1. Introduction 2. Types and Construction of Transformers. 3. The Ideal Transformer 4. Theory of...

1

TRANSFORMERS

Dr. Abdul Khaliq

2

Chapter Topics to be Covered

1. Introduction2. Types and Construction of Transformers.3. The Ideal Transformer4. Theory of Operation of Real Single Phase Transformer5. The Equivalent Circuit of a Transformer6. The Per Unit System of Measurements7. Transformer Voltage Regulation and Efficiency8. Transformer Taps and Voltage Regulation9. The Autotransformer10. Three Phase Transformers

3

2.1: Why Transformers are Important

• The first power distribution system in U.S. was 120V dc system invented by Thomas A. Edison to supply power for incandescent light bulbs.

• Edison’s first central power station went into operation in New York city in 1882.

• The power was transmitted at very low voltage level, thus requiring a very large current to supply significant amount of power resulting in high power losses.

• To avoid this problem central power stations were located every few blocks of the city.

• The invention of transformer and concurrent development of ac power sources eliminated these restrictions forever.

4

Introduction to Transformers

• A transformer is a device that changes ac electric power from one voltage level to another voltage level through the action of magnetic field. The transformer has two windings:i. Primary winding: The winding connected to the power

source is called the primary winding.ii. Secondary winding: The winding connected to the load is

called secondary winding.

2

5

2.2: Types of Transformers w. r. t. Core Design

Core Type Transformer Shell Type Transformer

6

2.2: Types of Transformers w.r.t. Core Design

• With respect to the core of the transformer there are two types of the transformers:i. Core type transformer: Winding wrapped around two

sides of the core.ii. Shell type transformer: This is three legged laminated

core, with the winding wrapped around the center leg.• Core is made up of laminations with the High voltage

winding outside & Low voltage winding inside.• Advantages:

i. Easy to insulate H.V. winding from coreii. Less leakage than if they are at distance from each other.

7

1. Unit transformer: To step up the voltage at generation.2. Substation transformer: To step down voltage from

transmission to distribution level.3. Distribution transformer: Takes distribution voltage and

steps it down to final voltage at which the power is used.4. Potential transformer: Is used to change the high voltage to

low voltage, but can handle very small current.5. Current transformer: Is used to change the primary current

to much smaller secondary current but the current is directly proportional to the primary current.Types 4 and 5 are used for instrumentation.

Transformer Types w.r.t. its Applications

8

Ideal Transformer

Ideal transformer is a lossless deviceNp: no of turns on primary sideNs: no of turns on the secondary side

Then Voltage eq. is:

Where a is the turn ratio, and the Current eq. is:

aN

N

ti

ti

tiNtiN

p

s

s

p

sspp

1

)(

)(

)()(

==

=

aN

N

tV

tV

S

P

S

P ==)(

)(

3

9

Phasor

• Vp. & Vs have same phase angle• Ip & Is have same phase angle• Turn ratio just effects only the magnitude of voltage &

current.

1. If Vp is +ve @ dotted end w.r.t un-dotted end then Vs will be also +ve @ dotted end.

2. If Ip flows into the dotted end of primary then Is it will flow out of dotted end of secondary.

aI

I

N

N

V

V

p

s

S

P

s

p===

Ideal Transformer

10

Power in Ideal Transformer

Primary side:

Where θp is the angle between Vp and Ip

Secondary side:

Where θs is angle between Vs and Is

Since voltage and current angles are unaffected for an ideal transformer

Power factor on primary and secondary side is the same

θcosssout IVP =

θcosPPin IVP =

0=−⇒ SP θθ

11

Power in Ideal Transformer

For an Ideal TransformerOutput power =Input powerLikewiseθcosssout

outin

IVP

PP

=

⇒

( )

inPPout

P

P

out

PS

P

S

PIVP

aIa

VP

aIIa

VV

==

=⇒

==

θ

θ

cos

cos

&

outSSPPin

outSSPPin

SIVIVS

QIVIVQ

==

===

=

and

θθ sinsin

12

Impedance Transformation Through a Transformer

• With a transformer it is possible to match the magnitude of the load impedance to a source simply by picking the proper turn ratio.

• To analyze the circuit replace the portion of circuit on one side of the transformer by an equivalent circuit with the same terminal characteristics. This process is known as referring the first to the second side.

