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 TRANSFORMER CONSTRUCTION

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

Primary coil

Secondary coil

• 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

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

• 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

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.

Faraday’s law: dt

dNe

Φ= where N = number of turns

=Φdt

drate of change of flux

e = instantaneous induced voltage

dtwtEN

t

N

e

dt

d

wtEte

rms

t

rms

)sin(21

)(

)sin(2)(

0∫=Φ

=Φ=

)cos()(

)cos(2

2)(

)cos(2

)(

max wtt

wtfN

Et

wtwN

Et

rms

rms

Φ−=Φ

−=Φ

−=Φ

π

Therefore maxmax

44.42

2 Φ=Φ= fNfN

Ermsπ

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

aNs

Np

Es

Ep == aN

N

E

E ==2

1

2

1

• 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

• 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=VpIp=VsIs

ThereforeaN

N

V

V

I

I

p

s

p

s

s

p 1===aN

N

V

V

I

I 1

1

2

1

2

1

2 ===

• 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

• From Ohm’s Law,

p

p

p

p

S

s

ss

Iaa

V

aIa

V

Z

I

VZ

)(==

=

2

2

2

==

=

s

p

s

p

ps

N

Na

Z

Z

a

ZZ

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

AV

kVA

V

SI

AV

kVA

V

SI

pp

ss

184.47967

33.33

8.277120

33.33

===

===

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

VN

NVV

N

N

V

V

p

sps

s

p

s

p

1201500

752400 =

==

=

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

AN

NII

N

N

I

I

s

pps

p

s

s

p

20875

15004.10 =

==

=

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

AV

kVA

V

SI

AV

kVA

V

SI

AV

kVA

V

SI

s

ratedVs

s

ratedVs

p

ratedp

5.312240

75

625120

75

42.107200

75

240240

120120

===

===

===

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) IRV =

AI

N

N

I

I

AV

R

VI

p

p

s

s

p

ss

82407200

240

2401

240

==

=

=Ω

==

(ii) kWRIP 6.57)1(24022 ===

(iii)

2

=

s

p

s

p

N

N

Z

Z

Ω=Ω

= 900)1(

240

72002

pZ

25

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

1.0m 1.0k 1.0k 1.0k 1.0m 1.0m

Rp jXp jXs Rs

RmjXm

a:1

Figure 2.11:Equivalent T-circuit

26

1.0m 1.0k 1.0k 1.0k 1.0m 1.0m

jXp jXs '=a2jXs Rs '=a

2Rs

jXm

a:1

Rm

Rp

Figure 2.12: T-circuit referred to the primary

1.0m 1.0k 1.0k 1.0k 1.0m 1.0mTR1

jXs

jXm Rm

Rp'=Rp

a2

jXp'=jXp

a2 Rsa:1

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.

1.0m 1.0k 1.0k 1.0m

Req p jXeq p

RmjXm

a:1

Figure 2.14: Cantilever circuit referred to the primary

Req,p = Rp + a2Rs = Rp+Rs’

Xeq,p = Xp + a2Xs = Xp+Xs’

27

1.0m 1.0k 1.0k 1.0m

Req s jXeq s

RmjXm

a:1

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 = 51840ΩXp = 14.6Ω Xs = 0.0216Ω Xm = 12960Ω

28

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:

Ω=+=

+=

===

839.5)00432.0()99.25(92.2

99.25277

7200

2,

2,

peq

sppeq

s

p

R

RaRR

V

Va

Ω+=

Ω=+=

+=

)2.29839.5(

19.29)0216.0()99.25(6.14

,

2,

2,

jZ

X

XaXX

peq

peq

sppeq

Equivalent impedance referred to the low/secondary side:

Ω=+=

+=

=

−32,

2,

10643.8)00432.0(99.25

92.2

99.25

xsR

Ra

RR

a

eq

sp

seq

Ω+=

Ω=+=

+=

−2,

2,

2,

10)3.486.0(

0432.0)0216.0(99.25

6.14

xjZ

X

Xa

XX

seq

seq

sp

seq

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 ΩXp = 20Ω Xs = 80Ω Xm = 12000Ω

29

Draw the equivalent winding impedance referred to the high voltage side and low voltage side.

