Electrical technology 9788131785935, 9789332517677, 9789332517684, 9332517681

Citation preview

ELECTRICAL TECHNOLOGY ELECTRICAL FUNDAMENTALS Volume I

S. P. Bali Former Faculty Member Military College of Electronics and Mechanical Engineering Secunderabad, India

Delhi • Chennai

Copyright © 2013 Dorling Kindersley (India) Pvt. Ltd. Licensees of Pearson Education in South Asia No part of this eBook may be used or reproduced in any manner whatsoever without the publisher’s prior written consent. This eBook may or may not include all assets that were part of the print version. The publisher reserves the right to remove any material in this eBook at any time. ISBN 9788131785935 eISBN 9789332517677 Head Office: A-8(A), Sector 62, Knowledge Boulevard, 7th Floor, NOIDA 201 309, India Registered Office: 11 Local Shopping Centre, Panchsheel Park, New Delhi 110 017, India

Dedicated to My late wife SUKSHAM BALI (10 April 1940 – 07 August 2007)

And departing leave behind us FOOTPRINTS on the sands of time

Preface

A celebrity is a person who works hard all his life to become well known, then wears dark glasses to avoid being recognised. With the rising momentum of development, the scope of text books dealing, in particular, with Electrical Engineering has expanded considerably. An important guiding objective in writing this book is to provide the students with a text book they can read, understand and study by themselves. Intended to serve as a text book for the subject of Electrical Engineering for BE/B.Tech Degree students it will also serve as a text-cum-reference for the students of Diploma Engineering. So also it will be useful to candidates appearing for AMIE, IETE, GATE, UPSC Engineering Services and IAS entrance examination. It will be equally helpful to practicing engineers to understand the theoretical aspects of their professions. Despite the publication of a large number of text books on this field, the students continue to remain perplexed. Keeping this fact in mind, this text book has been developed in a systematic manner, with emphasis on basic concepts. Written in a simple, easy to understand language, reinforced by illustrations which speak of themselves and are easy to understand and supplemented by selected worked examples based on step-by-step solutions the various chapters are interlinked, yet independent. The book can be read in the sequence in which it is written without facing any difficulty. The following features are intended to serve as learning aids: 1. More than enough worked examples are given in each chapter, wherever applicable to emphasize the practical utility of the results derived. 2. Detailed summary is given at the end of each chapter, as an aid to memory. 3. Multiple choice questions (MCQ) along with their answers are included in each chapter for the self-assessment of the student. 4. The illustrative method of treatment is used, each illustration bringing home a point. 5. Conventional questions are also given at the end of each chapter. Answers to numerical questions are also given. 6. Where possible, mechanical analysis is given. 7. Equivalent circuits are given for a better understanding of the problem. 8. The per unit system is discussed in detail with plenty of worked examples. The book has been designed in two volumes, Volume 1: Electrical Fundamentals and Volume 2: Machines and Measurements. Volume 1 comprises Chapters 1 to 30. More stress is given to Electrical Fundamentals which form the foundation for further study. Part A of the book on Electrical Fundamentals starts from the requirement of a system of units, how many of them are there, what does it consist of, and which one is prevalent. It further proceeds to explain electrical parameters like resistance R, inductance L and capacitance C as a separate entity and in collaboration as LCR Circuits. The characteristics of these components, their types and construction and control over their extent are explained in detail along with their colour coding and the effect of temperature on their performance. Topics like types of current, potential fall and rise of potential and equipotential surfaces are explained with the help of hydraulic analogy. Networks, series, parallel, and complex and associated terms are explained. Excellent analytical tools that are derived from just three simple scientific laws—Ohm’s law and Kirchhoff’s voltage and current laws are explained in detail with the help of worked examples. Circuit elements are defined in terms of their circuit equation. This forms the ground for node-pair voltage and mesh current analysis. The idea of field is introduced, the field forming the link between electrical and mechanical systems. Suitability of material for specific tasks is explained leading to the topic of losses and efficiency. The analysis and synthesis of complex waveforms is discussed. First order and second order circuits are discusses leading to Laplace Transform. The natural, forced, and steady state conditions of electrical networks are discussed. Networks a.c. and Networks d.c. are given in separate chapters to distinguish between the two. An in-depth study of network theorems is included and the steps involved in Thevenizing and

vi Electrical Technology Nortonizing a circuit are enumerated. Electrochemical energy conversion basics are discussed leading to constant current and constant voltage generators, and conversion from one to the other. After dealing with resonance (LCR Circuits) topics like filters and attenuators and transmission lines are covered. Mathematical topics like vectors matrices and determinants, circular and hyperbolic functions are introduced where suitable. After d.c. circuits, single phase a.c. circuits and relevant network theorems are discussed along with plenty of worked out examples; then three phase circuits and delta and star connection and conversion from one to the other followed by step-by-step worked out problems are given. The book also features —a Web-based circuit simulator, specially created to help students practice key circuits. The customized FREE version integrated with the book will enable students to build, analyze and learn the circuits. Besides being used as a practice/pre-lab tool by students, it can also serve as an exciting tool for instructors to teach the circuits. Appendices A to M provide extremely useful information concluding with a brief Glossary of terms have been made available as online resources. Last but not the least I would like to put on record the appreciation of the production and editorial staff in providing me unstained help in completing the project right from the day it was conceived till its completion. Suggestions for improvement of the book will be thankfully acknowledged.

S. P. Bali

Contents Preface

Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

v

PART A: ELECTRICAL FUNDAMENTALS 1. Systems of Units 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8

Introduction 3 Scientific Notation 3 Fundamental and Derived Units 4 Standards and Units 5 Systems of Units 6 The SI System of Units 6 Importance of SI System 7 Definitions 8 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

2. Electrons in Action 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10

Introduction 23 Electric Circuit 23 Current 24 Electromotive Force 25 Reference Zero 26 Safety Precautions While Handling Electric Circuits 26 3.7 Insulators 27 3.8 Semiconductors 28 3.9 Conductors 30

4. Simple d.c. Circuits 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10

12

Introduction 12 Conduction in Solids 12 Bonding in Atoms 14 Energy Bands 16 Electrons in Action 16 Direction of Current Flow 18 Diffusion Current Momentarily 18 Drift Velocity 18 The Nature of Electric Current 19 Effects of Electricity 20 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

3. Electric Circuit 3.1 3.2 3.3 3.4 3.5 3.6

3

4.11 4.12 4.13 4.14 4.15 4.16 4.17 4.18 4.19 4.20 4.21 4.22 4.23 4.24

23

Introduction 35 The Basic Circuit 35 Resistors 36 Resistivity (Specific Resistance) 36 Types of Resistors 40 Resistor Tolerance and Wattage 42 Ohm’s Law 43 Lumped Resistance and Distributed Resistance 45 Leakage Resistance 45 Temperature Coefficient of Resistance 45 Zero Ohm Resistors 48 Chip Resistors 48 Resistor Networks 48 Simulated Resistors 50 Adjustable Resistors 50 Variable Resistors 50 Types of Electric Circuits 51 Resistances in Series 51 Voltage Division Formula 55 Dominant Resistance 56 Resistors in Parallel 56 Current Division Formula 58 Dominant Resistance 59 Series-Parallel (Complex) Circuits 59 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

5. Networks (d.c.) 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8

35

Introduction 69 Ohm’s Law 69 Kirchhoff ’s Laws 71 Voltage Drop and Polarity 72 Equipotential Points 73 The Bridge Network 73 Networks 78 Superposition Theorem 78

69

viii Electrical Technology

5.9 5.10 5.11 5.12 5.13 5.14 5.15 5.16 5.17 5.18 5.19 5.20 5.21 5.22

Voltage and Current Sources 80 Dependent Voltage Sources 85 Millman’s Theorem 86 Thevenin’s Theorem 87 Thevenizing a Circuit 88 Norton’s Theorem 90 Nortonizing a Circuit 90 Maximum Power Transfer Theorem Efficiency 93 Y Transformation 94 Δ Balanced Networks 97 Network Reduction 97 Mesh Currents 97 Node-Voltages 98 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

6. Mesh-Current and Node-Voltage Analysis 6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9

113

Introduction 113 Matrices and Determinants 113 Network Analysis by Mesh Current 118 Network Analysis by Node-Pair Voltages 121 The Resistance Matrix 122 The Conductance Matrix 123 The Super Mesh 124 The Super Node 126 Nodal Analysis Vs Mesh Analysis—A Comparison 127 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

