1. CHAPTER 2
TRANSFORMER
2.1 INTRODUCTION
• Transformer allow voltage level to be changed throughout the electrical distribution
system so that the most economical voltage can be used in each part of system .
• Generators are limited to about 25kV due to the size of insulation required, but
transmission losses at 25kV would be unacceptable.
• Thus, voltage are stepped up for transmission.
• Transformer allow the voltage level to be changed to the most economical level.
• Compared to rotating machines, the transformer is relatively simple.
• Transformer comprises two or more electric circuit coupled by magnetic circuit.
Figure 2.1: An autotransformer with a sliding brush contact
17

2. 2.2 TRANSFORMER CONSTRUCTION
Primary Secondary coil
coil
Figure 2.2
Figure2.3: An ideal step-down transformer showing magnetic flux in the core
• Consists of two or more electrical windings that are linked together by a magnetic
field.
• Except for special-purpose transformers, the coupling is enhanced with a
ferromagnetic core.
• When AC voltage is applied to the primary winding, magnetic flux is established,
which links the secondary winding.
• If the flux is sinusoidal, a sinusoidal voltage will be induced in the secondary.
18

3. • The primary is the side that is connected to the source (generator or power system),
and the secondary is the side that is connected to the load.
• Either side may be the high-voltage or low-voltage side.
Figure 2.4: Step up transformer (hydro station)
Core Material
• The core constructed of a high-permeance (low-reluctance) material to minimize the
magnetixing current.
• To keep eddy current losses down, the core is made of laminations, the thickness of
which is inversely proportional to the rated frequency of the transformer.
• Eddy current losses are proportional to the lamination thickness squared.
• Thus, halving the thickness would reduce the eddy current losses by 75%.
Figure 2.5: Three-phase pole-mounted step-down transformer.
19

4. Core Configuration
• There are two types of transformers cores are used:-
(i) Core type
(ii) Shell type
• In the core type, there are winding on each leg of the core (the winding surround the
core).
• In the shell type, the winding are on the center leg of the core, and the core surrounds
the windings.
Figure 2.6. Two types of transformer core.
Conductors
• Copper provides the best conductivity and, therefore the minimum volume for the
coil.
• Aluminum is usually used to reduce the cost.
• The conductor must carry current without overheating.
• The conductors may be round, square, or rectangular, and there may be several
conductors in parallel to reduce the I2R losses.
20

5. • The operating temperature of the coil is extremely important because the insulation
may deteriorate at increased temperature and the resistance of the coil also increases
with temperature.
Figure 2.7: Three-phase oil-cooled transformer with cover cut away
2.3 IDEAL TRANSFORMER OPERATION
• A transformer makes use of Faraday's law and the ferromagnetic properties of an iron
core to efficiently raise or lower AC voltages.
• It of course cannot increase power so that if the voltage is raised, the current is
proportionally lowered and vice versa.
Figure 2.8
21

6. Faraday's Law
• Any change in the magnetic environment of a coil of wire will cause a voltage (emf)
to be "induced" in the coil.
• No matter how the change is produced, the voltage will be generated.
dΦ
Faraday’s law: e = N where N = number of turns
dt
dΦ
= rate of change of flux
dt
e = instantaneous induced voltage
e(t ) = 2 Erms sin( wt )
dΦ e
=
dt N
t
1
Φ (t ) = ∫ 2 Erms sin( wt )dt
N0
− 2 Erms
Φ (t ) = cos( wt )
wN
− 2 Erms
Φ (t ) = cos( wt )
2πfN
Φ (t ) = −Φ max cos( wt )
2πfNΦ max
Therefore Erms = = 4.44 fNΦ max
2
Voltage
• For an ideal transformer, all the flux is confined to the iron core and thus links both
the primary and secondary.
• Therefore: Ep = 4.44fNpΦmax
Es = 4.44fNsΦmax
Ep Np E1 N1
= =a = =a
Es Ns E2 N 2
• A transformer is called a step-down transformer if the primary side has more turns
than the secondary side (a>1).
• A transformer is called a step-up transformer if the primary side has fewer turns than
the secondary side (a<1).
• A step-down transformer has the high voltage facing the power system and low
voltage facing the load.
22

