer formed during heat treatment.
This prevents current from circulating from one
lamination across to its neighbors.
EDDY CURRENTS
Eddy current loss
Clearly, the current in each lamination will be less than the very large current we had with the solid core;
but there are more of these small currents. So have we really won? The answer is yes, for two reasons.
Power loss (the reduction of which is our aim) is proportional to the square of induced voltage. Induced
voltage is proportional to the rate of change of flux, and each of our laminations carries one quarter of
the flux. So, if the voltage in each of our four laminations is one quarter of what it was in the solid core
then the power dissipated in each lamination is one sixteenth the previous value. Hurrah!
But wait; it gets better. Look at the long thin path that the Eddy current takes to travel round the
lamination. Suppose we made the laminations twice as thin (we halved d1). The path length of the current
isn't much changed; it's still about 2×d2. However, the width of the path has halved and therefore its
resistance will double and so the current will be halved. The bottom line is that Eddy current loss is
inversely proportional to the square of the number of laminations.
This idea of dividing up the iron into thin sections is carried a
stage further in the iron dust cores. Here the iron is ground into a
powder, mixed with some insulating binder or matrix material
and then fired to produce whatever shape of core is required.
These cores can function at several megahertz but their
permeability is lower than solid iron.
EDDY CURRENTS
Note on expression for loss
2
I R
JE
E
2
f 2 B 2
When alternating field is present at frequency f
E
dB
 fB
dt
Then
 E  f B
2
2
2
As the frequency increases so do the power losses. Can fix this
using ferrites
2
2
P  kf B
e
Ferrites
EDDY CURRENTS
EDDY CURRENTS
Ferrites
Ferrites
EDDY CURRENTS
Ferrites
EDDY CURRENTS
EDDY CURRENTS
The following are examples of induced currents which
are normally not identified as Eddy Currents
• Opposing magnetic force due to induced currents in linear motors.
• Braking action due to Eddy Currents in a pendulum.
• Opposing force due to induced currents in MHD generator.
• Induced currents in plasma engine.
• Currents due to back emf in a motor.
• Currents produced by generator.
EDDY CURRENTS
• Opposing magnetic force due to induced currents in linear motors.
EDDY CURRENTS
• Braking action due to Eddy currents in a pendulum.
EDDY CURRENTS
• Opposing force due to induced
currents in MHD generator.
MHD = magnetohydrodynamic
• Induced currents in plasma engine.
EDDY CURRENTS
EDDY CURRENTS
• Currents due to back emf in a motor
EDDY CURRENTS
EDDY CURRENTS
• Currents produced by generator.
EDDY CURRENTS
Eddy Currents
A magnet falls more slowly through a metallic tube than it
does through a nonmetallic tube.
When a magnet is dropped down a metallic tube, the changing
magnetic field created by the falling magnet pushes electrons in
the metal tube around in circular, eddy-like currents. These eddy
currents have their own magnetic field that opposes the fall of the
magnet. The magnet falls dramatically slower than it does in
ordinary free fall in a nonmetallic tube.
Material
• A cow magnet or neodymium magnet.
• A nonmagnetic object, such as a pen or a pencil.
• One 3 foot (90 cm) length of aluminum, copper, or brass tubing (do not use
iron!) with an inner diameter larger than the cow magnet and with walls as thick as
possible.
• One 3 foot (90 cm) PVC or other nonmetallic tubing.
• Optional: 2 thick, flat pieces of aluminum (available at hardware and homerepair stores); cardboard; masking tape; rubber bands or cord.
Assembly
•None required.
EDDY CURRENTS
To do and notice:
Hold the metal tube vertically. Drop the cow magnet through the tube. Then drop a nonmagnetic object, such as a
pen or pencil, through the tube. Notice that the magnet takes noticeably more time to fall. Now try dropping both
magnetic and nonmagnetic objects through the PVC tube.
In addition to dropping these objects through the tubes, a very simple, visible, and dramatic demonstration can be
done by merely dropping the magnet between two thick, flat pieces of aluminum. The aluminum pieces should be
spaced just slightly farther apart than the thickness of the magnet. A permanent spacer can easily be made with
cardboard and masking tape if you don't want to hold the pieces apart each time. Rubber bands or cord can hold
the pieces all together. The flat surfaces need to be only slightly wider than the width of the magnet itself.
Thickness, however, is important. The effect will be seen even with thin pieces of aluminum, but a thickness of