7. Electrochemical Action 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.9 7.10 7.11 7.12 7.13 7.14

91

Introduction 135 Primary Cells 136 Electrolysis 136 Faraday’s Laws 137 Simple Voltaic Cell 139 e.m.f. of a Cell 140 Local Action 140 Polarization 141 Internal Resistance 141 Characteristics of a Good Cell 144 The Leclanche Cell 144 The Dry Cell 145 Secondary Batteries/Cells 146 Elements of Secondary Cells 147

135

7.15 The Electrolyte 148 7.16 Capacity of Cells 148 7.17 Internal Resistance of Secondary Cells 149 7.18 Makeup of Cells 149 7.19 Charging and Discharging of Lead-Acid Secondary Batteries 149 7.20 Constant Current Charging 150 7.21 Constant Voltage Charging 151 7.22 Efficiencies of a Cell 151 7.23 Faults 151 7.24 Alkaline Cells 151 7.25 Nife Nickel Cadmium Alkaline Cell 152 7.26 Mercury Cell 152 7.27 Silver-Oxide Cell 153 7.28 Grouping of Cells 153 7.29 Grouping Cells for Maximum Current 154 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

8. Electromagnetism 8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8 8.9 8.10 8.11 8.12 8.13 8.14 8.15 8.16 8.17 8.18 8.19 8.20 8.21 8.22 8.23 8.24 8.25

162

Introduction 162 Attraction and Repulsion 163 The Inverse Square Law 163 Lines of Force 164 Magnetic Flux 165 Permeability 165 Permeability (B–H) Curves 166 The Domain Theory of Magnetism 166 Electromagnetism 170 Direction of Magnetic Field 171 Magnetizing Force of Electromagnetic Fields 171 Indicating the Direction of Current Flow 172 Rule of Direction 172 Electrodynamic Forces 173 Forces between Magnetic Poles 174 Magnetic Moment 174 Flux Density of a Solenoid 175 Magnetic Circuit 175 Magnetic Induction 178 Magnetic Shields 180 Reluctance 180 Series Magnetic Circuits 181 Parallel Magnetic Circuit 183 Electromagnets 184 Electromagnetic Relays 186

Contents ix

Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

9. Inductors 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 9.10 9.11 9.12

211

Introduction 211 The B-H Curve 211 Hysteresis Loop 212 Hysteresis Loss 212 Determination of B-H Curve 213 Determination of Hysteresis Loop 215 Hysteresis Loss 216 Eddy Currents 219 Eddy Current Losses 220 Separation of Hysteresis and Eddy Current Losses 221 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

11. Magnetic Materials 11.1 11.2 11.3 11.4 11.5

190

Introduction 190 Inductance 191 Factors Determining Inductance 192 Energy Stored in the Magnetic Field of an Inductor 194 Losses in Inductors 194 Toroids 195 Inductor Types 196 Time-Constant 196 Graphical Derivation of the Transient Characteristics of an R-L Circuit 200 Universal Time Constant 203 Inductors in Series and Parallel 205 Transient Behaviour 207 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

10. Hysteresis 10.1 10.2 10.3 10.4 10.5 10.6 10.7 10.8 10.9 10.10

12. Electrostatics

226

Introduction 226 Magnetic Materials 226 Non-Magnetic Alloys 228 Ferrites 228 Magnetic Materials with Rectangular Hysteresis Loops 229 11.6 Grain-Oriented Magnetic Material 230 11.7 Permanent Magnets 230 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

235

12.1 12.2 12.3 12.4 12.5 12.6 12.7 12.8 12.9 12.10 12.11 12.12 12.13 12.14

Introduction 235 Electrification by Friction 235 Application of Electron Theory 236 Coulomb’s Law 237 Permittivity 237 Electrostatic Induction 238 The Gold-Leaf Electroscope 239 Electric Fields 240 Electric Flux 241 Potential 241 Equipotential Lines 241 Gauss’s Law 242 Dielectric Strength 244 The Electric Field Due to a Line of Charge 244 12.15 The Electric Field Due to a Charged Disk 245 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

13. Capacitors and d.c. Transients 249 13.1 13.2 13.3 13.4 13.5 13.6 13.7 13.8 13.9 13.10 13.11 13.12 13.13 13.14

Introduction 249 Capacitance 249 Capacitor Action 250 Permittivity 251 Factors Determining Capacitance 252 Energy Stored in the Electric Field between the Capacitor Plates 254 Power Factor (Capacitors) 254 Types of Capacitors 255 Capacitor Colourcode 258 Time Constant 262 Graphical Derivation of the Transient Characteristics of an R-C Circuit 263 Universal Time Constant 266 Connecting Capacitors in Series 269 Connecting Capacitors in Parallel 270 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

14. Dielectric Materials 14.1 14.2 14.3 14.4 14.5 14.6

277

Introduction 277 Dielectric Materials 277 Permittivity (Dielectric Constant) 277 Power Factor 278 Insulation Resistance (or Insulance) 278 Dielectric Absorption 278

x Electrical Technology

14.7 14.8 14.9 14.10 14.11 14.12

15. Field Theory 15.1 15.2 15.3 15.4 15.5 15.6 15.7 15.8 15.9 15.10 15.11 15.12 15.13 15.14 15.15 15.16 15.17 15.18

17. Three-phase Circuits and Systems

289

Introduction 289 The Electric Field 290 Vectors 291 Electric Field Lines 293 Field Plotting by Curvilinear Squares 294 Effect of Fringing 296 Capacitance of a Parallel Plate Capacitator 298 Capacitance of a Multiplate Capacitator 299 Capacitance between Concentric Cylinders 301 Dielectric Stress 302 Concentric Cable Field Plotting 302 Capacitance of an Isolated Twin Line 304 Energy Stored in an Electric Field (Alternate Method) 306 Induced e.m.f. and Inductance 307 Inductance Due to Internal Linkages at Low Frequency 308 Inductance of a Pair of Concentric Cylinders 309 Energy Stored in an Electromagnetic Field 312 Magnetic Energy Stored in an Inductor 312 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

16. Single Phase Alternating Voltage and Current 16.1 16.2 16.3 16.4 16.5 16.6 16.7

Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

Dielectric Strength 279 Thermal Effects 280 Loss Angle 280 Dielectric Materials (General) 281 The Dielectric Phenomenon 284 Dielectric Breakdown 285 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

316

Introduction 316 Comparison of a.c. and d.c. 317 The Sine Wave 318 Basic a.c. Generator 318 Phasor Diagrams 321 Addition of Sinusoidal Waveforms 325 Alternate Treatment: a.c. Values 330

334

17.1 17.2 17.3 17.4 17.5 17.6 17.7 17.8

Introduction 334 Why Three Phase? 334 Generating Three-Phase Voltage 335 Phase and Line Voltages 336 Star Connection 337 Delta Connection 339 Y– Δ Change Over Switch 341 Supply of Three-Phase Electrical Energy 341 17.9 Balanced and Unbalanced Loads 342 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

18. Complex Algebra

347

18.1 18.2 18.3 18.4 18.5 18.6 18.7

Introduction 347 The J-Operator 347 Inductive Reactance 348 Capacitive Reactance 349 Rectangular and Polar Notation 349 Rules of Complex Algebra 350 Admittance, Conductance, and Susceptance 352 18.8 Impedance and Admittance Triangles 353 Summary Conventional Questions (CQ)

19. Work, Power and Energy 19.1 19.2 19.3 19.4 19.5 19.6 19.7 19.8 19.9

356

Introduction 356 Work Done by an Electric Current 357 Methods of Heating Rooms 359 Heating Water 360 Power 361 Power in a Resistance 361 Power in a.c. Circuits 363 Three-Phase Power 365 Energy 366 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

20. Power Factor Correction 20.1 Introduction 372 20.2 The Need for Correction

372

372

Contents xi

20.3 Power Factor Correction 373 20.4 Types of Compensation 374 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

21. LCR Circuits 21.1 21.2 21.3 21.4 21.5 21.6 21.7 21.8 21.9 21.10

Introduction 376 Inductive Reactance 376 Capacitive Reactance 378 Filtering 379 Basic Series a.c. Circuits 380 The Concept of Impedance 381 Series Connected Impedances 385 Polar Notation 388 Parallel Connected Impedances 389 Components of Current 392 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

22. Resonance 22.1 22.2 22.3 22.4 22.5 22.6 22.7 22.8 22.9

376

403

Introduction 403 Series Resonance 403 Q-Quality Factor of a Series Circuit 410 Selectivity and Bandwidth 411 Parallel Resonance 412 Parallel Resonant Circuits 413 Quality Factor of a Parallel Network 415 Impedances in Parallel 416 Resonant Filters 417 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

