Transformers

Chapter Two
Chapter Two
Transformers
By Yimam A.(MSc.)
April 12, 2022
By Yimam A.(MSc.) Chapter Two April 12, 2022 1 / 94

Chapter Two
Outline
1 Introduction
2 Equivalent Circuits and phasor diagrams
3 Voltage regulation and Efficiency of Transformers
4 Autotrasformer
5 Parallel operation of Singlephase Transformers
6 Three phase transformer Connection
7 Transient Inrush Current in Transformer

Chapter Two
Learning Objectives
At the end of this chapter the students should be able to understand:
What is a transformer and its necessity in power system including its basic principle.
What is the necessity of connecting transformers in parallel and how to calculate the
load sharing between transformers operated in parallel?
How to draw phasor diagrams of a transformer to represent various alternating quan-
tities ?
Why is there sudden inrush of magnetising current when a transformer is connected to
the lines although it may be at no-load?
What is an auto-transformer and how can auto-transformer is different to a potential
divider?

Chapter Two
Introduction
Introduction
Transformer is considered to be a backbone of a power system.
For generation, transmission & distribution of electric power, AC system is adopted
instead of DC system b/c voltage level can be changed comfortably by using a trans-
former.
For economic reasons, high voltages are required for transmission where as ,for safety
reasons, low voltages are required for utilization. Hence transformer is an essential part
of power system.
A transformer is a static device that transfers AC electrical power from one circuit to
the other at the same frequency but the voltage level is usually changed.

Chapter Two
Introduction
Cont...
Commonly used transformer type, depending upon voltage
Step-up transformer: If the secondary winding has more turns than the primary wind-
ing or if the secondary voltage is higher than the primary voltage.
Step-down transformer: If the secondary winding has less turns than the primary
winding or if the secondary voltage is lower than the primary voltage.

Chapter Two
Introduction
Applications
Main applications of the transformers are given below:
To change the level of voltage and current in electric power systems.
As impedance-matching device for maximum power transfer in low-power electronic
and control circuits.
As a coupling device in electronic circuits
To isolate one circuit from another, since primary and secondary are not electrically
connected.
To measure voltage and currents; these are known as instrument transformers.

Chapter Two
Introduction
Cont...
Transformers are extensively used in AC power systems because of the following reasons:
Electric energy can be generated at the most economic level (11–33 kV)
Stepping up the generated voltage to high voltage, extra high voltage EHV (voltage
above 230 kV), or to even ultra high voltage UHV (750 kV and above) to suit the
power transmission requirement to minimise losses and increase transmission capacity
of lines.
The transmission voltage is stepped down in many stages for distribution and utilisation
for domestic, commercial and industrial consumers.

Chapter Two
Introduction
Working Principle of a Transformer
Transformer in its simplest form consists of two windings insulated from each other
and wound on a common core made up of magnetic material.
The transformer winding connected to the power source is called the primary winding
or input winding.
The transformer winding connected to the loads is called the secondary winding or
output winding.
The transformer with three windings is called triple or three winding transformer. The
third winding on the transformer is called the tertiary winding.
The primary and secondary windings are not connected electrically, but coupled mag-
netically.

Chapter Two
Introduction
Cont...
When primary winding is connected to an AC source, an exciting current I1 Amp flows
through the primary number of turns N1
It will produce an alternating flux in the core which will be linked with both the primary
and secondary windings.
The flux in the core will alternate at the same frequency as the frequency of the supply
voltage.
The alternating flux cuts the primary turns N1 and produces self induced emf of E1
volts in the primary winding according to Faraday’s law of electromagnetic induction.
The induced emf in the primary winding E1 is almost equal to the applied voltage V1
and it will oppose the applied voltage.
Similarly, mutually induced emf of E2 volts is indued in the secondary winding.

Chapter Two
Introduction
Cont...
This emf can be utilized to deliver the power to load which is connected across secondary
winding.
Thus, the power is transferred from the primary to the secondary circuit by electro-
magnetic induction.
The induced emfs in the windings will also have the same frequency as the frequency
of the supply voltage.
The magnitude of emf induced in the secondary winding depends on the number of
turns N2 in the secondary winding.
So, if the number of turns in the secondary winding N2 is greater than the number of
turns in the primary winding N1 , a higher voltage can be produced.Such a transformer
is called as step-up transformer.

Chapter Two
Introduction
Cont...