about 1/4 inch (6 mm) will produce a remarkably slow rate of fall. Allow at least a 6 inch (15 cm) fall.
What's going on
As the magnet falls, the magnetic field around it constantly changes position. As the
magnet passes through a given portion of the metal tube, this portion of the tube
experiences a changing magnetic field, which induces the flow of eddy currents in an
electrical conductor, such as the copper or aluminum tubing. The eddy currents create a
magnetic field that exerts a force on the falling magnet. The force opposes the magnet's
fall. As a result of this magnetic repulsion, the magnet falls much more slowly.
EDDY CURRENTS
Eddy currents are often generated in transformers and
lead to power losses. To combat this, thin, laminated strips of
metal are used in the construction of power transformers, rather than
making the transformer out of one solid piece of metal. The thin strips
are separated by insulating glue, which confines the eddy currents to
the strips. This reduces the eddy currents, thus reducing the power
loss.
With the new high-strength neodymium magnets, the effects of eddy
currents become even more dramatic. These magnets are now
available from many scientific supply companies, and the price has
become relatively affordable. (An excellent source is Dowling Miner
Magnetics Corp., P.O. Box 1829, Sonoma, CA 95476. )
ELEC 3105 BASIC EM AND POWER
ENGINEERING
Hysteresis Losses
Hysteresis
In a previous lecture it was
stated that the area of the
hysteresis loop represents the
energy dissipated as heat.
The hysteresis power loss for
one cycle can be expressed
as:
Ferromagnetic materials
P W f
SOFT and HARD Ferromagnetic materials
h
Soft ::: transformer cores, solenoids, ….
Hard :: permanent magnets
Area of hysteresis loop
is equivalent to energy
lost in one cycle.
h
where f is the cycle
frequency and Wh is the
energy dissipated in
each cycle.
49
We want to find an expression for Wh.
Hysteresis
First consider the work which
must be done by the power source
supplying the primary to increase
the magnetic field in the core by
dB.
We shall use the model device a toroidal coil with an iron core.
I  I  dI
H  H  dH
B  B  dB
t  t  dt
Increase the current in
time interval dt
Then by Lenz’s law an
electromotive force will be
induced in the winding tending to
oppose the increase in current.
Hysteresis
d
V  N
dt
To increase the current the generator must
furnish energy in the amount of.
W  VIdt  NId
I  I  dI
H  H  dH
B  B  dB
t  t  dt
Taking B as constant over the cross-section
of the toroid core, an expression for the
flux can be written.
  BA
d  AdB
Hysteresis
From Ampere’s law we can obtain
an expression for the magnetic
field inside the toroid core.
NI
H
2r
To increase the current the generator must
furnish energy in the amount of.
W  2rAHdB
where the volume of the toroid is:
I  I  dI
H  H  dH
B  B  dB
t  t  dt
vol  2rA
Hysteresis
Now we can sum “integrate” over
the initial magnetization part of the
hysteresis curve in order to obtain
the work done by the generator in
establishing
the
maximum
magnetic flux density in the core
Bmax
Bmax
W  vol  HdB
0
This integral is the shaded area of the
above curve multiplied by the volume
“vol” of the toroid’s core.
Hysteresis
The work done in one cycle, Wh,
is the striped area of the hysteresis
cycle. The cross hatched area is
the work returned to the source.
Consider area 0abc
In order for the external source to increase the H
field from a to a value corresponding at b, it must do work on
the ferromagnetic specimen equivalent to the area 0abc.
Consider area bcd
When H is reduced back to zero, the ferromagnetic
material will do work on the source equivalent to the area
bcd, because H remains positive while dB changes sign.
W
 w   HdB
vol
Bmax
0
Hysteresis
W
 w   HdB
vol
Bmax
0
W 
h
Ferromagnetic materials
SOFT and HARD Ferromagnetic materials
Area of Hysteresis loop
Power loss
P W f  f 
h
h
Area of Hysteresis loop
Soft ::: transformer cores, solenoids, ….
Hard :: permanent magnets
Area of hysteresis loop
is equivalent to energy
lost in one cycle.
49
FERROMAGNETIC MATERIALS
Hysteresis
ELEC 3105 BASIC EM AND POWER
ENGINEERING
START Losses in
Transformers again
!!!!
TRANSFORMER LOSSES
There are two dominant loss mechanisms in transformer
Eddy currents
P  kB f
2
2
Hysteresis
e
P W f
h
h
Total
core
loss
P  W f  kB f
2
he
2
h
k constant that depends on geometry
ELEC 3105 BASIC EM AND POWER
ENGINEERING
Ideal / Real Transformer
IDEAL TRANSFORMER
What would the ideal transformer look like?
• PERFECT IMPEDANCE TRANSFORMATION AT ALL FREQUENCIES
L ,L ,M  
• NO LOSS
1
2
• PERFECT FLUX COUPLING
• INFINITE CORE PERMEABILITY
• TRANSFORMER DRAWS NO CURRENT WHEN SECONDARY IS OPEN
Sorry: We must deal with real transformers.
REAL TRANSFORMER
Consider:

v
i
1  k L
2
1
i
1
1
1  k L
i
1
i
2
a
2
i
a

v
2
kL
1
2


Ideal transformer
Magnetization inductance
Leakage inductance
v  i j 1  k L   i  i  jkL
a

v  jL i  jMi
2
1
1
1
1
1 1
1
2
Equivalent circuit reproduces basic
transformer equation
1
REAL TRANSFORMER
Add losses:
1  k L
1
i
2

v
1

v


a
2
Resistance of secondary
Ideal transformer
Eddy currents and Hysteresis losses in core
Resistance of primary
TRANSFORMER EFFICIENCY
Ideal transformer
𝑃𝑜𝑢𝑡 = 𝑃𝑖𝑛
Efficiency = 100 %
Real transformer
𝑃𝑜𝑢𝑡 ≠ 𝑃𝑖𝑛
Efficiency < 100 %
𝜂 = 100%
𝜂 < 100%
𝑃𝑖𝑛 = 𝑃𝑜𝑢𝑡 + 𝑃𝑐 + 𝑃𝑒
Copper loss in winding resistance
𝜂=
Eddy current & core loss
𝑃𝑜𝑢𝑡
100%
𝑃𝑜𝑢𝑡 +𝑃𝑐 +𝑃𝑒
Real transformer
VOLTAGE REGULATION
Observed voltage drop when we draw a current from the secondary
% 𝑅𝑒𝑔𝑢𝑙𝑎𝑡𝑖𝑜𝑛 =
𝑉𝑠 𝑛𝑜 𝑙𝑜𝑎𝑑 −𝑉𝑠 𝑙𝑜𝑎𝑑𝑒𝑑
𝑉𝑠 𝑙𝑜𝑎𝑑𝑒𝑑
Real transformer
100%
ELEC 3105 BASIC EM AND POWER
ENGINEERING
Transformer core saturation
42
CORE SATURATION
Hysteresis
At a sufficiently high H
the core saturates and B
is essentially constant.
In Lecture 21 it was stated that the
area of the hysteresis loop represents
the energy dissipated as heat.
The hysteresis power loss for
one cycle can be expressed as:
Ferromagnetic materials
P W f
SOFT and HARD Ferromagnetic materials
h
h
where f is the cycle frequency
and Wh is the energy dissipated
in each cycle.
Soft ::: transformer cores, solenoids, ….
Hard :: permanent magnets
Area of hysteresis loop
is equivalent to energy
lost in one cycle.
The usual design principle is to
have B = Bsat at the voltage peaks
in the primary
We want to find an expression
for Wh.
49
(minimizes the amount of iron
needed in the core)
Recall the Hysteresis curve
B versus H
43
CORE SATURATION
Recall for the primary that:
d
v 
dt
1
1
Then at maximum
v
1 , max
v
1 , peak
d