23. The Fourier Series

421

23.1 Introduction 421 23.2 Complex Waveforms 421 23.3 Synthesis of Non-Sinusoidal Waveforms 424 23.4 The Fourier Series 425 23.5 Analyzing a Complex Waveform 426 23.6 Summary of Properties of Fourier Analysis 431 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

24. Networks (a.c.) 24.1 Introduction 434 24.2 Voltage Division 434

434

24.3 24.4 24.5 24.6 24.7 24.8 24.9 24.10 24.11 24.12 24.13 24.14 24.15 24.16

Current Division 436 Superposition Theorem 436 Thevenin’s Theorem 437 Constant Voltage Generator 439 Constant-Current Generator 439 Norton’s Theorem 440 Procedure for Solving a Network Using Thevenin’s Theorem 443 Procedure for Solving a Network Using Norton’s Theorem 443 Maximum Power Transfer Theorem 445 Millman’s Theorem 447 Reciprocity Theorem 449 Duality 450 A.c. Circuit Analysis 452 Mesh-Current and Nodal Analysis 454 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

25. Delta Wye Transformations

462

25.1 Introduction 462 25.2 Delta and Star Connections 462 Transformations 463 25.3 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

26. Attenuators and Filters 26.1 26.2 26.3 26.4 26.5 26.6 26.7 26.8 26.9 26.10 26.11 26.12 26.13 26.14 26.15 26.16 26.17 26.18 26.19 26.20

472

Introduction 472 The Decibel 473 Characteristic Impedance 479 Symmetrical T-Attenuator 481 Symmetrical Π-Attenuator 482 Insertion Loss 484 Asymmetrical T - and Π - Sections 486 The L-Section Attenuator 488 Cascading Two-Port Networks 489 Filters 490 Types of Filters 490 Active and Passive Filters 491 Frequency Response 492 Symmetrical Networks 494 Equivalence of Balanced and Unbalanced Sections 497 Maclaurin’s Theorem 497 Circular Functions 498 Hyperbolic Functions 499 Theorem Connecting α and Z0 502 Prototype (Constant K) Filter Sections 503

xii Electrical Technology

26.21 M-Derived Filters 507 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

27. Transmission Lines 27.1 27.2 27.3 27.4 27.5 27.6 27.7 27.8 27.9 27.10 27.11 27.12 27.13 27.14 27.15

Introduction 511 The Infinite Line 511 Short Line Terminated in Zo 512 Transmission Line Parameters 513 Phase Delay, Wavelength and Velocity of Propagation 514 Current and Voltage along an Infinite Line 515 Propagation Constant 516 Line Constants 517 Conditions for Minimum Attenuation 520 Distortion 520 Loading 521 Reflection 523 Open-Circuit Termination 525 Short-Circuit Termination 526 Standing Waves 526 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

28. First and Second Order Systems 28.1 28.2 28.3 28.4 28.5 28.6 28.7 28.8 28.9 28.10 28.11 28.12

511

Introduction 531 First Order Systems 531 Solving the Equation 532 General Procedure 534 Signal Waveforms 534 Second-Order Circuits 535 The Characteristic Equation 535 The Complex Plane 538 Impedance Concepts 539 Initial and Final Conditions 540 The Admittance Concept 544 Forced Response 545

531

28.13 Complete Response 548 28.14 Components of the Complete Response 548 28.15 Characteristics of the Components 548 28.16 Network Functions for the One-Port and Two-Port 549 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

29. Laplace Transform

553

29.1 Introduction 553 29.2 Flowchart for Mathematical Procedure 553 29.3 The Laplace Transformation 556 29.4 Basic Theorems of Linearity 556 29.5 Step Function 556 29.6 Exponential Function 557 29.7 Sine and Cosine Functions 558 29.8 Laplace Transform Operations 560 29.9 Inverse Laplace Transformation 562 29.10 Use of Partial Fractions for Inverse Laplace Transforms 565 29.11 Inverse Laplace Transforms and the Solution of Differential Equations 565 29.12 Circuit Analysis with Laplace Transforms 566 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

30. Coupled Circuits 30.1 30.2 30.3 30.4 30.5

Index

571

Introduction 571 Degree of Coupling 572 Classification of Coupled Circuits 572 Category (i) 572 Impedance of Coupled Circuits 577 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

587

Part A

ELECTRICAL FUNDAMENTALS Magnetic flux

Current direction (a) Straight isolated wire

(b) Loop

Simple magnetic fields

(c) Solenoid

1

Systems of Units OBJECTIVES

1 Inch

Inches

In this chapter you will learn about:   Scientific notation for expressing very large and very small   quantities   The need for a system of units     What makes up a system of units   The difference between fundamental and derived units     The existing standard of units     Systems of units     SI system of units     Definitions of the base units   Importance of the SI system

1 1 mm

2

2 3

4

6

7

8

10 11 12

9

Centimetres

2.54 cm 25.4 mm

cm1

4

3

5

2

3

6

7

8

9

1.1 INTRODUCTION From the earlier days of civilization, mankind found it necessary to adopt some system of units that would simplify his daily activities of barter, travel, and accumulation of knowledge. When selling, buying or exchanging goods, he had to arrive at some convenient unit of value; when travelling, he found it expedient to assign unit values to distance; and as he continued his activities from day to day, he found it necessary to relate past events, as well as future plans or commitments, to some definite time values. Therefore, it is not surprising to find that early in civilization, basic systems of measurement of length, mass and time were common. Although this fundamental concept of length, mass and time was universal, over the centuries, different nations used different standards. Consequently, the exchange of ideas among scholars, scientists and merchants was hindered by the confusion and time-consuming necessity of comparing the various systems of units. Finally, in the latter part of the eighteenth century, remedial steps were taken towards international agreements regarding standard systems of units.

1.2 SCIENTIFIC NOTATION In electrical and electronic practices, values often range from fractional values in millionths to unit values in excess of a million. The Standard System of Scientific Notation, also known as the engineer’s short hand, is employed because it simplifies the writing of numbers involving many zeros, both in fractional expressions as well as in unit value expressions, as has been illustrated in Table 1.1. This is the basic mathematical system of using exponents to indicate both very large as well as very small numbers.

4 Electrical Technology Table 1.1 Scientific Notation Prefix

Symbol

Numerical Value

tera

T

1,000,000,000,000

or 10 

12

giga

G

1,000,000,000

or 10 

9

mega

M

1,000,000

or 10 

kilo

k

1,000

or 10 

3

hecto

h

100

or 10 

2

deca

da

10

or 10 

1

deci

d

0.1

or 10 

–1

centi

c

0.01

or 10 

–2

milli

m

0.001

or 10 

–3

micro

m n p f a

0.000001

or 10 

–6

0.000000001

or 10

–9

0.000000000001

or 10 

–12

0.000000000000001

or 10 

0.000000000000000001

or 10 

nano pico femto atto

6

–15 –18

1.3 FUNDAMENTAL AND DERIVED UNITS The standard measure of any kind of physical quantity is the unit. The number of times the unit occurs in a given amount of the same quantity is the number of measures in that quantity. Without the unit the number of measures has absolutely no physical meaning. In the field of science and engineering, two kinds of units are used; fundamental units and derived units.

1.3.1 Fundamental Units The fundamental units in mechanics are length, mass, and time. The sizes of the fundamental units are arbitrary and can be selected to fit a certain set of circumstances. Length, mass, and time are called the primary fundamental units. Measures of certain physical quantities in thermal, electrical, and illumination disciplines are also represented by the fundamental units. These units are used only when the relevant classes are involved and they may, therefore, be termed as auxiliary fundamental units.

1.3.2 Derived Units Units which can be expressed in terms of fundamental units are called derived units (see Table 1.2). Every decimal unit originates from some physical law defining the unit. A derived unit is recognized by its dimensions, which can be defined as Table 1.2 Derived Units (Electrical) Unit

Quantity

Derivation

Coulomb (C)

Charge

Amperes × seconds (A-s.)

Farad (F)

Capacitance

Amperes × seconds, divided by volts (A-s/V)

Henry (H)

Inductance

Volts × seconds, divided by amperes (V-s/A)

Hertz (Hz)

Frequency

Reciprocal of seconds (s–1)

Joule (J)

Work

Newtons × metres (N-m)

Ohm (Ω)

Resistance

Volts divided by amperes (V/A)

Volt (V)

Voltage

Watts divided by amperes (W/A)

Watt (W)

Power

Joules per seconds (J/s)

Systems of Units 5

the complete algebraic formula for the derived unit. The dimensional symbols for the fundamental units of length, mass, and 3 time are L, M, T, respectively. The dimensional symbol for volume is L  . The dimensional formulae are useful for converting units from one system to another.

