KG Reddy College of Engineering &...

KG Reddy College of Engineering & Technology

(Approved by AICTE, New Delhi, Affiliated to JNTUH, Hyderabad)

Chilkur (Village), Moinabad (Mandal), R. R Dist, TS-501504

Accredited by NAAC

3

COURSE FILE CONTENTS

S.N. Topics Page No.

1 Vision, Mission, PEO’s, & PO’s & PSO’s

2 Syllabus (University Copy)

3 Course Objectives, Course Outcomes And Topic Outcomes

4 Course Prerequisites

5 CO’s. PO’s Mapping

6 Course Information Sheet (CIS)

a). Course Description

b). Syllabus

c). Gaps in Syllabus

d). Topics beyond syllabus

e). Web Sources-References

f). Delivery / Instructional Methodologies

g). Assessment Methodologies-Direct

h). Assessment Methodologies –Indirect

i). Text books & Reference books

7 Micro Lesson Plan

8 Teaching Schedule

9 Unit wise hand written notes

10 OHD/LCD SHEETS /CDS/DVDS/PPT (Soft/Hard copies)

11 University Previous Question papers

12 MID exam Descriptive Question Papers with Key

13 MID exam Objective Question papers with Key

14 Assignment topics with materials

15 Tutorial topics and Questions

16 Unit wise-Question bank

1 Two marks question with answers 5 questions

2 Three marks question with answers 5 questions

3 Five marks question with answers 5 questions

4 Objective question with answers 10 questions

5 Fill in the blanks question with answers 10 questions

17 Beyond syllabus Topics with material

18 Result Analysis-Remedial/Corrective Action

19 Record of Tutorial Classes




Accredited by NAAC

4

20 Record of Remedial Classes

21 Record of guest lecturers conducted

1. Vision, Mission, PEO’s, PO’s & PSO’s

Vision

To be renowned department imparting both technical and non-technical skills to the students

through implementing new engineering pedagogy and research to produce competent new age

electrical engineers

Mission

To transform the students into motivated and knowledgable new age electrical engineers.

To advance the quality of education to produce world class technocrats with an ability to

adapt to the academically challenging environment.

To provide a progressive environment for learning through organized teaching

methodologies, contemporary curriculum and research in the thrust areas of electrical

engineering.




Accredited by NAAC

5

Program Educational Objectives

PEO 1: Apply knowledge and skills to provide solutions to Electrical and Electronics Engineering problems

in industry and governmental organizations or to enhance student learning in educational institutions

PEO 2: Work as a team with a sense of ethics and professionalism, and communicate effectively to manage

cross-cultural and multidisciplinary teams

PEO 3: Update their knowledge continuously through lifelong learning that contributes to personal, global

and organizational growth

Program Outcomes

PO.1 Engineering knowledge: Apply the knowledge of mathematics, science, engineering fundamentals

and an engineering specialization to the solution of complex engineering problems.

PO.2 Problem analysis: Identify, formulate, review research literature, and analyze complex engineering

problems reaching substantiated conclusions using first principles of mathematics, natural science and

engineering sciences.

PO.3 Design/development of solutions: design solutions for complex engineering problems and design

system components or processes that meet the specified needs with appropriate consideration for the public

health and safety, and the cultural, societal and environmental considerations.




Accredited by NAAC

6

PO.4 Conduct investigations of complex problems: use research based knowledge and research methods

including design of experiments, analysis and interpretation of data, and synthesis of the information to

provide valid conclusions.

PO.5 Modern tool usage: create, select and apply appropriate techniques, resources and modern

engineering and IT tools including prediction and modeling to complex engineering activities with an

understanding of the limitations.

PO.6 The engineer and society: apply reasoning informed by the contextual knowledge to assess societal,

health, safety, legal and cultural issues and the consequent responsibilities relevant to the professional

engineering practice.

PO.7 Environment sustainability: understand the impact of the professional engineering solutions in the

societal and environmental contexts, and demonstrate the knowledge of, and need for sustainable

development.

PO.8 Ethics: apply ethical principles and commit to professional ethics and responsibilities and norms of

the engineering practice.

PO.9 Individual and team work: function effectively as an individual and as a member or leader in

diverse teams, and in multidisciplinary settings.

PO.10 Communication: communicate effectively on complex engineering activities with the engineering

community and with society at large, such as, being able to comprehend and write effective reports and

design documentation, make effective presentations, and give and receive clear instructions.

PO.11 Project management and finance: demonstrate knowledge and understanding of the engineering

and management principles and apply these to one’s own work, as a member and leader in a team, to

manage projects and in multidisciplinary environments.

PO.12 Lifelong learning: recognize the need for, and have the preparation and ability to engage in

independent and lifelong learning in the broader context of technological change.

PSO-1:Apply the engineering fundamental knowledge to identify, formulate, design and investigate complex

engineering problems of electric circuits, power electronics, electrical machines and power systems and to

succeed in competitive exams like GATE, IES, GRE, TOEFL, GMAT, etc.

PSO-2: Apply appropriate techniques and modern engineering hardware and software tools in power systems




Accredited by NAAC

7

and power electronics to engage in life-long learning and to get an employment in the field of Electrical and

Electronics Engineering.

PSO-3: Understand the impact of engineering solutions in societal and environmental context, commit to

professional ethics and communicate effectively.




Accredited by NAAC

8

2. SYLLABUS (University Copy)

EE304PC: ELECTRICAL MACHINES I

B.Tech. II Year I Sem. L T P C

3 1 0 4

Prerequisite: Basic electrical & Electronics Engineering

Course Objectives:

To study and understand different types of DC generators, Motors and Transformers, their

construction, operation and applications.

To analyze performance aspects of various testing methods.

Course Outcomes:

Identify different parts of a DC machine & understand its operation

Carry out different testing methods to predetermine the efficiency of DC machines

Explain different excitation and starting methods of DC machines

Control the voltage and speed of a DC machines

Analyze single phase and three phase transformer circuits

UNIT – I

D.C. Generators: Principle of operation – Action of commutator – constructional features – armature

windings – lap and wave windings – simplex and multiplex windings – use of laminated armature – E.

M.F Equation. Armature reaction – Cross magnetizing and de-magnetizing AT/pole – compensating

winding commutation – reactance voltage – methods of improving commutation. Methods of

Excitation – separately excited and self excited generators – build-up of E.M.F - critical field resistance

and critical speed - causes for failure to self excite and remedial measures. Load characteristics of

shunt, series and compound generators




Accredited by NAAC

9

UNIT – II

Motors: Principle of operation – Back E.M.F. - Torque equation – characteristics and application of

shunt, series and compound motors – Armature reaction and commutation. Speed control of D.C. Motors

- Armature voltage and field flux control methods. Motor starters (3 point and 4 point starters) Testing of

D.C. machines - Losses – Constant & Variable losses – calculation of efficiency – condition for maximum

efficiency.

UNIT - III

Methods of Testing – direct, indirect, and regenerative testing – Brake test – Swinburne’s test

Hopkinson’s test – Field’s test - separation of stray losses in a d.c. motor test.

UNIT - IV

Single phase transformers: Types - constructional details-minimization of hysteresis and eddy current

losses- EMF equation - operation on no load and on load - phasor diagrams Equivalent circuit - losses

and efficiency – regulation - All day efficiency - effect of variations of frequency & supply voltage on

iron losses.

UNIT - V

OC and SC tests - Sumpner’s test - predetermination of efficiency and regulation-separation of losses

test-parallel operation with equal and unequal voltage ratios - auto transformers- equivalent circuit -

comparison with two winding transformers. Polyphase transformers - Polyphase connections - Y/Y,

Y/ , /Y, / and open

TEXT BOOKS:

1. “I.J. Nagrath & D.P. Kothari”, “Electric Machines”, Tata Mc Graw Hill Publishers, 3rd edition,

2004.

2. “P.S. Bimbra”, “Electrical Machines”, Khanna Publishers, 7th Edition, 204.

REFERENCE BOOKS:

1. E. Clayton & N. M. Hancock “The Performance and Design Of Direct Current Machines” 3rd

Edition

Pitman, London 1959.




Accredited by NAAC

10

2. “A. E. Fritzgerald, C. Kingsley and S. Umans”, “Electric Machinary”, McGraw Hill Companies, 6th

edition, 2003.

3. “Abhijith Chakrabarthi & SubithaDebnath”, “Electrical Machines”, Mc Graw Hill, 2015.

11

3. COURSE OBJECTIVES AND COURSE OUTCOMES AND

TOPIC OUTCOMES

COURSE OBJECTIVES

Study and understand different types of DC generators, motors and transformers, their

construction, operation and applications.

Analyze performance aspects of various testing methods.

Explain different excitation and starting methods of DC machines

Control the voltage and speed of a DC machines

COURSE OUTCOMES

After this course, the student will be able to

CO1: Identify different parts of a DC machine & understand its operation

CO2: Carry out different testing methods to predetermine the efficiency of DC machines

CO3: Explain different excitation and starting methods of DC machines

CO4: Control the voltage and speed of a DC machines

CO5: Analyze single phase and three phase transformer circuits

12

TOPIC OUTCOMES

LECTURE

NO.

TOPIC TO BE COVERED TOPIC OUTCOME

(Upon the completion of this topic the student

will be able to)

L1 Over view of course Outline the topics of electrical machines

course

L2 Unit-I: Introduction Identify the importance of the dc generators

L3 Energy Conversion (Gaps in

syllabus)

Get knowledge on the energy conversions

L4 Principle of Operation Explain the operation of dc generators

L5 Constructional Features List and explain the parts of dc generator

L6 Armature windings Draw the lap and wave windings

L7 Armature windings Draw the lap and wave windings

L8 EMF equation Derive the EMF equation of dc generator

L9 Armature Reaction Explain the effect of armature reaction

L10 Methods of improving

commutation

List the methods of improving commutation

L11 Types of generators Classify different types of dc generators

L12 Critical resistance and critical

speed

Define critical resistance and critical speed

L13 Characteristics of generators Draw the characteristics of generators

L14 Unit-II: Principle of operation

of DC Motors

Demonstrate the working principle of dc

motors

L15 Back EMF Derive the expression of back EMF

L16 Torque Equation Derive torque equation

L17 Applications of Motors List the applications of dc motors

L18 Armature reaction and

Commutation

Explain Armature reaction and Commutation

L19 Speed control of DC motors Control speed of DC motors

L20 3 and 4 point starters Explain 3 and 4 point starters

L21 Testing of DC machines Test dc machines performance

L22 Losses of dc machine Study the losses of dc machine

L23 Constant and variable losses Study the constant and variable of dc

machine

L24 Calculation of efficiency Calculate the efficiency of dc machine

L25 Condition for maximum

efficiency

Obtain the condition for maximum efficiency

13

LECTURE

NO.



will be able to)

L26 Unit-III: Methods of testing Identify the methods of testing dc machines

L27 Methods of testing Explain the methods of testing dc machines

L28 Brake test Perform the brake test on dc machine

L29 Swinburnes test Perform the swinburnes test on dc machine

I MID EXAMINATION

L30 Hopkinsons test Perform the Hopkinsons test on dc machine

L31 Fields test Perform the Fields test on dc machine

L32 Seperation of stray loss Discuss the Seperation of stray loss

L33 Unit-IV: Constructional Details

of a transformer

Describe the Constructional Details of a

transformer

L34 Hysterisis and eddy current loss calculate the losses in transformer

L35 EMF Equation of transformer Derive the EMF equation of transformer

L36 Operation on no load and on

load

Explain the Operation on no load and on load

L37 Equivalent Circuit of 1-Ph

transformer

Draw the Equivalent Circuit of 1-Ph

transformer

L38 Regulation of single phase

transformer

Check the Regulation of single phase

transformer

L39 All day efficiency Calculate the all day efficiency

L40 Effect of variations of

frequency and supply voltage

on iron losses

Describe the effect of variations of frequency

and supply voltage

L41 Practical applications of the

transformers (Lecture beyond

syllabus)

Identify the practical applications of the

transformers

L42 Unit-V: Testing of 1-phase

transformers

Test the single phase transformers

L43 OC and SC tests Conduct the oc and sc tests on 1-ph T/F’s

L44 Sumpner’s test Conduct the sumpners test on 1-ph T/F

L45 Predetermination of efficiency

and regulation

Predetermination of efficiency and regulation

L46 Separation of losses test Conduct Separation of losses test

L47 Parallel operation with equal an

unequal voltage ratios

Conduct Parallel operation

14

LECTURE

NO.



will be able to)

L48 Auto-transformers &

Equivalent Ckt

Discuss about auto transformers & Draw the

equivalent circuit of 1-ph transformer

L49 Polyphase connections-y/y, y/∆,

∆/y, ∆/∆ and open∆

Illustrate Polyphase connections-y/y, y/∆,

∆/y, ∆/∆ and open∆

II MID EXAMINATION

4. COURSE PRE–REQUISITES

1. Fundamental laws

2. Electromagnetism in Basic Electrical and Electronics Engineering

5. CO’S, PO’S MAPPING

PO1 PO2 PO3 PO4 PO5 PO6 PO7 PO8 PO9 PO10 PO11 PO12

CO1 3 - - - 2 - - - - - - -

CO2 3 - - - 3 - - - - - - 1

CO3 3 - - - - - - - - - - -

CO4 3 - - 3 2 - - - - - - -

CO5 3 3 2

1-Low, 2-Medium, 3-High

6. COURSE INFORMATION SHEET

6.a). COURSE DESCRIPTION:

PROGRAMME: B. Tech. (EEE.) DEGREE: BTECH

COURSE: ELECTRICAL MACHINES - I YEAR: II SEM: I

CREDITS: 4

COURSE CODE: EE304PC

REGULATION: R18

COURSE TYPE: CORE

15

COURSE AREA/DOMAIN: CONTACT HOURS: 3+ 1(L+T))

hours/Week.

CORRESPONDING LAB COURSE CODE (IF ANY):

NIL

LAB COURSE NAME: NIL

6.b). SYLLABUS:

Unit Details Hours

I

D.C. Generators: Principle of operation – Action of commutator –

constructional features – armature windings – lap and wave windings –

simplex and multiplex windings – use of laminated armature – E. M.F

Equation.

Armature reaction – Cross magnetizing and de-magnetizing AT/pole –

compensating windingcommutation – reactance voltage – methods of

improving commutation. Methods of Excitation – separately excited and self

excited generators – build-up of E.M.F - critical field resistance and critical

speed - causes for failure to self excite and remedial measures. Load

characteristics of shunt, series and compound generators

11

II

Motors: Principle of operation – Back E.M.F. - Torque equation –

characteristics and application of shunt, series and compound motors –

Armature reaction and commutation. Speed control of D.C. Motors -

Armature voltage and field flux control methods. Motor starters (3 point and

4 point starters) Testing of D.C. machines - Losses – Constant & Variable

losses – calculation of efficiency – condition for maximum efficiency.

11

III

Methods of Testing – direct, indirect, and regenerative testing – Brake test –

Swinburne’s testHopkinson’s test – Field’s test - separation of stray losses in a

d.c. motor test. 6

IV

Single phase transformers: Types - constructional details-minimization of

hysteresis and eddy current losses- EMF equation - operation on no load

and on load - phasor diagrams.Equivalent circuit - losses and efficiency –

regulation - All day efficiency - effect of variations of frequency & supply

voltage on iron losses.

8

V

OC and SC tests - Sumpner’s test - predetermination of efficiency and

regulation-separation of losses test-parallel operation with equal and unequal

voltage ratios - auto transformers- equivalent circuit - comparison with two

winding transformers. Polyphase transformers - Polyphase connections -

Y/Y, Y/ , /Y, / and open

6

16

Contact classes for syllabus coverage 42

Lectures beyond syllabus 01

Tutorial classes 5

Classes for gaps& Add-on classes 01

Total No. of classes 49

6.c). GAPS IN THE SYLLABUS - TO MEET INDUSTRY/PROFESSION REQUIREMENTS:

S.NO. DESCRIPTION PROPOSED

ACTIONS

No. of Classes

1 Energy conversion PPT 01

6. d). TOPICS BEYOND SYLLABUS / ADVANCED TOPICS:

S.NO. DESCRIPTION PROPOSED ACTIONS No. of Classes

1 Practical applications of Transformers NPTEL 01

6. e). WEB SOURCE REFERENCES:

Sl. No. Name of book/ website

a. https://swayam.gov.in/nd1_noc19_ee60/preview

c. https://www.youtube.com/watch?v=g53tqrBjIgc&list=PL5105727DD6E8DE98&index=1

6. f). DELIVERY / INSTRUCTIONAL METHODOLOGIES:

CHALK & TALK STUD. ASSIGNMENT WEB RESOURCES

LCD/SMART BOARDS STUD. SEMINARS ☐ ADD-ON COURSES

https://u10190948.ct.sendgrid.net/wf/click?upn=FjujsDMP-2F-2BhPeKswV30OEgOF4dt5oax9BTLt04o2CaUwiNG4PhL3kHVidW2SdSQkZqwdAaZhmgaMf-2FryKlNHRA-3D-3D_88qnucqjhNfJl6N1vAFFjI1MgwLFKoHl1z8ZnVl13nrYRmq1pcq3viEq9-2FTxQoPE-2FqdhPoymtB-2FLdylgej3zu0aobri2RF6kNGd8MdV276iGPoFT8DHJvTo9f6EkRLqSskXdnbJ9k0gbyOMtFENnFQP1zRH-2B1k5v1tsnwinh5XHwYdCKt1XeM696AKo5LMbNZIiDbWia1q1Movatx3eW8BeTEI0W-2BkfoBoEV8dL9zUo-3D

17

6.g). ASSESSMENT METHODOLOGIES - DIRECT

ASSIGNMENTS STUD.

SEMINARS

TESTS/MODEL

EXAMS

UNIV.

EXAMINATION

STUD. LAB

PRACTICES

STUD. VIVA ☐ MINI/MAJOR

PROJECTS

☐

CERTIFICATIONS

☐ ADD-ON

COURSES

☐ OTHERS

6.h). ASSESSMENT METHODOLOGIES - INDIRECT

ASSESSMENT OF COURSE OUTCOMES

(BY FEEDBACK, ONCE)

STUDENT FEEDBACK ON

FACULTY (TWICE)

☐ASSESSMENT OF MINI/MAJOR

PROJECTS BY EXT. EXPERTS

☐ OTHERS

6.i). TEXT / REFERENCE BOOKS:

T/R BOOK TITLE/AUTHORS/PUBLICATION

Text Book 1.“I.J. Nagrath & D.P. Kothari”, “Electric Machines”, Tata

Mc Graw Hill Publishers, 3rd edition, 2004.

Text Book 2.“P.S. Bimbra”, “Electrical Machines”, Khanna Publishers, 7th Edition, 204.

Reference

Book

2.E. Clayton & N. M. Hancock “The Performance and

Design Of Direct Current Machines” 3rd

Edition

Pitman, London 1959.

Reference

Book “Abhijith Chakrabarthi & SubithaDebnath”, “Electrical

Machines”, Mc Graw Hill, 2015.

18

7. Topic wise Coverage [Micro Lesson Plan]

S.No.

Topic

Scheduled

date

Actual date

1 Over view of course 18/7/19

2 Unit-I: Introduction 19/7/19

3 Energy Conversion 20/7/19

4 Principle of Operation 23/7/19

5 Constructional Features 25/7/19

6 Armature windings 26/7/19

7 Armature windings 27/7/19

8 EMF equation 30/7/19

9 Armature Reaction 1/8/19

10 Methods of improving commutation 2/8/19

11 Types of generators 3/8/19

12 Critical resistance and critical speed 6/8/19

13 Characteristics of generators 8/8/19

14 Unit-II: Principle of operation of DC Motors 9/8/19

15 Back EMF 13/8/19

16 Torque Equation 16/8/19

17 Applications of Motors 17/8/19

18 Armature reaction and Commutation 20/8/19

19 Speed control of DC motors 22/8/19

20 3 and 4 point starters 23/8/19

19

21 Testing of DC machines 27/8/19

22 Losses of dc machine 29/8/19

23 Constant and variable losses 31/8/19

24 Calculation of efficiency 3/9/19

25 Condition for maximum efficiency 5/9/19

26 Unit-III: Methods of testing 6/9/19

27 Methods of testing 7/9/19

28 Brake test 17/9/19

29 Swinburnes test 19/9/19

30 Hopkinsons test 20/9/19

31 Fields test 24/9/19

32 Seperation of stray loss 26/9/19

33 Unit-IV: Constructional Details of a transformer 1/10/19

34 Hysterisis and eddy current loss 3/10/19

35 EMF Equation of transformer 4/10/19

36 Operation on no load and on load 5/10/19

37 Equivalent Circuit of 1-Ph transformer 15/10/19

38 Regulation of single phase transformer 17/10/19

39 All day efficiency 18/10/19

40 Effect of variations of frequency and supply voltage on

iron losses

19/10/19

41 Practical applications of transformers(Beyond Syllabus) 22/10/19

42 Unit-V: Testing of 1-phase transformers 29/10/19

43 OC and SC tests 31/10/19

20

8. TEACHING SCHEDULE

Subject UTILIZATION OF ELECTRICAL ENERGY

Text Books (to be purchased by the Students)

Book 1 1.“I.J. Nagrath & D.P. Kothari”, “Electric Machines”, Tata Mc Graw Hill

Publishers, 3rd

edition, 2004.