dt
1 , max
Area of the core
 B A
sat
Frequency
44
CORE SATURATION
 B A
v
1 , peak
sat
If v1 goes beyond this range,
however, H is above Hsat, and the
effective inductance seen at the
primary becomes small. The flux
is also not well-confined to the
core.
L
Because of the reduced inductance, the current in the primary becomes large in these peak
parts of the cycle.
t
i
1
45
CORE SATURATION
v
1 , peak
 B A
sat
If v1 goes beyond this range,
however, H is above Hsat, and the
effective inductance seen at the
primary becomes small. The flux
is also not well-confined to the
core.
L
The voltage in the secondary also ceases to be a clean sine wave, and acquires harmonics.
t
v
2
46
CORE SATURATION
Hysteresis
In Lecture 21 it was stated that the
area of the hysteresis loop represents
the energy dissipated as heat.
The hysteresis power loss for
one cycle can be expressed as:
Ferromagnetic materials
The saturation problem is
also the reason for not using
very low frequencies. The
impedance in the primary
becomes small as the
frequency becomes small, so
that the current becomes
large and the core saturates.
P W f
SOFT and HARD Ferromagnetic materials
h
h
where f is the cycle frequency
and Wh is the energy dissipated
in each cycle.
Soft ::: transformer cores, solenoids, ….
Hard :: permanent magnets
Area of hysteresis loop
is equivalent to energy
lost in one cycle.
At a sufficiently high H the
core saturates and B is
essentially constant.
We want to find an expression
for Wh.
49
Recall
47
Z  jL
L
ELEC 3105 BASIC EM AND POWER
ENGINEERING
Transformer and the power
grid
48
TRANSFORMER AND THE POWER GRID
Typical power requirement for a small city: 1000MW
Maximum voltage provided by typical generator: 30 KV
Thus need a current of 3.3*104 A
Suppose generator is 100 km from city.
  100km
a  2.5cm
Copper transmission line 2.5 cm in radius has resistivity
49
  1.7cm
TRANSFORMER AND THE POWER GRID
  100km
a  2.5cm
Copper transmission line 2.5 cm in radius has resistivity
  1.7cm
Resistance over length of transmission line length
R
Power loss in wire conduit

A


 0.86
a
2
I R  950MW
2
Leaves about 50 MW for city.
Need very large core to handle this current without saturation.
50
TRANSFORMER AND THE POWER GRID
  100km
a  2.5cm
Solution is to step up the voltage and thus reduce the current and
power loss in copper conduit.
Typical AC high voltage line: 765 kV
Current required: I = 1000 A
Power loss in wire conduit
I R  0.86MW
2
Leaves about 999.14MW for city.
51
GEOMAGNETIC STORMS AND CORE SATURATION
1989
http://www.geolab.emr.ca/geomag/e_gic_history.html - year1989
A great magnetic storm occurred on 13 March 1989, that caused a nine-hour blackout of the
21,000 MV Hydro Québec, power system. A vivid description of that failure has been
provided by G. Blais and P. Metsa (1993) of Hydro Québec:
"Telluric currents induced by the storm created harmonic voltages and currents of
considerable intensity on the La Grande network. Voltage asymmetry on the 735-kV
network reached 15%. Within less than a minute, the seven La Grande network static var
compensators on line tripped one after the other....With the loss of the last static var
compensator, voltage dropped so drastically on the La Grande network (0.2 p.u.) that all
five lines to Montréal tripped through loss of synchronism (virtual fault), and the entire
network separated. The loss of 9,450 MV of generation provoked a very rapid drop in
frequency at load-centre substations. Automatic underfrequency load-shedding controls
functioned properly, but they are not designed for recovery from a generation loss
equivalent to about half system load. The rest of the grid collapsed piece by piece in 25
seconds."
Many other power utilities in North America experienced problems ranging from minor
voltage fluctuations to tripping out of lines and capacitors. A summary of these effects and
the times of their occurrence is given by Cucchi and Ponder (1991).
52
GEOMAGNETIC STORMS AND
CORE SATURATION
Auroral electrojet 106 A
53
GEOMAGNETIC STORMS AND CORE SATURATION
Auroral electrojet fluctuates by 105 A in one minute
 I
 10 A
B 