4

13

Class Activity 1: Effect of Transformer on the losses

Example 2-1: A single phase power system consists of a 480-V 60 Hz generator supplying a load Zload=4+j3Ω through a transmission line of impedance Zline=0.18+j0.24Ω. Answer the following questions about this system.a) If the power system is exactly as described above what

will the line losses be.b) Suppose a 1:10 step up transformer is placed at the

generator end of transmission line and 10:1 step down transformer is placed at the load end of the line, What will the load voltage be now? what will the transmission line losses be now?

14


15


16

Real Single Phase Transformer

• Primary connected to an AC source and secondary is open circuited

• λ is the flux linkage in the coil across which voltage is being induced.

• Total flux linkage is the sum of the flux passing through each turn in the coil added over all turns of the coil.

• However, average flux can be given by:

dt

dNe

N

in

__

__

φ

λφ

=

=

thus

∑=

=N

i

i

1

φλ

dt

deind

λ=

5

17

The Voltage Ratio Across a Real Transformer

• When the voltage is applied on the primary side of the transformer then the flux produced by this voltage is:

∫= dttvN

P

P

)(1__

φ

18


• Some of this flux produced in the primary links with the secondary winding and some goes into the leakage.

• There is similar division of flux in the secondary side:

• Thus the Faraday’s law for the primary winding can be written as:

LPMP φφφ +=

LSMS φφφ +=

)()(

)(

tetedt

dN

dt

dN

Dt

dNtv

LPp

LPP

MP

PPp

+=

+=

=

φφ

φ

19

Like wise for secondary side:

)()()(

)(

tetetV

dt

dN

dt

dN

dt

dNtV

LSss

LSS

Ms

sss

+=

+=

=

φφ

φ

aN

N

te

te

N

te

dt

d

N

te

dt

dNte

dt

dNte

S

P

S

P

S

SM

P

P

MsS

MPP

==⇒

==⇒

=

=

)(

)(

)()(

)(

)(

φ

φ

φ

• From eqs. above the primary and secondary voltages due to mutual flux are given by:

• The ratio of the primary voltage producing the mutual flux to the secondary voltage induced by the mutual flux is equal to the turn ratio of the transformer.


20

• In a good design ØM >> ØLPand ØM >> ØLS. Therefore, Ratio of the total voltage on the primary of transformer to the ratio of the total voltage on the secondary of a transformer is approximately. Which was the case for the ideal transformer.

• The smaller the leakage flux, the closer the transformer will be to the ideal one.

aN

N

tV

tV

s

p

s

p==

)(

)(


6

21

The Magnetization Current in Real Transformer

• When AC power is connected to the transformer a current flows even when secondary is open circuited.

• This is the current required to produce the flux in real ferromagnetic material and it consists of two components:1. Magnetization current

iM, Current required to produce the flux in the core of transformer.

2. Core losses current ih+e, current required for hystersis and eddy current losses.

tN

V

tdtVN

tVtV

dt

NdetdtV

N

p

M

M

p

Mp

p

p

ωω

φ

ωφ

ω

φφ

sin

cos1

cos)(

)()()(1

=

=

=

==

∫

∫ Q

22


23


• If the value of the current required to produce a given flux is known then it is possible to construct the sketch of magnetizing current in the winding of the core.

Observations: 1) im is not sinusoidal2) Øm reaches max; a small increase in Ø requires very large increase in im.

3) im lags voltage applied by 90˚.

24

The Hysteresis and Eddy Current in Real

Transformer

• This is the current required to supply power to make up the hysteresis and eddy current losses in the core, known as core loss current.

• since the eddy current are proportional to the rate of change of flux. Therefore, the eddy current are largest when the flux is passing through the 0 wb.

7

25

The Total No Load (Excitation) Current in Real Transformer

• Total no load current is called excitation current which is sum of magnetization and core loss current.

ehmex iii ++=

26

The Current Ratio & Dot Convention

• Connect a load on the secondary side of the transformer.• A current flowing into the dotted end of the transformer

produces a positive mmf.

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• A current flowing into the dotted end produces +ve mmf, while a current flowing into the undotted end produce a –ve mmf.

• If both currents are entering the dotted end then the mmf will add to each other.

• If one current enters and the other one leaves then the mmf will subtract from each other.