Book 2 2.“P.S. Bimbra”, “Electrical Machines”, Khanna Publishers, 7th Edition, 204.

Reference Books

Book 3 1.“Abhijith Chakrabarthi & SubithaDebnath”, “Electrical Machines”, Mc

Graw Hill, 2015.

44 Sumpner’s test 1/11/19

45 Predetermination of efficiency and regulation 2/11/19

46 Separation of losses test 5/11/19

47 Parallel operation with equal an unequal voltage ratios 7/11/19

48 Auto-transformers 8/11/19

49 Polyphase connections-y/y, y/∆, ∆/y, ∆/∆ and open∆ 12/11/19

21

Book 4 2.E. Clayton & N. M. Hancock “The Performance and Design Of Direct

Current Machines” 3rd

Edition Pitman, London 1959.

Unit

Topic Chapters Nos No of

classes

Book 1

Book 2

Book 3 Book 4

I

Introduction 5 4 1

Working principle

and construction of

dc generators

3 4 5

characteristics 3

4 5

II

DC Motors types 4 5 5 5

Characteristics 2 4 5 6

III

Methods of testing 1 6 1 6

IV

Single phase Transformers 6 8 7 4

Phasor diagrams 8 7 3

V

Transformer tests 8 7 4

Poly phase Transformers

7 7 3

Contact classes for syllabus coverage 42

Tutorial classes, Gaps, Lecture beyond Syllabus 7

Total No. of classes 49

22

10.

PPTS

23

11. UNIVERSITY PREVIOUS QUESTION PAPERS

27

12. MID EXAM DESCRIPTIVE QUESTION PAPERS WITH KEY

K. G. Reddy College of Engineering &Technology

(Approved by AICTE, Affiliated to JNTUH)

Name of the Exam: Mid Examinations 2018

Year-Sem & Branch: II-I-EEE Duration: 60 Min

Subject: EM-I Date & Session

Answer ANY TWO of the following Questions 2X5=10

Answers (Key)

1. List and Explain the parts of a DC Generator A DC generator has the following parts

Yoke

Pole of generator

Field winding

Armature of DC generator

Brushes of generator and Commutator

Bearing

Yoke of DC Generator

Yoke are the outer frame of DC generator serves two purposes,

It holds the magnetic pole cores of the generator and acts as cover of the generator.

It carries the magnetic field flux.

Q.NO QUESTION Bloom’s level Course

Outcome

1 List and Explain the parts of a DC Generator Understanding CO1

2 Definition of Pole Pitch, Coil Span and Coil Span

Remembering CO1

3 Derive An Emf Equation Of Dc Generator Apply CO2

4 Explain the swinburnes test on DC Machine Understanding CO3

https://www.electrical4u.com/principle-of-dc-generator/

https://www.electrical4u.com/what-is-magnetic-field/

28

In small generator, yoke are made of cast iron. Cast iron is cheaper in cost but heavier than

steel. But for large construction of DC generator, where weight of the machine is concerned,

lighter cast steel or rolled steel is preferable for constructing yoke of DC generator.

Pole Cores and Pole Shoes of DC Generator

Let's first discuss about pole core of DC generator. There are mainly two types of construction

available.

One: Solid pole core, where it is made of a solid single piece of cast iron or cast steel.

Two: Laminated pole core, where it made of numbers of thin, limitations of annealed steel

which are riveted together.

The pole shoes are so typically shaped, that, they spread out themagnetic flux in the air gap and

reduce the reluctance of the magnetic path.

Armature Core of DC Generator

The purpose of armature core is to hold the armature winding and provide low reluctance path

for the flux through the armature from N pole to S pole. Although a DC generator provides

direct current but induced current in the armature is alternating in nature.

Armature Winding of DC Generator

Armature winding are generally formed wound. These are first wound in the form of flat

rectangular coils and are then pulled into their proper shape in a coil puller. Various conductors

of the coils are insulated from each other.

Commutator of DC Generator

The commutator plays a vital role in DC generator. It collects current from armature and sends

it to the load as direct current. It actually takes alternating current from armature and converts it

to direct current and then send it to external load.

Brushes of DC Generator

The brushes are made of carbon. These are rectangular block shaped. The only function of these

carbon brushes of DC generator is to collect current from commutator segments. The brushes

are housed in the rectangular box shaped brush holder or brush box. As shown in figure, the

brush face is placed on the commutator segment which is attached to the brush holder.

Bearing of DC Generator

https://www.electrical4u.com/what-is-magnetic-field/#Magnetic-Flux-or-Magnetic-Lines-of-Force

https://www.electrical4u.com/armature-winding-pole-pitch-coil-span-commutator-pitch/

https://www.electrical4u.com/what-is-flux-types-of-flux/

https://www.electrical4u.com/electric-current-and-theory-of-electricity/


29

For small machine, ball bearing is used and for heavy duty DC generator, roller bearing is used.

The bearing must always be lubricated properly for smooth operation and long life of generator.

2. Define Pole Pitch, Coil Pitch and Coil Span

The pole pitch is defined as peripheral distance between centers of two adjacent poles in DC

machine.

Coil Span or Coil Pitch

Coil of dc machine is made up of one turn or multi turns of the conductor. If the coil is made up

of single turn or single loop of conductor, it is called single turn coil.

Definition of Coil Span

Coil span is defined as peripheral distance between two sides of a coil, measured in terms of

number of armature slots between them.

3. Derive the EMF equation of DC generator

E. M.F Equation

The derivation of EMF equation for DC generator has two parts:

Induced EMF of one conductor

Induced EMF of the generator

Derivation for Induced EMF of One Armature Conductor

For one revolution of the conductor,

Let,Φ = Flux produced by each pole in weber (Wb)

And P = number of poles in the DC generator.

therefore,Total flux produced by all the poles

And, Time taken to complete one revolution

Where, N = speed of the armature conductor in rpm.

Now, according to Faraday’s law of induction, the induced emf of the armature conductor is

denoted by “e” which is equal to rate of cutting the flux.

Therefore,

Induced emf of one conductor is

https://www.electrical4u.com/electrical-conductor/




30

Derivation for Induced EMF for DC Generator

Let us suppose there are Z total numbers of conductor in a generator, and arranged in such a

manner that all parallel paths are always in series.

Here,Z = total numbers of conductor

A = number of parallel paths

Then,Z/A = number of conductors connected in series

We know that induced emf in each path is same across the line

Therefore,Inducedemf of DC generator

E = emf of one conductor × number of conductor connected in series.

Induced emf of DC generator is

Simple wave wound generator

Numbers of parallel paths are only 2 = A

Therefore,

Induced emf for wave type of winding generator is

Simple lap-wound generator

Here, number of parallel paths is equal to number of conductors in one path

i.e. P = A,

Therefore, Induced emf for lap-wound generator is

4. Explain the swinburnes test on DC machine

Swinburne Test of DC Machine

This method is an indirect method of testing a DC machine. It is named after Sir James

Swinburne. Swinburne's test is the most commonly used and simplest method of testing of

31

shunt and compound wound DC machines which have constant flux. In this test the efficiency

of the machine at any load is pre-determined. We can run the machine as a motor or as a

generator. In this method of testing no load losses are measured separately and eventually we

can determine the efficiency.

The circuit connection for Swinburne's test is shown in figure below. The speed of the

machine is adjusted to the rated speed with the help of the shunt regulator R as shown in figure.

Advantages of Swinburne's Test

The main advantages of this test are:

This test is very convenient and economical as it is required very less power from supply

to perform the test.

Since constant losses are known, efficiency of Swinburne's test can be pre-determined

at any load.

Disadvantages of Swinburne's Test

The main disadvantages of this test are :

Iron loss is neglected though there is change in iron loss from no load to full load due to

armature reaction.

We cannot be sure about the satisfactory commutation on loaded condition because the

test is done on no-load.

We can’t measure the temperature rise when the machine is loaded. Power losses can

vary with the temperature.

In DC series motors, the Swinburne’s test cannot be done to find its efficiency as it is a

no load test.

https://www.electrical4u.com/series-wound-dc-motor-or-dc-series-motor/

32

13. MID EXAM OBJECTIVE QUESTION PAPERS WITH KEY

36

14. ASSIGNMENT TOPICS WITH MATERIAL

1. Explain the working principle of DC Generators

There are two types of generators, one is ac generator and other is DC generator. Whatever may

be the types of generators, it always converts mechanical power to electrical power.

An AC generator produces alternating power. A DC generator produces direct power. Both of

these generators produce electrical power, based on same fundamental principle of Faraday's

law of electromagnetic induction. According to this law, when a conductor moves in a magnetic

field it cuts magnetic lines of force, due to which an emf is induced in the conductor.

The magnitude of this induced emf depends upon the rate of change of flux (magnetic line

force) linkage with the conductor. This emf will cause a current to flow if the conductor circuit

is closed.

Hence the most basic tow essential parts of a generator are

Magnetic field

Conductors which move inside that magnetic field.

Single Loop DC Generator

In the figure above, a single loop of conductor of rectangular shape is placed between two

opposite poles of magnet.

https://www.electrical4u.com/electric-power-single-and-three-phase/

https://www.electrical4u.com/faraday-law-of-electromagnetic-induction/









37

Let's us consider, the rectangular loop of conductor is ABCD which rotates inside the magnetic

field about its own axis ab. When the loop rotates from its vertical position to its horizontal

position, it cuts the flux lines of the field. As during this movement two sides, i.e. AB and CD

of the loop cut the flux lines there will be an emf induced in these both of the sides (AB and

BC) of the loop

Now the loop is opened and connected it with a split ring as shown in the figure below. Split

ring are made out of a conducting cylinder which cuts into two halves or segments insulated

from each other. The external load terminals are connected with two carbon brushes which are

rest on these split slip ring segments.

Working Principle of DC Generator

It is seen that in the first half of the revolution current flows always along ABLMCD i.e. brush

no 1 in contact with segment a. In the next half revolution, in the figure the direction of the

induced current in the coil is reversed. But at the same time the position of the segments a and b

are also reversed which results that brush no 1 comes in touch with the segment b. Hence, the

current in the load resistance again flows from L to M. The wave from of the current through

the load circuit is as shown in the figure. This current is unidirectional.

https://www.electrical4u.com/electrical-resistance-and-laws-of-resistance/

38

This is basic working principle of DC generator, explained by single loop generator model. The

position of the brushes of DC generator is so arranged that the change over of the segments a

and b from one brush to other takes place when the plane of rotating coil is at right angle to the

plane of the lines of force. It is so become in that position, the induced emf in the coil is zero.

2. Explain the parts of a DC Machine

A DC generator has the following parts

Yoke

Pole of generator

Field winding



Bearing










available.







https://www.electrical4u.com/construction-of-dc-generator-yoke-pole-armature-brushes-of-dc-generator/




39




















3. Explain the demagnetizing and cross magnetizing effects

Demagnetizing and Cross Magnetizing Conductors:

The conductors which are responsible for producing demagnetizing and distortion effects are

shown in the Fig.1.





http://2.bp.blogspot.com/-P9NOAu6GsdA/UDrcJilkfaI/AAAAAAAAGRE/kyWa--Q5mkM/s1600/ccc162.jpeg

40

Fig. 1

The brushes are lying along the new position of MNA which is at angle θ from GNA. The

conductors in the region AOC = BOD = 2θ at the top and bottom of the armature are carrying

current in such a direction as to send the flux in armature from right to left. Thus these

conductors are in direct opposition to main field and called demagnetizing armature conductors.

The remaining armature conductors which are lying in the region AOD and BOC carry current

in such a direction as to send the flux pointing vertically downwards i.e. at right angles to the

main field flux. Hence these conductors are called cross magnetizing armature conductors

which will cause distortion in main field flux.

Fig. 2

Calculation of Demagnetizing and Cross Magnetizing Amp-Turns

Let us the number of demagnetizing and cross magnetizing amp-turns.

Let Z = Total number of armature conductors

P = Number of poles

I = Armature conductor current in Amperes

= Ia/2 for simplex wave winding

= Ia/P for simplex lap winding

θm = Forward lead of brush in mechanical degrees.

http://3.bp.blogspot.com/-4Lw4SL-7PNY/UDrcSwBcOPI/AAAAAAAAGRM/cuUJN4rXx-Q/s1600/ccc163.jpeg

41

The voltage generated in the armature, placed in a rotating magnetic field, of a DC generator is

alternating in nature. The commutation in DC machine or more specifically commutation in

DC generator is the process in which generated alternating current in the armature winding of a

dc machine is converted into direct current after going through the commutator and the

stationary brushes.

Again in DC Motor, the input DC is to be converted in alternating form in armature and that is

also done through commutation.

This transformation of current from the rotating armature of a DC machine to the stationary

brushes needs to maintain continuously moving contact between the commutator segments and

the brushes. When the armature starts to rotate, then the coils situated under one pole (let it be N

pole) rotates between a positive brush and its consecutive negative brush and the current flows

through this coil is in a direction inward to the commutator segments.

Then the coil is short circuited with the help of a brush for a very short fraction of time (1⁄500

sec). It is called commutation period. After this short-circuit time the armature coils rotates

under S pole and rotates between a negative brush and its succeeding positive brush. Then the

direction is reversed which is in the away from the commutator segments. This phenomena of

the reversal of current is termed as commutation process.

We get direct current from the brush terminal.

The commutation is called ideal if the commutation process or the reversal of current is

completed by the end of the short circuit time or the commutation period. If the reversal of

current is completed during the short circuit time then there is sparking occurs at the brush

contacts and the commutator surface is damaged due to overheating and the machine is called

poorly commutated.

https://www.electrical4u.com/voltage-or-electric-potential-difference/





https://www.electrical4u.com/working-or-operating-principle-of-dc-motor/

42

4. Explain about the 3 point starter

A 3 point starter in simple words is a device that helps in the starting and running of a shunt

wound DC motor or compound wound DC motor. Now the question is why these types of DC

motors require the assistance of the starter in the first case.

The only explanation to that is given by the presence of back emfEb, which plays a critical role

in governing the operation of the motor. The back emf, develops as the motor armature starts to

rotate in presence of the magnetic field, by generating action and counters the supply voltage.

This also essentially means, that the back emf at the starting is zero, and develops gradually as

the motor gathers speed.

The general motor emf equation

at starting is modified to E = Ia.Ra as at starting Eb = 0.

Thus we can well understand from the above equation that the current will be dangerously high

at starting (as armature resistance Ra is small) and hence its important that we make use of a

device like the 3 point starter to limit the starting current to an allowable lower value.

Let us now look into the construction and working of three point starter to understand how

the starting current is restricted to the desired value. For that let’s consider the diagram given

https://www.electrical4u.com/shunt-wound-dc-motor-dc-shunt-motor/


https://www.electrical4u.com/compound-wound-dc-motor-or-dc-compound-motor/

https://www.electrical4u.com/types-of-dc-motor-separately-excited-shunt-series-compound-dc-motor/






43

below showing all essential parts of the three point starter.

Construction of 3 Point Starter

Construction wise a starter is a variable resistance, integrated into number of sections as shown

in the figure beside. The contact points of these sections are called studs and are shown

separately as OFF, 1, 2, 3, 4, 5, RUN. Other than that there are 3 main points, referred to as

'L' Line terminal. (Connected to positive of supply.)

'A' Armature terminal. (Connected to the armature winding.)

'F' Field terminal. (Connected to the field winding.)

And from there it gets the name 3 point starter. Now studying the construction of 3 point starter

in further details reveals that, the point 'L' is connected to an electromagnet called overload

release (OLR) as shown in the figure. The other end of OLR is connected to the lower end of

conducting lever of starter handle where a spring is also attached with it and the starter handle

contains also a soft iron piece housed on it. This handle is free to move to the other side RUN

against the force of the spring. This spring brings back the handle to its original OFF position

under the influence of its own force. Another parallel path is derived from the stud '1', given to

the another electromagnet called No Volt Coil (NVC) which is further connected to terminal 'F'.

The starting resistance at starting is entirely in series with the armature. The OLR and NVC acts

as the two protecting devices of the starter.

5. Explain the methods of testing DC machines

Methods of Testing – direct, indirect, and regenerative testing:

Testing of DC motor



44

Testing of machines is used for finding losses, efficiency and temperature rise. Direct method is

used, for small machines. Indirect method is used for large shunt machines. In practice,

seinburne,s test are mostly used.

1.Direct method of testing

In direct method of testing the generator or motor is put on full load and whole of the power

developed by it is wasted,. Brake test is a typical example of direct test. The direct tests can be

used only on small machines.

2. Indirect method of testing

This method consists of measuring the losses and then calculating the efficiency. The simplest

of the indirect test is Swinburne’s test. Hopkinson test is commonly used test under this method

on shunt motors. This method also enables the determination of losses without actually loading

the machine.

3. Swinburne’s test (No load test)

In this method (simplest indirect method) the losses are measured separately and efficiency at

any desired load is pre-determined.

The iron and friction losses are determined by measuring the input to the machine on no-load,

the machine being run as a motor at normal voltage and speed.

4. Hopkinson’s Test (Back-to-back test or Regenerative test)

Through this test full-load testing of two d.c. shunt machines can be carries out, mainly identical

ones. In this test, power drawn from the supply only corresponds to no load losses of the

machines. Electrically these two machines are mechanically connected in parallel and controlled

in such a way that one machine acts as a generator and the other as motor.

Brake test:

Brake Test of DC Machine

DC Machines can be tested by three different methods namely Direct Method, Indirect Method

and Regenerative Method. Direct Method of testing of DC Machine, also known as Brake Test

(if carried out for a DC Motor) will be discussed in this post.

Direct method is suitable for small DC machines. In Direct Method, the DC machine is

subjected to rated load and the entire output power is wasted. The ratio of output power to the

input power gives the Efficiency of DC Machine. For a DC Generator the output power is

wasted in resistor.

http://www.theelectricalportal.com/2015/07/generator-protection.html

http://www.theelectricalportal.com/2015/08/speed-control-of-dc-motordc-series.html

http://1.bp.blogspot.com/-phCJIvu7BKI/VjDZiOTyMaI/AAAAAAAAAbo/hQsIFNynYQ8/s1600/download+(2).jpg

45

Direct Method of testing when conducted on a motor is also known as Brake Test. Brake Test of

DC Motor is carried out as shown in figure below.

A belt around the air cooled pulley has its end attached to the spring balance S1 and S2. Using

belt tightening hand wheels H1 and H2, the load of motor is adjusted to its rated value.

Assuming the spring balance to be calibrated in kilogram, then rated load on the DC motor is

given as

Motor Output Power = Torque x Angular Speed

= (Force x Radius) x Angular Speed

6. Give the constructional details of a transformer

Two coils of wire (called windings) are wound on some type of core material. In some cases the

coils of wire are wound on a cylindrical or rectangular cardboard form. In effect, the core

material is air and the transformer is called an AIR-CORE TRANSFORMER. Transformers

used at low frequencies, such as 60 hertz and 400 hertz, require a core of low-reluctance

https://2.bp.blogspot.com/-tzwatVbF3eE/WD-B0lnaS8I/AAAAAAAACeo/Xg_jdYXqDfwOeNtKfPHiXa_24suozFT1gCLcB/s1600/Brake+Test+of+DC+Motor.jpg

46

magnetic material, usually iron. This type of transformer is called an IRON-CORE

TRANSFORMER. Most power transformers are of the iron-core type.

The principle parts of a transformer and their functions are:

The CORE, which provides a path for the magnetic lines of flux.

The PRIMARY WINDING, which receives energy from the ac source.

The SECONDARY WINDING, which receives energy from the primary winding and

delivers it to the load.

The ENCLOSURE, which protects the above components from dirt, moisture, and mechanical

damage.

(i) CORE

There are two main shapes of cores used in laminated-steel-core transformers.

One is the HOLLOWCORE, so named because the core is shaped with a hollow square

through the center. This shape of core. Notice that the core is made up of many

laminations of steel it shows how the transformer windings are wrapped around both

sides of the core.

(ii) WINDINGS

As stated above, the transformer consists of two coils called WINDINGS which are wrapped

around a core. The transformer operates when a source of ac voltage is connected to one of the

windings and a load device is connected to the other. The winding that is connected to the

source is called the PRIMARY WINDING. The winding that is connected to the load is called

the SECONDARY WINDING. The primary is wound in layers directly on a rectangular

cardboard form.

47

7. Obtain the equivalent circuit of a single phase transformer

Equivalent impedance of transformer is essential to be calculated because the electrical power

transformer is an electrical power system equipment for estimating different parameters of

electrical power system which may be required to calculate total internal impedance of an

electrical power transformer, viewing from primary side or secondary side as per requirement.

This calculation requires equivalent circuit of transformer referred to primary or equivalent

circuit of transformer referred to secondary sides respectively. Percentage impedance is also

very essential parameter of transformer. Special attention is to be given to this parameter during

installing a transformer in an existing electrical power system. Percentage impedance of

different power transformers should be properly matched during parallel operation of power

transformers. The percentage impedance can be derived from equivalent impedance of

transformer so, it can be said that equivalent circuit of transformer is also required during

calculation of % impedance.