7. • The opposite is true for the step-up transformer.
Current
• Because the losses are zero in the ideal transformer, the apparent power in and out of
transformer must be the same:
│Sin│=│Sout│=│Vp││Ip│=│Vs││Is│
Ip Vs Ns 1 I1 V2 N2 1
Therefore = = = = = =
Is Vp Np a I2 V1 N1 a
• The ratio of the currents is the inverse of the voltage ratio.
• If we raise the voltage level to a load with a step-up transformer, then the secondary
current drawn by the load would have to be less than the primary current, since the
apparent power is constant.
Impedance
Vs
Zs =
Is
• From Ohm’s Law, Vp
Vp
ZS = a =
aI p a (a ) I p
Zp
Zs =
a2
2
Zp  Np 
=a =
N
2


Zs  s 
Example 1
The nameplate on a single-phase, step-down transformer indicates that is rated
33.33kVA,
7967V/120V. Find the rated current on the primary and secondary sides.
Solution
23

8. S 33.33kVA
Is = = = 277.8 A
Vs 120V
S 33.33kVA
Ip = = = 4.184 A
Vp 7967V
Example 2
An ideal, step-down transformer has 1500 turns in the primary coil and 75 turns in the
secondary coil. If 2400 V is applied to the primary, what is the voltage on the secondary?
Solution
Vp Np
=
Vs Ns
Ns  75 
Vs = V p = 2400  = 120V
Np  1500 
Example 3
An ideal, step-down transformer has 1500 turns in the primary coil and 75 turns in the
secondary coil. If 10.4 A flows in the primary of the transformer, calculate the load
current.
Solution
Ip Ns
=
Is Np
Np  1500 
Is = I p = 10.4  = 208 A
Ns  75 
Example 4
A transformer is rated 75kVA, 7200/240 or 120V. Find the rated current on each side of
the transformer. Find the exciting current if it is 1.7% of rated current.
Solution
The rated current on each side of the transformer :-
24

9. S rated 75kVA
Ip = = = 10.42 A
Vp 7200V
S rated 75kVA
I s120V = = = 625 A
Vs120 120V
S rated 75kVA
I s 240V = = = 312.5 A
Vs 240 240V
The exciting current is on the primary side, so:-
I ex = 0.017Ip = 0.017(10.42) = 0.177A
Example 5
A transformer is rated 75kVA, 7200/240V. A load having an impedance of 1.0Ω is
connected to the secondary.
(i) What is the current on each side?
(ii) What is the power delivered to the load?
(iii) What is the input impedance on the primary side?
Solution
(i) V = IR
V 240V
Is = s = = 240 A
R 1Ω
Ip N
= s
Is N p
240
Ip = 240 = 8 A
7200
(ii) P = I 2 R = 240 2 (1) = 57.6kW
2
Zp  Np 
(iii) = 
Zs  Ns



2
 7200 
Zp =   (1Ω) = 900Ω
 240 
25

10. 2.4 PRACTICAL TRANSFORMER
The Equivalent Circuit
• All transformer have winding resistance, a core with finite permeability, leakage flux
and hysterisis and eddy current losses and are thus nonideal.
• The transformer as Figure 2.9 which represented by the core and coils, is still not
ideal because it has magnetizing current flowing into it.
Figure 2.9
• Adding the magnetizing branch as shown in Figure 2.10, the transformer is now
become an ideal transformer.
Figure 2.10
Rp jXp a:1 jXs Rs
jXm Rm 1.0k
1.0m
1.0m
1.0m
Figure 2.11:Equivalent T-circuit
26

11. 2 2
Rp jXp jXs '=a jXs Rs '=a Rs a:1
jXm Rm 1.0k
1.0m
Figure 2.12: T-circuit referred to the primary
Rp jXp
Rp'= 2
jXp'= 2
a a jXs Rs
a:1
TR1 jXm Rm 1.0k
1.0k
1.0m
1.0m
1.0m
Figure 2.13: T-circuit referred to the secondary
Reflection In Transformer
• One simplification that introduces only a small error is to move the magnetizing
branch to the primary terminals and then combine the primary and secondary
resistance and leakage reactance.
• The combination of the winding resistance is called the equivalent resistance, and the
combination of the leakage reactance is called the equivalent reactance.
• The combination can be referred to one side only, either primary or secondary.
• The following Figure 2.14 and 2.15, illustrates transformer reflection.
Re q p jXe q p
a:1
1.0k
1.0m
1.0k
1.0m
jXm Rm
Figure 2.14: Cantilever circuit referred to the primary
Req,p = Rp + a2Rs = Rp+Rs’
Xeq,p = Xp + a2Xs = Xp+Xs’
27