• For a good designed transform, R should be very small nearly zero, as long as the core is operating in unsaturated region.

• In order for mmf to be zero, current must flow in to the one dotted end and out of the other dotted end.

RiNiNF

iNiNFFF

iNFiNF

ssppnet

ssppspnet

sssppp

φ=−=

−=−=

−== &

aN

N

i

i

iNiN

iNiNF

p

s

s

p

sspp

ssppnet

1

0

=≈

≈

≅−=

28


Assumptions to convert a real transformer into an ideal transformer.1. The core must have no

hysteresis or eddy current.

2. The Magnetization curve must be an ideal one.

3. The leakage flux in the core must be zero, implying that all the flux in the core couples both windings.

4. The resistance of the transformer winding must be zero.

8

29

2.5: The Equivalent Circuit of a Transformer

An accurate model of the transformer, have to account for all the losses of the transformer.1. Copper Losses( ): The resistive heating losses in the

primary and secondary winding of the transformer. They are proportional to the square of the current in the winding.

2. Leakage Flux( ): The flux which escapes the core and pass only through one of the transformer winding. These escaped fluxes produce a self inductancein the primary and secondary coil.

3. Eddy Current Losses: These are resistive heating losses in the core and are proportional to the square of the voltage applied to the transformer.

4. Hysteresis loses: These are associated with the rearrangement of the magnetic domains in the core during each half cycle and are nonlinear function of applied voltage.

RI2

LSLP φφ &

30

Driving The Equivalent Circuit Model

1. Copper Losses: The resistive copper losses are modeled by placing the a resistance RP in the primary and RS in the secondary winding of the transformer.

2. Leakage Flux: which escapes the core and pass only through one of the transformer winding.• LP: self inductance of

primary coil• LS: self inductance of

secondary coilTherefore, Leakage flux will be

modeled by primary and secondary inductances.

& SSLSPPLP

LSsLS

t

LPPLP

iNiN

dt

dNte

d

dNte

)()(

)(

)(

ρφρφ

φ

φ

==

=

=

ρ

ρρ

ρ

ρρ

2

2

2

2

)(

)()(

)(

)()(

SSS

SLS

SSSSSLS

PP

p

pLP

p

pPPpLP

NLdt

diLte

dt

diNiN

dt

dNte

NLdt

diLte

dt

diNiN

dt

dNte

==

==

==

==

Q

Q

31

3. (Eddy Current +Hysteresis Losses): The core loss current, ih+e, is proportional to voltage applied to the core and is in phase with voltage. Therefore, can be modeled by Rcacross primary.

4. The Magnetization Current: Is proportional to the voltage applied to the core (in unsaturated region) and is lagging the applied voltage by 90o. So it can be modeled by a reactance connected across the winding, represented by XM.

The XM & RC represent the excitation effect which includes the core loss current (eddy+hysteresis) and the magnetization current.

The XM & RC are placed inside, after LP and RP, because the voltage applied to the core is input voltage.

Driving The Equivalent Circuit Model

32

The Final Equivalent Circuit Model

• To analyze practical circuits containing transformers, it is important to convert the entire circuit to a single voltage level.

• Therefore the circuit must be referred either to primary or to its secondary side.

9

33

The Final Equivalent Circuit Model Referred to

One Side of Transformer

I. Referred to Primary Side

II. Referred to Secondary Side

34

Approximate Equivalent Circuit of a Transformer

• Excitation branch adds complexity. However, current through this branch is very small, as compared to the load current.

• It causes negligible voltage drop in the Rp and Xp and hence can be neglected.

speqspeq XaXXRaRRPP

22 +=+=s

p

eqs

p

eq Xa

XXR

a

RR

SS+=+=

22

35

The Per-Unit Analysis

• The advantage of per unit analysis is that by properly specifying the base quantities, the transformer equivalent circuit can be simplified.

• The ideal transformer winding can be eliminated, such that the voltages, currents, and external impedances and admittances expressed in per-unit do not change when they are referred from one side of transformer to the other.

• This can be a significant advantage even in a power system of moderate size, where hundreds of transformers may be encountered. 36

The Per-Unit Analysis

• The following two rules are adopted for the base

quantities:

1. The value of Sbase is the same for entire power system

of concern.