Equivalent Circuit of Transformer Referred to Primary

48

For drawing equivalent circuit of transformer referred to primary, first we have to establish

generalequivalent circuit of transformer then, we will modify it for referring from primary

side. For doing this, first we need to recall the complete vector diagram of a transformer which

is shown in the figure below.

Let us consider the transformation ratio be,

In the figure right, the applied voltage to the primary is V1 and voltage across the primary

winding is E1. Total current supplied to primary is I1. So the voltage V1 applied to the primary is

partly dropped by I1Z1 or I1R1 + j.I1X1 before it appears across primary winding.

The voltage appeared across winding is countered by primary induced emf E1.

http://www.electrical4u.com/voltage-or-electric-potential-difference/


http://www.electrical4u.com/electric-current-and-theory-of-electricity/



49

The equivalent circuit for that equation can be drawn as below,

From the vector diagram above, it is found that the total primary current I1 has two components,

one is no - load component Io and the other is load component I2′. As this primary current has

two a component or branches, so there must be a parallel path with primary winding of

transformer. This parallel path of currentis known as excitation branch of equivalent circuit of

transformer. The resistive and reactive branches of the excitation circuit can be represented as

The load component I2′ flows through the primary winding of transformer and

induced voltage across the winding is E1 as shown in the figure right. This induced voltage

E1transforms to secondary and it is E2 and load component of primary current I2′ is transformed

to secondary as secondary current I2. Current of secondary is I 2. So the voltage E2 across





50

secondary winding is partly dropped by I2Z2 or I2R2 + j.I2X2 before it appears across load. The

load voltage is V2.

From above equation, secondary impedance of transformer referred to primary is,

So, the complete equivalent circuit of transformer referred to primary is shown in the figure

below,

51

8. Explain about the Auto Transformers

Auto transformers:

Auto transformer is kind of electrical transformer where primary and secondary shares same

common single winding. So basically it’s a one winding transformer.

Theory of Auto Transformer

In Auto Transformer, one single winding is used as primary winding as well as secondary

winding. But in two windings transformer two different windings are used for primary and

secondary purpose. A diagram of auto transformer is shown below.

The winding AB of total turns N1 is considered as primary winding. This winding is tapped

from point ′C′ and the portion BC is considered as secondary. Let's assume the number of turns

in between points ′B′ and ′C′ is N2.

If V1 voltage is applied across the winding i.e. in between ′A′ and ′C′.

Hence, the voltage across the portion BC of the winding, will be,

https://www.electrical4u.com/electrical-power-transformer-definition-and-types-of-transformer/


52

As BC portion of the winding is considered as secondary, it can easily be understood that value

of constant ′k′ is nothing but turns ratio or voltage ratio of that auto transformer.

When load is connected between secondary terminals i.e.between ′B′ and ′C′, load current I2starts

flowing. The current in the secondary winding or common winding is the difference

Copper Savings in Auto Transformer

Now we will discuss the savings of copper in auto transformer compared to conventional two

winding transformer.

We know that weight of copper of any winding depends upon its length and cross-sectional

area. Again length of conductor in winding is proportional to its number of turns and cross-

sectional area varies with rated current.

So weight of copper in winding is directly proportional to product of number of turns and rated

https://www.electrical4u.com/emf-equation-of-transformer-turns-voltage-transformation-ratio-of-transformer/


53

current of the winding.

Therefore, weight of copper in the section AC proportional to,

and similarly, weight of copper in the section BC proportional to,

Hence, total weight of copper in the winding of auto transformer proportional to,

In similar way it can be proved, the weight of copper in two winding transformer is proportional

to,

N1I1 + N2I2

⇒ 2N1I1 (Since, in a transformer N1I1 = N2I2)

Let's assume, Wa and Wtw are weight of copper in auto transformer and two winding

transformer respectively,

54

∴ Saving of copper in auto transformer compared to two winding transformer,

9. Draw the phasor diagram of Transformer which operates on Load

The transformer is said to be loaded, when its secondary circuit is completed through an

impedance or load. The magnitude and phase of secondary current (i.e. current flowing through

secondary) I2 with respect to secondary terminals depends upon the characteristic of the load

i.e. current I2 will be in phase, lag behind and lead the terminal voltage V+2+ respectively when

the load is non-inductive, inductive and capacitive. The net flux passing through the core

remains almost constant from no-load to full load irrespective of load conditions and so core

losses remain almost constant from no-load to full load. Vector diagram for an ideal transformer

supplying inductive load is shown

55

Resistance and Leakage Reactance In actual practice, both of the primary and secondary

windings have got some ohmic resistance causing voltage drops and copper losses in the

windings. In actual practice, the total flux created does not link both of the primary and

secondary windings but is divided into three components namely the main or mutual flux Ø

linking both of the primary and secondary windings, primary leakage flux ØL1 linking with

primary winding only and secondary leakage flux ØL2 linking with secondary winding only.

The primary leakage flux ØL1 is produced by primary ampere-turns and is proportional to

primary current, number of primary turns being fixed. The primary leakage flux ØL1 is in phase

with I1 and produces self inducedemf ØL1 is in phase with I1 and produces self inducedemf

EL1 given as 2f L1 I1 in the primary winding.

The self inducedemf divided by the primary current gives the reactance of primary and is

denoted by X1.

56

i.e. X1 = EL1/I1 = 2πfL1I1/I1 = 2FL1,

Similarly leakage reactance of secondary X2 = EL2/E2 = 2fπL2I2/I2 = 2πfL2

Equivalent Resistance and Reactance. The equivalent resistances and reactance’s of transformer

windings referred to primary and secondary sides are given as below Referred to primary side

Equivalent resistance,

Equivalent resistance, = X'1 = Referred to secondary side Equivalent resistance,

Equivalent resistance, = X2 + K2X1 Where K is the transformation ratio.

57

15. TUTORIAL TOPICS AND QUESTIONS

UNIT-I

1 What is the necessity of laminating the armature core of a DC generator?

2 What do you mean by “back e.m.f” in DC Machine?

3 Mention the types of armature winding and their specifications.

4 Why electromagnets are preferred other than permanent magnets in large DC

machines?

5 Mention the reasons, why armature of a DC machine is made of laminated

silicon steel?

6 Write the basic equation of induced e.m.f in DC Generator.

7 For which kind of machines lap winding is preferred?

8 Define commutation Process in DC generators.

9 What is the main function of compensating winding?

10 What is the use of equalizer rings?

58

UNIT-II

1

i) With the help of speed torque characteristics, explain the motoring function of DC

compoundmotor.

ii) Which type of speed control techniques used in DC motor? Explain each one ofthem

2 Explain the principle of operation of dc motor with neat sketch.

3 Derive the expression for torque in dc motor.

4 Derive the terminal voltage and current expressions for the self and separately excited dc

motors

5 Explain the applications of self and separately excited dc motors.

6 Explain the following speed control methods.

a) Armaturecontrol b) fluxcontrol

7 Explain the Ward-Leonard system of speed control.

8 Explain the principle of operation of 3- point starter.

9 Explain the principle of operation of 4- point starter.

10 Explain the necessity of starters in dc machine and compare the 3-point and 4-point

starters. UNIT-III

1

Explain the experimental procedure to conduct ‘Retardation Test’ on a dc shunt machine

with the help of connection diagram. How the different losses are estimated from the test

results?

2

i) With neat circuit diagram, explain the procedure to conduct Swinburne’s test.

ii) List the calculations to be made to predetermine the efficiency of DC motor by using

Swinburne’s testresults.

3 Derive the expression for condition for maximum efficiency.

4 Explain how many losses are there in dc machine with equations.

59

UNIT-IV

5 Explain the procedure to conduct Hopkinson’s test with neat sketches.

6 Explain the procedure separate the losses in dc machine with neat sketches.

7 Classify the methods of testing? And compare them.

8 Indirect test is superior to the direct test justify this statement with proof.

9 With neat circuit diagram Calculate the efficiency by break test.

10 Draw and explain the internal and external characteristics of dc motor.

1 Give the concept of single phase ideal transformer. Describe its performance with the help of neat phasor diagram

2 Explicate in detail with a neat diagram about the constructional details of single

phasetransformers.

3 Derive the EMF equation of transformer? Hence derive the voltage ratio.

4 What is the efficiency of transformer? How the efficiency of transformer can

becalculated?

5 Discuss the effect of variable frequency and supply voltage on iron loss and

performance of the transformer?

6 Define voltage regulation of a transformer & enumerate the factors which influence

the magnitude of this change?

7 Draw the exact equivalent circuit of a transformer and describe briefly the various

parameters involved in it?

8 Define ‘efficiency’ and ‘all-day efficiency’ of a transformer. Mention how these are

affected by the power factor?

9 Draw the complete phasor diagram for a transformer, when the load power factor is

i) Lagging ii) Leading.

10 Discuss the different losses taking place in the transformer and their variation with the

load current.

60

UNIT-V

1 Show that an auto-transformer will result in saving copper in place of two winding

transformer.

2 With neat diagram, discuss the Construction of a three-phase transformer.

3 What are the disadvantages of current & voltage harmonics in transformers?

Discuss how these harmonics can be eliminated.

4 Describe the two possible ways of connections of 3-phase transformers with

relevant relations amongst voltage and currents

5 Discuss T-T connection of transformer with the help of neat phasor diagrams.

6 With neat diagram describe the Scott connection of a 3 phase transformer.

7 Describe the tertiary winding connection of transformer with the help of neat

diagrams.

8 With a neat sketch discuss the constructional details of a three phase transformer.

9 Justify thatpowerhandling capacity is reduced by 42.3%inOpen-deltaconnection.

10

Describe the advantage of using tertiary in a bank of star-star transformers and List

out the merits and demerits of a delta/star connected three phase transformer

61

16. UNIT WISE QUESTION BANK

a) Two and Three marks questions

Unit-I

1.What is the function of Generator?

An electric generator is a machine that converts mechanical energy into electrical energy

2.What is the basic principle of DC Generators?

This principle is nothing but the Faraday's law of electromagnetic induction. It states that,

'whenever the number of magnetic lines of force i.e. flux linking with a conductor or a coil

changes, an electromotive force is set up in that conductor or coil.

3. Give the Fleming's Right Hand Rule :

If three fingers of a right hand, namely thumb, index finger and middle finger are outstretched

so that everyone of them is at right angles with the remaining two, and if in this position index

finger is made to point in the direction of lines of flux, thumb in the direction of the relative

motion of the conductor with respect to flux then the Outstretched middle finger gives the

direction of the e.m.f. induced in the conductor.

4. What are the essential parts of DC Machines

The essential parts of the DC machine are

1.Magnetic Frame or Yoke 2. Pole-Cores and Pole-Shoes

3.Pole Coils or Field Coils 4. Armature

5.Armature Windings or Conductors 6. Commutator 7. Brushes and Bearings

5. What is the function of Yoke?

The outer frame or yoke serves two main purposes. They are

(i) It provides mechanical support for the poles and acts as a protecting cover for the whole

machine.

(ii)It carries the magnetic flux produced by the poles.

6. What is the function of Armature?

1. Armature core provides house for armature winding i.e. armature conductors.

62

2. To provide a path of low reluctance to the magnetic flux produced by the field winding.

7. What do mean by Separately Excited DC Generators ?

A DC generator whose field winding is supplied from an independent external DC source (e.g.,

a battery etc.) is called a separately excited generator.

8. Define the term Pole-pitch ?

It may be variously defined as :

a) The periphery of the armature divided by the number of poles of the generator i.e. the

distance between two adjacent poles.

b) It is equal to the number of armature conductors (or armature slots) per pole.

EX: If there are 48 conductors and 4 poles, the pole pitch is 48/4 = 12.

:

9. Define the term Conductor?

The length of a wire lying in the magnetic field and in which an e.m.f. is induced, is called a

conductor (or inductor).

10. What is meant by armature reaction?

The armature reaction means that the effect of magnetic field set up by armature current on the

distribution of flux under main poles of a generator. The armature magnetic field has two effects

: (i) It demagnetises or weakens the main flux (ii) It cross-magnetises or distorts it.

Unit-II

1.What is the function of Motor?

A machine that converts Electrical power into mechanical power is known as a DC motor.

2. Give the operating principle of DC Motor.

Its operation is based on the principle that when a current carrying conductor is placed in a magnetic

field, the conductor experiences a mechanical force and it is shown in the fig. The direction of this

force is given by Fleming’s left hand rule and magnitude is given by;

F = BIl newtons

3.Why commutator is employed in D.C.machines?

Conduct electricity between rotating armature and fixed brushes, convert

alternating emf into unidirectional emf (mechanical rectifier).

63

4.Distinguish between shunt and series field coil construction? Shunt field coils are wound with wires of small section and have more no of turns. Series field coils are wound with wires of larger cross section and have less no of turns.

5.How will you change the direction of rotation of d.c.motor? Either the field direction or direction of current through armature conductor is reversed.

6.What is back emf in D.C.motor?

As the motor armature rotates, the system of conductor come across alternate north

and South Pole magnetic fields causing an emf induced in the conductors. The

direction of the emf induced in the conductor is in opposite to current. As this emf

always opposes the flow of current in motor operation it is called as back emf.

7.What is the function of no-voltage release coil in D.C. motor starter?

As long as the supply voltage is on healthy condition the current through the NVR

coil produce enough magnetic force of attraction and retain the starter handle in ON

position against spring force. When the supply voltage fails or becomes lower than a

prescribed value then electromagnet may not have enough force to retain so handle

will come back to OFF position due to spring force automatically.

8.Enumerate the factors on which speed of a d.c.motor depends?

N= (V-IaRa)/Ф

So speed depends on air gap flux, resistance of armature, voltage

applied to armature.

9.What are the conditions to be fulfilled by for a dc shunt generator to

build back emf?

The generator should have residual flux, the field winding should be connected in such a manner that the flux setup by field in same direction as residual flux, the field resistance should be less than critical field resistance, load circuit resistance should be above critical resistance.

10.Define armature reaction in dc

machines?

The interaction between the main flux and armature flux cause disturbance called

as armature reaction.

64

Unit-III

1. Define a transformer?

A transformer is a static device which changes the alternating voltage from one level to

another.

2.What is the turns ratio and transformer ratio of transformer?

Turns ratio =

N2/ N1

Transformer =

E2/E1 = I1/

I2=K

3.Mention the difference between core and shell type transformers?

In core type, the windings surround the core considerably and in shell type the core

surrounds the windings i.e winding is placed inside the core

4.What is the purpose of laminating the core in a transformer? In order to minimise eddy current loss.

5.Give the emf equation of a transformer and define each term?

Emf induced in primary coil E1= 4.44fФmN1

volt Emf induced in secondary Coil E2 =4.44

fФmN2. f-----------freq of AC input

Ф---------maximum value of flux in thecore

N1, N2----Number of primary & secondary turns.

6.Does transformer draw any current when secondary is open? Why?

Yes, it (primary) will draw the current from the main supply in order to magnetize

the core and to supply for iron and copper losses on no load. There will not be any

current in the secondary since secondary is open.

7.Define voltage regulation of a transformer?

When a transformer is loaded with a constant primary voltage, the secondary voltage

decreases for lagging PF load, and increases for leading PF load because of its internal

resistance and leakage reactance. The change in secondary terminal voltage from no

load to full load expressed as a percentage of no load or full load voltage is termed as

regulation.

%regulation =E2-V2/E2 *100

V2>E2 for leading p.f load

V2<E2 for lagging p.f load

8.Define all day efficiency of a transformer?

It is computed on the basis of energy consumed during a certain period,

usually a day of 24 hrs. All day efficiency=output in kWh/input in kWh for

65

24 hrs.

9.Why transformers are rated in kVA?

Copper loss of a transformer depends on current & iron loss on voltage. Hence total

losses depend on Volt-Ampere and not on PF. That is why the rating of

transformers is in kVA and not in kW.

10.What determines the thickness of the lamination or stampings?

a. F

requ

ency

b. I

ronl

oss

66 Mr.Khamruddin Syed

Associate Professor,

Dept. of EEE

Unit - IV 1. What are the typical uses of autotransformer?

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

2. Explain on the material used for core construction?

The core is constructed by sheet steel laminations assembled to provide a continuous

magnetic path with minimum of air gap included. The steel used is of high silicon content

sometimes heat treated to produce a high permeability and a low hysteresis loss at the usual

operating flux densities. The eddy current loss is minimized by laminating the core, the

laminations being used from each other by light coat of coreplate vanish or by oxide layer

on the surface. The thickness of lamination varies from 0.35mm for a frequency of 50Hz

and 0.5mm for a frequency of25Hz.

3. How does change in frequency affect the operation of a given transformer?

With a change in frequency, iron and copper loss, regulation, efficiency & heating varies so

the operation of transformer is highly affected.

4. What is the angle by which no-load current will lag the ideal applied voltage?

In an ideal transformer, there are no copper & core loss i.e. loss free core. The no load

current is only magnetizing current therefore the no load current lags behind by angle 900.

However the winding possess resistance and leakage reactance and therefore the no load

current lags the applied voltage slightly less than900

. 5. Why are breathers used in transformers?

Breathers are used to entrap the atmospheric moisture and thereby not allowing it to pass on

to the transformer oil. Also to permit the oil inside the tank to expand and contract as its

temperature increases and decreases.

6. What is the function of transformer oil in a transformer? i. It provides good insulation

ii. Cooling.

7. Can the voltage regulation goes –ive? If so under what condition? Yes, if the load has leading PF.

8. Name the factors on which hysteresis loss depends?

1. Frequency 2. Volume of the core 3. Maximum flux density

9. Why the open circuit test on a transformer is conducted at rated voltage?

The open circuit on a transformer is conducted at a rated voltage because core loss depends

upon the voltage. This open circuit test gives only core loss or iron loss of the transformer.



Dept. of EEE

10. What is the purpose of providing Taps in transformer and where these are provided?

In order to attain the required voltage, taps are provided, normally at high voltages side(low

current).

Unit-V

1. What are the necessary tests to determine the equivalent circuit of the

transformer?

a. Open circuittest b. Short circuittest

2. Define efficiency of the transformer?

Transformer efficiency n= (output power/input power) x 100

3. Mention the difference between core and shell type transformers?

In core type, the windings surrounded the core considerably and in shell type the core

surround the windings i.e winding is placed inside the core

4. Full load copper loss in a transformer is 1600W. What will be the loss at half load?

If n is the ratio of actual load to full load then copper loss = n2 (F.L copper loss) Pc = (0.5)2

– 1600=400W.

5. Define all day efficiency of a transformer?

It is computed on the basis of energy consumed during a certain period, usually a day of 24hrs.

All day efficiency=output in kWh/input in kWh tor 24 hrs.

6. List the advantage of stepped core arrangement in a transformer?

c. To reduce the space effectively

d. To obtain reduce length of mean turn of the winding 3.To reduce I

2Rloss.

7. Why are breathers used in transformers? Breathers are used to entrap the atmospheric moisture and thereby not allowing it to

pass on to the transformer oil. Also to permit the oil inside the tank to expand and contract as its temperature increases and decreases

8. List the arrangement of stepped core arrangement in a transformer? To reduce the space effectively

To obtain reduced length of mean turn of the winding



Dept. of EEE

To reduce I2Rloss.

9. What are the applications of step-up & step-down transformer? Step-up transformers are used in generating stations. Normally the generated voltagewill be either 11kV. This voltage (11kV) is stepped up to 110kV or 220kV or 400Kv and transmitted through transmission lines (simply called as sending end voltage). Step-down transformers are used in receiving stations. The voltage are stepped down to 11kV or 22kV are stepped down to 3phase 400V by means of a distribution transformer and made available at consumer premises. The transformers used at generating stations are called power transformers.

10. How transformers are classified according to their construction?

1. Core type 2.shell type.

In core type, the winding (primary and secondary) surround the core and in shell type,

the core surround the winding.

b) Five marks question

Unit-I

1.Explain the principle of operation of DC Generator.

There are two types of generators, one is ac generator and other is DC generator. Whatever may

be the types of generators, it always converts mechanical power to electrical power.

An AC generator produces alternating power. A DC generator produces direct power. Both of

these generators produce electrical power, based on same fundamental principle of Faraday's

law of electromagnetic induction. According to this law, when a conductor moves in a magnetic

field it cuts magnetic lines of force, due to which an emf is induced in the conductor.

The magnitude of this induced emf depends upon the rate of change of flux (magnetic line

force) linkage with the conductor. This emf will cause a current to flow if the conductor circuit

is closed.

Hence the most basic tow essential parts of a generator are

https://www.electrical4u.com/electric-power-single-and-three-phase/










Dept. of EEE

Magnetic field

Conductors which move inside that magnetic field.

Single Loop DC Generator

In the figure above, a single loop of conductor of rectangular shape is placed between two

opposite poles of magnet.