12. Re q s jXe q s
a:1
jXm Rm 1.0m
1.0m
1.0k
1.0k
Figure 2.15: Cantilever circuit referred to the secondary
Req,s = (Rp/a2 ) + Rs = R’p + Rs
Xeq,s = (Xp/a2 ) + Xs = X’p+ Xs
Figure 2.16
Example 6
A step-down transformer is rated 100kVA, 7200/277 V and has the following equivalent
circuit parameter:
Rp = 2.92Ω Rs = 0.00432Ω Rm = 51840Ω
Xp = 14.6Ω Xs = 0.0216Ω Xm = 12960Ω
28

13. Find the equivalent winding impedance referred to the high voltage side. Repeat for the
low voltage side.
Solution
Equivalent impedance referred to the high/primary side:
Vp 7200
a= = = 25.99
Vs 277
Req , p = R p + a 2 Rs
Req , p = 2.92 + (25.99) 2 (0.00432) = 5.839Ω
X eq , p = X p + a 2 X s
X eq , p = 14.6 + (25.99) 2 (0.0216) = 29.19Ω
Z eq , p = (5.839 + j 29.2)Ω
Equivalent impedance referred to the low/secondary side:
a = 25.99
Rp
Req ,s = 2 + Rs
a
2.92
Req ,s = 2
+ (0.00432) = 8.643x10 −3 Ω
25.99
Xp
X eq ,s = + Xs
a2
14.6
X eq ,s = + (0.0216) = 0.0432Ω
25.99 2
Z eq , s = (0.86 + j 4.3) x10 −2 Ω
Example 7
A step-down transformer is rated 140kVA, 400/1000 V and has the following equivalent
circuit parameter:
Rp = 10Ω Rs = 50Ω Rm = 5000 Ω
Xp = 20Ω Xs = 80Ω Xm = 12000Ω
29

14. Draw the equivalent winding impedance referred to the high voltage side and low voltage
side.
Solution
N1 400
a= = = 0.4
N 2 1000
Refer to the high voltage side/secondary side
R1 10
Req = 2 + R2 = + 50 = 112.5Ω
a 0.4 2
X 20
X eq = 21 + X 2 = + 80 = 205Ω
a 0.4 2
Z eq = 112.5 + j 205Ω
112.5 ohm j205 ohm
1,0m
1,0k
Refer to the low voltage side/primary side
Req = R1 + a 2 R2 = 10 + 0.4 2 (50) = 18Ω
X eq = X 1 + a 2 X 2 = 20 + 0.4 2 (80) = 32.8Ω
Z eq = 18 + j 32.8Ω
18 ohm j32.8 ohm
1,0m
1,0k
30

15. 2.5 DETERMINING CIRCUIT PARAMETER
There are 2 simple test that can provide the data required to calculate values for the
elements of the transformer equivalent circuit:-
(i) Short-Circuit Test
(ii) Open-Circuit Test
(i) Short-Circuit Test (normally at low voltage side)
Rp jXp jXs ' Rs '
HV side jXm Rm 1.0k
1.0m
1.0k
Figure 2.17: Short-circuit test arrangement
• One-side of the transformer is shorted, and voltage is applied on the other side until
rated current flows in the windings.
• Measured the
1. applied voltage (Vsc),
2. winding current (Isc),
3. input power (Psc)
• Generally, the low-voltage side of the transformer is shorted and voltage is applied to
the high-voltage side.
• The measurements are used to calculate Req and jXeq which referred to the high
voltage side.
• With the low-voltage winding shorted, the input impedance is:-
Zin = (Rp + jXp) + [(R’s + jX’s)║[Rm║jXm]] ,without ignored Rm and jXm
• If Rm and jXm>>>R’s and jX’s, therefore Rm and jXm ignored.
Zin = (Rp + jXp) + [(R’s + jX’s)
• From short circuit test, Vsc, Isc and Psc are measured. Therefore:-
Vsc
│Z eq │= │Z sc │=
Isc
R eq = Psc/(I2sc)
X eq =√ (│Z eq│2 – R2eq)
31

