Download ELECTRIC CIRCUITS I

ELECTRICAL TECHNOLOGY
EET 103/4
• Define and analyze the principle of
transformer, its parameters and structure.
• Describe and analyze Ideal transformer,
equivalent circuit, and phasor diagram
• Calculate and justify efficiency, losses,
performance
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TRANSFORMERS
(CHAPTER 22)
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FUNCTION OF A TRANSFORMER
The main function of an electrical power transformer
is to transfer electrical energy from one side
(primary) to the other side (secondary). The
secondary current and voltage may or may not be at
the same level as that of the primary current and
voltage. The energy is transferred by means of
magnetic coupling. The magnetic flux produced by
the current in primary winding links the secondary
winding. Since the flux varies with time, this flux
linkage results in an induced voltage in the
secondary winding. If the secondary winding is
terminated with a load, the induced voltage will drive
a secondary current through the load.
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22.1 Introduction
 This chapter covers:
 Mutual inductance that exists between
coils of the same or different dimensions.
 Basic to the operation of a transformer
 Three of the basic operations of a
transformer.
 step up/down
 impedance matching
 isolation
 The dot convention.
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22.2 Mutual Inductance
 Transformers are constructed of two coils
placed so that the charging flux developed
by one will link the other.
 The coil to which the source is applied is called
the primary coil.
 The coil to which the load is applied is called the
secondary coil.
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22.2 Mutual Inductance
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22.2 Mutual Inductance
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22.2 Mutual Inductance
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22.2 Mutual Inductance
• Primary transformer formula for voltage using
Faraday’s Law:
• Voltage induced across the primary and selfinductance is directly related to the number of
turns in the coil and the rate of change of
magnetic flux linking the primary coil:
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22.2 Mutual Inductance
• The magnitude of eS , the voltage inducted
across the secondary is determined by;
where NS is the number of turns in the
secondary winding and m is the portion of
the primary flux  P that links the
secondary winding
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22.2 Mutual Inductance
If all of the flux linking the primary links the
secondary:
The coefficient of coupling (k) between
two coils is determined by;
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22.2 Mutual Inductance
Since the maximum level of m is P , the
coefficient of coupling between two coils can
never be greater than 1.
Coils with low coefficients of coupling are
termed loosely coupled.
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22.2 Mutual Inductance
Mutual inductance between two coils is
proportional to the instantaneous change in
flux linking one coil due to an instantaneous
change in the current through the other coil.
In terms of inductance of each coil and the
coefficient of coupling, the mutual inductance
is:
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22.3 The Iron-core Transformer
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22.3 The Iron-core Transformer
The iron core will serve to increase the
coefficient of coupling between the coils by
increasing the mutual flux m..
The magnetic flux lines will always take the path of
least reluctance, which in this case is the iron core.
When the current Ip through the primary circuit of the
ion-core transformer is a maximum, the flux m
linking both coils is also a maximum.
The flux is directly proportional to the current
through the winding.
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22.3 The Iron-core Transformer
If;
i p  2 I p sin t
then;
m   m sin t
By Faraday’s law;
ep  N p
Or;
d p
dm
d
 Np
 N p  m sin t 
dt
dt
dt

e p  N p m cos t  N p m sin t  90

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22.3 The Iron-core Transformer
The effective value of ep is;
Ep 
N p  m
2
 4.44 fN p  m
Since the flux linking the secondary flux equals that of the
primary;
Es 
N s  m
2
 4.44 fN s  m
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22.3 The Iron-core Transformer
Ep
Dividing;
Es
Np
Ns

Np
Ns
 a  Transforma tion ratio
For and ideal transformer;
Hence;
Vg
VL

E p  Vg
and
Es  VL
Np
Ns
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22.3 The Iron-core Transformer
• The ratio of the magnitudes of the induced
voltages is the same as the ratio of the
corresponding turns:
• The ratio Np/Ns , usually represented by the
lowercase letter a, is referred to as the
transformation ratio.
– If a < 1, the transformer is a step-up transformer.
– If a > 1, the transformer is a step-down
transformer.
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22.3 The Iron-core Transformer
Example22.2
(a) Find the maximum flux m.
(b) Find the secondary turns Ns.
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22.3 The Iron-core Transformer
Example22.2 – Solution
(a)
E p  4.44 fN p  m
m 
Ep
4.44 fN p
200

 15.02 mWb
4.44  60  50
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22.3 The Iron-core Transformer
Example22.2 – Solution (cont’d)
(b)
Ep
Es

Ns 
Np
Ns
N p Es
Ep
50  2400

 600 turns
200
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22.3 The Iron-core Transformer
• The primary and secondary current of a
transformer are related by the inverse ratio of
the turns:
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22.4 Reflected Impedance and Power
The impedance of the primary circuit of an
ideal transformer is related to the impedance
load by the transformation ratio
If the load is capacitive or inductive, the reflected
impedance will also be capacitive or inductive.
For an ideal iron-core transformer;
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22.4 Reflected Impedance and Power
Example22.3
(a) Find Ip and Vg.
(b) Find Zp.
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22.4 Reflected Impedance and Power
Example22.3 – Solution
(a)
Ip
Ns