Chapter Two
Introduction
Cont...
If the secondary turns are less than the primary, it leads to a lower output voltage.
Such a transformer is called step-down transformer.
However, a transformer can be used as step-up or step-down depending on the way it
is connected.
If the low voltage winding is used as primary it can be used as step-up, while it can be
used as step-down if the high voltage winding is used as primary.

Chapter Two
Equivalent Circuits and phasor diagrams
Equivalent Circuit of a Transformer
To model a real transformer accurately, we need to account for the following losses:
Copper losses(I2R) – resistive heating losses in the windings.
Eddy current losses – resistive heating losses in the core, proportional to the square
of voltage applied to the transformer.
Hysteresis losses – energy needed to rearrange magnetic domains in the core: non-
linear function of the voltage applied to the transformer.
Leakage flux – the fluxes ϕLP
and ϕLS
that escapes from the core and flux that
passes through one winding only. These escaped fluxes produce a self-inductance in
the primary and secondary coils.

Chapter Two
Exact Equivalent Circuit of a Real Transformer
Copper losses are modeled by the resistors RP and RS.
Leakage flux in a primary winding produces the voltage.
Since much of the leakage flux pass through air, and air has a constant reluctance that
is much higher than the core reluctance, the primary coil’s leakage flux is:
eLp (t) = Np
dϕLp
dt
ϕLp = ρNP ip where ρ is permeance of flux path.Therefore
eLP
(t) = Np
d
dt
(ρNP iP ) = N2
P ρ
dip
dt

Chapter Two
Cont...
Recognizing that the self-inductance of the primary coil is
LP = N2
P ρ
The induced voltages in the primary coil and secondary coil are:
eLp (t) = Lp
dip
dt
, and eLs (t) = Ls
dis
dt
The leakage flux can be modeled by primary and secondary inductors.
The magnetization current (IM ) is a current proportional (in the unsaturated region)
to the voltage applied to the core and lagging the applied voltage by 90◦, so it can be
modeled by a reactance XM connected across the primary voltage source.

Chapter Two
Cont...
The core- loss current(ih+e) is a current proportional to the voltage applied to the core
that is in phase with the applied voltage, so it can be modeled by a resistance RC
connected across the primary voltage source.
Both currents are nonlinear; therefore, XM and RC are just approximations.
However, the exact circuit is not very practical.
Therefore, the equivalent circuit is usually referred to the primary side or the secondary
side of the transformer.
Io ≪ IL
Io
∼
= 2 − 3% of full load current for typical power transformer.

Chapter Two
Cont...
Figure 1: Exact equivalent circuit of real transformer

Chapter Two
Cont...
Figure 2: Equivalent circuit referred to the primary side

Chapter Two
Cont...
Figure 3: Equivalent circuit referred to the secondary side

Chapter Two
Approximate equivalent circuit of a transformer
For many practical applications, approximate models of transformers are used.
a) Referred to the primary side.
b) Referred to the secondary side.
c) Without an excitation branch referred to the primary side.
d) Without an excitation branch referred to the secondary side.
The values of components of the transformer model can be determined experimentally by:
An open-circuit test
A short-circuit test.

Chapter Two
Cont...
Figure 4: Approximate equivalent circuit of a transformer

Chapter Two
Cont...
Figure 5: Approximate equivalent circuit with no excitation branch

Chapter Two
Determining the values of components
Open-circuit test
Full line voltage is applied to the primary side of the transformer. The input voltage,
current, and power are measured.
From this information, the power factor of the input current and the magnitude and
the angle of the excitation impedance can be determined.
To evaluate RC and XM , we determine the conductance of the core-loss resistor is:
GC =
1
RC
The susceptance of the magnetizing inductor is:
BM =
1
XM

Chapter Two
Cont...
Figure 6: Connection for transformer open circuit test

Chapter Two
Cont...
Since both elements are in parallel, their admittances add. Therefore, the total exci-
tation admittance is:
YE = GC − jBM =
1
RC
− j
1
XM
The magnitude of the excitation admittance in the open-circuit test is:
|YE| =
IOC
VOC
The angle of the admittance in the open-circuit test can be found from the circuit
power factor (PF):
cos θ = PF =
POC
VOCIOC

Chapter Two
Cont...
In real transformers, the power factor is always lagging, so the angle of the current
always lags the angle of the voltage by θ degrees. The admittance is:
YE =
IOC
VOC
< −θ =
IOC
VOC
< − cos−1
PF
The power factor angle (θ) is given by
θ = cos−1 POC
VOCIOC
Therefore, it is possible to determine values of RC and XM in the open-circuit test.