2. The ratio of voltage bases on either side of

transformer is selected to be the same as the ratio of

transformer voltage ratings.

quantitytheofvaluebase

valueactualquantiyunitper

=−

10

37

Per-unit Impedance• A single phase two-winding transformer is rated 20

kVA, 480/120 volts, 60 Hz. The equivalent leakage

impedance of transformer referred to the 120-volt

winding, denoted winding 2, is . Using

the transformer ratings as base values, determine

the per-unit leakage impedance referred to the

winding 2 and the winding 1.

Ω∠= o

eqZ 13.780525.02

38

Per-unit Impedance• Three zones of a single phase circuit are identified in the figure

below. These zones are connected by transformers T1 and T2

whose ratings are also shown. Using base values of 30 KVA and

240 volts in zone 1, draw the per unit circuit and determine the

per unit impedances and per unit source voltage. The calculate

the load current both in per unit and in amperes. Transformer

winding resistances and shunt admittances are neglected.

39

The Open Circuit Test

• Rp & Xp are too small as compared to Rc and Xm. Therefore, significant voltage drops in Rc and Xm, approximately all the voltage drop across the excitation impedance

• Easy to look at the admittance and conductance to calculate Rc and Xm. [Parallel equivalent of Gc and BM].

MCE

M

M

C

C

jBGY

XB

RG

−=

==11

.(i)11

……−=Mc

EX

jR

Y

40

The Open Circuit Test

• Magnitude of YE referred to primary side.

• The P.F. for real transformer is always lagging. Therefore: I lags V by θ

• Finally , compare (i) and (ii) and you can obtain the values of RC and XMdirectly.

)(cos

cos

1

OCOC

OC

OCOC

OC

OC

OCE

IV

P

IV

PPF

V

IY

−=θ

=θ=

=

(ii) jB...... -A Y

Fcos

E

1

=

−∠=

−∠=

−P

V

IY

V

IY

OC

OC

E

OC

OC

E θ

11

41

Short Circuit Test

• In this test the secondary terminals of the transformer are short circuited.

• Apply a fairly low voltage to primary side, till current in short circuited winding is equal to rated current.

• Caution: Make sure to keep the primary voltage low, otherwise you could burn the transformer’s winding.

42

Short Circuit Test

• In this test the input voltage is so low. Therefore, no current flows though the excitation branch.=> All the voltage drop is in the series elements in the circuit.

• The magnitude and angle of the series impedance referred to primary side are given by:

SCSC

sc

SCSC

SC

SC

SC

SE

VI

P

IV

PPF

I

VZ

1cos

cos

−=

==

=

θ

θ

θθ

+∠=−∠

∠=

SC

SC

SC

SC

SEI

V

I

VZ

0

)()(

22

SPsPSE

eqeqSE

XaXjRaRZ

jXRZ

+++=

+=

43

Class Activity 2: Determining Transformer Model’s Parameters

Example 2-2: The equivalent circuit impedance of a 2-kVA, 8000/240 V. 60 HZ transformer are to be determined. The open circuit test and short circuit test were performed on the primary side of the transformer, and the following data were taken. Find the impedance of the approximate equivalent circuit referred to primary side, and sketch the circuit.

Open-circuit test (on primary) Short-circuit test (on primary)

Voc=8000 V Vsc=489 V

Ioc=0.214 A Isc=2.5 A

Poc= 400 W Psc= 240 W44

2.7: Transformer Voltage Regulation and Efficiency

• Due to series impedance in the transformer, the output voltage of the transformer varies with the load even if input voltage remains constant.

• The voltage regulation can be calculated at any load. However, we define full load voltage regulation as a quantity which compares the output voltage of transformer at no load with the output voltage at full load to conveniently compare various transformers.

%100/,

,

×−

=⇒=flS

flSP

PSV

Va

V

VRaVV

%100,

,,×

−=

flS

flSnlS

V

VVVR

Since at no load:

12

45

Transformer Voltage Regulation

• Low voltage regulation means less impedance of the winding. So is mostly desirable.

• However, some times high impedances are used to reduce the short circuit current.

• Figure 2-18(b) and 2-18(d)

46

VR with Lagging P.F.

seqseqs

pIjXIRV

a

V++= s

pV

a

V>

The voltage regulation of a transformer with lagging loads must be greater than zero.