Let's us consider, the rectangular loop of conductor is ABCD which rotates inside the magnetic

field about its own axis ab. When the loop rotates from its vertical position to its horizontal

position, it cuts the flux lines of the field. As during this movement two sides, i.e. AB and CD

of the loop cut the flux lines there will be an emf induced in these both of the sides (AB and

BC) of the loop

Now the loop is opened and connected it with a split ring as shown in the figure below. Split

ring are made out of a conducting cylinder which cuts into two halves or segments insulated

from each other. The external load terminals are connected with two carbon brushes which are

rest on these split slip ring segments.

Working Principle of DC Generator





Dept. of EEE

It is seen that in the first half of the revolution current flows always along ABLMCD i.e. brush

no 1 in contact with segment a. In the next half revolution, in the figure the direction of the

induced current in the coil is reversed. But at the same time the position of the segments a and b

are also reversed which results that brush no 1 comes in touch with the segment b. Hence, the

current in the load resistance again flows from L to M. The wave from of the current through

the load circuit is as shown in the figure. This current is unidirectional.

This is basic working principle of DC generator, explained by single loop generator model. The

position of the brushes of DC generator is so arranged that the change over of the segments a

and b from one brush to other takes place when the plane of rotating coil is at right angle to the

plane of the lines of force. It is so become in that position, the induced emf in the coil is zero.

2. Explain about function of Action of commutator


https://www.electrical4u.com/construction-of-dc-generator-yoke-pole-armature-brushes-of-dc-generator/



Dept. of EEE

The voltage generated in the armature, placed in a rotating magnetic field, of a DC generator is

alternating in nature. The commutation in DC machine or more specifically commutation in

DC generator is the process in which generated alternating current in the armature winding of a

dc machine is converted into direct current after going through the commutator and the

stationary brushes.

The commutation is called ideal if the commutation process or the reversal of current is

completed by the end of the short circuit time or the commutation period. If the reversal of

current is completed during the short circuit time then there is sparking occurs at the brush

contacts and the commutator surface is damaged due to overheating and the machine is called

poorly commutated.

Physical Concept of Commutation in DC Machine

For the explanation of commutation process, let us consider a DC machine having an armature

wound with ring winding. Let us also consider that the width of the commutator bar is equal to

the width of the brush and current flowing through the conductor is IC.










Dept. of EEE

Let the commutator is moving from left to right. Then the brush will move from rightto left.At

the first position, the brush is connected the commutator bar b (as shown in fig 1). Then the total

current conducted by the commutator bar b into the brush is 2IC.

When the armature starts to move right, then the brush comes to contact of bar a. Then the

armature current flows through two paths and through the bars a and b (as shown in fig 2). The

total current (2IC) collected by the brush remain same.

When the brush totally comes under the bar a (as shown in fig 5) and disconnected with the bar

b then current IC flows through the coil B in the counter-clockwise direction and the short

circuit is removed.

In this process the reversal of current or the process of commutation is done.

3.Give the constructional features of DC Machines

A DC generator has the following parts

Yoke

Pole of generator

Field winding



Bearing












Dept. of EEE



available.




























Dept. of EEE






4.Describe the types of armature windings

Definition of Pole Pitch

The pole pitch is defined as peripheral distance between centers of two adjacent poles in DC

machine.

Coil Span or Coil Pitch

Coil of dc machine is made up of one turn or multi turns of the conductor. If the coil is made up

of single turn or single loop of conductor, it is called single turn coil.

Definition of Coil Span

Coil span is defined as peripheral distance between two sides of a coil, measured in terms of

number of armature slots between them.



Dept. of EEE

Back Pitch (Yb)

A coil advances on the back of the armature. This advancement is measured in terms of

armature conductors and is called back pitch. It is equal to the number difference of the

conductor connected to a given segment of the commutator.

Front Pitch (Yf)

The number of armature conductors or elements spanned by a coil on the front is called front

pitch. Alternatively, the front pitch may be defined as the distance between the second

conductor of the next coil which are connected together at the front i.e. commutator end of the

armature. In other words, it is the number difference of the conductors connected together at the

back end of the armature. Both front and back pitches for lap and wave windings are shown in

the figure below.

Resultant Pitch (Y)

It is the distance between the beginning of one coil and the beginning of the next coil to which it

is connected. As a matter of precautions, it should be kept in mind that all these pitches, though

normally stated in terms of armature conductors, are also times of armature slots or commutator

bars.

Commutator Pitch

Commutator pitch is defined as the distance between two commutator segments which two ends

of same armature coil are connected. Commutator pitch is measured in terms of commutator

bars or segment.

Single Layer Armature Winding

Armature coil sides are placed in the armature slots in different manner. In some arrangement,

each slot is occupied by one side of an armature coil. In other words one coil sides is placed in



Dept. of EEE

each armature slot. This arrangement is referred as single layer winding.

Two Layer Armature Winding

In other types of armature Winding, arrangement every armature slot is occupied by two coil

sides, one on upper half and other on lower half of the slot. The coils in two layers winding are

so placed, that if one side is placed on upper half of the slot then other side is placed on the

lower half of some other slot at a distance of one coil pitch away.

5.Differentiate the lap winding and wave winding.

Armature windings are mainly of two types – lap winding and wave winding. Here we are

going to discuss about lap winding.

Lap winding is the winding in which successive coils overlap each other. It is named "Lap"

winding because it doubles or laps back with its succeeding coils.



Dept. of EEE

In this winding the finishing end of one coil is connected to one commutator segment and the

starting end of the next coil situated under the same pole and connected with same commutator

segment.

Here we can see in picture, the finishing end of coil - 1 and starting end of coil - 2 are both

connected to the commutator segment - 2 and both coils are under the same magnetic pole that

is N pole here.

Lap winding are of two types –

Simplex Lap Winding

Duplex Lap Winding

Let us start from 1st conductor,

Back connections Front connections

1 to (1+YB) = (1+5) = 6 6 to (6-YF) = (6-3) = 3

3 to (3+5) = 8 8 to (8-3) = 5

5 to (5+5) = 10 10 to (10-3) = 7

7 to (7+5) = 12 12 to (12-3) = 9

9 to (9+5) = 14 14 to (14-3) = 11

11 to (11+5) = 16 16 to (16-3) =13



Dept. of EEE

13 to (13+5) = 18 = (18-16) = 2 2 to (18-3) = 15

15 to (15+5) = 20 = (20-16) = 4 4 to (20-3) = 17 = (17-16) = 1

Advantages of Lap Winding

This winding is necessarily required for large current application because it has more

parallel paths.

It is suitable for low voltage and high current generators.

Disadvantages of Lap Winding

It gives less emf compared to wave winding. This winding requires more no. of

conductors for giving the same emf, it results high winding cost.

It has less efficient utilization of space in the armature slots

6.Briefly explain about Simplex and Multiplex windings





Dept. of EEE

In simplex wave winding

Back pitch (YB) and front pitch (YF) are both odd and are of same sign. Back pitch and front

pitch are nearly equals to the pole pitch and may be equal or differ by ±2. + For progressive

winding, - for retrogressive winding.

Here, Z is the no of conductors in the winding. P is the no of poles. Average pitch (YA) must be

an integer no. because it may close itself. ±2 is taken because after one round of the armature

the winding falls sort of two conductors. If average pitch is taken Z/P then after one round the

winding will close itself without including all coil sides. Since average pitch must be an integer,

this winding is not possible with any no. of conductors. Let us take 8 conductors in a 4 pole

machine.

Being fractional no the wave winding is no possible but if there was 6 conductors then the

winding can be done. Since,

For this problem the DUMMY COILS are introduced.

Construction of Wave Winding

Let us develop a simplex and progressive wave winding diagram of a machine having 34

conductor in 17 slots and 4 poles. Average pitch:

Now we have to construct a table for the connection diagram:



Dept. of EEE

7.Give the importance of laminated armature and E.M.F Equation

Use of laminated armature

Eddy current losses are directly proportional to area of armature or more precisely the path of

motion. Now consider an armature with single piece of iron.

In this case with single piece of armature. In this Eddy current losses are represented by white

lines, Now consider laminated armature with some laminations (in practical we have each

lamination of around 0.4 mm).

In laminated armature eddy current losses are reduced to very less or '0' quantity. That is why

armature of DC machines (either motor or generator) is laminated.

E. M.F Equation

The derivation of EMF equation for DC generator has two parts:

Induced EMF of one conductor

Induced EMF of the generator

Derivation for Induced EMF of One Armature Conductor




Dept. of EEE

For one revolution of the conductor,

Let,Φ = Flux produced by each pole in weber (Wb)

And P = number of poles in the DC generator.

therefore,Total flux produced by all the poles

And, Time taken to complete one revolution

Where, N = speed of the armature conductor in rpm.

Now, according to Faraday’s law of induction, the induced emf of the armature conductor is

denoted by “e” which is equal to rate of cutting the flux.

Therefore,

Induced emf of one conductor is

Derivation for Induced EMF for DC Generator

Let us suppose there are Z total numbers of conductor in a generator, and arranged in such a

manner that all parallel paths are always in series.

Here,Z = total numbers of conductor

A = number of parallel paths

Then,Z/A = number of conductors connected in series

We know that induced emf in each path is same across the line

Therefore,Inducedemf of DC generator






Dept. of EEE

E = emf of one conductor × number of conductor connected in series.

Induced emf of DC generator is

Simple wave wound generator

Numbers of parallel paths are only 2 = A

Therefore,

Induced emf for wave type of winding generator is

Simple lap-wound generator

Here, number of parallel paths is equal to number of conductors in one path

i.e. P = A,

Therefore, Induced emf for lap-wound generator is

8.Briefly explain the Armature reaction

In a DC machine, the carbon brushes are always placed at the magnetic neutral axis. In no load

condition, the magnetic neutral axis coincides with the geometrical neutral axis. Now,

when the machine is loaded, the armature flux is directed along the inter polar axis (the

axis in between the magnetic poles)and is triangular in wave shape. This results an

armature current flux directed along the brush axis and causes cross magnetization of

the main field. This cross magnetization effect results in the concentration of flux at the




Dept. of EEE

trailing pole tip in generator action and at the leading pole tip in motor action.

The armature reaction is the effect of the armature flux on the main flux. In case of

a DC motor the resultant flux is strengthened at the leading pole and weakened at the

trailing pole tips.

Inter polar windings are always kept in series with armature, so inter polar winding

carries the armature current; therefore works satisfactorily irrespective of load, the

direction of rotation or the mode of operation. Inter poles are made narrower to ensure

that they influence only the coil undergoing commutation and its effect does not spread

to the other coils. The base of the inter poles is made wider to avoid saturation and to

improve response.

Compensating Winding

Commutation problem is not the only problem in DC machines. At heavy loads, the cross

magnetizing armature reaction may cause very high flux density in the trailing pole tip in

generator action and leading pole tip in the motor action.

Consequently, the coil under this tip may develop induced voltage high enough to cause a flash

over between the associated adjacent commutator segments particularly, because this coil is

physically close to the commutation zone (at the brushes) where the air temperature might be

already high due to commutation process.

This flash over may spread to the neighboring commutator segments, leading ultimately to a

complete fire over the commutator surface from brush to brush. Also, when the machine is

subjected to rapidly fluctuating loads, then the voltage L× di/dt, that appears across the adjacent

commutator segments may reach a value high enough to cause flash over between the adjacent

commutator segments. This would start from the center of pole as the coil below it possesses the

maximum inductance. This may again cause a similar fire as described above. This problem is

more acute while the load is decreasing in generating action and increasing in motor action as


https://www.electrical4u.com/dc-motor-or-direct-current-motor/



Dept. of EEE

then, the induced e.m.f and voltage L× di/dt will support each other. The above problems are

solved by use of compensating winding.

9.Describe the importance of Cross magnetizing and de-magnetizing AT/pole

Demagnetizing and Cross Magnetizing Conductors

The conductors which are responsible for producing demagnetizing and distortion effects are

shown in the Fig.1.

Fig. 1

The brushes are lying along the new position of MNA which is at angle θ from GNA. The

conductors in the region AOC = BOD = 2θ at the top and bottom of the armature are carrying

current in such a direction as to send the flux in armature from right to left. Thus these

conductors are in direct opposition to main field and called demagnetizing armature conductors.

The remaining armature conductors which are lying in the region AOD and BOC carry current

in such a direction as to send the flux pointing vertically downwards i.e. at right angles to the

main field flux. Hence these conductors are called cross magnetizing armature conductors

which will cause distortion in main field flux.

These conductors are shown in the Fig. 2

http://2.bp.blogspot.com/-P9NOAu6GsdA/UDrcJilkfaI/AAAAAAAAGRE/kyWa--Q5mkM/s1600/ccc162.jpeg



Dept. of EEE

Fig. 2

Calculation of Demagnetizing and Cross Magnetizing Amp-Turns

Let us the number of demagnetizing and cross magnetizing amp-turns.

Let Z = Total number of armature conductors

P = Number of poles

I = Armature conductor current in Amperes

= Ia/2 for simplex wave winding

= Ia/P for simplex lap winding

θm = Forward lead of brush in mechanical degrees.

10.Explain the compensating winding

Compensating Windings

The cross-magnetizing effect of armature reaction may cause trouble in d.c. machines subjected

to large fluctuations in load. In order to neutralize the cross magnetizing effect of armature

reaction, a compensating winding is used.

The compensating windings consist of a series of coils embedded in slots in the pole faces.

These coils are connected in series with the armature. The series-connected compensating

windings produce a magnetic field, which varies directly with armature current. Because the

compensating windings are wound to produce a field that opposes the magnetic field of the

armature, they tend to cancel the cross magnetizing effect of the armature magnetic field.

http://3.bp.blogspot.com/-4Lw4SL-7PNY/UDrcSwBcOPI/AAAAAAAAGRM/cuUJN4rXx-Q/s1600/ccc163.jpeg



Dept. of EEE

The neutral plane will remain stationary and in its original position for all values of armature

current. Because of this, once the brushes have been set correctly, they do not have to be moved

again.

Interpoles

Another way to reduce the effects of armature reaction is to place small auxiliary poles

called"interpoles" between the main field poles. The interpoles have a few turns of large wire

and are connected in series with the armature.

Interpoles are wound and placed so that each interpole has the same magnetic polarity as the

main pole ahead of it, in the direction of rotation. The field generated by the interpoles produces

the same effect as the compensating winding.

This field, in effect, cancels the armature reaction for all values of load current by causing a

shift in the neutral plane opposite to the shift caused by armature reaction. The amount of shift

caused by the interpoles will equal the shift caused by armature reaction since both shifts are a

result of armature current.

UNIT-II

https://images-blogger-opensocial.googleusercontent.com/gadgets/proxy?url=http://1.bp.blogspot.com/-6EzF4Xd4dcw/VHFQgjqlmtI/AAAAAAAABOY/m-Zvo2OFqHM/s400/Reliance_Field_Windings.jpg&container=blogger&gadget=a&rewriteMime=image/*



Dept. of EEE

1.Describe the principle of operation of DC Motor

A DC motor in simple words is a device that converts electrical energy (direct current system)

into mechanical energy. It is of vital importance for the industry today, and is equally important

for engineers to look into the working principle of DC motor in details that has been discussed

in this article. In order to understand the operating principle of DC motor we need to first look

into its constructional feature.

The very basic construction of a DC motor contains a current carrying armature which is

connected to the supply end through commutator segments and brushes. The armature is placed

in between north south poles of a permanent or an electromagnet as shown in the diagram

above.

As soon as we supply direct current in the armature, a mechanical force acts on it due to

electromagnetic effect of the magnet. Now to go into the details of the operating principle of

DC motor its important that we have a clear understanding of Fleming’s left hand rule to

determine the direction of force acting on the armature conductors of DC motor.

Then the force on the left hand side armature conductor

,


https://www.electrical4u.com/construction-of-dc-motor-yoke-poles-armature-field-winding-commutator-brushes-of-dc-motor/


https://www.electrical4u.com/fleming-left-hand-rule-and-fleming-right-hand-rule/



Dept. of EEE

Similarly force on the right hand side conductor

Therefore, we can see that at that position the force on either side is equal in magnitude but

opposite in direction. And since the two conductors are separated by some distance w = width

of the armature turn, the two opposite forces produces a rotational force or a torque that results

in the rotation of the armature conductor.

2.Determine the EMF Equation and Torque equation of DC Motor

Consider a DC generator with the following parameters,

P = number of field poles

Ø = flux produced per pole in Wb (weber)

Z = total no. of armature conductors

A = no. of parallel paths in armature

N = rotational speed of armature in revolutions per min. (rpm)

Now,

Average emf generated per conductor is given by dΦ/dt (Volts) ... eq. 1

Flux cut by one conductor in one revolution = dΦ = PΦ ….(Weber),

Number of revolutions per second (speed in RPS) = N/60

Therefore, time for one revolution = dt = 60/N (Seconds)

From eq. 1, emf generated per conductor = dΦ/dt = PΦN/60 (Volts) …..(eq. 2)

Above equation-2 gives the emf generated in one conductor of the generator. The conductors

are connected in series per parallel path, and the emf across the generator terminals is equal to

the generated emf across any parallel path.

http://www.electricaleasy.com/2012/12/basic-construction-and-working-of-dc.html

http://www.electricaleasy.com/2012/12/armature-winding-of-dc-machine.html



Dept. of EEE

Therefore, Eg = PΦNZ / 60A

For simplex lap winding, number of parallel paths is equal to the number of poles (i.e. A=P),

Therefore, for simplex lap wound dc generator, Eg = PΦNZ / 60P

For simplex wave winding, number of parallel paths is equal to 2 (i.e P=2),

Therefore, for simplex wave wound dc generator, Eg = PΦNZ / 120

Torque Equation Of A DC Motor

When armature conductors of a DC motor carry current in the presence of stator field flux, a

mechanical torque is developed between the armature and the stator. Torque is given by the

product of the force and the radius at which this force acts.

Torque T = F × r (N-m) …where, F = force and r = radius of the armature

Work done by this force in once revolution = Force × distance = F × 2πr (where, 2πr =

circumference of the armature)

Net power developed in the armature = word done / time

= (force × circumference × no. of revolutions) / time

= (F × 2πr × N) / 60 (Joules per second) .... eq. 2.1

But, F × r = T and 2πN/60 = angular velocity ω in radians per second. Putting these in the above

equation 2.1

Net power developed in the armature = P = T × ω (Joules per second)

Armature Torque (Ta)

The power developed in the armature can be given as, Pa = Ta × ω = Ta × 2πN/60

The mechanical power developed in the armature is converted from the electrical power,

Therefore, mechanical power = electrical power

http://www.electricaleasy.com/2014/01/basic-working-of-dc-motor.html



Dept. of EEE

That means, Ta × 2πN/60 = Eb.Ia

We know, Eb = PΦNZ / 60A

Therefore, Ta × 2πN/60 = (PΦNZ / 60A) × Ia

Rearranging the above equation,

Ta = (PZ / 2πA) × Φ.Ia (N-m)

The term (PZ / 2πA) is practically constant for a DC machine. Thus, armature torque is directly

proportional to the product of the flux and the armature current i.e. Ta ∝Φ.Ia

Shaft Torque (Tsh)

Due to iron and friction losses in a dc machine, the total developed armature torque is not

available at the shaft of the machine. Some torque is lost, and therefore, shaft torque is always

less than the armature torque.

Shaft torque of a DC motor is given as,

Tsh = output in watts / (2πN/60) ....(where, N is speed in RPM)

3.Describe the Characteristics Of DC Shunt Motors

Torque Vs. Armature Current (Ta-Ia)

In case of DC shunt motors, we can assume the field flux ɸ to be constant. Though at heavy

loads, ɸ decreases in a small amount due to increased armature reaction. As we are neglecting

the change in the flux ɸ, we can say that torque is proportional to armature current. Hence, the

Ta-Ia characteristic for a dc shunt motor will be a straight line through the origin.

Since heavy starting load needs heavy starting current, shunt motor should never be started

on a heavy load.

Speed Vs. Armature Current (N-Ia)

As flux ɸ is assumed to be constant, we can say N ∝Eb. But, as back emf is also almost

constant, the speed should remain constant. But practically, ɸ as well as Eb decreases with

increase in load. Back emfEb decreases slightly more than ɸ, therefore, the speed decreases

slightly. Generally, the speed decreases only by 5 to 15% of full load speed. Therefore, a shunt

motor can be assumed as a constant speed motor. In speed vs. armature current characteristic in

http://www.electricaleasy.com/2014/01/losses-in-dc-machine.html

http://www.electricaleasy.com/2013/01/armature-reaction-in-dc-machines.html



Dept. of EEE

the following figure, the straight horizontal line represents the ideal characteristic and the actual

characteristic is shown by the dotted line.

4.Describe the Characteristics Of DC Compound and series Motor

DC compound motors have both series as well as shunt winding. In a compound motor, if series

and shunt windings are connected such that series flux is in direction as that of the shunt flux

then the motor is said to be cumulatively compounded. And if the series flux is opposite to the

direction of the shunt flux, then the motor is said to be differentially compounded.

Characteristics of both these compound motors are explained below.

(a) Cumulative compound motor

Cumulative compound motors are used where series characteristics are required but the load is

likely to be removed completely. Series winding takes care of the heavy load, whereas the shunt

winding prevents the motor from running at dangerously high speed when the load is suddenly

removed. These motors have generally employed a flywheel, where sudden and temporary loads

are applied like in rolling mills.