Chapter Two
Short Circuit Test
Fairly low input voltage is applied to the primary side of the transformer.
This voltage is adjusted until the current in the secondary winding equals to its rated
value. The input voltage, current, and power are again measured.
Since the input voltage is low, the current flowing through the excitation branch is
negligible; therefore, all the voltage drop in the transformer is due to the series elements
in the circuit.
The magnitude of the series impedance referred to the primary side of the transformer
is |Zsc| = Vsc
Isc
The power factor of the current is given by: PF = cos θ = Psc
VscIsc
Lagging

Chapter Two
Cont...
Figure 7: Connection for transformer short circuit test.

Chapter Two
Cont...
Therefore:
|ZSE| =
VSC < 0◦
ISC < −θ◦
=
VSC < θ◦
ISC
Since the series impedance ZSE is equal to
ZSE = Req + jXeq
ZSE = (Rp + a2
Rs) + (jXp + a2
Xs)
It is possible to determine the total series impedance referred to the primary side of
the transformer. But there is no easy way to split the series impedance into primary
and secondary components.
@ Examples

Chapter Two
Voltage regulation and Efficiency of Transformers
Voltage Regulation of Transformers
The major performance measures of a single-phase transformer are
Voltage Regulation, and
Efficiency
The choice of a transformer for a specific application depends upon the above two
performance indices.
The loads connected to the secondary of the transformers are designed to operate at
constant voltage and hence it is desired to keep the secondary terminal voltage constant.
Voltage Regulation is defined as the percentage change in the secondary terminal volt-
age when full load is thrown off at a specified power factor, keeping the primary voltage
constant.
The allowed voltage regulation in the transmission network is ±12.5% while in distri-
bution network is ±5%.

Chapter Two
Cont...
The voltage regulation is defined mathematically as
V R =
Vs,nl − Vs,fl
Vs,fl
× 100% =
VP /a − Vs,fl
Vs,fl
× 100%
In a per-unit system:
V R =
Vp,pu − Vs,fl,pu
Vs,fl,pu
× 100%
Where Vs,nl and Vs,fl are the secondary no load and full load voltages.
Note, the voltage regulation of an ideal transformer is zero.

Chapter Two
The transformer phasor diagram
To determine the VR of a transformer, it is necessary to understand the voltage drops
within it.
Usually, the effects of the excitation branch on transformer VR can be ignored and,
therefore, only the series impedances need to be considered.
The VR depends on the magnitude of the impedances and on the current phase angle.
A phasor diagram is often used in the VR determinations.
The phasor voltage Vs is assumed to be at 0◦ and all other voltages and currents are
compared to it.

Chapter Two
Cont...
Considering the diagram referred to the
secondary side and by applying the KVL,
the primary voltage is:
Vp
a
= Vs + ReqIs + jXeqIs

Chapter Two
Cont...
A transformer operating at a lagging power factor.
It is seen that
Vp
a Vs, V R 0 .
Figure 8: at a lagging power factor

Chapter Two
Cont...
A transformer operating at a unity power factor. It is seen that V R 0.
Figure 9: at a unity power factor

Chapter Two
Cont...
A transformer operating at a leading power factor. If the secondary current is leading,
the secondary voltage can be higher than the referred primary voltage; V R 0
Figure 10: at a leading power factor

Chapter Two
Derivation of the approximate equation for Vp/a
For lagging loads the vertical component of Req and Xeq will partially cancel each
other. Due to that the angle of Vp/a will be very small, hence we can assume that Vp/a
is horizontal. The approximation will be

Chapter Two
The Transformer Efficiency
The efficiency of a transformer is defined as:
η =
Pout
Pin
× 100% =
Pout
Pout + Ploss
× 100%
Note: the same equation describes the efficiency of motors and generators.
Considering the transformer equivalent circuit, we notice three types of losses:
Copper (I2
R) losses – are accounted for by the series resistance
Hysteresis losses – are accounted for by the resistor RC
Eddy current losses – are accounted for by the resistor RC