Is N p
Ns
Ip 
Is
Np
5
  0.1  12.5 mA
40
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22.4 Reflected Impedance and Power
Example22.3 – Solution (cont’d)
(b)
Z p  a2ZL
Np
40
a

8
Ns
5
Z p  82  2 k  128 k
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22.4 Reflected Impedance and Power
Example 22.4
For the residential supply,
determine (assuming a
totally resistive load) the
following :
a. The value of R to
ensure a balanced load
b. the magnitude of I1 and
I2
c. The line voltage VL
d. The turn ratio a=Np/Ns
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22.4 Reflected Impedance and Power
Example22.4 - Solution
a. PT = (10) (60) W+ 200 W + 2000 W
= 600 W + 200 W + 2000 W = 2800 W
Pin = Pout
VpIp=VsIs=2800 W (purely resistive load)
(2400 V) Ip= 2800 W and Ip=1.17 A
R=Vphase/Ip=2400 V/1.17 A = 2051.28 ohms
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22.4 Reflected Impedance and Power
Example 22.4 – Solution (cont’d)
b. P1=600 W = VI1= (120V) I1
And I1= 5 A
P2= 2000 W = VI2= (240W) I2
And I2= 8.33 A
c. VL  3V  1.73(2400V )  4152 V
d. a 
Np
Ns

Vp
Vs

2400 V
 10
240 V
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22.6 Equivalent Circuit (Ironcore Transformer)
For the non ideal or practical iron-core
transformer, the equivalent circuit appears
below.
Part of the equivalent circuit includes an ideal
transformer.
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22.6 Equivalent Circuit (Ironcore Transformer)
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22.6 Equivalent Circuit (Ironcore Transformer)
All elements in the iron-core transformer other
than the ideal transformer are the elements of
the transformer that contribute to the nonideal
characteristics of the device.
Resistance Rp and Rs are simply the geometric
resistance of the primary and secondary windings.
Leakage flux Lp and Ls is a small amount of flux that
links each coil but does not pass through the core.
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22.6 Equivalent Circuit (Ironcore Transformer)
The resistance RC represents the hysteresis and
eddy current losses (core losses) within the core
due to an ac flux through the core.
The inductance Lm (magnetizing inductance) is the
inductance associated with the magnetization of the
core, that is, the establishing of the flux m in the
core.
The capacitances CP and CS and the lumped
capacitances of the primary and secondary circuits,
respectively, and Cw represent the equivalent
lumped capacitance between the windings of the
transformer.
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22.6 Equivalent Circuit (Ironcore Transformer)
For power transformer;
• Winding capacitances have negligible effects
because of the low operating frequency.
• Under normal operating condition, normally i’p >> im.
Hence, Cp, Cs, Cw, RC and Lm may be omitted.
The approximate equivalent circuit becomes
as follows;
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22.6 Equivalent Circuit (Ironcore Transformer)
Reduced equivalent circuit for the nonideal ironcore transformer.
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22.6 Equivalent Circuit (Ironcore Transformer)
Normally, Lp and Ls are represented by their equivalent
reactances;
X p  L p
and
Xp
X s  Ls
Xs
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22.6 Equivalent Circuit (Ironcore Transformer)
Xp
Xs
Xs and Rs may be reflected into the primary circuit
using the relationship;
2
Rp ' a Rs
and
X p ' a 2 X s
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22.6 Equivalent Circuit (Ironcore Transformer)
The equivalent circuit now becomes;
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22.6 Equivalent Circuit (Ironcore Transformer)
Combining the series resistances and reactances,
results in the following circuit;
Re  R p  R p '
Xe  X p  X p'
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22.6 Equivalent Circuit (Ironcore Transformer)
The load resistance RL may also be reflected into the
primary circuit;
RL ' a 2 RL
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22.6 Equivalent Circuit (Ironcore Transformer)
By voltage divider rule;


Ri
aVL  
 Vg
 Re  Ri   jX e 
Ri  a 2 RL
is the load resistance reflected into the
primary circuit of the transformer
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22.6 Equivalent Circuit (Ironcore Transformer)
Example 22.7
(a) Determine Re and Xe.
(b) Determine the magnitude VL and Vg.
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22.6 Equivalent Circuit (Ironcore Transformer)
Example 22.7 – Solution
(a)
Re  Rp  a 2 Rs  1  22 1  5 
X e  X p  a 2 X s  2  22  2  10 
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22.6 Equivalent Circuit (Ironcore Transformer)
Example 22.7 – Solution
(b)
aVL  I p a 2 RL  2400 V
2400
VL 
 1200 V
a

Vg  I p Re  a 2 RL  jX e

 100 5  240  j10  2450  j100 V
Vg  2452 V2.34
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