(b) Differential compound motor

Since in differential field motors, series flux opposes shunt flux, the total flux decreases with

increase in load. Due to this, the speed remains almost constant or even it may increase slightly



Dept. of EEE

with increase in load (N ∝Eb/ɸ). Differential compound motors are not commonly used, but

they find limited applications in experimental and research work.

Characteristics Of DC Series Motors

Torque Vs. Armature Current (Ta-Ia)

This characteristic is also known as electrical characteristic. We know that torque is directly

proportional to the product of armature current and field flux, Ta ∝ɸ.Ia. In DC series motors,

field winding is connected in series with the armature, i.e. Ia = If. Therefore, before magnetic

saturation of the field, flux ɸ is directly proportional to Ia. Hence, before magnetic saturation Ta

α Ia2. Therefore, the Ta-Ia curve is parabola for smaller values of Ia.

After magnetic saturation of the field poles, flux ɸ is independent of armature current Ia.

Therefore, the torque varies proportionally to Ia only, T ∝Ia.Therefore, after magnetic

saturation, Ta-Ia curve becomes a straight line.

The shaft torque (Tsh) is less than armature torque (Ta) due to stray losses. Hence, the curve

Tsh vs Ia lies slightly lower.

In DC series motors, (prior to magnetic saturation) torque increases as the square of armature

current, these motors are used where high starting torque is required.

Speed Vs. Armature Current (N-Ia)

We know the relation, N ∝Eb/ɸ

For small load current (and hence for small armature current) change in back emfEb is small

and it may be neglected. Hence, for small currents speed is inversely proportional to ɸ. As we

know, flux is directly proportional to Ia, speed is inversely proportional to Ia. Therefore, when

armature current is very small the speed becomes dangerously high. That is why a series motor

should never be started without some mechanical load.




Dept. of EEE

But, at heavy loads, armature current Ia is large. And hence, speed is low which results in

decreased back emfEb. Due to decreased Eb, more armature current is allowed.

Speed Vs. Torque (N-Ta)

This characteristic is also called as mechanical characteristic. From the above

two characteristics of DC series motor, it can be found that when speed is high, torque is low

and vice versa.

5.Determine the concept of Armature reaction

In a DC machine, the carbon brushes are always placed at the magnetic neutral axis. In no load

condition, the magnetic neutral axis coincides with the geometrical neutral axis. Now, when the

machine is loaded, the armature flux is directed along the inter polar axis (the axis in between

the magnetic poles)and is triangular in wave shape. This results an armature current flux

directed along the brush axis and causes cross magnetization of the main field. This cross

magnetization effect results in the concentration of flux at the trailing pole tip in generator

action and at the leading pole tip in motor action.

The armature reaction is the effect of the armature flux on the main flux. In case of a DC

motor the resultant flux is strengthened at the leading pole and weakened at the trailing pole

tips.







Dept. of EEE

Where,

Ia = armature current,

Z = total number of conductors,

P = total number of poles,

β = angular shift of carbon brushes (in electrical Degrees).

6.Explain the speed control methods for DC Motor

Speed control means intentional change of the drive speed to a value required for performing

the specific work process. Speed control is a different concept from speed regulation where

there is natural change in speed due change in load on the shaft. Speed control is either done

manually by the operator or by means of some automatic control device.One of the important

features of DC motor is that its speed can be controlled with relative ease. We know that the

emf equation of DC motor is given as,

N = 60A E / PZØ

N = E / kØ

where, k = PZ/60A

N = V - Ia Ra / kØ

Therefore speed (N) of 3 types of DC motor – SERIES, SHUNT and COMPOUND can be

controlled by changing the quantities on RHS of the expression. So speed can be varied by

changing

Terminal voltage of the armature V.

External resistance in armature circuit Ra.

Flux per pole φ.

The first two cases involve change that affects armature circuit and the third one involves

change in magnetic field. Therefore speed control of DC motor is classified as

Armature control methods









Dept. of EEE

Field control methods.

Speed Control of DC Series Motor

Speed control of DC series motor can be done either by armature control or by field control.

Armature Control of DC Series Motor

Speed adjustment of DC series motor by armature control may be done by any one of the

methods that follow,

Armature Resistance Control Method:

This is the most common method employed. Here the controlling resistance is connected

directly in series with the supply of the motor as shown in the fig.

The power loss in the control resistance of DC series motor can be neglected because this

control method is utilized for a large portion of time for reducing the speed under light load

condition. This method of speed control is most economical for constant torque. This method of

speed control is employed for DC series motor driving cranes, hoists, trains etc.

Shunted Armature Control:

The combination of a rheostat shunting the armature and a rheostat in series with the armature is

involved in this method of speed control. The voltage applied to the armature is varies by

varying series rheostat R1. The exciting current can be varied by varying the armature shunting

resistance R2. This method of speed control is not economical due to considerable power losses

in speed controlling resistances. Here speed control is obtained over wide range but below

normal speed.




Dept. of EEE

Armature terminal voltage control:

7.Explain the operating principle of 3 point starter .

A 3 point starter in simple words is a device that helps in the starting and running of a shunt

wound DC motor or compound wound DC motor. Now the question is why these types of DC

motors require the assistance of the starter in the first case.

The only explanation to that is given by the presence of back emfEb, which plays a critical role

in governing the operation of the motor. The back emf, develops as the motor armature starts to

rotate in presence of the magnetic field, by generating action and counters the supply voltage.

This also essentially means, that the back emf at the starting is zero, and develops gradually as

the motor gathers speed.

The general motor emf equation

at starting is modified to E = Ia.Ra as at starting Eb = 0.

Thus we can well understand from the above equation that the current will be dangerously high

at starting (as armature resistance Ra is small) and hence its important that we make use of a

device like the 3 point starter to limit the starting current to an allowable lower value.

Let us now look into the construction and working of three point starter to understand how

the starting current is restricted to the desired value. For that let’s consider the diagram given












Dept. of EEE

below showing all essential parts of the three point starter.

Construction of 3 Point Starter

Construction wise a starter is a variable resistance, integrated into number of sections as shown

in the figure beside. The contact points of these sections are called studs and are shown

separately as OFF, 1, 2, 3, 4, 5, RUN. Other than that there are 3 main points, referred to as




And from there it gets the name 3 point starter. Now studying the construction of 3 point starter

in further details reveals that, the point 'L' is connected to an electromagnet called overload

release (OLR) as shown in the figure. The other end of OLR is connected to the lower end of

conducting lever of starter handle where a spring is also attached with it and the starter handle

contains also a soft iron piece housed on it. This handle is free to move to the other side RUN

against the force of the spring. This spring brings back the handle to its original OFF position

under the influence of its own force. Another parallel path is derived from the stud '1', given to

the another electromagnet called No Volt Coil (NVC) which is further connected to terminal 'F'.

The starting resistance at starting is entirely in series with the armature. The OLR and NVC acts

as the two protecting devices of the starter.

8.Explain the Working Principle of Four Point Starter





Dept. of EEE

The 4 point starter like in the case of a 3 point starter also acts as a protective device that helps

in safeguarding the armature of the shunt or compound excited DC motor against the high

starting current produced in the absence of back emf at starting.

The 4 point starter has a lot of constructional and functional similarity to a three point starter,

but this special device has an additional point and a coil in its construction. This naturally brings

about some difference in its functionality, though the basic operational characteristic remains

the same. The basic difference in circuit of 4 point starter as compared to 3 point starter is that

the holding coil is removed from the shunt field current and is connected directly across the line

with current limiting resistance in series.

Now to go into the details of operation of 4 point starter, lets have a look at its constructional

diagram, and figure out its point of difference with a 3 point starter.

Construction and Operation of Four Point Starter

A 4 point starter as the name suggests has 4 main operational points, namely




Like in the case of the 3 point starter, and in addition to it there is, A 4th point N. (Connected to

the No Voltage Coil NVC)

The remarkable difference in case of a 4 point starter is that the No Voltage Coil is connected

independently across the supply through the fourth terminal called 'N' in addition to the 'L', 'F'

and 'A'. As a direct consequence of that, any change in the field supply current does not bring

about any difference in the performance of the NVC. Thus it must be ensured that

no voltage coil always produce a force which is strong enough to hold the handle in its 'RUN'

position, against force of the spring, under all the operational conditions. Such a current is

adjusted through No Voltage Coil with the help of fixed resistance R connected in series with

the NVC using fourth point 'N' as shown in the figure above.

Apart from this above mentioned fact, the 4 point and 3 point starters are similar in all other

ways like possessing is a variable resistance, integrated into number of sections as shown in the

figure above. The contact points of these sections are called studs and are shown separately as

OFF, 1, 2, 3, 4, 5, RUN, over which the handle is free to be maneuvered manually to regulate

https://www.electrical4u.com/3-point-starter-working-principle-and-construction-of-three-point-starter/



https://www.electrical4u.com/3-point-starter-working-principle-and-construction-of-three-point-starter/






Dept. of EEE

the starting current with gathering speed.

Now to understand its way of operating lets have a closer look at the diagram given above.

Considering that supply is given and the handle is taken stud No.1, then the circuit is complete

and line current that starts flowing through the starter. In this situation we can see that the

current will be divided into 3 parts, flowing through 3 different points.

1 part flows through the starting resistance (R1+ R2+ R3…..) and then to the armature.

A 2nd part flowing through the field winding F.

And a 3rd part flowing through the no voltage coil in series with the protective resistance R.

So the point to be noted here is that with this particular arrangement any change in the shunt

field circuit does not bring about any change in the no voltage coil as the two circuits are

independent of each other. This essentially means that the electromagnet pull subjected upon the

soft iron bar of the handle by the no voltage coil at all points of time should be high enough to

keep the handle at its RUN position, or rather prevent the spring force from restoring the handle

at its original OFF position, irrespective of how the field rheostat is adjusted. This marks the

operational difference between a 4 point starter and a 3 point starter. As otherwise both are



Dept. of EEE

almost similar and are used for limiting the starting current to a shunt wound DC

motor or compound wound DC motor, and thus act as a protective device.

9.Determine the Testing of D.C. machines - Losses – Constant & Variable losses

The testing of dc machine is needed for proper fabrication and smooth trouble free operation.

The tests which are mainly needed for these purposes are…

Open circuit test

Short circuit test

Load test

Determination of efficiency

Open Circuit Test

The open circuit test is needed to determine the open circuit characteristic or magnetizing

characteristic of a dc machine. The open circuit test gives the mmf and hence the

excitation current or field current needed to generate the required voltage on no load at a fixed

speed. The open circuit characteristic curve shows the variation of induced emf as a function of

field current at constant speed and zero load current.

This curve is practically determined by running the machine as a separately excited generator on

on-load. This curve is also called no load saturation curve as it gives the saturation characteristic

of the generator.

Short Circuit Test

The short circuit test is needed to determine the voltage drop across the armature at any load

current. In this testing of DC machine the armature is short circuited with an ammeter to get

the short circuit current. Short circuit test gives the short circuit characteristic curve which

shows the variation of short circuit current as a function of excitation current.

Load Test

The load testing of DC machine is needed to determine the rating of a machine. When we run a

machine, then some energy is lost in the machine, which converts into the heat and cause

temperature rise. If a machine produces too much heat then it can affect the insulation of the

machine and ultimately it can cause the breakdown of the machine. Therefore, the load must be

set to a value that it can operate within the temperature limit. The maximum value of the load

that can be delivered by the machine without any harm is called the continuous rating of that

machine.

Determination of Efficiency

The efficiency of DC machine like any other machine is determined by the ratio of output

power to that of the input power.






https://www.electrical4u.com/types-of-dc-generators/#Separately_Excited_DC_Generator

https://www.electrical4u.com/voltage-drop-calculation/

https://www.electrical4u.com/ammeter/



Dept. of EEE

There are three methods of determining the efficiency of a machine.

Direct method

Indirect method

Regenerative method

The 1st equation is giving an idea about the direct estimation of the efficiency. In this method

the machine is fully loaded and the output is directly measured. This method of measurement is

only applied for the small machines.

The 2nd and 3rd equations are giving an idea about the indirect estimation of the efficiency.

Indirect method is helpful of determining the efficiency of shunt wound

generator and compound wound generators. In this method it is required to determine to

determine the losses only. So, power supply is required to supply the losses only without

loading the machine. For the regenerative method of determining efficiency, it is required to

have two identical machines. One machine is used as motor and drives the other and the other is

used as generator and feedback the power into the supply. Two machines are mechanically

coupled. Therefore the losses can be determined because the internal power drawn is only to

supply losses of the two machines. Except these testes, the insulation test and the test for

making the commutation satisfactory is done while building up the machine.

Swinburne Test of DC Machine









https://www.electrical4u.com/types-of-dc-generators/#Shunt_Wound_DC_Generators

https://www.electrical4u.com/types-of-dc-generators/#Shunt_Wound_DC_Generators

https://www.electrical4u.com/types-of-dc-generators/#Compound_Wound_DC_Generator



Dept. of EEE

Advantages of Swinburne's Test

The main advantages of this test are:

This test is very convenient and economical as it is required very less power from supply

to perform the test.

Since constant losses are known, efficiency of Swinburne's test can be pre-determined

at any load.

Disadvantages of Swinburne's Test

The main disadvantages of this test are :

Iron loss is neglected though there is change in iron loss from no load to full load due to

armature reaction.

We cannot be sure about the satisfactory commutation on loaded condition because the

test is done on no-load.

We can’t measure the temperature rise when the machine is loaded. Power losses can

vary with the temperature.

In DC series motors, the Swinburne’s test cannot be done to find its efficiency as it is a

no load test.

Losses in DC Machine

As we know “Energy neither can be created nor it can be destroyed, it can only be transferred

from one form to another”. In DC machine, mechanical energy is converted into the electrical

energy. During this process, the total input power is not transformed into output power. Some

part of input power gets wasted in various forms. The form of this loss may vary from one

machine to another. These losses give in rise in temperature of machine and reduce the




Dept. of EEE

efficiency of the machine. In DC Machine, there are broadly four main categories of energy

loss.

Copper Losses or Electrical Losses in DC Machine or Winding Loss

The copper losses are the winding losses taking place during the current flowing through the

winding. These losses occur due to the resistance in the winding. In DC machine, there are only

two winding, armature and field winding.

Thus copper losses categories in three parts; armature loss, field winding loss, and brush contact

resistance loss. The copper losses are proportional to square of the current flowing through the

winding.

Armature Copper Loss in DC Machine

Armature copper loss = Ia2Ra

Where, Ia is armature current and Ra is armature resistance.

These losses are about 30% of the total full load losses.

Field Winding Copper Loss in DC Machine

Field winding copper loss = If2Rf

Where, If is field current and Rf is field resistance.

These losses are about 25% theoretically, but practically it is constant.

Brush Contact Resistance Loss in DC Machine

Brush contact loss attributes to resistance between the surface of brush and commutator. It is not

a loss which could be calculated separately as it is a part of variable losses. Generally, it

contributes in both the types of copper losses. So, they are factor in the calculation of above

losses.

Core Losses or Iron Losses in DC Machine or Magnetic Losses

As iron core of the armature is rotating in magnetic field, some losses occurs in the core which

is called core losses. Normally, machines are operated with constant speed, so these losses are

almost constant. These losses are categorized in two form; Hysteresis loss and Eddy current

loss.

Hysteresis Loss in DC Machine




http://www.electrical4u.com/hysteresis-eddy-current-iron-or-core-losses-and-copper-loss-in-transformer/

https://www.electrical4u.com/hysteresis-eddy-current-iron-or-core-losses-and-copper-loss-in-transformer/

https://www.electrical4u.com/hysteresis-eddy-current-iron-or-core-losses-and-copper-loss-in-transformer/



Dept. of EEE

Hysteresis losses occur in the armature winding due to reversal of magnetization of the core.

When the core of the armature exposed to magnetic field, it undergoes one complete rotation of

magnetic reversal. The portion of armature which is under S-pole, after completing half

electrical revolution, the same piece will be under the N-pole, and the magnetic lines are

reversed in order to overturn the magnetism within the core. The constant process of magnetic

reversal in the armature, consume some amount of energy which is called hysteresis loss. The

percentage of loss depends upon the quality and volume of the iron.

The Frequency of Magnetic Reversal

Where,

P = Number of poles

N = Speed in rpm

Steinmetz Formula

The Steinmetz formula is for the calculation of hysteresis loss.

Where,

η = Steinmetz hysteresis co-efficient

Bmax = Maximum flux Density in armature winding

F = Frequency of magnetic reversals

V = Volume of armature in m3.

Eddy Current Loss in DC Machine

According to Faraday’s law of electromagnetic induction, when an iron core rotates in the

magnetic field, an emf is also induced in the core. Similarly, when armature rotates in magnetic

field, small amount of emf induced in the core which allows flow of charge in the body due to

conductivity of the core. This current is useless for the machine. This loss of current is called

eddy current. This loss is almost constant for the DC machines. It could be minimized by

selecting the laminated core.

Mechanical Losses in DC Machine





Dept. of EEE

The losses associated with mechanical friction of the machine are called mechanical losses.

These losses occur due to friction in the moving parts of the machine like bearing, brushes etc,

and windage losses occurs due to the air inside the rotating coil of the machine. These losses are

usually very small about 15% of full load loss.

Stray Load Losses in DC Machine

There are some more losses other than the losses which have been discussed above. These

losses are called stray-load losses. These miscellaneous losses are due to the short-circuit

current in the coil undergoing commutation, distortion of flux due to armature and many more

losses which are difficult to find. These losses are difficult to determine. However, they are

taken as 1% of the whole load power output.

10.Derive the expression for Efficiency of Dc Machines.

Calculation of Efficiency

Let, I0 is the no load current (it can be measured by ammeter A1)

Ish is the shunt field current (it can be measured by ammeter A2)

Then, no load armature current

=

Also let, V is the supply voltage. Therefore, No load power input = VI0 watts.

In Swinburne's test no load power input is only required to supply the losses. The losses occur

in the machine mainly are:

Iron losses in the core

Friction and windings losses

Armature copper loss.

Since the no load mechanical output of the machine is zero in Swinburne's test, the no load

input power is only used to supply the losses.

The value of armature copper loss

=

Here, Ra is the armature resistance.







Dept. of EEE

Now, no to get the constant losses we have to subtract the armature copper loss from the no load

power input.

Then,

After calculating the no load constant losses now we can determine the efficiency at any load.

Let, I is the load current at which we have to calculate the efficiency of the machine.

Then, armature current (Ia) will be (I - Ish),

When the machine is motoring.

And , when the machine is generating.

Calculation of Efficiency When the Machine is Motoring on Load

Power input = VI

Armature copper loss,

Constant losses,

∴ Efficiency of the motor:

Calculation of Efficiency When the Machine is Generating on Load

Power input = VI


Constant losses,



Dept. of EEE

∴ Efficiency of the generator:

The DC generator efficiency is perpetual but varies with load. Think through a shunt

generator supplying a load current IL at a terminal voltage V.

Then Generator output = VIL

Generator input = Output + Losses

= VIL + Variable losses + Constant losses

= VIL + I2a Ra + Wc

= VIL + (IL + Ish2)Ra + Wc

( ∵ Ia = IL + Ish)

The shunt field current Ish is generally small as compared to IL and, therefore, can be neglected.

Generator input = VIL + I2a Ra + Wc

Now Efficiency η = output / Input

= VIL / (VIL + I2a Ra + Wc

= 1 / {1+[(ILRa/V)+(Wc/VIL)]}

UNIT-III

1.Determine the concept of methods of Testing?

Testing of DC motor

http://www.mytech-info.com/2014/04/dc-shunt-generator.html

http://www.mytech-info.com/2014/04/dc-shunt-generator.html



Dept. of EEE

Testing of machines is used for finding losses, efficiency and temperature rise. Direct method is

used, for small machines. Indirect method is used for large shunt machines. In practice,

seinburne,s test are mostly used.

a.Direct method of testing

In direct method of testing the generator or motor is put on full load and whole of the power

developed by it is wasted,. Brake test is a typical example of direct test. The direct tests can be

used only on small machines.

b. Indirect method of testing

This method consists of measuring the losses and then calculating the efficiency. The simplest

of the indirect test is Swinburne’s test. Hopkinson test is commonly used test under this method

on shunt motors. This method also enables the determination of losses without actually loading

the machine.

c. Swinburne’s test (No load test)

In this method (simplest indirect method) the losses are measured separately and efficiency at

any desired load is pre-determined.

The iron and friction losses are determined by measuring the input to the machine on no-load,

the machine being run as a motor at normal voltage and speed.

d. Hopkinson’s Test (Back-to-back test or Regenerative test)

Through this test full-load testing of two d.c. shunt machines can be carries out, mainly identical

ones. In this test, power drawn from the supply only corresponds to no load losses of the

machines. Electrically these two machines are mechanically connected in parallel and controlled

in such a way that one machine acts as a generator and the other as motor.