1.4 STANDARDS AND UNITS If all the clocks are off time, how do we know the right time? If all readings differ somewhat from the actual values, how do we know the true value of anything? The responsibility of establishing basic standard values to maintain reliability and uniformity in measurements rests with the International Bureau of Weights and Measures, that has defined measurement units used in Electrical Engineering in terms of the basic units of length, mass and time. The definitions of the basic (fundamental) units are as follows:

Length

a eb

n ele av w 3

tr

Me

s

gth

r

The metre was defined in 1950 by an International commission as 1,650,763.73 wavelengths of the radiation of the orange-red line of Krypton 86 as shown in Figure 1.1. This is a definition in terms of wavelength of light, which makes it more permanent than the previously used definition. A standard metre is defined as 1,650,763.73 wavelengths in vacuum of the orange-red line of the spectrum of krypton 86.

3.7

6 0,7

5 1,6 1 Wavelength

Mass

The kilogram is the mass of a platinum-iridium alloy of cylindrical shape kept at the International Bureau of Weights and Measures at Sevres, in France. It is approximately equal to the mass of a cube (measuring 0.1 m on each side) of pure water at 4°C.

Krypton 86 atom

Figure 1.1 A Standard Metre

Time

The second is now defined as 1/31,556,925.9747 of the tropical year 1900. The tropical year is the time between successive arrivals of the apparent sun at the vernal equinox. Refer Figure 1.2. A fourth fundamental unit must be introduced for the study of electricity and magnetism. For practical reasons, the additional fundamental quantity is chosen to be ‘current’ while the corresponding fundamental unit is called ampere. The ampere serves as the link between electrical, magnetic, and mechanical quantities and is more readily measurable.

Ampere

Pole of the ecliptic

Fixed star (sidereal time)

Pole of the earth Mean ecliptic (apparent path of the sun)

Mean sun (mean solar time) Earth Apparent sun (ephemeris time)

g Vernal equinox

Celestial equator

(a projection of the The ampere is that current which, when flowing equator of the earth through two infinitely long parallel straight wires of on the celestial sphere) negligible cross-sectional area and placed 1m apart Figure 1.2 Basis for Celestial Time in a vacuum, produces a force per unit length of –7 2 × 10 N/m. Refer Figure 1.3. An ampere is defined as the magnitude of current which, when flowing through each of two long parallel wires separated by one metre in 1A free space, results in a force between the two wires due to their magnetic fields of 2 × 10–7 newton for each metre of length. Force = 2 × 10 –7 N 1m Although it is not possible to get two straight parallel conductors of infinite length and negligible cross-section, yet the definition shows how the ampere is derived mathematically from the basic 1m units of length, mass and time. The other electrical units are also derived units related to the protype units and such other units as Figure 1.3 An Ampere Defined shown in Table 1.2.

1A

6 Electrical Technology

1.5 SYSTEMS OF UNITS The two principal systems of units are: (1) British System and (2) Metric System. The metric system is subdivided into CGS System (centimetre, gram, second) and MKS system (metre, kilogram, second). The MKS system is shown in Table 1.3. The modern form of the MKS system is the SI system of units. Table 1.3 The MKS System of Units Quantity of Fundamental Units

Definition

Derived Unit

The legal metre

Metre, m

Mass

The legal kilogram

Kilogram, kg

Time

Same as in the CGS (i.e.) second

Second, s

Area

Square metre

m2

Volume

Cubic metre

m3

Density

Kilogram per cubic metre

kg m–3

Velocity

Metre/second

ms–1

Metre/second/second

ms–2

accelerates l kg by 1 m/s/s

Newton, m × kg × s–2

One kilogram moving with velocity of 1 m/s

m × kg × s–1

One newton acting through 1 m

1 J or m2 × kg × s–2

Joule/second

Watt, m2 × kg × s–3

Newton/square metre

m–1 × kg × s–2

Kilogram square metre

kg × m2

Stress

Newton/square metre

m–1 × kg × s–2

Modulus of elasticity

Newton/square metre

m–1 × kg × s–2

One newton acting at one metre

m–2 × kg × s–2

Newton/1 m

kg × s–2

Kilogram/metre/second

m–1 kg × s–1

Length

Acceleration Force Momentum Work and energy Power Pressure Moment of inertia

Torque Surface tension Viscosity

1.6 THE SI SYSTEM OF UNITS The initials SI are an abbreviation for Systems International d’ Units (International System of Units). The modern form of the metric system, finally agreed upon at an International conference in 1960. SI is now being adopted throughout the world and is likely to become the primary world standard of units and measurement. The system rationalizes the main metric units of measurement and standardizes their names and symbolic representation. In the field of engineering, we must be able to describe physical phenomenon quantitatively in terms that will mean the same to everyone. We need a standard set of units that are consistent among themselves and reproducible in any place in the world. In electrical engineering, we use the SI (System International) system in which the metre is the unit of length, the kilogram the unit of mass and the second the unit of time. Another basic quantity is the temperature which, in the SI system, is measured in kelvins. To define electrical quantities, an additional unit is needed; taking the ampere as the unit of electric current satisfies this requirement. The candela (Table 1.4) is needed to define illumination quantities. The mole is the amount of substance of a system which contains as many elementary entities as there are atoms in 0.012 kg carbon 12. When the mole is used, the elementary entities must be specified and may comprise atoms, molecules, ions, electrons or other particles, or specified groups of such particles. (1 g atomic weight) = 1 mol of atom 23 = Avogadro’s Number (6.02 × 10  ) of atoms. In order to determine the luminous intensity of a source of light, it is necessary to compare it with a source of known intensity. The most recent primary standard is known as a black body radiator (see Figure 1.5). The modern preferred term is full radiator or Planckian radiator.

Systems of Units 7

Table 1.4 The Seven Base Units of the SI System Quantity

Unit

Symbol

Length

Metre

m (Figure 1.1)

Mass

Kilogram

kg

Time

Second

s (Figure 1.2)

Intensity of electric current

Ampere

a (Figure 1.3)

Thermodynamic temperature

Kelvin

k (Figure 1.4)

Luminous intensity

Candela

cd (Figure 1.5)

Mol

mol

Amount of substance °Fahrenheit

°Celsius

Kelvin

°C 0

10

100

373 Boiling point of water

212

90 80 70 60 50

100°

100°

180°

40

Platinum

30 20

25

77

298

0

32

273 Freezing point of water

–17.8

0

255.2

– 273.16

– 459.72

0 Absolute zero

Fused thorium oxide crucible Fused thorium oxide tube Powdered thorium oxide

10 0 10

Figure 1.4 Reference Points on Various Temperature Scales

Figure 1.5 Planckian Radiator

The SI system, besides seven base units (Table 1.4), has the following supplementary units. (Table 1.5) When data are specified in other units, they are first converted to SI units and then substituted with applicable equations. Three conversion factors frequently needed are: 1 metre = 39.37 inches; 1 kilogram = 2.205 pounds; and 1 newton = 0.2248 pound Table 1.5 Supplementary Units in the SI System Quantity

Unit

Symbol

Plane angle

Radian

rad

Solid angle

Steradian

Sr

1.7 IMPORTANCE OF SI SYSTEM The SI system has a number of unique features. These are as follows: 1. It is a coherent system. Coherent system means that the units defined are independent. They are known as the basic units and all other units are known as derived units. 2. It is a decimal system.

8 Electrical Technology 3. 4. 5. 6. 7.

Many units defined are used in commerce and industry, viz., volt, ampere, kilogram, etc. It can be used by technicians, scientists, engineers as well as researchers. The choice of newton avoids the confusion due to the introduction of g in equations in some branches of engineering. The use of watt, as the unit of power, has avoided the use of calorie, BTU, British and metric horse power. It is very convenient for the electrostatic and electromagnetic units, both of which had to be used in electricity till the SI was adopted on an international basis. 8. All practical units in electricity have to become absolute units in the SI system and, thus, a lot of inconvenience has been avoided in industry and research laboratories of the world which had been using these units for a long time. Note: Any system of units expressible in terms of fundamental units is called an absolute system and both the fundamental and derived units are called absolute units. The fundamental units of classical physics are the centimetre, the gram and the second. The absolute system of units is called the CGS system. Table 1.6 Conversion Factors 1 cm

=0.3937 in

1 in

=2.54 cm

1m

=39.37 in

1 in

=0.0254 m

1m

=3.28 ft

1 ft

=0.3048 m

1n

=0.2248 Ib

1 Ib

=4.45 n

1n

0.1019 kg 1.1 b

=0.4536 kg

1 ft lb

=1.356 Joules

1 kg

=9.81 N

1 kg

=2.205 Ib

1 Joule

=0.7374 ftlb

1 Joule

=2.388 × 10 fcal

1 kcal

=4187 Joules

1 kcal

=3.97 BTU

1 BTU

=0.252 kcal

1 kW

=1.34 h.p.