16. (ii) Open-Circuit Test(normally at high voltage side)
Rp jXp jXs ' Rs '
LV side jXm Rm 1.0k
1.0m
1.0k
Figure 2.18: Open-circuit test arrangement
• High-voltage side of the transformer is opened and rated voltage is applied to the
low-voltage side.
• Readings of Voc, Ioc and Poc are taken on the low voltage side. Then Rm and Xm
can be calculated.
• Measured the
1. applied voltage (Voc),
2. winding current (Ioc),
3. input power (Poc)
• Therefore Rm and Xm which referred to the low voltage side can be calculate.
• The input impedance during the open-circuit test is the primary winding in series with
the exciting branch:
Zin = (Rp + jXp) + (Rm║jXm)
• The leakage reactance and winding resistance on the primary side can be negligible as
they are too low compare to Rm and Xm,
Rm and jXm>>>Rp and jXp,
Therefore, Zin = (Rm║jXm)
• From open-circuit test, Voc, Ioc and Poc are measured. Therefore:-
Poc = VocIoccosθoc
θoc = cos-1(Poc/IocVoc)
IR = Ioccos θoc
IX = Iocsin θoc
32

17. Rm and Xm can be calculated by:
Rm = Voc/IR
Xm = Voc/IX
Example 8
Short-circuit and open-circuit tests were performed on a 100kVA transformer, rated
7200V/277V, with the results listed below. Assuming step-down operation, determine the
equivalent circuit parameters of the transformer referred to the high voltage side.
Vsc = 414 V Voc = 277 V
Isc = 13.89 A I oc = 14.88 A
Psc = 1126 W Poc = 1000 W
Solution
Short circuit(HV) (Req,Xeq)
Vsc 414
Zeq = = = 29.8Ω
I sc 13.89
Psc 1126W
Req = = = 5.836Ω
I sc 13.89 2
2
2 2
Xeq = Z eq − Req = 29.23Ω
Open circuit(LV) (Rm,Xm)
 P   1000W 
θ oc = cos −1  oc  = cos −1 
V I   277(14.88)  = 75.96°

 oc oc   
I R = I oc cos θ oc = 14.88(cos 75.96°) = 3.61A
I x = I oc sin θ oc = 14.88(sin 75.96°) = 14.435 A
V 277
Rm = oc = = 76.73Ω
I R 3.61
V 277
X m = oc = = 19.19Ω
I x 14.435
33

18. Referred to the high voltage side/primary side
Req ,hv = 5.836Ω
X eq ,hv = 29.23Ω
Rm ,lv = 76.73Ω
2
 7200 
Rm ,hv =   (76.73Ω) = 51.84kΩ
 277 
X m ,lv = 19.19Ω
2
 7200 
X m ,hv =   (19.19Ω) = 12.97 kΩ
 277 
Example 9
Short-circuit and open-circuit tests were performed on a 100kVA transformer, 50 Hz,
rated at 120V/2400V, and the results are listed as follows:
Vsc = 40 V Voc = 120 V
Isc = 41.67 A I oc = 6 A
Psc = 380 W Poc = 40 W
(i) Draw the equivalent circuit with the necessary parameters of the transformer referred
to the low voltage side.
(ii) Draw the equivalent circuit with the necessary parameters of the transformer referred
to the high voltage side
Solution
Short circuit(HV) (Req,Xeq)
Vsc 40
Zeq = = = 0.96Ω
I sc 41.67
Psc 380W
Req = = = 0.22Ω
I sc 41.67 2
2
2 2
Xeq = Z eq − Req = 0.96 2 − 0.22 2 = 0.93Ω
34