2.Determine the Brake Test of DC Machine

http://www.theelectricalportal.com/2015/07/generator-protection.html

http://www.theelectricalportal.com/2015/08/speed-control-of-dc-motordc-series.html

http://1.bp.blogspot.com/-phCJIvu7BKI/VjDZiOTyMaI/AAAAAAAAAbo/hQsIFNynYQ8/s1600/download+(2).jpg



Dept. of EEE

DC Machines can be tested by three different methods namely Direct Method, Indirect Method

and Regenerative Method. Direct Method of testing of DC Machine, also known as Brake Test

(if carried out for a DC Motor) will be discussed in this post.

Direct method is suitable for small DC machines. In Direct Method, the DC machine is

subjected to rated load and the entire output power is wasted. The ratio of output power to the

input power gives the Efficiency of DC Machine. For a DC Generator the output power is

wasted in resistor.

Direct Method of testing when conducted on a motor is also known as Brake Test. Brake Test of

DC Motor is carried out as shown in figure below.

A belt around the air cooled pulley has its end attached to the spring balance S1 and S2. Using

belt tightening hand wheels H1 and H2, the load of motor is adjusted to its rated value.

Assuming the spring balance to be calibrated in kilogram, then rated load on the DC motor is

given as

https://2.bp.blogspot.com/-tzwatVbF3eE/WD-B0lnaS8I/AAAAAAAACeo/Xg_jdYXqDfwOeNtKfPHiXa_24suozFT1gCLcB/s1600/Brake+Test+of+DC+Motor.jpg



Dept. of EEE

Motor Output Power = Torque x Angular Speed

= (Force x Radius) x Angular Speed

As the torque because of force F1 and F2 are opposing each other, therefore net torque will be

subtraction of torque because of F1 and F2.

Therefore,

Motor Output = ω (S1 – S2) x r x9.8 Watt

Now assuming the terminal voltage of DC Motor to be Vt and IL to be the load current then,

Power input to the DC Motor = VtIL

Thus the efficiency of DC Motor can be calculated as below.

Efficiency = Output / Input

https://3.bp.blogspot.com/-7KWwDIWSK8Q/WD-AVE6z11I/AAAAAAAACec/Dc9KMmEdQg80mE9hoD1R08iTZfxWNqTcgCLcB/s1600/Brake+Test+of+DC+Motor-2.jpg



Dept. of EEE

= [ω (S1 – S2) x r x9.8 Watt] / VtIL

For conducting Brake Test on DC Series Motor, it must be ensured that belt is sufficiently tight

before the motor is switched on to the sully as DC Series Motor shall not be started at no load.

Disadvantages of Brake Test of DC Motor:

1) The Spring Balance Readings are not stable rather it fluctuates.

2) Output power is wasted.

3) The frictional torque at a particular setting of Hand wheel H1 and H2 do not remain

constant.

3.Determine the test processes of Swinburne Test of DC Machine









http://electricalbaba.com/why-dc-series-motor-should-not-be-started-at-no-load/



Dept. of EEE

Calculation of Efficiency

Let, I0 is the no load current (it can be measured by ammeter A1) Ish is the shunt

field current (it can be measured by ammeter A2)

Then, no load armature current

=

Also let, V is the supply voltage. Therefore, No load power input = VI0 watts.

In Swinburne's test no load power input is only required to supply the losses. The losses occur

in the machine mainly are:

Iron losses in the core

Friction and windings losses

Armature copper loss.

Since the no load mechanical output of the machine is zero in Swinburne's test, the no load

input power is only used to supply the losses.

The value of armature copper loss

=

Here, Ra is the armature resistance.

Now, no to get the constant losses we have to subtract the armature copper loss from the no load

power input.

Then,

After calculating the no load constant losses now we can determine the efficiency at any load.

Let, I is the load current at which we have to calculate the efficiency of the machine.

Then, armature current (Ia) will be (I - Ish), when the machine is motoring.

And






Dept. of EEE

, when the machine is generating.

Calculation of Efficiency When the Machine is Motoring on Load

Power input = VI


Constant losses,

∴ Efficiency of the motor:

Calculation of Efficiency When the Machine is Generating on Load

Power input = VI


Constant losses,

∴ Efficiency of the generator:

4.Expaline the testing procedure of Hopkinson's Test or Regenerative Test On DC Motor

In the previous post we have seen how to determine the efficiency of DC machines using brake

test. Hopkinson's test is also a test of finding the efficiency of a dc motor. Hopkinson's test

or regenerative test is a full load test and it requires two identical machines which are coupled to

each other.In this test two identical d.c. machines mechanically coupled to each other and

simultaneously tested.One is operated as generator another one as motor,hence we can find

http://www.electricaledition.com/2016/02/brake-test-on-dc-shunt-motor.html

http://www.electricaledition.com/2016/02/brake-test-on-dc-shunt-motor.html



Dept. of EEE

efficiency of two dc machines simultaneously.So output power of dc machines are going to be

wasted.The mechanical output of motor given to generator through shaft to shaft mechanical

coupling.And generator's electrical power supplied to run the motor,where losses will

be supplied by external power source.

If there are no losses in the motor-generator set,the electrical power from the generator and

mechanical output from motor are enough to run motor,generatorrespectively.So no need of any

external power supply to the motor.But due to losses, the generator output is not sufficient to

drive the motor. Thus motor takes current from the supply to account for losses.

Observe circuit diagram of Hopikinson's test. The two shunt dc machines are connected in

parallel. In that two machines,one is started as a motor another one operated as generator.Here

the only rotor connections are mentioned,stator connections are not shown for simplicity.

Connection Diagram of Hopkinson's Test

First switch S is kept open. The other machine which is coupled to first will act as load on first

which is acting as motor. Thus second machine will act as a generator.With the help of field

rheostat speed of the motor is adjusted to normal value.Note down the observed voltmeter

readings.With the help of generator field rheostat voltage of the generator is adjusted up

to voltmeter reading is zero.This is to make sure generator voltage is having same magnitude

and polarity of that of supply voltage.By making this we can prevent heavy circulating current

flowing in the local loop of armatures on closing the switch.

http://www.electricaledition.com/2015/09/dc-machine-construction-motor-generator.html

http://www.electricaledition.com/2016/01/dc-generator-working-principle.html

http://www.electricaledition.com/2016/01/dc-generator-working-principle.html

http://www.electricaledition.com/2016/02/parallel-operation-of-dc-generators.html

https://3.bp.blogspot.com/-f0ZbCeqoR1U/VsraUXCbQ6I/AAAAAAAAA4c/VsnIeySyp_E/s1600/brake-test-on-dc-machines.jpeg



Dept. of EEE

Now close the switch S. The two machines can be put into any load by adjusting their field

rheostats. The generator current I2 can be adjusted to any value by increasing the excitation of

generator or by reducing the excitation of motor. The various reading shown by different

ammeters are noted for further calculations.

The input to the motor is nothing but the output of the generator and small power taken from

supply. The mechanical output given by motor after supplying losses will in turn drive the

generator.

Calculation of Efficiency by Hopkinson's Test

Let V = Supply voltage

Motor power Input = V(I1 + I2)

Generator power Input = VI1

We can determine the efficiency of DC machines in two cases.

Case 1: Assuming that the efficiency of both the machines are same.

Case 2: Assuming both the machines has same iron loss, friction loss and windage loss.

Case 1:

Assuming that the efficiency of both the machines are same.

Motor output power = η x Motor Input power

= η V(I1+I2)

i.e., Motor Input power = Generator Input

Now the Generator output = η x generator Input

= η x ηV(I1 +I2)



Dept. of EEE

= η2 V(I1+I2)

VI1 = η2V(I1+I2)

∵ Generator output η = √ {I1 / (I1+I2)}

Note: The above expression is used to determine the efficiency satisfactorily perfect for a rough

test. If case need to find more accuracy then the efficiency of the two machines can be

determined separately using the below expressions.

Case

Assuming both the machines has same iron loss, friction loss and windage loss.

However the iron loss, friction loss and windage loss of both the machines will be same due to

both the machines are identical. On this notion we can find the efficiency of each machine.

5.Determine the test method of Field’s Test

This is one of the methods of testing the d.c. series motors. Unlike shunt motors, the series

motor can not be tested by the methods which area available for shunt motors as it is impossible

to run the motor on no load. It may run at dangerously high speed on no load. In case of small

series motors brake test may be employed.

The series motors are usually tested in pairs. The field test is applied to two similar series

motors which are coupled mechanically. The connection diagram for the test is shown in the

Fig. 1.



Dept. of EEE

Fig. 1 Field test

As shown in the Fig. 1 one machine is made to run as a motor while the other as a generator

which is separately excited. The field of the two machines are connected in series so that both

the machines are equally excited. This will make iron losses same for the two machines. The

two machines are running at the same speed. The generator output is given to the variable

resistance R.

The resistance R is changed until the current taken by motor reaches full load value. This

will be indicated by ammeter A1. The other readings of different meters are then recorded.

Let V = Supply voltage

I1 = Current taken by motor

I2 = Load current

V2 = Terminal p.d. of generator

Ra, Rse = Armature and series field resistance of each machine

Power taken from supply = VI1

Output obtained from generator = V2 I2

Total losses in both the machines, WT = VI1 - V2 I2

Armature copper and field losses, WCU = ( Ra + 2 Rse ) I12 + I22 Ra

Total stray losses = WT - WCU

http://1.bp.blogspot.com/-nwFVwtCM7KE/UCcIgP0ycUI/AAAAAAAAF4I/Ye2cND9aGo8/s1600/ccc1.jpeg



Dept. of EEE

Since the two machines are equally excited and are running at same speed the stray loses

are equally divided.

For Motor ;

Input to motor = V1 I1

Total losses = Aramture Cu loss + Field Cu loss + Stray loss

= I12 ( Ra + Rse) + Ws

Output of motor = Input - Total losses = V1 I1 - [ I12 ( Ra + Rse) + Ws ]

For Generator :

Efficiency of generator is of little importance because it is running under conditions of

separate excitation. Still it can be found as follows.

Output of generator = V2 I2

Field Cu loss = I12 Rse

Armature Cu loss = I22 Ra

Total losses = Armature Cu loss + Field Cu loss + Stray loss

= I22 Ra + I12 Rse + Ws

Input to generator = Output + Total losses = V2 I2 + [ I22 Ra + I12 Rse + Ws ]

http://2.bp.blogspot.com/-lcrMkcULrNI/UCcOJ7RYXRI/AAAAAAAAF40/QuVKzAHh50Q/s1600/ccc11.jpeg

http://4.bp.blogspot.com/--RyOQ4JO8jE/UCcOQN0T5AI/AAAAAAAAF48/vddLo4uihw4/s1600/ccc12.jpeg



Dept. of EEE

The important point to be noted is that this is not regenerative method though the two

machines are mechanically coupled because the generator output is not fed back to the motor as

in case of Hopkinson's test but it is wasted in load resistance.

6.Explan the testing procedure of Separation of stray losses in a d.c. motor test

There are some more losses other than the losses which have been discussed above. These

losses are called stray-load losses. These miscellaneous losses are due to the short-circuit

current in the coil undergoing commutation, distortion of flux due to armature and many more

losses which are difficult to find. These losses are difficult to determine. However, they are

taken as 1% of the whole load power output.

UNIT-IV

1. Briefly explain about single phase transformer

A transformer is a static machine used for transforming power from one circuit to another

without changing frequency. This is a very basic definition of transformer. Since there is no

rotating or moving part so transformer is a static device. Transformer operates on ac supply.


https://www.electrical4u.com/what-is-transformer-definition-working-principle-of-transformer/

http://2.bp.blogspot.com/-WSz0CB_mvhg/UCcWU56_f3I/AAAAAAAAF5o/sbjjoRFzuEU/s1600/ccc13.jpeg



Dept. of EEE

Transformer works on the principle of mutual induction

Generally, the name associated with the construction of a transformer is dependant upon

how the primary and secondary windings are wound around the central laminated steel core.

The two most common and basic designs of transformer construction are the Closed-core

Transformer and the Shell-core Transformer

In core type transformer it has two vertical legs or limbs with two horizontal sections named

yoke. Core is rectangular in shape with a common magnetic circuit. Cylindrical coils (HV

and LV) are placed on both the limbs.

Shell type transformer: It has a central limb and two outer limbs. Both HV, LV coils are

placed on the central limb. Double magnetic circuit is present.

Berry type transformer: The core looks like spokes of wheels. Tightly fitted metal sheet

tanks are used for housing this type of transformer with transformer oil filled inside.

Transformer Core Construction

Shell type transformer cores overcome this leakage flux as both the primary and secondary

windings are wound on the same centre leg or limb which has twice the cross-sectional area of

the two outer limbs. The advantage here is that the magnetic flux has two closed magnetic paths



Dept. of EEE

to flow around external to the coils on both left and right hand sides before returning back to the

central coils.

This means that the magnetic flux circulating around the outer limbs of this type of transformer

construction is equal to Φ/2. As the magnetic flux has a closed path around the coils, this has the

advantage of decreasing core losses and increasing overall efficiency

2.Give the constructional features of single phase transformer

Two coils of wire (called windings) are wound on some type of core material. In

some cases the coils of wire are wound on a cylindrical or rectangular cardboard form.

In effect, the core material is air and the transformer is called an AIR-CORE

TRANSFORMER. Transformers used at low frequencies, such as 60 hertz and 400

hertz, require a core of low-reluctance magnetic material, usually iron. This type of

transformer is called an IRON-CORE TRANSFORMER. Most power transformers are

of the iron-core type.

The principle parts of a transformer and their functions are:

The CORE, which provides a path for the magnetic lines of flux.

The PRIMARY WINDING, which receives energy from the ac source.

The SECONDARY WINDING, which receives energy from the primary winding and

delivers it to the load.

The ENCLOSURE, which protects the above components from dirt, moisture, and mechanical

damage.

(i) CORE

There are two main shapes of cores used in laminated-steel-core transformers.

One is the HOLLOWCORE, so named because the core is shaped with a hollow square

through the center. This shape of core. Notice that the core is made up of many



Dept. of EEE

laminations of steel it shows how the transformer windings are wrapped around both

sides of the core.

(ii) WINDINGS

As stated above, the transformer consists of two coils called WINDINGS which are wrapped

around a core. The transformer operates when a source of ac voltage is connected to one of the

windings and a load device is connected to the other. The winding that is connected to the

source is called the PRIMARY WINDING. The winding that is connected to the load is called

the SECONDARY WINDING. The primary is wound in layers directly on a rectangular

cardboard form.

3.What are the ways to minimize hysteresis and eddy current losses

Core loss of a transformer consists of two parts 1)Hysteresis losses 2)eddy current losses. In

order to reduce eddy current losses , the magnetic core of the transformer is not made from a

single magnetic material;because in this case the circulating eddy current flowing will be

higher.Instead the magnetic core is a stack of thin silicon steel lamination and the laminations



Dept. of EEE

are insulated from one an other by thin layer of varnish in order to reduce eddy current and

hence eddy current losses.

In order to minimize hysteresis losses , soft magnetic materials eg: Si steel, steel alloys ,Mn-Zn

ferrite are used because they have high saturation magnetization, Low coercivity ,High

magnetic permeability etc. which reduce losses due to hysteresis.

4.Determine the transformer operation on no load and give its phasor

diagrams

Let the applied voltage V1 applied to the primary of a transformer, with secondary open-

circuited, be sinusoidal (or sine wave). Then the current I1, due to applied voltage V1, will also

be a sine wave. The mmf N1 I1 and core flux Ø will follow the variations of I1 closely. That is

the flux is in time phase with the current I1 and varies sinusoidally.

Let,

NA = Number of turns in primary ;NB = Number of turns in secondary

Ømax = Maximum flux in the core in webers = Bmax X A f = Frequency of alternating current

input in hertz (HZ)

As shown in figure above, the core flux increases from its zero value to maximum value

Ømax in one quarter of the cycle , that is in ¼ frequency second.



Dept. of EEE

Therefore, average rate of change of flux = Ømax/ ¼ f = 4f ØmaxWb/s

Now, rate of change of flux per turn means induced electro motive force in volts.

Therefore,

average electro-motive force induced/turn = 4f Ømaxvolt

If flux Ø varies sinusoidally, then r.m.s value of induced e.m.f is obtained by multiplying the

average value with form factor.

Form Factor = r.m.s. value/average value = 1.11 Therefore, r.m.s value of e.m.f/turn = 1.11 X

4f Ømax = 4.44f Ømax Now, r.m.s value of induced e.m.f in the whole of primary winding.=

(induced e.m.f./turn) X Number of primary turns

Therefore,

EA = 4.44f NAØmax = 4.44fNABmA

Similarly, r.m.s value of induced e.m.f in secondary is

EB = 4.44f NB Ømax = 4.44fNBBmA

In an ideal transformer on no load, VA = EA and VB = EB , where VB is the terminal

voltage

Voltage Transformation Ratio.

The ratio of secondary voltage to primary voltage is known as the voltage

transformation ratio and is designated by letter K. i.e.

Voltage transformation ratio, K = V2/V1 = E2/E1 = N2/N1

Current Ratio.

The ratio of secondary current to primary current is known as current ratio and is

reciprocal of voltage transformation ratio in an ideal transformer.

Transformer on No Load.



Dept. of EEE

When the primary of a transformer is connected to the source of an ac supply and the

secondary is open circuited, the transformer is said to be on no load. The Transformer on

No Load alternating applied voltage will cause flow of an alternating current I0 in the

primary

winding, which will create alternating flux Ø. No-load current I0, also known as excitation or

exciting current, has two components the magnetizing component Im and the energy component

Ie. Im is used to create the flux in the core and Ie is used to overcome the hysteresis and eddy

current losses occurring in the core in addition to small amount of copper losses occurring in the

primary only (no copper loss occurs in the secondary, because it carries no current, being open

circuited.)

From vector diagram shown in above it is obvious that

1. Induced emfs in primary and secondary windings, E1 and E2 lag the main flux Ø by and are

in phase with each other.

2. Applied voltage to primary V1 and leads the main flux Ø by and is in phase opposition to

E1.

3. Secondary voltage V2 is in phase and equal to E2 since there is no voltage drop in

secondary.



Dept. of EEE

4. Im is in phase with Ø and so lags V1 by

5. Ie is in phase with the applied voltage V1.

6. Input power on no load = V1Ie = V1I0 cos Ø0 where Ø0 = tan-1

5.Determine the transformer operation on load and give its phasor diagrams

The transformer is said to be loaded, when its secondary circuit is completed through an

impedance or load. The magnitude and phase of secondary current (i.e. current flowing through

secondary) I2 with respect to secondary terminals depends upon the characteristic of the load

i.e. current I2 will be in phase, lag behind and lead the terminal voltage V+2+ respectively

when the load is non-inductive, inductive and capacitive. The net flux passing through the core

remains almost constant from no-load to full load irrespective of load conditions and so core

losses remain almost constant from no-load to full load. Vector diagram for an ideal

transformer supplying inductive load is shown



Dept. of EEE

Resistance and Leakage Reactance In actual practice, both of the primary and secondary

windings have got some ohmic resistance causing voltage drops and copper losses in the

windings. In actual practice, the total flux created does not link both of the primary and

secondary windings but is divided into three components namely the main or mutual flux Ø

linking both of the primary and secondary windings, primary leakage flux ØL1 linking with

primary winding only and secondary leakage flux ØL2 linking with secondary winding only.

The primary leakage flux ØL1 is produced by primary ampere-turns and is proportional to

primary current, number of primary turns being fixed. The primary leakage flux ØL1 is in phase

with I1 and produces self inducedemf ØL1 is in phase with I1 and produces self inducedemf

EL1 given as 2f L1 I1 in the primary winding.

The self inducedemf divided by the primary current gives the reactance of primary and is

denoted by X1.

i.e. X1 = EL1/I1 = 2πfL1I1/I1 = 2FL1,

Similarly leakage reactance of secondary X2 = EL2/E2 = 2fπL2I2/I2 = 2πfL2

Equivalent Resistance and Reactance. The equivalent resistances and reactance’s of transformer

windings referred to primary and secondary sides are given as below Referred to primary side

Equivalent resistance,

Equivalent resistance, = X'1 = Referred to secondary side Equivalent resistance,

Equivalent resistance, = X2 + K2X1 Where K is the transformation ratio.

6.Describe the equivalent circuit of single phase transformer

Equivalent impedance of transformer is essential to be calculated because the electrical power

transformer is an electrical power system equipment for estimating different parameters of

electrical power system which may be required to calculate total internal impedance of an

electrical power transformer, viewing from primary side or secondary side as per requirement.

This calculation requires equivalent circuit of transformer referred to primary or equivalent

circuit of transformer referred to secondary sides respectively. Percentage impedance is also

very essential parameter of transformer. Special attention is to be given to this parameter during

installing a transformer in an existing electrical power system. Percentage impedance of



Dept. of EEE

different power transformers should be properly matched during parallel operation of power

transformers.

Equivalent Circuit of Transformer Referred to Primary

For drawing equivalent circuit of transformer referred to primary, first we have to establish

generalequivalent circuit of transformer then, we will modify it for referring from primary

side. For doing this, first we need to recall the complete vector diagram of a transformer which

is shown in the figure below.