Solution

4.01000

400

2

1 ===N

Na

Refer to the high voltage side/secondary side

Ω+=

Ω=+=+=

Ω=+=+=

2055.112

205804.0

20

5.112504.0

10

2221

2221

jZ

Xa

XX

Ra

RR

eq

eq

eq

1,0k 1,0m

112.5 ohm j205 ohm

Refer to the low voltage side/primary side

Ω+=

Ω=+=+=

Ω=+=+=

8.3218

8.32)80(4.020

18)50(4.0102

22

1

22

21

jZ

XaXX

RaRR

eq

eq

eq

1,0k 1,0m

18 ohm j32.8 ohm

30

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)

1.0m 1.0k 1.0k 1.0k 1.0m 1.0m

jXp jXs ' Rs '

jXm Rm

Rp

HV side

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)[RmjXm]] ,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:-

Z eq = Z sc =sc

sc

I

V

R eq = Psc/(I2sc)

X eq =√ (Z eq2 – R2eq)

31

(ii) Open-Circuit Test(normally at high voltage side)

1.0m 1.0k 1.0k 1.0k 1.0m 1.0m

jXp jXs ' Rs '

jXm Rm

Rp

LV side

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) + (RmjXm)

• 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 = (RmjXm)

• 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

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 VIsc = 13.89 A I oc = 14.88 APsc = 1126 W Poc = 1000 W

Solution

Short circuit(HV) (Req,Xeq)

Zeq = Ω== 8.2989.13

414

sc

sc

I

V

Req = Ω== 836.589.13

112622

W

I

P

sc

sc

Xeq = Ω=− 23.2922eqeq RZ

Open circuit(LV) (Rm,Xm)

Ω===

Ω===

=°===°==

°=

=

= −−

19.19435.14

277

73.7661.3

277

435.14)96.75(sin88.14sin

61.3)96.75(cos88.14cos

96.75)88.14(277

1000coscos 11

x

ocm

R

ocm

ococx

ococR

ococ

ococ

I

VX

I

VR

AII

AII

W

IV

P

θθ

θ

33

Referred to the high voltage side/primary side

Ω=Ω

=

Ω=

Ω=Ω

=

Ω=

Ω=

Ω=

kX

X

kR

R

X

R

hvm

lvm

hvm

lvm

hveq

hveq

97.12)19.19(277

7200

19.19

84.51)73.76(277

7200

73.76

23.29

836.5

2

,

,

2

,

,

,

,

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 VIsc = 41.67 A I oc = 6 APsc = 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)

Zeq = Ω== 96.067.41

40

sc

sc

I

V

Req = Ω== 22.067.41

38022

W

I

P

sc

sc

Xeq = Ω=−=− 93.022.096.0 2222eqeq RZ

34

Open circuit(LV) (Rm,Xm)