19. Open circuit(LV) (Rm,Xm)
 P   40W 
θ oc = cos −1  oc  = cos −1 
V I   120(6)  = 86.82°

 oc oc   
I R = I oc cos θ oc = 6(cos 86.82°) = 0.33 A
I x = I oc sin θ oc = 6(sin 86.82°) = 5.99 A
V 120
Rm = oc = = 363.64Ω
I R 0.33
V 120
X m = oc = = 20.03Ω
Ix 5.99
(i) Referred to the low voltage side/primary side
Rm ,lv = 363.64Ω
X m,lv = 20.03Ω
Req ,hv = 0.22Ω
2
 120 
Req ,lv =  (0.22Ω) = 0.55mΩ
 2400 
X eq ,hv = 0.93Ω
2
 120 
X eq ,lv =  (0.93Ω) = 2.325mΩ
 2400 
The equivalent circuit:-
0.55m ohm j2.325m ohm
1,0m
1,0k
363.64 ohm 1,0k
j20.03 ohm 1,0m
35

20. (ii) Referred to the high voltage side/secondary side
Req ,hv = 0.22Ω
X eq ,hv = 0.93Ω
Rm,lv = 363.64Ω
2
 2400 
Rm,hv =  (363.64Ω) = 145456Ω
 120 
X m,lv = 20.03Ω
2
 2400 
X m,hv =   (20.03Ω) = 8012Ω
 120 
0.22 ohm j0.93 ohm
1,0m
1,0k
145 456 ohm 1,0k
j8012 ohm 1,0m
36

21. 2.6 TRANSFORMER LOSSES
Generally, in any machine there will be two types of losses namely:
(i) iron losses
P iron = IRm2Rm = VocIoccosθoc
(ii) copper losses
P copper = I12R1 + I22R2
= I12R eq,p = I22Req,s
2.7 TRANSFORMER EFFICIENCY
η = OutputPower =
InputPower
Pout
=
Pin −Ploss
Pin Pin
Output Power , P out = VAcosθ2
Input Power , P in = Output Power + Total Losses
Total Losses , P loss = P copper losses + P core losses
Energy dissipated in Hysteresis and
the resistance of eddy current losses
winding
2.8 VOLTAGE REGULATION
The purpose of voltage regulation is actually to see what is the voltage being dropped in
the secondary winding between no-load and full-load condition.
Vnl − Vfl
VR =
Vnl
37

22. If transformer equivalent circuit referred to:-
(i) primary side,
R 01 X01
1 2
aV2
V 1 − aV 2
VR =
V1
(ii) secondary side,
R 02 X02
1 2
V2
V1 − V 2
a
VR =
V1
a
Example 10
By referring to the Example 9,calculate the terminal voltage, V1 and voltage
regulation,
VR if a load at 0.8 power factor lagging is connected to 2400 V side. (neglect the
magnetizing impedance).
Solution
Load is connected to secondary side:-
38

23. S 100kVA
I2 = = = 41.67∠ − cos −1 (0.8) = 41.67∠ − 36.87° = 33.34 − j 25 A
V2 2400
120
a= = 0.05
2400
If referred primary side:
I1 0.55m ohm j2.325m ohm I2'=I2/a
V1 V2'=aV 2
1,0m
1,0k
I2
V 1 = I 2' (0.55m + j 2.325m) + V 2' = (0.55m + j 2.325m) + aV 2
a
(33.34 − j 25)
= (0.55m + j 2.325m) + 0.05(2400)
0.05
= 121.53 + j1.28V = 121.54∠0.6°V
V 1 − aV 2 121.54 − 0.05(2400)
VR = = x100% = 1.27%
V1 121.54
If referred to secondary side:
I1'=aI1 0.22 ohm j0.93 ohm I2
V1'=V1/a V2
1,0m
1,0k
V 1' = I 2(0.22 + j 0.93) + V 2
= (33.34 − j 25)(0.22 + j 0.93) + 2400
= 2430.58 + 25.51
V 1 = V 1' (a) = 0.05( 2430.58 + 25.51)
= 121.53 − j1.28V = 121.54∠0.6°V
V 1 − V 2 121.54 − 2400
a 0.05
VR = = x100% = 1.27%
V1 121.54
a 0.05
39