Let us consider the transformation ratio be,

In the figure right, the applied voltage to the primary is V1 and voltage across the primary

winding is E1. Total current supplied to primary is I1. So the voltage V1 applied to the primary is

partly dropped by I1Z1 or I1R1 + j.I1X1 before it appears across primary winding.

The voltage appeared across winding is countered by primary induced emf E1.

The equivalent circuit for that equation can be drawn as below,

From the vector diagram above, it is found that the total primary current I1 has two

components, one is no - load component Io and the other is load component I2′. As this

primary current has two a component or branches, so there must be a parallel path with primary

winding of transformer. This parallel path of currentis known as excitation branch of equivalent











Dept. of EEE

circuit of transformer. The resistive and reactive branches of the excitation circuit can be

represented as

The load component I2′ flows through the primary winding of transformer and

induced voltage across the winding is E1 as shown in the figure right. This induced voltage

E1transforms to secondary and it is E2 and load component of primary current I2′ is transformed

to secondary as secondary current I2. Current of secondary is I 2. So the voltage E2 across

secondary winding is partly dropped by I2Z2 or I2R2 + j.I2X2 before it appears across load. The

load voltage is V2.

From above equation, secondary impedance of transformer referred to primary is,




Dept. of EEE

So, the complete equivalent circuit of transformer referred to primary is shown in the figure

below,

7., Determine the losses and efficiency of single phase transformer

In any electrical machine, 'loss' can be defined as the difference between input power and output

power. An electrical transformer is an static device, hence mechanical losses (like windage or

friction losses) are absent in it. A transformer only consists of electrical losses (iron losses and

copper losses). Transformer losses are similar to losses in a DC machine, except that

transformers do not have mechanical losses.

Losses in transformer are explained below -

(I) Core Losses Or Iron Losses

Eddy current loss and hysteresis loss depend upon the magnetic properties of the material used

for the construction of core. Hence these losses are also known as core losses or iron losses.

http://www.electricaleasy.com/p/electrical-machines.html

http://www.electricaleasy.com/2014/03/electrical-transformer-basic.html




Dept. of EEE

Hysteresis loss in transformer: Hysteresis loss is due to reversal of magnetization in the

transformer core. This loss depends upon the volume and grade of the iron, frequency of

magnetic reversals and value of flux density. It can be given by, Steinmetzformula:

H=ηBmax1.6fV (watts)

where,η=Steinmetzhysteresisconstant

V = volume of the core in m3

Eddy current loss in transformer: In transformer, AC current is supplied to the primary

winding which sets up alternating magnetizing flux. When this flux links with secondary

winding, it produces induced emf in it. But some part of this flux also gets linked with

other conducting parts like steel core or iron body or the transformer, which will result in

induced emf in those parts, causing small circulating current in them. This current is called

as eddy current. Due to these eddy currents, some energy will be dissipated in the form of

heat.

(Ii) Copper Loss In Transformer

Copper loss is due to ohmic resistance of the transformer windings. Copper loss for the primary

winding is I12R1 and for secondary winding is I2

2R2. Where, I1 and I2 are current in primary and

secondary winding respectively, R1 and R2 are the resistances of primary and secondary

winding respectively. It is clear that Cu loss is proportional to square of the current, and current

depends on the load. Hence copper loss in transformer varies with the load.

Efficiency Of Transformer

Just like any other electrical machine, efficiency of a transformer can be defined as the output

power divided by the input power. That is efficiency = output / input .

Transformers are the most highly efficient electrical devices. Most of the transformers have full

load efficiency between 95% to 98.5% . As a transformer being highly efficient, output and

input are having nearly same value, and hence it is impractical to measure the efficiency of

transformer by using output / input. A better method to find efficiency of a transformer is

using, efficiency = (input - losses) / input = 1 - (losses / input).



Dept. of EEE

Condition For Maximum Efficiency

Let,

Copper loss = I12R1

Iron loss = Wi

Hence, efficiency of a transformer will be maximum when copper loss and iron losses are

equal.

That is Copper loss = Iron loss.

8.Determine the regulation of single phase transformer

http://2.bp.blogspot.com/-vPFBtCslCIk/Uz1Ld8zvH-I/AAAAAAAAArs/HTMe4G_BwC0/s1600/efficiency+of+transformer.png



Dept. of EEE

The voltage regulation is the percentage of voltage difference between no load and full load

voltages of a transformer with respect to its full load voltage.

Explanation of Voltage Regulation of Transformer

Say an electrical power transformer is open circuited, means load is not connected with

secondary terminals. In this situation, the secondary terminalvoltage of the transformer will be

its secondary induced emf E2. Whenever full load is connected to the secondary terminals of the

transformer, ratedcurrent I2 flows through the secondary circuit and voltage drop comes into

picture. At this situation, primary winding will also draw equivalent full load current from

source. The voltagedrop in the secondary is I2Z2 where Z2 is the secondary impedance of

transformer. Now if at this loading condition, any one measures the voltage between secondary

terminals, he or she willgetvoltage V2 across load terminals which is obviously less than no load

secondary voltage E2 and this is because of I2Z2 voltage drop in the transformer.

Expression of Voltage Regulation of Transformer, represented in percentage, is

9.Explain about all day efficiency

Large capacity transformers used in power systems are classified broadly into Power

transformers and Distribution transformers. The former variety is seen in generating stations

and large substations. Distribution transformers are seen at the distribution substations.

The basic difference between the two types arises from the fact that the power transformers are

switched in or out of the circuit depending upon the load to be handled by them. Thus at 50%

load on the station only 50% of the transformers need to be connected in the circuit. On the

other hand a distribution transformer is never switched off. It has to remain in the circuit



Dept. of EEE

irrespective of the load connected. In such cases the constant loss of the transformer continues

to be dissipated. Hence the concept of energy based efficiency is defined for such transformers.

It is called ’all day’ efficiency.

The all day efficiency is thus the ratio of the energy output of the transformer over a day to the

corresponding energy input. One day is taken as duration of time over which the load pattern

repeats itself. This assumption, however, is far from being true. The power output varies from

zero to full load depending on the requirement of the user and the load losses vary as the square

of the fractional loads. The no-load losses or constant losses occur throughout the 24 hours.

Thus, the comparison of loads on different days becomes difficult. Even the load factor, which

is given by the ratio of the average load to rated load, does not give satisfactory results.

The calculation of the all day efficiency is illustrated below with an example. The graph of

load on the transformer, expressed as a fraction of the full load is plotted against time. In an

actual situation the load on the transformer continuously changes. This has been presented by a

stepped curve for convenience. For the same load factor different average loss can be there

depending upon the values of xi and ti. Hence a better option would be to keep the constant

losses very low to keep the all day efficiency high. Variable losses are related to load and are

associated with revenue earned.

The constant loss on the other hand has to be incurred to make the service available. The

concept of all day efficiency may therefore be more useful for comparing two transformers

subjected to the same load cycle. The concept of minimizing the lost energy comes into effect

right from the time of procurement of the transformer.

The constant losses and variable losses are capitalized and added to the material cost of the

transformer in order to select the most competitive one, which gives minimum cost taking

initial cost and running cost put together. Obviously the iron losses are capitalized more in the

process to give an effect to the maximization of energy efficiency. If the load cycle is known at

this stage, it can also be incorporated in computation of the best transformer.

10.Explain the effect of variations of frequency and voltage on iron losses



Dept. of EEE

The iron losses of a transformer includes two types of losses,

1. Hysteresis loss and 2. Eddy current loss

For a given volume and thickness of laminations, these losses depend on the operating

frequency, maximum flux density and the voltage.

It is known that for a transformer,

V = 4.44 f Φm N = 4.44 f Bm A N

Where A = area

... Bm α (V/f) .......... For constant A and N

Thus as voltage changes, the maximum flux density changes and both eddy current and

hysteresis losses also changes. As voltage increases, the maximum flux density in the core

increases and total iron loss increases.

As frequency increases, the flux density in the core decreases but as the iron loss is directly

proportional to the frequency hence effect of increased frequency is to increase the iron losses.

Key Point : Thus iron loss increases as the voltage and frequency increases for the transformer.

Unit-v

1. Determine the open circuit and short circuit tests.

In this test secondary (usually high voltage) winding is left open, all metering instruments

(ammeter, voltmeter and wattmeter) are connected on primary side and normal rated voltage is

applied to the primary (low voltage) winding, as illustrated below

http://3.bp.blogspot.com/-BhB4hySKZJg/TijBGq8cE3I/AAAAAAAAA7E/MGhlYfwacXY/s1600/ABB135.jpeg



Dept. of EEE

Iron loss = Input power on no-load

W0 watts (wattmeter reading) No-load current = 0 amperes (ammeter reading) Angle of lag, =

/Io Ie = and Im = √o - Caution: Since no load current I0 is very small, therefore, pressure coils

of watt meter and the volt meter should be connected such that the current taken by them

should not flow through the current taken by them should not flow through the current coil of

the watt meter.

V0 = Rated Voltage

Wo = Input power

I0 = Input current = no oad current

Im = I0sinϕ0

Ic = I0cos ϕ0

cos ϕ0 = No load power factor

Hence power input can be written as,

Wo = V0 I0 cos ϕ0



Dept. of EEE

W0 = Pi = Iron losses

W0 = V0I0 cos ϕ0

cos ϕ0 = W0 / V0I0 = no load power factor

Ro = V0/Ic Ω

X0 = V0 /Im

2. Short-circuit or Impedance Test.

This test is performed to determine the full-load copper loss and equivalent resistance and

reactance referred to secondary side. In this test, the terminals of the secondary (usually the

low voltage) winding are short circuited, all meters (ammeter, voltmeter and wattmeter) are

connected on primary side and a low voltage, usually 5 to 10 % of normal rated primary

voltage at normal frequency is applied to the primary, as shown in fig below.

The applied voltage to the primary, say Vs’ is gradually increased till the ammeter A indicates

the full load current of the side in which it is connected. The reading Ws of the wattmeter gives

total copper loss (iron losses being negligible due to very low applied voltage resulting in very

small flux linking with the core) at full load. Le the ammeter reading be Is.



Dept. of EEE

Equivalent impedence referred to primary= Commercial Efficiency and All day Efficiency (a)

Commercial Efficiency. Commercial efficiency is defined as the ratio of power output to

power input in kilowatts.(b) All-day Efficiency. The all day efficiency is defined as the ratio of

output in kwh to the input in kwh during the whole day. Transformers used for distribution are

connected for the whole day to the line but loaded intermittently. Thus the core losses occur for

the whole day but copper losses occur only when the transformer is delivering the load current.

Hence if the transformer is not used to supply the load current for the whole day all day

efficiency will be less than commercial efficiency. The efficiency (commercial efficiency) will

be maximum when variable losses (copper losses) are equal to constant losses (iron or core

losses).sign is for inductive load and sign is for capacitive load Transformer efficiency, Where

x is the ratio of secondary current I2 and rated full load secondary current.

Wsc = (Pcu) F.L. = full Load copper loss

Wsc = VscIsc cosϕsc

cosϕsc = VscIsc / Wsc

Wsc = I2sc R1e = Copper loss

Z1e = Vsc /Isc = Root(R21e+X21e)



Dept. of EEE

X1e = Root(Z21e – R21e)

2. Briefly explain about the sumpner’s test

Sumpner's test or back to back test can be employed only when two identical transformers are

available. Both transformers are connected to supply such that one transformer is loaded on

another. Primaries of the two identical transformers are connected in parallel across a supply.

Secondaries are connected in series such that emf's of them are opposite to each other. Another

low voltage supply is connected in series with secondaries to get the readings, as shown in the

circuit diagram shown below.

In above diagram, T1 and T2 are identical transformers. Secondaries of them are connected in

voltage opposition, i.e. EEF and EGH. Both the emf's cancel each other, as transformers are

identical. In this case, as per superposition theorem, no current flows through secondary. And

thus the no load test is simulated. The current drawn from V1 is 2I0, where I0 is equal to no load

current of each transformer.

http://www.electricaleasy.com/2014/03/electrical-transformer-basic.html

http://3.bp.blogspot.com/-vxsi3ZS6Oyw/U01AhM2N22I/AAAAAAAAAvk/ONrMJXqunKc/s1600/transformer+sumpners+test.png



Dept. of EEE

Thus input power measured by wattmeter W1 is equal to iron losses of both transformers.i.e.

iron loss per transformer Pi = W1/2.

Now, a small voltage V2 is injected into secondary with the help of a low voltage transformer.

The voltage V2 is adjusted so that, the rated current I2 flows through the secondary. In this case,

both primaries and secondaries carry rated current. Thus short circuit test is simulated and

wattmeter W2 shows total full load copper losses of both transformers.i.e. copper loss per

transformer PCu = W2/2.

From above test results, the full load efficiency of each transformer can be given as -

3.Describe the separation of losses test in transformers.

Hysteresis loss and eddy current loss are the components of the iron losses. For the applied flux

density Bmax to the core, we have

Hysteresis loss = Afand Eddy current loss = Bf2

The no load loss can be expressed as

Wc=Af+Bf2 (1.36)where A and B are constants.

Therefore,

http://4.bp.blogspot.com/-68aGv4bw8ew/U01HYh_y0dI/AAAAAAAAAv0/FYcsF1o6oIA/s1600/transformer+full+load+efficiency.png



Dept. of EEE

Figure 1.38 shows the graph, which is a straight line when and f are plotted along the y-axis and x-

axis, respectively. The intercept on the y-axis gives the value of A, whereas the slope of the line gives

the value of B. Now the hysteresis and eddy current loss can be determined at any desired frequency.The

experimental circuit arrangement for determining and f is shown in Figure 1.39.In Figure 1.39, a

variable frequency alternator supplies to the transformer under the test, which is driven by DC shunt

motor whose speed can be varied over a wide range. The switches S1 and S2 are opened and the alternator

is started with the help of the DC shunt motor. The speed is adjusted to the

value of the required frequency. The excitation of the field coil (X-XX) is varied until the voltmeter on

the secondary side of the transformer achieves the rated value. If E2 is the transformer emf on the

secondary, we have

E2=4.44ΦmfN

i.e.

For constant , the flux density in the transformer remains constant. To achieve this, the frequency of

the alternator emf is varied so that remains constant. The necessary f can be adjusted to vary E2so

that is kept constant. For different values of frequencies above and below the rated value, the reading

of wattmeter (W) is noted. The graph and f is drawn to get the constants A and B. After getting the

value of A and B, the hysteresis loss and eddy current loss is obtained.

4.Determine the transformers Polyphase connections - /Y, /

Getting this phasing correct when the windings aren’t shown in regular Y or Δ configuration

can be tricky. Let me illustrate, starting with Figure below.

Inputs A1, A2, A3 may be wired either “Δ” or “Y”, as may outputs B1, B2, B3.



Dept. of EEE

Now, let’s examine a Δ-Y system in Figure below.

Phase wiring for “Δ-Y” transformer.

Such a configuration (Figure above) would allow for the provision of multiple voltages (line-

to-line or line-to-neutral) in the second power system, from a source power system having no

neutral.

And finally, we turn to the Δ-Δ configuration: (Figure below)

Phase wiring for “Δ-Δ” transformer.

When there is no need for a neutral conductor in the secondary power system, Δ-Δ connection

schemes (Figure above) are preferred because of the inherent reliability of the Δ configuration.

5.Determine the transformer poly phase connections open

https://www.allaboutcircuits.com/textbook/alternating-current/chpt-10/three-phase-transformer-circuits/#02209.png



Dept. of EEE

Considering that a Δ configuration can operate satisfactorily missing one winding, some

power system designers choose to create a three-phase transformer bank with only two

transformers, representing a Δ-Δ configuration with a missing winding in both the primary

and secondary sides: (Figure below)

“V” or “open-Δ” provides 2-φ power with only two transformers.

This configuration is called “V” or “Open-Δ.” Of course, each of the two transformers has to

be oversized to handle the same amount of power as three in a standard Δ configuration, but

the overall size, weight, and cost advantages are often worth it. Bear in mind, however, that

with one winding set missing from the Δ shape, this system no longer provides the fault

tolerance of a normal Δ-Δ system. If one of the two transformers were to fail, the load voltage

and current would definitely be affected.

The following photograph (Figure below) shows a bank of step-up transformers at the Grand

Coulee hydroelectric dam in Washington state. Several transformers (green in color) may be

https://www.allaboutcircuits.com/textbook/alternating-current/chpt-10/three-phase-transformer-circuits/#02210.png



Dept. of EEE

seen from this vantage point, and they have grouped in threes: three transformers per

hydroelectric generator, wired together in some form of three-phase configuration. The

photograph doesn’t reveal the primary winding connections, but it appears the secondaries are

connected in a Y configuration, is that there is only one large high-

voltage insulator protruding from each transformer. This suggests the other side of each

transformer’s secondary winding is at or near ground potential, which could only be true in a

Y system. The building to the left is the powerhouse, where the generators and turbines are

housed. On the right, the sloping concrete wall is the downstream face of the dam

d) & e) Objective questions and Fill in the blanks

UNIT-I

01) The emf induced in the dc generator armature winding is

AC

DC

AC and DC

None of the above

The emf induced in the dc generator armature winding is AC, but we need DC current from DC

generator, so to convert this AC current to DC current mechanical rectifier called as

commutator is used.

02) Commutator in DC generator is used for

collecting of current

reduce losses

https://www.allaboutcircuits.com/textbook/direct-current/chpt-1/conductors-insulators-electron-flow/



Dept. of EEE

increase efficiency

convert AC armature current in to DC

Commutator is used as mechanical rectifier in DC machines. It converts AC armature current

in to DC current.

03) A DC generator without commutator is a

AC generator

DC motor

DC generator

induction motor

Commutator is used to convert AC current into DC current. If commutator is not present, then it

will acts as an AC generator with huge losses.

04) In DC generators brushes are used for

collecting of current without any sparkings

collecting of voltage

reduce eddy current loss

convert ac armature current in to dc

In case of DC generator the brushes need to collect current with minimum sparking, which is

known as successful commutation.

05) Which of the following bearings and their uses are correct

ball bearings → small machines

roller bearings → large machines

neither 1 nor 2

both A and B



Dept. of EEE

These are different applications of different bearings. Since the ball is a sphere, it only contacts

the inner and outer race at a very small point, as contact area is very small, so if the ball

bearing is overloaded, the balls can deform or squish the running bearing. So it has to be used

for small rating machines. Since the roller is a cylinder, so the contact between the inner and

outer race is not a point but a line. This spreads the load over more area, allowing the bearing

to handle much greater loads than ball bearings. Hence it is used for large machines.

06) Lap winding is prefered for which type of machines?

low current and low voltage

high current and high voltage

high current and low voltage

low current and high voltage

Number of parallel paths are more in case of lap winding than wave winding. So summation of

currents at the output is more. Therefore it is preferred for high current and low voltage rated

machines.

07) Wave winding is prefered for which of the following rating?

low current and low voltage

high current and high voltage

high current and low voltage

low current and high voltage

Number of parallel paths in wave winding is only 2. Therefore is used for low current and high

voltage rated machines. Summation of currents in parallel paths is less than wave winding.

08) Equalizer rings are used in which of the followings?

lap winding

wave winding

both 1 and 2

none of the above



Dept. of EEE

Equalizer rings are essential in lap winding than wave winding because of more no of parallel

paths in lap winding. As in lap winding no of parallel path is more, there would be severe

sparking at brushes due to difference in currents in different parallel paths. But in wave

winding no of parallel paths is two and the sparking at commutation is less. So, equalizer rings

are used in lap winding to avoid any unequal distribution.

09) A 4 pole lap wound dc generator has 4 brushes, if one of the brush is damaged, what will be

the change in V, I and P ratings

V, I and P

V/2, I/2 and P/4

V/2, I and P/2

V, I/2 and P/2

If one brush is damaged then two parallel paths will be damaged. So only two parallel paths

will provide the I/2 current and voltage is same for parallel paths. As the current is halved,

delivered power is also halved for same terminal voltage.

10) For a dc machine shunt resistance and armature resistance values are

high and high

high and low

low and low

low and high

For a DC machine the values of armature resistance is very low and shunt resistance is high.

The power delivered by the DC Machine depends upon the armature current Ia. Ia should be

high to deliver maximum power. The Shunt field is parallel to the armature field, so its

resistance must be high for minimum value of shunt field current. If its value will be low then

armature current will be lower and power delivered will be less. The series field is connected in

series. So if series resistance is more then there would considerable series drop and again

armature current will be lower.

01) Shunt field of DC generators consists of--------number of turns and ---------- conductors

respectively



Dept. of EEE

large and thick

large and thin

less and thick

less and thin

To deliver maximum power armature current must be high and the shunt field current is

minimum. So, shunt field resistance of DC Machine is very high value around 50Ω to 500 Ω.

, it can be concluded that for high resistance length of the coil is to be large and area

to be small. So shunt field must have large no of turns and thinner wire than the series field.

02) Series field consists of--------number of turns and ---------- conductors respectively

large and thick

large and thin

less and thick

less and thin

The power delivered by a DC machine depends upon armature current. The series field is in

series with the armature so they are carrying same current through them. Series field are kept

at low resistance for minimum drop. , we can conclude that area should be high and no

of turns should be less.