Ω===

Ω===

=°===°==

°=

=

= −−

03.2099.5

120

64.36333.0

120

99.5)82.86(sin6sin

33.0)82.86(cos6cos

82.86)6(120

40coscos 11

x

ocm

R

ocm

ococx

ococR

ococ

ococ

I

VX

I

VR

AII

AII

W

IV

P

θθ

θ

(i) Referred to the low voltage side/primary side

Ω=Ω

=

Ω=

Ω=Ω

=

Ω=

Ω=Ω=

mX

X

mR

R

X

R

lveq

hveq

lveq

hveq

lvm

lvm

325.2)93.0(2400

120

93.0

55.0)22.0(2400

120

22.0

03.20

64.363

2

,

,

2

,

,

,

,

The equivalent circuit:-

1,0m

1,0

k 1

,0m 1,0k

0.55m ohm j2.325m ohm

363.64 ohm j20.03 ohm

35

(ii) Referred to the high voltage side/secondary side

Ω=Ω

=

Ω=

Ω=Ω

=

Ω=

Ω=

Ω=

8012)03.20(120

2400

03.20

145456)64.363(120

2400

64.363

93.0

22.0

2

,

,

2

,

,

,

,

hvm

lvm

hvm

lvm

hveq

hveq

X

X

R

R

X

R

1,0m

1,0

k 1

,0m 1,0k

0.22 ohm j0.93 ohm

145 456 ohm j8012 ohm

36

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 + I2

2R2

= I12R eq,p = I2

2Req,s

2.7 TRANSFORMER EFFICIENCY

=η =InputPower

rOutputPowe

in

lossin

in

out

P

PP

P

P −=

Output Power , P out = VAcosθ2

Input Power , P in = Output Power + Total Losses

Total Losses , P loss = P copper losses + P core losses

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

VflVnlVR

−=

37

Energy dissipated in the resistance of

winding

Hysteresis and eddy current losses

If transformer equivalent circuit referred to:-(i) primary side,

a V 2

R 0 1 X 0 11 2

1

21

V

aVVVR

−=

(ii) secondary side,

V 2

R 0 2 X 0 21 2

aV

VaV

VR1

21 −=

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

05.02400

120

2534.3387.3667.41)8.0(cos67.412400

100 1

22

==

−=°−∠=−∠=== −

a

AjkVA

V

SI

If referred primary side:

1,0k 1,0mV1

I1 I2'=I2/a

V2'=aV2

0.55m ohm j2.325m ohm

VVj

mjmj

aVmjma

IVmjmIV

°∠=+=

++−=

++=++=

6.054.12128.153.121

)2400(05.0)325.255.0(05.0

)2534.33(

)325.255.0(')325.255.0(' 22

221

%27.1%10054.121

)2400(05.054.121

1

21=−=

−= x

V

aVVVR

If referred to secondary side:

1,0k 1,0mV1'=V1/a

I1'=aI1 I2

V2

0.22 ohm j0.93 ohm

VVj

aVV

jj

VjIV

°∠=−=+==

+=++−=

++=

6.054.12128.153.121

)51.2558.2430(05.0)('

51.2558.2430

2400)93.022.0)(2534.33(

)93.022.0('

11

221

%27.1%100

05.054.121

240005.054.121

1

21

=−

=−

= x

aV

VaV

VR

39

2.9 AUTOTRANFORMERS 1

.0m

1.0

m

1.0

m 1

.0m

LOAD LOAD

OR

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

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 1

,0m

1,0

m 1

,0m

V1=660V

V2=220V

N1

N2

I2

I2-I1

(i) 3220

660

2

1 ===N

Na

(ii) AkVA

I 87.37660

251 ==

kVAkVAauto 251000

)87.37(660 ==

(iii) AkVA

I 64.113220

252 ==

AII 77.7587.3764.11312 =−=−

41

I1

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

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

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.

44

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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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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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 = 50000ΩXp = 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 VIsc = 10kVA/400V=25A I oc = 4 APsc = 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 ttVs 377sin8.282)( = V. The turn ratio of the transformer is 50/200. If the secondary

current of the transformer is )87.36377sin(07.7)( 0−= ttis A, calculate:(i) the primary current of the transformer(ii) voltage regulation

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The impedances of this transformer referred to the primary side are

Ω=

Ω=

225.0

05.0

eq

eq

X

R

Ω=Ω=

20

75

m

c

X

R

7. A 20kVA 8000/277V distribution transformer has the following equivalent circuit parameter:

Ω=

Ω=

Ω=

kR

X

R

c

p

p

250

45

32

Ω=Ω=Ω=

kX

X

R

m

s

s

30

06.0

05.0

(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 VIsc = 4.35 A I oc = 0.45 APsc = 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

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