Chapter Two
Cont...
Since the output power is
Pout = VsIs cos θ
The efficiency of the transformer can be expressed by:
Pout =
VsIs cos θ
VsIs cos θ + Pcu + Pcore
× 100%
Where Pcu + Pcore + Pout = Pin
Pin − Pout = Pcu + Pcore = total losses in transformer
Pcu = copper losses ; Pcore = core losses(Hysteresis eddy current losses)
@ Examples

Chapter Two
Autotrasformer
Autotrasformer
Sometimes, it is desirable to change the voltage by a small amount (for instance, when
the consumer is far away from the generator and it is needed to raise the voltage to
compensate for voltage drops).
In such situations, it would be expensive to wind a transformer with two windings of
approximately equal number of turns.
An autotransformer (a transformer with only one winding) is used instead.
Autotransformers are single-winding transformers that output variable voltage.
Output (up) or input (down) voltage is a sum of voltages across common and series
windings.

Chapter Two
Autotrasformer
Cont...
(a) A step-up autotransformer
connection.
(b) A step-down autotransformer connec-
tion

Chapter Two
Autotrasformer
Voltage and Current relationships in an Autotransformer
There is no isolation between the two input and output windings. The two windings
are electrically connected
The voltage across the common winding is called a common voltage VC, and the current
through this coil is called a common current IC.
The voltage across the series winding is called a series voltage VSE, and the current
through that coil is called a series current ISE.
The voltage and current on the low-voltage side are called VL and IL;
The voltage and current on the high-voltage side are called VH and IH.

Chapter Two
Autotrasformer
Cont...
For the autotransformers:
VC
VSE
=
NC
NSE
=
ISE
IC
NCIC = NSEISE
VL = VC; IL = IC + ISE
VH = VC + VSE; IH = ISE
.
The voltage on the high side of the auto-
transformer is given by
VH = VC + VSE
But
VC
VSE
=
NC
NSE
and VC = VL

Chapter Two
Autotrasformer
Cont...
VH = Vc +
NSE
NC
VC
= VL +
NSE
NC
VL
=
(NSE + NC)
NC
VL
Therefore
VL
VH
=
NC
NC + NSE
The current relationship will be:
IL = IC + ISE
IL =
NSE
NC
ISE + ISE
=
NSE
NC
IH + IH
=
NSE + NC
NC
IH
IL
IH
=
NSE + NC
NC

Chapter Two
Autotrasformer
The apparent power advantage of autotransformers
Not all the power traveling from the primary to the secondary winding of the auto-
transformer goes through the windings.
As a result, an autotransformer can handle much power than the conventional trans-
former (with the same windings).
Considering a step-up autotransformer, the apparent input and output powers are:
Sin = VLIL, Sout = VHIH
It is easy show that
Sin = Sout = SIO
Where SIO is the input and output apparent powers of the autotransformer.

Chapter Two
Autotrasformer
Cont...
However, the apparent power in the au-
totransformer’s winding is
SW = VCIC
= VL(IL − IH)
= VLIL − VLIH
= VLIL − VLIL
NC
NSE + NC
= VLIL
NSE + NC − NC
NSE + NC
= SIO
NSE
NSE + NC

Chapter Two
Autotrasformer
Cont..
Therefore, the ratio of the apparent power in the primary and secondary of the auto-
transformer to the apparent power actually traveling through its windings is
apparent power rating advantage
SIO
SW
=
NSE + NC
NC
This equation describes the apparent power rating advantage of an autotransformer over
a conventional transformer.
SW is the apparent power actually passing through the windings. The rest passes from
primary to secondary parts without being coupled through the windings.
Note that the smaller the series winding, the greater the advantage!

Chapter Two
Autotrasformer
Cont...
For example, a 5 MVA autotransformer that connects a 110 kV system to a 138 kV
system would have a turns ratio of NC/NSE 110:28. Such an autotransformer would
actually have windings rated at:
SW = SIO
NSE
NSE + NC
= 500KV A
28
28 + 110
= 1015kV A
The autotransformer would have windings rated at only about 1015 kVA, while a
conventional transformer doing the same job would need windings rated at 5000 kVA.
The autotransformer could be 5 times smaller than the conventional transformer and
also would be much less expensive.
For this reason, it is very advantageous to build transformers between two nearly equal
voltages as autotransformers.