03) What is/are the necessary conditions for voltage build up in self excited generator?

Poles should contain residual flux.

Field winding should be properly connected to armature winding.

Field winding resistance should be less than critical resistance.

All of the above.



Dept. of EEE

The necessary condition for voltage build-up process in a self-excited DC generator are

1. The poles should retain some residual magnetism. If the poles lost its residual magnetism it

can't start voltage build up process. It may be started by a separate DC source at shunt field

when armature is at rest.

2. The field winding should be properly connected to armature winding. If the field connection is

reversed then the field flux would oppose residual flux.

3. The field winding resistance should be less than critical resistance. Voltage will not build up if

the field resistance is greater than critical resistance.

4. The speed of the generator should be greater than the critical speed. This can be remedied by

increasing prime mover speed above critical speed.

04) No load saturation characteristics are plotted between

no load voltage and field current

no load voltage and armature current

short circuit current and field current

short circuit current and armature current

This operating characteristics is also known as saturation curve or open circuit characteristics

or magnetization curve or no load magnetizing curve of DC Generator. This curve is drawn

between no-load armature generated voltage with the field or exciting current, keeping the

speed constant by prime mover. This curve starts slightly above the zero due to residual

magnetism. It also determines the design of flux per pole under linear magnetization or

saturation curve.

05) The effect of ------------------ on main field flux is armature reaction?

armaturemmf

armature current

armature flux

all of the above

On no load armature flux is negligible as armature current is small but with load the rotating

armature produces a flux due to sufficient armature current or load current. The effect of this

armature flux on the main field flux is known as armature reaction. Due to armature reaction

the main field flux distribution is weakened and distorted.

06) Armature flux is.............with respect to main field flux or main field poles?



Dept. of EEE

rotates opposite direction

rotates same direction

stationary

none of the above

The rotating armature produce a rotating armature flux with respect to armature and there is a

working flux which is also under the pole distributed uniformly. Therefore armature flux is

stationary with respect to main filed flux. Armature mmf is stationary w.r.t. field poles but

rotating w.r.t. the armature.

07) Cross magnetization effects which of the following?

commutation

reduction in main field flux

reduces the terminal voltage

both 2 and 3

Cross magnetizing is one of the effect of armature reaction. By vector addition it is found that it

distorts the main field flux. As a result it shifts the MNA (magnetic neutral axis). There would be

sparking at the time of commutation if the brushes are not shifted to the MNA.

08) Demagnetization effects which of the following?

commutation

reduction in main field flux

reduces the terminal voltage

both B and C

Demagnetization is one of the effect of armature reaction. By vector addition it is found that it

reduces or weakens the main field flux. So it reduces the induced emf or terminal voltage in

case of generator because Eg is directly proportional to the flux (φ). In case of a motor it

reduces the torque and increases the speed because Te is proportional to flux and speed(N) is

inversely proportional to the flux.



Dept. of EEE

09) Brushes are always placed on--------------- , in order to achieve sparkles commutation?

MNA

Brushes should be placed where the direction of current are changes or production of zero

e.m.f., under no load condition MNA and GNA coincides with each other. At this axis current

direction is reversed or no emf is produced. But due to armature flux the main field flux gets

distorted and MNA does not coincides with GNA under loaded condition. That means neutral

zone is shifted. In order to achieve sparkless commutation brushes is placed on MNA. So,

brushes are always placed in MNA in loaded or unloaded condition.

10) Flux density under trailing pole tips in case of generator will

increase

The pole tip towards which the armature conductor leaves the influence of pole is called trailing

pole tips (symbol by dot sign). In a generator the main field flux and armature flux are aiding

each other under trailing pole tips region. Therefore total flux density will increase under

trailing pole tips.

UNIT-II

01) If field current is decreased in shunt dc motor, the speed of the motor

remains same.

increases.

decreases.

none of the above.

As a shunt field current If decreases, φ also decreases and the speed rises as speed is inversely

proportional to flux.

02) What is the mechanical power developed by a DC series motor is maximum?



Dept. of EEE

Back emf is equal to half the applied voltage.

Back emf is equal to applied voltage.

Back emf is equal to zero.

None of above.

03) In Ward-Leonard system, the lower limit of the speed imposed by

Field resistance.

Armature resistance.

Residual magnetism of the generator.

None of above.

In Ward-Leonard method of speed control, the lower limit of speed is imposed by residual

magnetism of the generator.

04) Ward-Leonard control is basically a ___________ control method.

Field control.

Armature resistance control.

Armature voltage control.

Field diverter control.



Dept. of EEE

Ward-Leonard speed control method consists a motor with a constant excitation and applying a

variable voltage to its armature to provide the required speed. Hence it is armature voltage

control method.

05) For very sensitive and wide speed control, the preferable control method is

Armature control.

Ward-Leonard control.

Multiple voltage control.

Field control.

In ward-Leonard method, very fine speed control over the whole range from zero to normal

speed in both directions can be obtained. The motor-generator set can provide speed both

below and above the rated speed and in both direction.

06) Commutator pitches of duplex and simplex lap windings are respectively

4 and 2

2 and 1

4 ang 1

2 and 2

Commutator pitches of duplex winding is 2 and Commutator pitches of simplex winding is 1. It

is defined as the distance between two ends of same armature coil are connected. So, difference

in duplex winding are 2 as there are two turns in the duplex winding.

07) In DC machine yoke offers

mechanical protection to the machine

flux path completion

produce working flux

both A and B

In case of DC machines yoke is used to



Dept. of EEE

1. Mechanical protection to the machine

2. Flux path completion.

Poles will produce working flux.

08) In a dc machine 72 number of coils are used. Find the number of commutator segments

required?

36

37

72

74

In DC machines, Number of coils = Number of commutator segments

09) which of the following type of brush and their application is/are correct

carbon brush → normal ratings

electro graphite → large ratings

copper graphite → low voltage high current density

all of the above

These are the different applications for different types of brushes in DC machines.

10) Which of the following windings are necessary in case of all dc machines?

closed winding

lap winding

wave winding

open type winding

In case of all DC machines we must use closed winding. Lap and wave windings are used based

on the requirement of voltage and current levels, but winding of all DC machines is closed

winding. It is never short pitched.



Dept. of EEE

01) DC machine windings are

full pitched

shortpiched

either of these

none of the above

DC machine wingdings are full pitched winding only because it offers maximum induced emf

within the coil. No requirement of short pitching of coils like alternator.

02) Which of the following represents the commutator pitch?

Number of conductors spanned by one coil at the back end of the armature.

Number of conductors spanned by one coil at the front end of the armature.

The distance between the staring of first coil and its next successive coil

Number of commutator segments between two successive coils.

The definitions of different types of pitches are

1. Back pitch: Number of conductors spanned by one coil at the back end of the armature.

2. Front pitch: Number of conductors spanned by one coil at the front end of the armature.

3. Resultant pitch: The distance between the staring of first coil and its next successive coil.

4. Commutator pitch: Number of commutator segments between two successive coils.

For lap winding YC is the difference of YB and YF where as for wave winding it is the sum of the

two.

03) In a dc machine 4 pole lap winding is used. The number of parallel paths are?

2

4

1

8



Dept. of EEE

In lap winding number of parallel paths = number of poles = 4. For Wave winding it will be

equal to 2.

04) In a dc machine 6 pole wave winding is used. The number of parallel paths are?

6

4

2

1

In wave winding / simplex wave winding number of parallel paths = always 2 (irrespective of

number of poles).

05) Inter pole winding is connected in-------------------------- ?

series with armature

series with main poles

parallel with armature

parallel with main poles

Interpolars are small narrow commutating poles connected in series with armature. It produce

an extra flux to neutralize the reactance m.m.f. produced at the time of commutation process.

06) In a lap winding dc machine number of conductors are 100 and number of parallel paths are

10. Find the average pitch

10

100

50

1

(? front pitch = Yb+1, back pitch = Yb -1 )



Dept. of EEE

07) In a 2 pole lap winding dc machine , the resistance of one conductor is 2Ω and total number

of conductors is 100. Find the total resistance

200Ω

100Ω

50Ω

10Ω

Total resistance depends upon no of parallel path. In lap winding parallel path is no of poles

and here it is two. Half of conductor are in series i.e. 50 in series and rest of 50 in series and

they are parallel together. 50 no 2Ω in series = 100Ω. When two such paths are parallel their

equivalent will be 50 Ω. Numerically it can also be stated,

X = Resistance of one conductor Z = Total number of

conductors A = No of parallel paths

08) Dummy coils are used for

increasing efficiency.

reducing armature reaction.

mechanical balancing.

all of the above

Dummy coils are required under wave winding only to make the machine mechanically

balanced. Consider a problem to understand the dummy coils.

Design wave winding for 60 conductors, 15 slots, 4 poles

For designing of wave winding we should find average pitch (Ya) and that should be integer.

For some set of data Ya will not be integer.

Therefore the nearest possible number of conductors are considered to make Ya integer. But this

will make the armature mechanically unbalance. For example one of the 15 slots will contain

only 2 conductors as 2 are eliminated in the design itself. To make the machine mechanically

balanced those two conductors were cut shot, insulated and placed in the slots as dummy coils.

In this way dummy coils will mechanically balance the machine.

09) Which of the following is/are the advantages of carbon brush over the copper brush?



Dept. of EEE

They are not hard as copper brush

They are self lubricating in nature

In case of any sparking they will be less damaged than copper brushes

All of the above

Advantages of carbon brush over the copper brushes are 1. They are not hard as copper brush.

2. They are self lubricating in nature which ensure excellent mechanical conditions with

rotating commutator. 3. In case of any sparking they will be less damaged than copper brushes.

But they has less current density than copper brushes.

10) Maximum power will be developed when back ems is

equals to supply voltage

The maximum power developed is depends on the back emf value. i.e.,Eb = Vt/2.

UNIT-III

01) Eddy current loss will depends on

Frequency

Flux density

Thickness

All of the above

. So eddy current loss will depend upon frequency, flux

density and the area of the eddy current loop.

02) Hysteresis loss will depends on



Dept. of EEE

f

f²

f³

f1.6

. So, hysteresis loss will depends on frequency.

03) Thin laminations are used in a machine in order to reduce

Eddy current losses

Hysteresis losses

Both 1 and 2

Copper losses

Thin laminations are used in order to reduce the eddy current losses only. Due to laminations

the area of the eddy currents loops are minimized and the losses due to eddy current losses are

minimized.

04) Hysteresis loop represents the area of

copper loss

eddy current loss

hysteresis loss

total iron losses

Hysteresis loop will represents only hysteresis losses. It is found out by area of B-H loop curve

of a magnetic material.

05) Total core loss is also called as -------------?

Eddy current loss

Hysteresis loss



Dept. of EEE

Magnetic loss

Copper loss

As iron loss is proportional to flux density or flux, these are also called as magnetic loss. The

total core loss or magnetic loss consists of eddy current loss and hysteresis loss.

06) Which of the following are variable losses?

eddy current loss

hysteresis loss

shunt field copper loss

armature copper loss

Armature copper loss is directly proportional to square of armature current. Therefore as the

load varies these will also vary.

07) Maximum efficiency will occur, when copper loss is_______to iron loss?

greater than

less than

equals to

any of the above

Condition for maximum efficiency is, copper loss = iron loss or variable losses is equals to

fixed losses.

08) The noise resulting from vibrations of lamination set by magnetic forces, is termed as

magnetostriction.

boo.

hum.

zoom.



Dept. of EEE

The hum is generated by the magnetic field that happens due to the continuous reversing of the

frequency of the supply or it is due to magnetostriction. The magnetic field in the AC machine

or transformers has coils which are still able to move slightly due to the vibration. The

laminations of the armature are treated in a similar way but also vibrate at the line frequency

and it is almost impossible to stop. We can only reduce it by good design.

09) Autotransformer makes effective saving on copper and copper losses, when its

transformation ratio is equal to

very low

less than one.

greater than one.

approx to one.

To quantify the saving the total quantity of copper used in an auto transformer is expressed as a

fraction of that used in a two winding transformer as,

This

means that an auto transformer requires the use of lesser quantity of copper given by the ratio

of turns. Hence, if the transformation ratio is approximately equal to one, then the copper

saving is good and the copper loss is less.

10) Which of the following statements is/are correct?

High frequency power supplies are light weight

Transformer size get reduced at high frequency

Both 1 and 2

None of the above

From the induced emf equation of transformer E ∝φfFor same emf, φf = constant φ1f1 =

φ2f2 B1A1f1 = B2A2f2For constant flux density B1 = B2 A1f1 = A2f2 For high frequency f2 > f1 A2 <

A1 Therefore at high frequencies transformer size get reduced and also light weight.



Dept. of EEE

UNIT-IV

01) The basic function of a transformer is to change

the power level

the power factor

the level of the voltage

the frequency

A transformer can be used for either step up or step down the voltages. Its basic application in

power system is to transform the voltage from one level to another level without changing its

frequency and power.

02) The frequency of a voltage at the secondary is

greater than the primary

equal to primary

less than primary

any of the above

The basic function of a transformer is step up or step down voltages without changing its supply

frequency. Therefore the frequency of voltage at the secondary is equal to the primary side.

03) Transformer action requires a

constant magnetic flux

increasing magnetic flux

alternating magnetic lux



Dept. of EEE

alternating electric flux

A basic transformer contains two magnetically coupled coils on a common core, if an

alternating voltage is given to one winding is known as primary due to electromagnetic and

action mutual induction, there is a voltage induced in the secondary winding where load can be

connected known as secondary. The energy transfer from one winding to another is entirely

through magnetic medium it is known as transformer action based on the principle of

electromagnetic induction proposed by faraday. Therefore, transformer action requires an

alternating or time varying magnetic flux to transfer power from primary side to secondary

side. Since induced emf in the winding is due to flux linkage.

04) For an ideal transformer the winding should have

maximum resistance on primary side and least resistance on secondary side

minimum resistance on primary side and maximum resistance on secondary side

equal resistance on primary and secondary side

noohmic resistance on either side

For an ideal transformer the losses should be zero on both sides. Therefore the ohmic

resistance on either side of the transformer should be equal to zero.

05) Which of the following statements is/are correct statements?

EMF per turn in HV winding is more than EMF per turn in LV winding

EMF per turn in HV winding is less than EMF per turn in LV winding

EMF per turn in both the windings are equal

all of the above

In a transformer primary volt-ampere is equal to secondary volt-ampere and primary ampere

ampere turns are also equal. So, EMF per turn in both the winding are equal. Total induced emf

on both sides depends on the number of turns, flux and frequency. If number of turns on

secondary more than primary, then emf induced in the secondary will more than primary side

and vice versa, but the emf per turn in both the winding are equal.



Dept. of EEE

06) The core flux in transformer depends mainly on

supply voltage

supply voltage and frequency

supply voltage, frequency and load

supply voltage and load

Therefore core flux in transformer depends mainly on

supply voltage and frequency.

07) If the applied voltage of a certain transformer is increased by 50% and the frequency is

reduced by 50%, the maximum flux density will (assuming that the magnetic circuit remains

unsaturated)

changes to three times the original value

changes to 1.5 times the original value

changes to 0.5 times the original value

remain the same as the original

.

08) The low voltage winding of a 400/230 V single phase 50 Hz transformer is to be connected

to keep the magnetization current at the same level in both the cases the voltage at 25 Hz should

be

230 V

460 V



Dept. of EEE

115 V

65 V

To maintain the magnetization current at the same level, flux Ï† should be same i.e V/f ratio

should be same.

09) A single phase transformer has specifications as 250 KVA, 11000 V/415 V, 50 Hz. What

are the values of primary and secondary currents?

primary = 22.7 A, secondary = 602.4 A

primary = 602 A, secondary = 22.7 A

primary = 301 A, secondary = 22.7 A

primary = 11.4 A, secondary = 301 A

10) R1 is the resistance of the primary winding of the transformer. The turn ratio in terms of

primary to secondary is K. Then the equivalent resistance of the primary referred to secondary

is

R1 /K

K²R1

R1 /K²

K×R1

By changing of primary parameters to secondary or vice versa it does not change the

performance of circuit. So that we are taking power rating are equal in before and after



Dept. of EEE

referring. Therefore, the equivalent resistance of the primary

referred to secondary = R1 /K²

UNIT-V

1) The most commonly used connections for power systems as a step - up and step - down transformers are

a. Star - delta, star - star

b. Delta - star, star - delta

c. Star - star, delta - delta

d. Star - delta, delta - star

ANSWER: Star - delta, delta - star

2) A transformer when connected to a 230V, 50Hz supply, under no load draws a current of 4A at a power

factor of 0.2 lagging. The magnetizing current (Im) and core loss (Pc) is equal to

a. 3.919A, 184W

b. 1.84A, 391.9W

c. 39.19A, 184W

d. 3A, 180W



Dept. of EEE

ANSWER: 3.919A, 184W

3) A differential relay comparator used for the protection of three phase transformers has

a. One comparator

b. Two comparator

c. Three comparator

d. Six comparator

ANSWER: Three comparator

4) If a two winding transformer is converted into an autotransformer by applying additive polarity and

subtractive polarity which results in the secondary voltages of 1840 and 1810 volts. Then the primary and

secondary voltages of transformer are

a. 1800V, 50V

b. 1810V, 40V

c. 1820V, 30V

d. 1825V, 15V

ANSWER: 1825V, 15V

5) For a single phase no load transformer, which among the following losses will be minimum?

a. hysteresis losses

b. eddy current losses



Dept. of EEE

c. copper losses

d. mechanical losses

ANSWER: copper losses

6) In T-T connection, the ratio of actual capacity to the available capacity is

a. 1

b. 0.928

c. 1.928

d. 0.5

ANSWER: 0.928

7) In double delta transformation, a double delta refers to the case where there are two delta transformations in

a. Parallel

b. Series

c. Both series and parallel

d. Neither series nor parallel

ANSWER: Series

8) Most familiar application of zig - zag transformer is as

a. Ground reference on an ungrounded system

b. Converting single phase to two phase

c. Reducing harmonics

d. All of these



Dept. of EEE

ANSWER: Ground reference on an ungrounded system

9) In a single phase, full wave bridge circuit and in three phase, delta full wave bridge circuit, the ripple

voltage frequency is always

a. Twice the line frequency, six times the line frequency

b. Both will be twice the line frequency

c. Both will be six times the lines frequency

d. None of these

ANSWER: Twice the line frequency, six times the line frequency

10) In Scott connection, the voltage across the teaser leads the mains by

a. 30 degree

b. 60 degree

c. 90 degree

d. 120 degree

ANSWER: 90 degre

21) In star-star connection of three phase transformer, if VL is the line voltage and IL is the line current

phase voltage and phase current is given by



Dept. of EEE

a. VL / √3, IL

b. VL , IL

c. √3VL , IL / √3

d. VL, IL / √3

ANSWER: VL / √3, IL

22) Due to presence of third harmonic component in the star-star connection of three phase transformer,

the frequency of the circuit component becomes

a. Three times of the circuit frequency

b. One third of the circuit frequency

c. Remains same

d. None of these

ANSWER: Three times of the circuit frequency

23) For the parallel operation of three phase transformers, which among the following connection is

not applicable?

a. Δ - Δ to Y – Y

b. Y - Δ to Δ - Y

c. Y - Y to Y – Y

d. None of these



Dept. of EEE

ANSWER: None of these

24) While connecting three phase transformer in star-star connection, the terminals of secondaries are

wrongly joined. Then the mutual phase angle and magnitude of line to line voltages of secondaries will

a. Equal to 120 degree and equal as before

b. Not equal to 120 degree and different

c. Equal to 120 degree and different

d. None of these

ANSWER: Not equal to 120 degree and different

25) For the star - star connection of three phase transformer, the phase angle between the phase voltages

and line voltages on both primary and secondary side is

a. 0 degree

b. 30 degree

c. 60 degree

d. 120 degree

ANSWER: 30 degree

26) A bank of three single phase transformer can be used for obtaining the three phase output. Three

magnetic circuits produced in case of a bank of three single phase transformer and in case of single phase

transformer are



Dept. of EEE

a. Linked, independent

b. Independent, linked

c. Linked, linked

d. Both are independent

ANSWER: Independent, linked

27) In a three phase transformer, the current flowing in three primaries produces three corresponding

fluxes. The sum of these three fluxes at any instant is

a. Zero

b. Three times of any individual flux

c. One third of any individual flux

d. None of these

ANSWER: Zero

28) In a three phase transformer, the angle between two consecutive cores is

a. 30 degree

b. 60 degree

c. 120 degree

d. 150 degree

ANSWER: 120 degree

29) Pulse transformers are small in size. The leakage inductance and permeability of alloy used is

a. Low, high



Dept. of EEE

b. Low, low

c. High, low

d. High, high

ANSWER: Low, high

30) Small iron core transformers used in certain frequency range is also called audio frequency

transformer.

The audio frequency range is

a. 20 t0 2000 Hz

b. 20 to 20000 Hz

c. 2 to 2000 Hz

d. 200 to 20000 Hz

ANSWER: 20 to 20000 Hz

KG Reddy College of Engineering &...

Documents

Transcript of KG Reddy College of Engineering &...