47

VR with Unity P.F.

seqseqs

pIjXIRV

a

V++= s

pV

a

V>

The voltage regulation of a transformer with unity loads must be greater than zero. However, it is a smaller number compared to lagging P.F. load.

48

VR with Leading P.F.

seqseqs

pIjXIRV

a

V++=

If the secondary current is leading , the secondary voltage can be actually higher than the referred primary voltage. Which means a -ve voltage regulation.

0<

−⇒< s

p

s

pV

a

VV

a

V

13

49

Simplified Voltage Regulation

• Considering a lagging p.f. because this is the practical case mostly, we will drive an approximate equation for calculating the primary voltage.

θθ sinIjXcosIRVa

Vseqseqs

p++=

50

Transformer Efficiency

In general:

From equivalent circuit we can calculate η easily.1. Copper losses -in series resistance, Rp and Rs2. Hystersis losses -accounted for by Rc3. Eddy current -accounted for by Rc

Therefore, at any given load efficiency can be calculated as:

100 & ×++

==θ

θηθ

CosIVPP

CosIVCosIVP

sscorecu

sss

sssout

%100%100 ×+

=×=lossesout

out

in

out

PP

P

P

Pη

51

2.8: Transformer Taps

• So far we considered turn ratio to be fixed. Practically, almost all the distribution transformers have taps in the winding which allows small changes in transformer’s turns ratio in the field.

• Typically 4 taps are designed, in addition to nominal setting with spacing of ± 2.5 percent of full load voltage between them.

• Taps accommodate the variations in local voltages. These taps normally can not be changed while the power system is operating and power is being supplied to the load.

• Tap Changing Transformer Under Load (TCUL): It is transformer which has the ability to change taps while power is connected to it.

52

Class Activity 4: Transformer Taps

Example 2-6: A 500 KVA, 13,200/480 V distribution transformer has four 2.5% taps on the primary winding. What are the voltage ratios of transformer at each tap setting.

+ 5.0% taps 13,860/480

+2.5% taps 13,530/480

Nominal rating 13,200/480

-2.5% taps 12,870/480

-5.0% taps 12,540/480

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53

Class Activity 3:

Example 2-5: A 15 kVA, 2300/230-V transformer is to be tested to determine its excitation branch components, its series impedance and its voltage regulation. The following test data have been taken from the primary side of the transformer.

Open-circuit test (on primary)

Short-circuit test (on primary)

Voc=2300 V Vsc=47 V

Ioc=0.21 A Isc=6.0 A

Poc= 50 W Psc= 160 W

a) Eq ckt. Referred to HV.b) Eq ckt. Referred to LV.c) Full load VR at 0.8 pf

Lagging, 1.0, and 0.8 leading.

d) Perform the above using approximate Eq.

e) Plot VR versus load from no load to full load at 0.8 pf Lagging, 1.0, and 0.8 leading.

f) Efficiency at full load and 0.8 pf.

54

2.9: The Auto Transformer

• Some times it is desirable to change voltage levels by only small amount. For example from 110 V to 120 V or 13.2 kv to 13.8 kv.

• In this case it is wasteful and excessively expensive to wind a transformer with two full windings each rated at about the same voltage and a special purpose transformer called Autotransformer is used.

• The voltage at the output of the whole transformer is the sum of the voltage on the first winding and voltage on the second winding.

• The winding across which both the primary and secondary voltages appear is called Common Winding and the smaller winding which is connected in series with the common winding and across which only one voltage (primary/secondary) appears is called Series Winding.

55

The Auto Transformer

56

A Step-down Auto Transformer

15

57

The Auto Transformer

SECL III +=

SESE

C

SEL II

N

NI +=

SE

SEC

H

L

N

NN

I

I +=

SECH VVV +=

C

C

SECH V

N

NVV +=

L

C

SELH V

N

NVV +=

SE

SEC

H

L

N

NN

V

V +=

HH

C

SEL II

N

NI +=

Voltage Relationship between two sides of the transformer

Current Relationship between two sides of the transformer

58

Application of Auto Transformers

• Autotransformer is used when the turn ratio is nearly equal to 1 and where there is no objection to electrical connection between the primary and secondary winding. Hence such transformers are used:

1) To give small boost to a distribution cable to correct the voltage drop.