Chapter Two
Autotrasformer
Advantages of Auto-transformer over Two-winding Transformer
Quantity of conducting material required is less
Quantity of magnetic material required is less
Operate at higher efficiency
Operate at better voltage regulation

Chapter Two
Autotrasformer
Disadvantages of Auto-transformers
There is a direct electrical connection between two windings.
If there is an open circuit in the common winding, full primary voltage will be applied
to the load which may damage an equipment connected to the secondary.
There is no isolation between the primary and secondary circuit.
The effective per unit impedance is small as compared to a conventional two winding
transformer. This can produce a serious problem where series impedance is used to
limit the fault current.

Chapter Two
Autotrasformer
Applications of Autotransformer
It can be used as a voltage booster at receiving end of transmission line to compensate
the voltage drop.
It is used for starting of three-phase induction motor.
It can be used for interconnecting power systems of different voltage levels.
It is used as a variac ( variable ac) in laboratory where variable single-phase or three-
phase output is required.
@ Examples

Chapter Two
Parallel operation of Singlephase Transformers
Parallel Operation of Singlephase Transformers
The transformers are said to be connected in parallel when their primaries are connected
to a common supply and the secondaries are connected to a common load.
Figure 12: parallel operation of transformers

Chapter Two
Reasons for operating transformers in parallel
1 In case of large loads, it is not practical and economical if all the load is supplied by a
single large transformer.
2 If the load becomes grater than the capacity of already installed transformer, capacity
of the substation can be expanded by connecting a transformer in parallel with the
existing transformer.
3 The total load requirement can be supplied by number of transformers of standard size.
This reduces the spare capacity requirement.
4 There will be no interruption of supply in case of breakdown in a transformer or when
the transformer is disconnected for maintenance.

Chapter Two
Conditions for parallel operation of transformers:
The following are the essential and desirable conditions to be satisfied for parallel
operation of transformers
1 Essential conditions
1 Same polarity for single phase transformers
2 Same terminal voltage rating
3 Same phase sequence and zero phase displacement for three phase transformers
2 Desirable conditions
1 Same percentage impedance
2 Same rectance to resistance (X/R) ratio
The parallel operation between two or more transformers can be established only if the
essential conditions are satisfied.
Satisfying the necessary conditions will enable the transformers to share the common
load in proportion to their kVA ratings.

Chapter Two
1) Connection with regard to Polarity
(a) Correct connection (b) Wrong connection
Figure 13: Connection of transformers with respect to polarity

Chapter Two
2)Same Voltage Rating and Voltage Ratio
If this condition is not met, the secondary e.m.f.s will not be equal and there will be
circulating current in the loop formed by the secondaries.
This will result in the unsatisfactory parallel operation of transformers.
Let EA and EB be their no-load secondary voltages and ZA and ZB be their impedances
referred to the secondary. Assume EA EB , then at no-load, the circulating current
in the loop formed by the secondaries is
IC =
EA − EB
ZA + ZB
When the load is connected to the system, this circulating current will tend to produce
unequal loading conditions i.e., the transformers will not share the load according to
their kVA ratings. It is because the circulating current will tend to make the terminal
voltages of the same value for both transformers.

Chapter Two
3) Equal Percentage Impedance
This condition is also desirable for proper parallel operation of single-phase transform-
ers.
The parallel operation is still possible. But the power factor at which transformers
operate will be different from the power factor of load.
Therefore, transformers will not share the load in proportion to their kVA ratings.

Chapter Two
4)Same Reactance/Resistance Ratio
If the reactance/resistance ratios of the two transformers are not equal, the power
factor of the load supplied by the transformers will not be equal.
Equal Percentage Impedance (Condition 3) is much more important than Same Reac-
tance/Resistance Ratio (condition 4).

Chapter Two
Equal voltage ratio and unequal impedances
The voltage ratios of two transformers are same but their impedance triangles are not
identical in size and shape.
Hence, no-load secondary voltage of each transformer E = EA = EB
Common terminal voltage = V2
Current supplied by transformer A = IA
Current supplied by transformer B = IB
Total current = I which is lagging behind V2 by ϕ
The equivalent impedance of transformer A referred to secondary = ZA
The equivalent impedance of transformer B referred to secondary = ZB

Chapter Two
Cont...
with equal voltage ratio in the two transformers their no load emfs EAEB will be
equal.
EA = EB = E
under this condition the equivalent circuit will get simplified as shown in Figure 14
I = IA + IB
IAZA = IBZB