 6.6 10 T
2r 2 150 10 m
5
o
6
o
3
emf 

 110V
t
Change in flux through circuit
  6.6 10 T 100 10 m 1000 10 m 
6
3

B
Auroral electrojet 106 A
High voltage line
Generating station
100km
Consuming station
54
1000km
3
GEOMAGNETIC STORMS AND CORE SATURATION
Auroral electrojet fluctuates by 105 A in one minute
emf 

 110V
t
At this frequency transformer coils look like short circuits
R 1
A DC current of

B
100 A
is produced
Auroral electrojet 106 A
High voltage line
Generating station
100km
Consuming station
55
1000km
GEOMAGNETIC STORMS AND CORE
SATURATION
Geomagnetic Effects on Power Systems - English
Geomagnetic Storms
Sun article
Sun
56
97.315 BASIC EM AND POWER ENGINEERING
Lecture 27
END Transformer core saturation
START Transformer application
57
TRANSFORMER 120V AC TO 5V DC
N
1
N
2
R
G
5V DC
v
20 : 1
G
In order to convert AC line power to low voltage DC, a
step down transformer is used first, followed by a
rectifier bridge, followed by coarse filtering, followed
by limiting.
R
L
58
A Thin Film Piezoelectric Transformer for Microelectronic
Applications
With the onset of miniaturization, many applications in the electronics industry now require small, low profile components with a high
efficiency of operation. Electromagnetic transformers, having thousands of wire turns around a ferrite core, have become an obstacle
to the progress of miniaturization, as they are among the most bulky devices on a circuit board. Piezoelectric transformers have
recently received some attention as a possible alternative. A piezoelectric transformer essentially consists of two mechanically coupled
and electrically insulated piezoelectric resonators (Fig. 1). When an electrical signal near the frequency of mechanical resonance is
applied to the input section of the transformer, strong mechanical vibration occurs due to the piezoelectric converse effect. This
vibration is transferred to the output section, inducing a voltage on its electrodes due to the piezoelectric direct effect, with a
consequent voltage gain.
It is now possible to manufacture piezoelectric films of sufficient thickness and quality to make thin film piezoelectric transformers
through the use of a composite film technology developed here at Queen's University. A ring transformer with vibrational motion in
the radial direction has been established as the best candidate for a transformer through use of Mason's model for piezoelectric
transformers, predicting a operating range of 1-10 MHz with voltage gains of 0.1-100 and a maximum efficiency of operation of 90%
(depending on the dimensions and quality of the film). Current work has been focused on optimizing the material parameters and the
poling procedure of the piezoelectric composite films, as well as developing a method of characterization for the piezoelectric
parameters.
59
60
Slides not used after this one
TRANSFORMER APPLICATIONS
• The transformer is usually the most efficient way to convert AC
voltages.
• This can be done by impedance matching.
• Impedance matching is the term used for achieving maximum
power transfer to a given load from a generator.
• Quite often the generator is set as is the load.
• For maximum power transfer we can use a transformer to convert
the power from the generator to the load as shown below.
N
i
1
1
v
G
R
Consider:
G
2

v

v


1
62
N
2
i
2
R
L
Consider:
TRANSFORMER
APPLICATIONS
N
i
1
R
v
G
G
2

v

v


2
i 
and
2
2
1
1
2
2
1
L
1
L
We also know that
we can write
from Ohm’s law
2
2
1
2
1
equate
1
R
L
v v
i 
R
G
1
1
2
1
2
v
N v