24. 2.9 AUTOTRANFORMERS
OR
1.0m
1.0m
LOAD LOAD
Figure 2.19: Autotransformer schematic
• Figure 2.19 show the autotransformer schematic.
• Autotransformer is a transformer with only one winding.
• The low-voltage coil is essentially placed on top of the high-voltage coil and is called
the series coil.
• The connection is called an autotransformer and can be used as a step-up or a step-
down transformer.
• The advantages of autotransformer are:-
(i) cheaper
(ii) more efficient, because losses stay the same while the rating goes up
compared to a conventional transformer
(iii) lower exciting current
(iv) better voltage regulation
• The disadvantages of autotransformer are:-
(i) larger short circuit current available
(ii) no isolation between the primary and secondary
40

25. Example 11
A 220/440V, 25kVA and 50 Hz transformer is connected as an autotransformer to
transform 660V to 220V.
(i) Determine the ratio ‘a’.
(ii) Determine the kVA rating of the auto transformer.
(iii) With a load of 25 kVA, 0.8 lagging power factor connected to 220 V terminals,
determine the currents in the load and the two transformer windings.
Solution
I1
1,0m
1,0m
1,0m
N1 I2
V1=660V
N2 I2-I1 V2=220V
N 1 660
(i) a= = =3
N 2 220
25kVA
(ii) I1 = = 37.87 A
660
660(37.87)
kVAauto = = 25kVA
1000
25kVA
(iii) I2 = = 113.64 A
220
I 2 − I 1 = 113.64 − 37.87 = 75.77 A
41

26. 2.10 THREE PHASE TRANSFORMERS
Introduction:
• Three phase transformers are used throughout industry to change values of three
phase voltage and current.
• Since three phase power is the most common way in which power is produced,
transmitted, an used, an understanding of how three phase transformer
connections are made is essential.
Construction:
• A three phase transformer is constructed by winding three single phase
transformers on a single core.
• These transformers are put into an enclosure which is then filled with dielectric
oil.
• The dielectric oil performs several functions.
• Since it is a dielectric, a nonconductor of electricity, it provides electrical
insulation between the windings and the case.
• It is also used to help provide cooling and to prevent the formation of moisture,
which can deteriorate the winding insulation.
Connections:
• There are only 4 possible transformer combinations:
1. Delta to Delta - use: industrial applications
2. Delta to Wye - use : most common; commercial and industrial
3. Wye to Delta - use : high voltage transmissions
4. Wye to Wye - use : rare, don't use causes harmonics and balancing problems.
• Three-phase transformers are connected in delta or wye configurations.
• A wye-delta transformer has its primary winding connected in a wye and its
secondary winding connected in a delta (see Figure 2.20).
• A delta-wye transformer has its primary winding connected in delta and its
secondary winding connected in a wye (see Figure 2.21).
42

27. Figure 2.20:Wye-Delta connection
Figure 2.21:Delta-Wye connection
Delta Conections:
• A delta system is a good short-distance distribution system.
• It is used for neighborhood and small commercial loads close to the supplying
substation.
• Only one voltage is available between any two wires in a delta system.
• The delta system can be illustrated by a simple triangle.
• A wire from each point of the triangle would represent a three-phase, three-wire
delta system.
• The voltage would be the same between any two wires (see Figure 2.22).
Figure 2.22
43