Chapter Two
Cont...
(a) Equivalent circuit of two transformers sup-
plying a common load (b) Simplified parallel connection
Figure 14: Parallel Connection of transformers

Chapter Two
Cont...
IAZA = IBZB = IZAB ,where ZAB is the
combined impedance of parallel combina-
tion of ZA and ZB
ZAB =
ZAZB
ZA + ZB
IAZA = IBZB = IZAB = I
ZAZB
ZA + ZB
IA = I
ZB
ZA + ZB
, IB = I
ZA
ZA + ZB
Figure 15: Phasor diagram of two transformers op-
erating in parallel

Chapter Two
Cont..
multiplying both sides of the equation for
IA by V2
V2IA = V2I
ZB
ZA + ZB
Now V2IA = SA = load kVA supplied by
transformer A and
V2I = S = total load kVA
Therefore
SA = S
ZB
ZA + ZB
and Similarly
SB = S
ZA
ZA + ZB

Chapter Two
Unequal Voltage Ratios
For unequal voltage turns ratio, if the primary is connected to the supply, a circulating
current will flow in the primary even at no load. Let
V1 be the primary supply voltage, a1 be the turns ratio of transformer A, ZA be the
equivalent impedance of transformer A (ZA = RA + jXA) referred to as secondary and
IA be the output current of transformer A
a2 be the turns ratio of transformer B, ZB be the equivalent impedance of transformer B
(ZB = RB +jXB) referred to as secondary, and IB be the output current of transformer
B.

Chapter Two
Cont...
The induced emf in the secondary of transformer A is
EA =
V1
a1
= V2 + IAZA
The induced emf in the secondary of transformer B is
EB =
V1
a2
= V2 + IBZB
Again, V2 = IZL where ZL is the impedance of the load
V2 = (IA + IB)ZL
From the above 3 Equations , we have
EA = IAZA + (IA + IB)ZL and EB = IBZB + (IA + IB)ZL

Chapter Two
Cont...
Therefore
EA − EB = IAZA − IBZB
i.e
IA =
(EA − EB) + IBZB
ZA
substituting IA in equation EB = IBZB + (IA + IB)ZL
EB = IBZB +
(EA − EB) + IBZB
ZA
ZL + IBZL
IB

ZB + ZL +
ZB
ZA
× ZL

=
EBZA − (EA − EB)ZL
ZA

Chapter Two
Cont...
i.e
IB =
EBZA − (EA − EB)ZL
ZAZB + ZL(ZA + ZB)
Similarly
IA =
EAZB + (EA − EB)ZL
ZAZB + ZL(ZA + ZB)
When the load is connected to the system, this circulating current will tend to produce
unequal loading conditions i.e., the transformers will not share the load according to
their kVA ratings. It is because the circulating current will tend to make the terminal
voltages of the same value for both transformers.
Therefore, the transformer with a smaller voltage ratio will tend to carry more than
its proper share of the load.

Chapter Two
Conclusion
If the two transformers have equal impedances (i.e., equal resistance and equal reactance)
They will share a load equally.
If their per unit impedance is same, they will share a load in proportion to their ratings.
Both the transformers will operate at the same power factor.
Currents IA and IB will be in phase and the total current I will be the arithmetic sum
of two currents.
If the two transformers have unequal impedances
Transformers will not share the load in proportion to their ratings.
Transformers will not operate at the same power factor.
Total current I will be the vector or phasor sum of two currents IA and IB .
@ Examples

Chapter Two
Three phase transformer Connection
Three phase transformer connection
The majority of the power generation/distribution systems in the world are 3-phase
systems.
Transformers for 3-phase circuits can be constructed in two ways:
1 Connect 3 single phase transformers
2 Three sets of windings wrapped around a common core.
A single three-phase transformer is lighter, smaller, cheaper, and slightly more efficient,
but using three separate single-phase transformers has the advantage that each unit in
the bank could be replaced individually in the event of trouble.
But,it is more difficult and costly to repair three-phase transformers and to transport
single large unit of three phase transformer than to transport three single phase trans-
formers individually.

Chapter Two
Three phase transformer connection
There are four major three-phase transformer connections
1 Wye-Wye (Y − Y )
2 Delta-Delta (∆ − ∆)
3 Wye-Delta (Y − ∆)
4 Delta-Wye (∆ − Y )
Three-phase transformers are less expensive than 3-single-phase transformers
because less total core material is needed for the three-phase transformer and the
packaging cost is reduced.
Additionally they take up less space, are lighter, require less on site external wiring for
installation, and more efficient than three single-phase transformers.
Single phase transformer has one voltage ratio which agrees with the turns ratio.