1 h.p.

=746 watts

–4

Since electrostatic and electromagnetic phenomena are affected quantitatively by the properties of the media in which they occur, a fourth fundamental unit is required in any absolute system of electrical units.

1.8 DEFINITIONS For quantitative work in circuits, we need to define the following quantities.

Force

A force of 1 newton is required to cause a mass of 1 kilogram to change its velocity at the rate of 1 metre per second.

Energy

An object requiring a force of 1 newton to hold it against the force of gravity (i.e., an object weighing 1 newton) receives 1 Joule of potential energy when it is raised to 1 metre. A mass of 1 kilogram moving with a velocity of 1 metre per second possesses ½ Joule of kinetic energy.

Power

Power measures the rate at which energy is transformed. The transformation of 1 Joule of energy in one second represents an average power of 1 watt. In general, instantaneous power p and average power P are defined by dw W (1.1) p= and P= dt T

Charge

The quantity of electricity is electric charge. Charge is said to be conservative since it can neither be created nor destroyed. It is said to be quantized because the charge on 1 electron (1.602 × 10–19 C) is the smallest amount of charge that can exist.

Systems of Units 9

The coulomb can be defined as the charge on 6.24 × 10 electrons, or as the charge experiencing a force of 1 newton in an electric field of 1 volt per metre, or as the charge transferred in 1 second by a current of 1 ampere. 18

Note: Milikan verified quantization of charge The charge on a body is an integral multiple of the electronic charge. Q=n×e (1.2) The charge in coulombs is Q = It (1.3) The law of conservation of charge states that the total charge of an isolated system is always conserved. If two identical bodies,   carrying charges Q1 and Q2, are brought in contact then the charge on each body will be  Q1 + Q2  .   2

Voltage

The energy-transfer capability of flow of electric charge is determined by the electric potential difference or voltage through which the charge moves. A charge of 1 coulomb receives or delivers energy of 1 Joule in moving through a voltage of 1 volt. In general, instantaneous voltage is defined by dw u= (1.4) dq volts

(1.5)

Electric Field Strength

The field is a convenient concept in calculating electric and magnetic forces. Around a charge we visualize a region of influence called an electric field. The electric field strength ε, a vector, is defined by the magnitude of force f on a unit positive charge in the field. In vector notation, the defining equation is (1.6) where magnitude e can be measured in newtons per coulomb. However, bearing in mind the definition of energy and voltage, we note that force force × distance energy voltage = = = charge charge × distance charge × dis t ance distance And electric field strength in newtons per coulomb is just equal and opposite to voltage gradient or e=−

dv in volts per metre dt

(1.7)

Magnetic Flux Density

Around a moving charge or current, we can visualize a region of influence called a magnetic field. In a bar magnet, the current consists of spinning electrons in the atoms of iron; the effect of this current on the spinning electrons of an unmagnetized piece of iron results in the familiar force of attraction. The intensity of the magnetic effect of this current is determined by the magnetic flux density B, a vector defined by the magnitude and direction of the force f exerted on a charge q moving in the field with velocity u. In vector notation, the defining equation is f = qu × B

(1.8)

A force of 1 newton is experienced by a charge of 1 coulomb moving with a velocity of 1 metre per second normal to a flux density of 1 tesla.

Magnetic Flux

Historically, magnetic fields were first described in terms of lines of force or flux. The flux lines (so called because of their similarity to flow lines in a moving fluid) are convenient abstractions that can be visualized in the familiar iron-filing patterns. Magnetic flux f (phi) in webers is the total quantity obtained by integrating magnetic flux density over area A. The defining equation of magnetic flux is f = ∫ B. dA

(1.9)

10 Electrical Technology Due to this background, magnetic flux density is frequently considered as a derived unit and expressed in webers per square metre. Example 1.1 If a current of 2 A flows for 2 minutes find the quantity of electricity transferred. Solution: Q = It coulombs Q = 2 × 2 × 60 = 240 C Example 1.2 A motor develops 60 horse powers with a speed of 800 rpm. Calculate its torque in newton metres. Solution: 60 h.p. = 60× 746 = 44760 W 800 = 83.776 radius per second 60 44760 Torque = = 534 newton - metres 83.776

Angular velocity = 2p ×

Example 1.3 A source e.m.f of 5 V supplies a current of 3 A for a period of 10 minutes. How much energy is provided in this time? Solution: Energy = V I t = 5×3×10× 60 = 9000 J = 9 kJ Example 1.4 An electronic heater consumes 1.8 MJ when connected to a 250 V supply for 30 minutes. Find the power rating of the heater and the current taken from the supply. Solution: Power rating of heater = Current taken from t he supply = I =

1.8 MJ = 1 kW 30 × 60 VI V 1 kW =4A 250 V

S UM M A RY 1. 2. 3. 4. 5. 6. 7. 8. 9.

The standard measure of any quantity is called its unit. The fundamental units in mechanics are measures of length, mass and time. All other units which can be expressed in terms of fundamental units are called derived units. A derived unit can be recognized by its dimensions, which can be defined as the complete algebraic formula for the derived unit. The standard unit should be reproducible. A numerical figure should be attached to each unit. The protype units are based on physical standards that never change. The fourth fundamental unit is the ampere. The rationalized MKS system is an absolute system in itself in which all the conversion factors are unity and derived units are identical with practical units.

Systems of Units 11

M U LT IP LE C H O I C E Q UE S TI O NS (M C Q ) 1. The units whose sizes cannot be chosen independently are called (a) Derived units (c) Absolute units

(b) Fundamental units (d) Auxiliary fundamental units

2. The basic units in SI system are (a) (b) (c) (d)

metre, kilogram, second metre, kilogram, second, ampere metre, kilogram, second, ampere, Kelvin, candela, mole metre, kilogram, second, ampere, Kelvin, candela

3. The symbol nm stands for (a) newton metre (c) nano milli

(b) Velocity (d) Force

ANSWERS (MCQ) 1. (a) 2. (c) 3. (b) 4. (c)

10. The unit of electric field strength is (a) Volts/coulomb (c) Dynes/ampere

(b) Newtons/ampere (d) Newtons/coulomb

11. Dimensions of power are

(b) M2L1T–2 (d) M1L1T–2

12. The ratios L/R and RC (L = inductance, R = resistance and C = capacitance) have the dimensions as those of (b) Acceleration (d) Force

(a) ML2T3A–1 (c) ML2T–3A–1

(b) Momentum (d) Power

7. (b)

(b) ML2T–3A–2 (d) ML2T–3A1

15. The dimensions of electrical conductivity are

(b) 1 micron = 10–6 cm (d) 1 micron = 10–4 cm

6. (d)

(b) Joule (d) Kilowatt

14. The dimensions of voltage in terms of M, L, T, A are

(b) Length (d) Velocity

5. (d)

(a) Kilowatt-hour (c) Dyne

(a) Energy (c) Angular momentum

7. A micron is related to the centimetre as (a) 1 micron = 10–8 cm (c) 1 micron = 10–9 cm

9. The unit of power is

13. Joule × sec is the unit of

(b) Force (d) Electric current

6. Which of the following is a derived unit? (a) Mass (c) Time

(b) 10–4 cm (d) 102 cm

(a) Velocity (c) Time

5. Out of the following the only scalar quantity is (a) Velocity (c) Momentum

(a) 10–8cm (c) 104 cm

(a) M1L2T–3 (c) M1L2T–1

(b) nano metre (d) number of moles

4. Which of the following is a fundamental quantity? (a) Volume (c) Time

8. The radius of an atom is approximately equal to

(a) M–1L–3T3A2 (c) ML3T–3A–2

8. (a)

9. (d)

10. (d)

11. (a)

(b) ML3T3A2 (d) M2L3T–3A2

12. (c)

13. (d)

14. (c)

15. (a).

CON V E NTI O NA L Q UE S TI O NS (C Q ) 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

Through what steps is a science developed? What is meant by scientific notation? Describe the SI system of units. Define the seven base units of the SI system. List five advantages and five disadvantages of the metric system. What do you consider the most important characteristics of a standard used to find the value of a unit? Why is it important to have a universally accepted set of units? Define the terms (a) Units (b) Absolute Units (c) Fundamental Units and (d) Derived Units. Cite suitable examples to support your answer. Describe the International Standards of mass and length. What is the number of base SI units?