28. Wye Connections:
• In a wye system the voltage between any two wires will always give the same
amount of voltage on a three phase system.
• However, the voltage between any one of the phase conductors (X1, X2, X3) and
the neutral (X0) will be less than the power conductors.
• For example, if the voltage between the power conductors of any two phases of a
three wire system is 208v, then the voltage from any phase conductor to ground
will be 120v.
• This is due to the square root of three phase power.
• In a wye system, the voltage between any two power conductors will always be
1.732 (which is the square root of 3) times the voltage between the neutral and
any one of the power phase conductors.
• The phase-to-ground voltage can be found by dividing the phase-to-phase voltage
by 1.732 (see Figure 2.23).
Figure 2.23
Connecting Single-Phase Transformers into a Three-Phase Bank:
• If three phase transformation is need and a three phase transformer of the proper
size and turns ratio is not available, three single phase transformers can be
connected to form a three phase bank.
• When three single phase transformers are used to make a three phase transformer
bank, their primary and secondary windings are connected in a wye or delta
connection.
• The three transformer windings in Figure 2.24 are labeled H1 and the other end is
labeled H2.
• One end of each secondary lead is labeled X1 and the other end is labeled X2.
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29. Figure 2.24
• Figure 2.25 shows three single phase transformers labeled A, B, and C.
• The primary leads of each transformer are labeled H1 and H2 and the secondary
leads are labeled X1 and X2.
• The schematic diagram of Figure 2.24 will be used to connect the three single
phase transformers into a three phase wye-delta connection as shown in Figure
2.26.
Figure 2.25
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30. Figure 2.26
• The primary winding will be tied into a wye connection first.
• The schematic in Figure 2.24 shows, that the H2 leads of the three primary
windings are connected together, and the H1 lead of each winding is open for
connection to the incoming power line.
• Notice in Figure 2.26 that the H2 leads of the primary windings are connected
together, and the H1 lead of each winding has been connected to the incoming
primary power line.
• Figure 2.24 shows that the X1 lead of the transformer A is connected to the X2
lead of transformer c.
• Notice that this same connection has been made in Figure 2.26.
• The X1 lead of transformer B is connected to X1, lead of transformer A, and the
X1 lead of transformer B is connected to X2 lead of transformer A, and the X1
lead of transformer C is connected to X2 lead of transformer B.
• The load is connected to the points of the delta connection.
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31. Tutorial 2
1. A step-down transformer is rated 25kVA, 660-240 V and has the following equivalent
circuit parameter:
Rp = 15Ω Rs = 10Ω Rm = 50000Ω
Xp = 20Ω Xs = 8Ω Xm = 13000Ω
Find the equivalent winding impedance referred to the high side and the low side.
2. Short-circuit and open-circuit tests were performed on a 10kVA transformer,50 Hz,
rated 400V-240V, with the results listed below.
Vsc = 10 V Voc = 240 V
Isc = 10kVA/400V=25A I oc = 4 A
Psc = 120 W Poc = 80 W
(i) Draw the equivalent circuit parameters of the transformer referred to the high
side.
(ii) Draw the equivalent circuit parameters of the transformer referred to the low side.
3. The coil possesses 400 turns and links an ac flux having peak value of 2mWb. If the
frequency is 60Hz, calculate the induced voltage E.
4. A transformer having 90 turns on the primary and 2250 turns on the secondary is
connected to a 120V, 60Hz source. Calculate:
(i) The effective voltage for the secondary side.
(ii) The peak voltage for the secondary side.
5. An ideal-transformer having 90 turns on the primary and 2250 turns on the secondary
is connected to a 200V, 50 Hz source. The load across the secondary draws a current of
2A at a power factor of 0.8 lagging. Calculate:
(i) The effective value of the primary current.
(ii) The primary current if the secondary current is 100mA.
(iii) The peak flux linked by the secondary winding.
6. The secondary winding of a transformer has a terminal voltage of
Vs (t ) = 282.8 sin 377t V. The turn ratio of the transformer is 50/200. If the secondary
current of the transformer is is (t ) = 7.07 sin(377t − 36.87 ) A, calculate:
0
(i) the primary current of the transformer
(ii) voltage regulation
47

32. The impedances of this transformer referred to the primary side are
Req = 0.05Ω Rc = 75Ω
X eq = 0.225Ω X m = 20Ω
7. A 20kVA 8000/277V distribution transformer has the following equivalent circuit
parameter:
R p = 32Ω Rs = 0.05Ω
X p = 45Ω X s = 0.06Ω
Rc = 250kΩ X m = 30kΩ
(i) Draw the equivalent circuit of this transformer referred to the high voltage side.
(ii) Calculate the voltage regulation if:-
(a) the transformer is supplying rated load at 277V and 0.8 pf lagging,
(b) the transformer is supplying load at 200V,100W and 0.85 pf lagging.
(c) the transformer is supplying load at 18kVA, 250V and 0.8 pf leading.
8. Short-circuit and open-circuit tests were performed on a 1000VA transformer, 50 Hz,
rated at 230V/115V, and the results are listed as follows:
Vsc = 13.2 V Voc = 115 V
Isc = 4.35 A I oc = 0.45 A
Psc = 20.1 W Poc = 30 W
(i) Draw the equivalent circuit with the necessary parameters of the transformer referred
to the low voltage side.
(ii) Find the input voltage of the transformer if the transformer is connected to a 140Var,
110V at 0.75 pf lagging
48