Chapter Two
1.Star-Star(Y-Y) Connection
The primary phase voltage is
VϕP =
VLP
√
3
The secondary phase voltage is
VLS =
√
3VϕS
The overall voltage ratio is
VLP
VLS
=
√
3VϕP
√
3VϕS
= a

Chapter Two
2.Star-Delta (Y − ∆) connection
VϕP =
VLP
√
3
VLS = VϕS
VLP
VLS
=
√
3VϕP
VϕS
=
√
3a

Chapter Two
3.Delta- Star (∆ − Y )
VϕP = VLP
VLS =
√
3VϕS
VLP
VLS
=
VϕP
√
3VϕS
=
a
√
3

Chapter Two
4.Delta-Delta (∆ − ∆)
VϕP = VLP
VϕS = VLS
VLP
VLS
=
VϕP
VϕS
= a

Chapter Two
Applications
Star-Star connections are employed in small current and high voltage transformers
Delat-Delta connected transformer is used when neither primary nor secondary requires
neutral terminal and the voltages are low and moderate.
Star-Delta connected transformer is mainly used in step down transformers, which are
located at the substation end of the transmission line
Delta-Star connected transformers are employed where it is necessary to step up the
voltage. For example, at the beginning of HV transmission system.
These connections are also very popular with distribution transformers where voltages
are stepped down to 400 V with three-phase,four-wire system.
@ Examples

Chapter Two
Waveform of magnetization current and harmonics in transformer
Magnetizing current in transformer
The core of a transformer is made of ferromagnetic material which has a non linear
magnetic property. Thus its B-H curve is non linear.
If it is assumed that magnetic saturation does not occur, the waveforms of the magne-
tization current Iµ and flux ϕ are identical.
Thus if the core is unsaturated, the magnetizing current Iµ and flux ϕ both are sinu-
soidal and in phase with each other.
However, modern transformers are designed to operate at high saturation level to reduce
the weight, size and cost of transformer.Unfortunately, this practice introduces some
troubles.
Due to the saturation of magnetic circuit of transformer, the magnetizing current can-
not be sinusoidal and it contains odd harmonics.

Chapter Two
Cont...
As the applied voltage to the primary of transformer is sinusoidal, the flux ϕ must also
be sinusoidal.
From the B-H curve of core material, the waveform of magnetization current can be
obtained graphically.
At time t1, the instantaneous value of flux is ϕ1. The corresponding value of magnetiz-
ing force is H1 Value of magnetization current I1 corresponding to H1 can be obtained.
At time t2, the instantaneous value of flux is ϕ2 and the corresponding values of mag-
netizing force and current are H2 and I2.
In this way,complete waveform of the magnetizing current can be obtained.

Chapter Two
Cont...
Figure 16: Wave form of magnetizing current

Chapter Two
Cont...
The magnetization current is symmetrical but non sinusoidal.
Fourier analysis of the current waveform reveals that the magnetizing current Iµ con-
sists of fundamental sine component along with some odd harmonic components.
Magnetizing current, Iµ = Iµ1 sin ωt + Iµ3 sin 3ωt + Iµ5 sin 5ωt + Iµ7 sin 7ωt
The third harmonic component is predominant (5 - 10 % of fundamental wave).
Neglecting higher order harmonics and considering third order harmonics, the magne-
tizing current can be written as, Iµ = Iµ1 sin ωt + Iµ3 sin 3ωt
Thus if the flux is sinusoidal, the magnetizing current contains harmonics.

Chapter Two
Cont...
(a) Magnetizing current and flux (b) Magnetizing current with harmonics

Chapter Two
Cont...
If the transformer is connected to the
sinusoidal current source, the harmonic
currents cannot be supplied by the
source.
In this case, the flux wave will fail to
reach its sinusoidal peak value and will
become flat topped.
As the flux is non sinusoidal, the induced
emf will be non sinusoidal.
Figure 18: Wave form of magnetizing current

Chapter Two
Conclusion
When the applied voltage is purely sinusoidal, flux waveform is sinusoidal and magne-
tizing current is non sinusoidal due to the presence of odd harmonic components.
When magnetizing current is sinusoidal, the flux wave is flat topped and the induced
emf in primary and secondary windings will be non-sinusoidal due to the presence of
odd harmonic components.