11. What are the dimensions of the following? (a) Velocity (b) Acceleration (c) Work 12. What are the dimensions of the following? (a) Energy (b) Momentum (d) Force (c) Surface tension (e) Power (f) Torque 13. What is the time taken by light to travel from the sun to the earth? 14. A certain brand of cigarettes is labeled 100’s by their manufacturer. This refers to the fact that the length of each cigarette is 100 mm. What is the length of the cigarette in inches? 15. Convert a density of 7.45 g/cm2 to MKS units.

ANSWERS (CQ) 10. Seven 11. (a) LT–1 (b) LT–2 (c) ML2T–2 12. (a) ML2T–2 (b) MLT–1 (c) MT–2 (d) MLT–2 (e) ML2T–3 (f) ML2T–2

13. Approximately 8 minutes 14. 3.94 inches 15. 7.45 × 103 kg/m3.

2

Electrons in Action OBJECTIVES In this chapter you will learn about:    Conduction in solids    Energy bands and energy levels  Effect of an electric field on energy levels    Resistivity of conductors, semiconductors and insulators   Bonds in atoms—metallic, ionic, and covalent      Electrons in action—the electric current    Direction of current flow—conventional, electron, ionic, and diffusion  The effects of electricity

+ 13

+ 13

+ 13

Construction of aluminium atom Nucleus—13+ positive charges First shell— 2-bound electrons First ring or shell Second ring or shell Second shell— 8-electrons Third shell— 3-electrons Third ring or shell Possible to be free.

2.1 INTRODUCTION Electrical engineering itself can be described as a study of the movement of electrons. For a current to flow within a medium, the presence of charge carriers is necessary, and these charge carriers must be mobile. In its normal state, an atom is electrically neutral, the positively charged nucleus being surrounded by electrons which collectively carry an equal and opposite charge. However, charge carriers are created if electrons are detached from the Potential atoms. To achieve this, work must be done to separate the electrons from the energy nuclei against the electrostatic attraction. When an electron is separated from its atom, two kinds of charge carriers are created: the negatively charged electron, and the positively charged ionized atom. Because of their size and mass, the ionized atoms are far less mobile than the electrons which, therefore, make a major contribution to conduction. The electrons nearer the nucleus are more tightly bound to the atom than those in the more distant positions, and more energy is required to detach them.The situation has been Distance from nucleus graphically represented in Figure 2.1, in which the nucleus is considered Nucleus to be at the bottom of a potential energy well for electrons. The deeper an Figure 2.1 The Attraction of the Nucleus electron is in the well, the more is the energy required to separate it from the atom. The motion of an electron about the nucleus prevents it from falling Acts as a Potential Well for right down into the nucleus. Electrons

2.2 CONDUCTION IN SOLIDS According to the quantum theory, an atom has well-defined energy levels (orbits) which the electrons can occupy but they cannot possess energies between these values. An atom has more energy levels than electrons to occupy them and the electrons tend to occupy the lower energy levels, leaving the upper energy levels vacant, which correspond to positions further from the nucleus. The energy levels may be represented by circular orbits which the electrons can occupy, as has been represented in Figure 2.2. The energy needed to ionize an atom is often expressed in terms of the potential in volts, which would extract an electron, i.e., in electron volts. 1 electron volt = 1.6 × 10–19 Joules (2.1)

Electrons in Action

The ionization potential represented in the potential diagram in Figure 2.3 is the distance from the outermost occupied orbit to the top of the potential well. When atoms assemble together in a solid, each single energy level of the isolated atoms is split into a range of levels owing to the influence of the other atoms, as can be seen in Figure 2.4. The potential barriers preventing electrons from passing from atom to atom in the solid are less than those preventing the escape of electrons from isolated atoms. The outer electron orbits overlap to such an extent in the solid that they may be considered to be common to all the atoms. Electrons which are excited into these orbits are not confined to a single atom but move randomly about the solid. These common levels are represented on the potential diagram by a continuous band of energy levels called the conduction band. In metals, the occupied levels are continuous with the band of vacant levels and the potential barrier is zero. This is shown in Figure 2.4. In insulators, the outer orbits are far from the occupied orbits and on the potential diagram, as has been shown in Figure 2.5, this is represented by a large gap between the occupied levels and the conduction band.

13

Vacant levels

Potential energy + Radius of orbit

+ The energy levels represented in two dimensions

Plan view of orbits

Figure 2.2 The Permitted Energy Levels in an Atom may be Represented by Circular Orbits

Conduction band Ionization potential 7.6 V

10.6 V

Occupied levels Magnesium

Sulphur

Figure 2.3 The Ionization Potential of the Metal Magnesium and the Insulator Sulphur Represented on a Potential Well Diagram

Figure 2.4 Energy Levels and Potential Barriers in a Metal with Occupied Levels Adjacent to the Conduction Band

Conduction band

Energy gap

+

+

+

+

Figure 2.5 Energy Levels in an Insulator with an Energy Gap which is Greater than the Energy of Thermal Vibration of the Atoms At very low temperatures, the electrons within a solid will tend to occupy the lower energy orbits which are confined to individual atoms and, therefore, few charge carriers will be available. At higher temperatures, the atoms vibrate more energetically and shake the electrons into one of the bands of energy levels which continue through the material. Thus, the number of electrons free to carry charge through the solid increases with the temperature. If an external potential is applied to the solid then the electrons in the conduction band flow down the potential gradient, as can be seen in Figure 2.6. In this case, even the smallest field would cause a very large current to flow and the electrons would continually accelerate without loss of energy. However, the potential diagram is imaginary and does not represent the atoms in three dimensions,

14 Electrical Technology and the electrons frequently collide with the energy levels, converting their kinetic energy into heat energy. The frequency of these collisions increases with the thermal vibration of the atoms, i.e., with the temperature. Thus, an increase in the temperature of a solid reduces the mobility of the electrons. Ele ctr on In metals, the occupied levels are so near to the upper vacant levels flo w that, at normal temperatures, a large proportion of the outer electrons is in the conduction band. The electrons move at random about the metal as the molecules of a gas move inside a container. Any increase in the temperature of a metal reduces the mobility proportionally more than it increases the number of conducting electrons and, therefore, there is a net increase in resistance. In insulators, the conduction band is far from the energy level of the outermost occupied orbit. The energy of the thermal vibration of the atoms is not sufficient to elevate more than a few of the electrons to the Figure 2.6 Effect of an Electric Field on conduction band, and the resistance of the material is very large. Any Energy Levels in a Solid increase in the temperature increases the number of electrons free to move more than it reduces their mobility, so there is a net reduction in resistance. Insulators subject to high potentials sometimes break down at high temperature and allow a substantial current to flow. Carbon, silicon and germanium belong to a class of metals called semiconductors, which are intermediate in resistance between metals and insulators, as has been graphically represented in Figure 2.7. The atomic structures of these materials are similar to each other (they are in the same group in the periodic table of elements). Transistor germanium Pure silicon Mica

Glass

Wood

Nichrome

Pure germanium

1012 1011 1010 109 108 107 106 105 104 103 102

Impure germanium

Platinium Copper

101 10–1 10–2 10–3 10–4 10–5 10–6 10–7 10–8

Figure 2.7 Resistivity (Ohm-metre) of Conductors, Semiconductors, and Insulators at 20°C In a semiconductor, the energy difference between the upper occupied orbits and the conduction band is only about twenty times as much as the energy of thermal vibration of the atoms. This range can be easily understood with the help of Figure 2.8. Consequently, any increase in the temperature produces a large increase in the number of electrons in the conduction band. The resistance of semiconductors drops rapidly as the temperature rises. Comparative figures are given in Table 2.1 and also in Figure 2.7.

