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Chapter 9
Ideal Transformer
Lecture 04
Electro Mechanical System
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Ideal Transformer
Introduction
 Transformers are one of the most useful electrical
devices ever invented.
 It can raise or lower the voltage or current in an ac
circuit.
 It can isolate circuits from each other.
 It can change the apparent value of impedance.
 The transformer enables us to transmit electrical
energy over great distances.
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Eelements of transformer
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Transformer application
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Voltage Induction
 For a coil consisting of N turns placed in a variable
flux , the flux alternates sinusoidally at a frequency f,
periodically reaching positive and negative max. The
flux induces a sinusoidal ac voltage.
d
e(t )  N
dt
The rms value of voltage E is given by
E  2fN max  E  4.44 fN max
Where:
E = effective voltage induced [V]
f = the frequency of flux [Hz]
N = number of turns.
max is the peak value of flux [Wb]
(The reason for using peak flux is that it is proportional to peak flux
density Bmax which in iron cores, determine the value of saturation.)
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Applied Voltage & Induced Voltage
 Consider a coil connected across an AC voltage source Eg.
 The coil and source resistances are negligible.
 The induced voltage E must equal the source voltage Eg.
 A sinusoidal AC flux  must exist to generate the induced voltage
on the N turns of the coil.
 max varies in proportion to Eg.
 Placing an iron core in the coil will not change the flux .
E g  4.44 fN max
 max 
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Eg
4.44 fN
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Applied Voltage & Induced Voltage
 Magnetization current Im drives the AC flux
 The current is 90° out-of-phase and lagging with respect to the voltage.
 To produce the same flux a smaller magnetomotive force is needed
with an iron core than an air core. Less magnetizing current is
required. So with an iron core, less current is needed to drive the
AC flux.
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Applied Voltage & Induced Voltage
Example
A coil, having 4000 turns, links an AC flux with a
peak value of 2mWb at a frequency of 60 Hz
Calculate the rms value of the induced voltage
What is the frequency of the induced voltage?
E = 4.44 f N max
E = 4.44 x 60 x 4000 x 0.002
E = 2131 V
The induced voltage has rms value of 2131 V and a
frequency of 60 Hz.
The peak voltage is: 2131 x √2 = 3014 V
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Applied Voltage & Induced Voltage
Example:
A coil, having 90 turns, is connected to a 120 V, 60 Hz
source the rms magnetization current is 4 A. Find:
a) Peak value of the flux.
c) Inductive reactance of the coil.
a)
Φmax 
b)
I max 
Eg
4.44 fN
2I 

b) Peak value the mmf.
d) Inductance of the coil.
120
 0.005  5mWb
4.44  60  90
2  4  5.66 A
Peak mmf U  NIm  90  5.66  509.1A (The flux is equal to
5mWb at an instant wh en coil mmf is 509.1 ampere - turns)
E g 120
 30

c) X m 
4
Im
Xm
30
 0.0796  79.6mH

d )L 
2 60
2f
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Elementary Transformer
 Consider an air-core coil,
excited by an AC source Eg,
draws a magnetization current
Im, produces a total flux 
 A second coil is brought close
to the first
 a portion  m1 of the flux
couples the second coil, the
mutual flux
 an AC voltage E2 is induced
 the flux linking only the first
coil is called the leakage
flux,  f1
 Improved flux coupling
 concentric windings, iron core
 weak coupling causes small E2
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Elementary Transformer
 The magnetization current
Im produces both fluxes
m1 and f1
 The fluxes are in-phase
 The voltages Eg and E2
are in-phase
 Terminal orientation such
that the
 coil voltages are inphase and are said to
possess the same
polarity
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