Chapter Two
Transient Inrush Current in Transformer
Switching Transient
The magnetization current Iµ of a transformer is 3 to 5% of its rated full load current.
This is the steady-state current.
In steady state operation, the voltage applied to primary V and flux ϕ both are sinu-
soidal.
However, when the transformer at no load is switched on to supply, an initial transient
current of high magnitude may flow through the primary winding.
This transient inrush current flows for a short duration but it may be as high as 5 to
7 times the rated current of transformer.

Chapter Two
Cont...
(a) Switching transient (b) Magnetizing current in transformer

Chapter Two
Magnetizing current in transformer:
The magnitude of inrush current depends on the instant of the voltage wave at which
the transformer is connected to the supply.
The inrush current is maximum when the transformer is switched on at the instant of
zero input voltage.
The inrush current will be minimum if the switching is done at maximum input voltage
However, it is not possible to switch on the transformer at a particular instant of time.

Chapter Two
Steady state flux in transformer
Let the supply voltage is V = Vm sin ωt
e= emf induced in primary winding and
N = the number of turns of the primary winding
Neglecting core losses and resistance of primary winding,V = −e = N dϕ
dt
dϕ =
1
N
vdt or
ϕ =
1
N
Z
vdt =
1
N
Z
Vm sin ωtdt =
Vm
N
Z
sin ωtdt
ϕ = −
Vm
Nω
cos ωt =
Vm
Nω
sin(ωt − 90◦
)

Chapter Two
Cont...
Therefore, the steady state flux is sinusoidal and its maximum value
ϕm =
Vm
Nω
However, The magnetizing current Iµ is not sinusoidal and it contains third harmonic
components due to non linear B-H curve.

Chapter Two
Steady state flux in transformer
(a) Flux and applied voltage waveform (b) Phenomenon of inrush

Chapter Two
Flux for switching of transformer at zero supply voltage
ϕ =
Vm
N
Z t
0
sin ωtdt =
Vm
Nω
(1 − cos ωt)
ϕ =
Vm
Nω
−
Vm
Nω
cos ωt = ϕm − ϕm cos ωt
Here, −ϕm cos ωt = ϕm sin(ωt − 90◦) is the steady state flux (ϕss).
ϕm is the transient flux (ϕt), which will decay according to the circuit time con-
stant(L/R).
Resultant flux, ϕ = ϕm − ϕm cos ωt
When ωt = π, resultant flux, ϕ = 2ϕm. The resultant flux at the instant of switching
has zero value.
The resultant flux will reach at a maximum value of 2ϕm (doubling effect).

Chapter Two
Cont...
It is assumed that the initial flux at the
instant of switching has zero value.
However, the initial flux may not be zero
but it will have some residual value ϕr
because of retentivity.
Therefore, the transient flux ϕt will now
be even more severe and ϕt = ϕm + ϕr
The resultant flux will now go through a
maximum value of (2ϕm + ϕr)
Figure 21: Flux for switching of transformer at
zero supply voltage

Chapter Two
Exciting current for switching of transformer at zero voltage
As the peak value of flux is 2ϕm , the core goes to deep saturation region of magneti-
zation.
Thus the corresponding value of magnetizing current has a very high peak value.
It can be noted that the transient flux and current in the initial stage of transient are
unidirectional.
However, this large inrush current will quickly decay because of the winding resistance.
The normal magnetizing current under steady state will be small(3 to 5% of the full
load current).

Chapter Two
Cont...
(a) Exciting current for switching of transformer (b) Inrush current waveshape

Chapter Two
Factors affecting inrush current and its effects
Various factors affecting the magnitude of inrush current are:
The value of residual flux in the transformer core.
The non-linear magnetizing characteristics of transformer core.
The phase of supply voltage at the instant of switching of transformer.
Effects of inrush current:
The inrush current may be as large as 100 times the normal magnetizing current or 5
to 10 times the rated load current.
This high inrush current will produce large electromagnetic forces on the windings
which can be 25 times more than normal.
Therefore, the windings of large transformer must be strongly braced.
Improper operation of protective devices like unwarranted tripping of relays,
It can also introduce power quality issues.

Chapter Two
Question
Thank You!

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