2.3 BONDING IN ATOMS

Conduction band

Energy gap

No energy gap

Localized orbits Metal

Semiconductor

Insulator

Figure 2.8 The Energy Levels in a Metal, Semiconductor and Insulator Compared Schematically with the Energy of Thermal Vibration of the Atom

The adhesion of atoms occurs through the action of electrons in the outer ring of the atom. These electrons (called valence electrons) enter the orbit of an adjacent outer ring atom and, thus, form a cohesive factor in the bonding together of atoms. There are three general categories of such atomic bonding; ionic, covalent, and metallic. The ionic bonds, as can be seen in Figure 2.9, are those which exist in compounds. When an atom becomes ionized, it indicates that the atom no longer has a net neutral charge, but is now either primarily positive or primarily negative. Such a

Electrons in Action

15

Table 2.1 Comparison of the Resistance of Metals, Semiconductors, and Insulators, at Different Temperatures (Figures are approximate because they vary widely with the method of preparing the sample.) Material

Resistance in ohms between faces of a centimetre cube

0°C

500°C

1000°C

Copper

1.6 × 10

Carbon Germanium Silicon

3.5 × 10–3 9 × 10–2 6 × 10–5

2.7 × 10–3 – –

2.1 × 10–3 – –

1016

109

107

Silica

–6

–6

5 × 10

9 × 10–6

condition occurs when an electron is removed from the structure of the atom or when an electron is added to the structure. An unbalanced condition is set up between the nucleus and the planetary electrons. Both the negative and positive ionization can be frequently encountered and are prevalent in gas-filled tubes. The diagram above represents a small fragment of sodium chloride, which forms cubic crystals. Each sodium ion is surrounded by six chloride ions, and each chloride ion is surrounded by six sodium ions. Ionized gases are sometimes referred to as the fourth state of matter. When a gas is ionized, it is in a highly agitated state and the ionization is conductive and also affected by magnetic fields. Na+ Cl– Na+ Cl– The term plasma, from the Greek meaning mold or matrix, is Figure 2.9 Sodium Chloride Crystal used by electrical and electronic engineers to refer to this state. Great studies have been made in recent years in plasma engineering. Plasma can be considered to be a highly agitated ionized state. It is Single atom of Cu equivalent to an electrical conducting fluid which can be acted upon by magnetic fields. The ionic bond between the sodium and chlorine atoms results in the outer shell of each atom possessing a full quota of electrons, which makes it extremely difficult to move the electrons from atom to atom in the normal free-electron movement found in metals. Each atom, in conjunction with its neighbouring ones, forms the cubic arrangement as shown in Figure 2.10, resulting in a crystal structure known as a crystal lattice network. Figure 2.10 Crystalline Structure The metallic bonds which form the elements gold, silver, copper, etc., are of Copper Atoms formed because the valence electron, as it orbits, may easily enter an adjacent atom’s orbit, particularly at the time when the free electron is at an equal distance from the nucleus of each atom. If one free electron thus enters the orbit of an adjacent atom, the atom from which it left momentarily becomes a positive ion. This positive ion will now have an attraction from an electron from an adjacent atom and will capture such an electron. The condition occurs repeatedly to form the metallic bond. When electric pressure is applied, the free electrons are compelled to run in sequential fashions from one atom to the next in a direction established by the polarity of the electric pressure applied. A graphic representation, of the same can be seen in Figure 2.11. Germanium and silicon are the two most important semiconductors used in semiconductor devices. The crystal structure of theses materials consists of a regular repetition in three dimensions of a unit Figure 2.11 In the Copper Atom, 29 cell having the form of a tetrahedron with an atom at each vertex. Electrons Whirl Around the This structure is illustrated symbolically in Figure [2.12 (a) and (b)]. Nucleus in Seven Different Germanium has a total of 32 electrons in its atomic structure arranged Orbits

16 Electrical Technology in shells. Each atom in a germanium crystal contributes four valence electrons, so that the atom is tetravalent. The inert ionic core of the germanium atom carries a positive charge of +4 measured in units of the electronic charge. The binding forces between neighbouring atoms result from the fact that each of the valence electrons of a germanium atom is shared by one of its four nearest neighbours, as shown in Figure 2.13. In spite of the availability of four valence electrons, the crystal has a low conductivity. When a valence electron is detached from an atom as, for example, due to thermal agitation, it is elevated to the conduction band of energies, as can be seen in Figure [2.14 (a) and (b)]. It can now move freely under the influence of an electric field and contributes to the flow of current. The electron leaves behind a positively charged atom with a vacancy for a valence electron and valence electrons from other atoms can easily transfer to (a) (b) the gap. The effect of repeated movement of electrons in one direction is to move the positively charged vacancy Figure 2.12 (a) Germanium has a Tetrahedral Strucin the opposite direction (Figure 2.15). Such a vacancy is ture in Three Dimensions but called a positive hole and it behaves in some ways like a (b) the Fact that Each Atom Shares one positively charged electron. Electron with one of its Four Neighbours can be Represented in Two Dimensions Semiconductor atomic nucleus with inner electrons

Ge

Ge

Conduction band Ge

Ge

Ge

Ge

Conduction band Bound valence electron

Valence band (a)

Figure 2.13 Electron Pairs or Covalent Bonds in a Germanium Crystal

Forbidden band Valence band (b)

Figure 2.14 Conduction and Valence Bands for (a) Conductors and (b) Insulators

2.4 ENERGY BANDS Between the two extremes of conductors and insulators is the group of substances, for example, germanium, silicon, and selenium, known as semiconductors; the forbidden zone between the valence band and the conduction band is less wide than for insulators, yet the two bands do not overlap as they do in conductors Figure [2.14 (a) and (b)]. Although electronic currents do not easily flow in them, a smaller quantum of energy than that for insulators is required to make an electron jump to the conduction band. This is shown in Figure 2.15.

2.5 ELECTRONS IN ACTION Electric current in a wire travels at nearly the speed of light (3 × 108 m/s), i.e, the effect of current is almost instantaneous. An individual electron moves much more slowly than the effect of the current. It may take minutes for an individual electron to travel a few feet in the wire. This is illustrated in Figure 2.16. Suppose you have a very long card board tube with a diameter just long enough to pass a tennis ball. You lay the tube on the floor and fill it full of tennis balls. When you push an extra ball into one end of the tube, another ball immediately comes out of the other end of the tube (Figure 2.16). If you did not know that the tube was full of tennis balls, you might think that the tennis ball pushed in one end of the tube travelled very quickly all along the tube and out the other end.

Electrons in Action

The effect is very fast. Yet each tennis ball moves only a short distance. Now suppose you stacked up six tubes filled with balls (Figure. 2.17) and pushed the balls first into one and then into another. You will have a steady stream of balls appearing at one end of the tubes, yet only balls in only one tube would be moving at any one time. Even within that tube each ball would move only a short distance. This is comparable to the way in which charge carriers (electrons) move through a wire when current is flowing in the wire. Let us assume that you could look inside an aluminium wire and see the atoms and their particles. For simplicity, only the valence electrons are shown. Now, suppose, the ends of the wire are connected to a flash light cell. The cell provides an electric field through the wire. The electric field frees some of the valence electrons of the aluminium atoms, as shown in Figure 2.18, by giving them additional energy. At the moment that an individual electron is freed, it may be travelling in a direction opposite to that of the main current. However, in the presence of the electric field, it soon changes its direction of movement. For every electron that is freed, a positive ion is created. This positive ion has an attraction for an electron. Eventually, one of the free electrons will migrate close to the positive ion. The electron will be captured by that positive ion, which then, of course, will become neutral atom.

17

Figure 2.15 Conduction and Valence Bands for n type and p Type Semiconductors; (a) Germanium Atom with Trivalent impurity (b) Energy Band Diagram of p-type Semiconductor (c) Germanium Atom with Pentavalent Impurity (d) Energyband Diagram of n-type Semiconductor

Hidden balls

Ball entering tube

Ball leaving tube

Figure 2.17 Electron Movement Illustrated

Figure 2.16 Apparent Speed Illustrated. A Ball Exits the Instant Another Ball Enters the Tube Direction of current

– –

Electron

–

– + – ++

+ ++ – – – –

–

+ ++

–

–

–

– –

–

+ ++ – – –

+ ++

+ ++ –

– –

+ –

+ – ++ –

+ ++ – – –

+ ++

+ ++ –

–

+ – ++

–

–

+ – – ++

–

Positive ion

Figure 2.18 Current in a Solid. A Free Electron Travels Only a Short Distance Before it Combines with a Positive Ion

18 Electrical Technology A free electron does not remain a free electron and travels the full length of the wire; rather, it travels a short distance down the wire and is captured by one of the positive ions. At some later time, this particular electron may again gain enough energy to free itself of its new parent atom; it then travels further down the wire as a free electron. We can think of the free electrons as hopping down the conductor from atom to atom to atom. As long as there is a new free electron created every time a free electron is captured, the number of electrons moving down the wire remains constant. Current continues to flow. It continues to flow in the same direction through the conducting wire. Current that flows in the same direction all the time is called direct current (abbreviated as d.c.). It is the type of current you get from flashlight cells and batteries. Alternating current (abbreviated as a.c.) is the type of current you have in your home. It is the type of current that periodically reverses the direction in which it is moving. The current reverses its direction every 1/100th of a second. Currents that reverse in direction are easier to visualize if you think of the individual electrons as surging back and forth between several atoms.