2) As autotransformer to give 50 to 60 % of full voltage to an induction motor during starting.

3) As furnace transformers for giving convenient supply to suit the furnace winding from a 230 V supply.

59

Since there is only one winding both for the primary and the secondary side. Therefore, the winding has to be designed based on the higher voltage and current ratings.

The autotransformer can’t be used for the applications where the electrical isolation is desired because there is a physical connection between the primary and the secondary circuits, so the electrical isolation of the two windings is lost.

The common practice is to use autotransformer whenever the two voltages are fairly close to each other.

The autotransformer is also used as variable transformer, where the low voltage taps move up and down the winding.

Issues to Consider while using the Auto Transformer

60

Three Phase Transformers

The major power generation, transmission, and distribution systems are three phase systems. Therefore, a it is necessary to understand the design and operation of a three phase transformer. A three phase transformer can be constructed in one of the two ways:i. Simply take three single phase transformers and connect

them in three phase bank.ii. Make a three phase transformer consisting of three sets of

windings wrapped around a common core. A single three phase transformer is the preferred practice

today because it is lighter, smaller, cheaper, and slightly more efficient.

16

61

Three Phase Transformer Bank

Three phase transformers bank with independent transformers.

Three phase transformers bank on a single three legged core.

62

Three Phase Transformer Connections

The Primary and secondary winding of a three phase transformer can be independently connected in either wye (Y) or delta (∆). This gives a total of four possible connections for a three phase transformer.

1. Y-Y2. Y-∆3. ∆-Y4. ∆-∆

Any single transformer in the bank behaves exactly like a stand alone single phase transformer. Thus the impedance, voltage regulation, efficiency, and similar calculations for three phase transformer can be done on per phase basis using the technique established for single phase transformer.

63

Three Phase Transformer Y Connections

64

Three Phase Transformer ∆∆∆∆ Connections

17

65

Three Phase Transformer Connections

1) Y-Y

2) Y-∆∆∆∆

3) ∆∆∆∆-Y

4) ∆∆∆∆-∆∆∆∆

aV

Va

V

V

VVVV

LS

LP

S

P

SLSPLP

3

3

==

==

φ

φ

φφ

3

3

a

V

Va

V

V

VVVV

LS

LP

S

P

SLSPLP

==

==

φ

φ

φφ

aV

Va

V

V

VVVV

LS

LP

S

P

SLSPLP

==

==

φ

φ

φφ

aV

Va

V

V

VVVV

LS

LP

S

P

SLSPLP

==

==

φ

φ

φφ 33

66

Instrument Transformers: CT & VT/PT

Two special purpose transformers used in the power system are Potential Transformer (PT) and Current Transformer (CT).

I. A potential transformer is specially wound transformer with the high voltage primary and low secondary winding. It has very low power rating because its sole purpose is to provide a sample of power system voltage to the instruments monitoring it. It must be very accurate to get the true value of primary voltage because based on that protection and control decisions are taken.

II. A Current Transformer samples the current in the line and reduces it to a safe and measurable level. The current transformer consists of a secondary winding wrapped around a ferromagnetic ring, with the single primary line running through the center of the coil.

67

Instrument Transformers: CT & VT/PT

I. Current Transformer (CT) II. Voltage /Potential Transformer (VT/PT)

68

Characteristics of a CT

• Thus the winding in a CT is loosely coupled and mutual flux is smaller than the leakage flux.

• The conventional voltage and current equation does not apply for a CT. However, the secondary current is proportional to the much larger primary current and the device can provide the sample of the line’s current for measurement, monitoring, protection, and control purposes.

• Typical CT ratio might be 600:5, 800:5, 1000:5. A 5 amps rating is standard on the secondary of the transformer.

• It is important to keep the CT short-circuited all the time. Because

• Extremely high voltage can appear across the open terminals, mostly there is an interlock on the secondary side.

18

69

Questions to Ponder

Why the discovery of the transformer accelerated the development of the use of electricity ?

Why the iron or magnetization losses are more important than the losses caused by the winding resistance ?

Why the transformer has to be cooled ? How it is done ? What is the connection of the transformer that supplies your

house and where is it ? What is an Isolation transformer?

1. Introduction 2. Types and Construction of Transformers. 3. The Ideal Transformer 4. Theory of...

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