R N R
Assuming large self inductance,
N  v
N
i  i   
N
N  R
i
2
1
N
v 
v
N
We have
N
1
L
G
63
Consider:
TRANSFORMER
APPLICATIONS
N
i
1
R
v
G
G
N
1
2

v

v


i
2
R
L
2
1
Thus
N

N
 v v v


R
 R
2
2
1
1
G
1
L
vR
v 
N 
  R  R
N 
G
2
1
 N

 N
G
 1 1
v

 v 
R
 R R 
2
2
G
1
1
L
G
L
2
G
In terms of
L
this is ……
v
2
1
64
G
Consider:
TRANSFORMER
APPLICATIONS
N
i
1
R
v
G
G
In terms of
this is ……
N
1

v

v


v 
2
2
i
2
R
L
2
1
v
2
N
v
N
2
1
1
v 
2
vR
G
L
N
N
R 
R
N
N
2
1
G
1
We want to design a transformer to have a maximum power transfer to the load.
Thus:
L
2
2
v
P
 maximum
2R
2
L
L
65
and …….
Consider:
TRANSFORMER
APPLICATIONS
N
i
1
R
v
G
G
N
1
2

v

v


i
2
R
L
2
1
2
v
P
 maximum
2R
2
L
L




1 
vR

P 
2R  N R  N R 
N

N


G
2
2
1
kR v
P 
2 k R  2kR R  R
L
L
L
L
L
2
1
2
G
1
G
2
G
G
L
L
2
with
66
N
k  
N
2
1



2
…….
2
L
TRANSFORMER
APPLICATIONS
Consider:
N
i
1
R
v
G
G
N
1
2

v

v


i
2
R
L
2
1
Differentiating with respect to k and setting to zero will give the condition for maximum transfer to the load.
2
Rv
k R  2kR R  R   kR v
dP
00 2
kR  R 
dk
L
G
2
2
G
2
G
L
L
L
2
G
kR
2
G
R R
G
L
4
G
with
67
L
N
k  
N
2
1



2
…….
L

TRANSFORMER
APPLICATIONS
Consider:
N
i
1
1
R
v
G
G
N
2

v

v


1
i
2
R
L
2
Simplifying
2
Rv
kR  R   kR v
0
2
Rv
kR R v

2
2
L
G
G
G
L
R
k
R
L
G
L
2
2
2
G
L
G
with
L
N
k  
N
2
G
2
1
R 
G
N
R

N
R
2
L
1
G
This is the transformer turns ratio which gives maximum
power transfer to the load.



2
gives
68
TRANSFORMER APPLICATIONS
Consider:
N
i
1
1
R
v
G
G
N
2

v

v


1
If we convert the load resistance to the apparent generator side resistance, we
get:
i
2
R
L
2
Power transformation
Input admittance
P 
2
v
R
rms
P
L
N 
 v 
N 

P  
N

2
R 
N


2
2
2
v
2 R
1
Lecture 25 Slide 34
L
1
2
2
2
2
1
L
2
2
N
R  R
N
1
L
2
Equivalent circuit for the primary
2
Gives
P 
2
v
P
2R
2
j L
1
2
L
N
R 
R  R
L
N
1
1
L
2
2
L
L
2
34
L
L
2
 R 

 R  R
 R 
2
Maximum power transfer condition to load
v
G
L
G
G
L
69
R
G
R  R
L
G