2.6 DIRECTION OF CURRENT FLOW Prior to the discovery of the electron, it was necessary to postulate the direction of flow of current. The electric current was assumed to be a flow of positive electricity from a positive to a negative terminal. This is still the accepted convention in practical applications, and the terminals of electrical instruments are, where necessary, marked positive and negative accordingly. The flow of current by holes being a positive current is naturally in the same direction as positive current. On the other hand, the electron theory shows that the normal electric current is an electron flow, i.e., a flow of negative electricity from a negative ion to a positive ion. Holes These two conceptions may at first appear to be conflicting but it is not + so—a positive flow from A to B, as represented in Figure 2.19, is mathA B ematically the same as a negative flow from B to A. No confusion should – arise if it is remembered that the conventional view assumes a flow of posiElectrons tive electricity, whereas the electron theory concerns the flow of negative Figure 2.19 Direction of Current Flow electrical charges. Both electrons and holes can contribute to electric conduction. This has been illustrated in Figure 2.20. Row of unionized atoms

+

2.7 DIFFUSION CURRENT MOMENTARILY

– Ionized atom –

+

Electron conduction

Positive hole conduction by shuffling of electrons

+

Figure 2.20 Both Electrons and Holes can Contribute to Conduction

Electrical potential difference is not the only cause of electric current. Free electrons in semiconductors, like the molecules in a gas, try to spread themselves out uniformly. If, for example, as has been shown in Figure 2.21, there are many electrons at point A, and only a few at point B, then more electrons move from A towards B in the opposite direction. This movement of charge is an electric current, which is called diffusion current and differs from that caused by an electric field, which is called conduction current.

2.8 DRIFT VELOCITY A

Figure 2.21 Diffusion Current

B

The electric current in a conductor is the axial drift of the free electrons along the length of the conductor. Let there be n free electrons per unit volume and let the mean velocity of the axial drift be u. Then, in time, dt the mean distance of translation is u  dt. If the area of cross-section is a, then an elementary volume of length u  dt, as seen in Figure 2.22, will contain nau  dt electrons, The number of electrons crossing a normal section in dt seconds is = nau   dt. Current i = dq/dt = nau

Electrons in Action

The total quantity of electricity crossing normal section in dt seconds = naeu  dt. \        current i = dq/dt = naeu and current density J = i/a = neu (2.2)

19

– – υ –

Example 2.1

–

Given the data that there are 8.5 × 1028 electrons in one cubic metre of copper, calculate the velocity of the axial drift when a copper conductor is worked at 1000 A/in2. Solution: u = J / ne

υ dt

Figure 2.22 Drift Velocity

J = 1000 A / in 2 = (1000 / 6.45)×104 = 1.55×106 A / m 2 e = 1.6×10−19 coulomb

(

∴ u = 1.55×106

) (8.5×1028 ×1.6×10−19 )

= 1.14×10−4 m/sec Note: The velocity of electron drift is very small, even when conductors are worked at a high current density.

2.9 THE NATURE OF ELECTRIC CURRENT The flow of charge carriers caused by an electric field may, in the case of a gas or liquid, consist of a flow of positive ions in the direction of the field, or of negative ions or electrons opposite to that direction, or of both at once. In a metal, the flow is known to consist largely of a movement of electrons opposite to the direction of the field. This is illustrated in Figure 2.23. dq i= (2.3) dt

q = N1 p + N 2 e

(2.4)

where, i denotes the instantaneous electric current and q the net charge. N1 p denotes the net positive charge and N2e the total negative charge. The symbol e is used to denote the charge of an electron, so that e = 1.602 × 10–19 C/electron There are 6.24 × 1018 electrons in 1C. Metal wire

Electrochemical cell

p -type semiconductor

Neon lamp

Holes

Electrons Positive ions Negative ions +

Source of e.m.f.

Positive ions Electrons –

Figure 2.23 The Clockwise Conventional Current in this Circuit Represents the Flow of Different Kinds of Charge Carriers in Different Parts of the Circuit The coulomb is that unit of charge which, when placed 1m apart from an identical particle charge, is repelled by a force of 10–7c2, where c is the velocity of light in metres/second. Example 2.2 How many electrons pass through a conductor in one minute if the current through it is 0.32 mA? Solution: Let n be the number of electrons. Q = ne = It

n = It / e =

0.32 × 10−3 × 60 1.6 × 10−19

n = 1.2 ×1017

20 Electrical Technology

Q = ne = It n = It / e =

0.32 × 10−3 × 60 1.6 × 10−19

n = 1.2 ×1017 Example 2.3 In copper the number of free electrons per metre3 is 8.4 × 1023. Calculate the drift velocity of electrons in a copper wire of cross-sectional area 2 × 10–6 m2, if a current of 0.63A is flowing through it. Solution: I = neAud ud = I / neA =

8.4 × 1023

0.63 × 1.6 × 10−19 × 2 × 10−6

ud = 2.3 × 10−5 ms−1 Example 2.4 What current must flow if 0.24 coulombs is to be transferred in 15 ms? Solution: Q = It I =Q t 0.24 0.24 × 103 = 15 15 × 10−3 = 240 / 15 =

I = 16 A

2.10 EFFECTS OF ELECTRICITY What is electricity? Electricity is known by the results it produces—the laws which govern its behaviour have been investigated and are well established. The effects of an electric current are threefold, namely, (1) magnetic, (2) heating and (3) chemical, as shown in Figure 2.24. Conversely, electric current is produced from the expenditure of magnetic heat or chemical energy. C

L

R

b Resistance

Lamp

N

a Z Electrolytic cell

Compass

Figure 2.24 Ths Effects of an Electric Current The magnetic effect has perhaps the widest applications in industry. It is the basis of electrical power generating systems as also of electric traction; it has also made possible, the early microphone, the telephone receiver, the electromagnet and the relay which are the foundations of telecommunication equipment. The heating effect has a wide use in electric furnaces as well as in welding and in heating and lighting appliances. The chemical effect is applied mainly to the use of storage battery installations for use where emergency and portable supplies are required and also in electroplating industry. The flow of an electric current in a conductor normally results in—(1) the creation of a magnetic field, and (2) the production of heat. The chemical effect results only under certain favourable conditions, requiring the presence of certain chemical solutions. Precision instruments for measuring electrical quantities are designed from knowledge of the laws governing these effects, though meters depending on electro-chemical effects are not practicable outside the laboratory.

Electrons in Action

21

S UM M A RY 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24.

Electrical engineering is a study of the movement of electrons. For current to flow in a medium charge, carriers must be present and these charge carriers must be mobile. When an electron is separated from its atom, two kinds of charge carriers are created. The ionized atoms are far less mobile than the electrons. The deeper an electron is in the well, the more the energy required to separate it from the atom. The electrons tend to occupy the lower energy levels. The outer electron orbits overlap to such an extent in solids that they may be considered to be common to all the atoms. In metals, the potential barrier is zero. In insulators, there is a large gap between the occupied bands and the conduction band. An increase in the temperature of a solid reduces the mobility of the electrons. In metals, there is a net increase in resistance with increasing temperature. In insulators, there is a net reduction in resistance at higher temperatures. Semiconductors are intermediate in resistance between metals and insulators. Valence electrons form a cohesive force in the bonding of atoms. Ionic bonds exist in compounds. Plasma is sometimes referred to as the fourth state of matter. Metallic bonds form the elements gold, silver copper, etc. Covalent bonds are formed in semiconducting materials. Conduction band and valence bands of conductors overlap. Free electrons hop down a conductor from atom to atom. Conventional current is a flow of positive electricity from the positive to the negative terminal. Electron current is a flow of negative electricity from the negative to the positive terminal. Both electrons and holes contribute to current in semiconductors. Electricity is known by the results (magnetic, heating and chemical) it produces.

M U LT IP LE C H O I C E Q UE S TI O NS ( M C Q ) 1. Under what conditions can an electric current flow in a material? (a) There must be charged particles available (b) There must be movable particles available (c) There must be particles available which are both charged and movable

2. What particles cannot be current carriers? (a) Those in the atomic nucleus (b) Valence electrons (c) Inner electrons

4. Why is a metal a good electrical conductor?

(a) Because there are valence electrons in a metal (b) Because all the valence electrons in a metal are free electrons

5. Which type of conduction is very dependent on temperature? (a) Conduction caused by p-doping (b) Conduction caused by n-doping

6. What special property is shown by a semiconductor?

3. Which electrons are important in holding a crystal together?

(a) It insulates (b) It conducts (c) Both (a) and (b)

(a) The valence electrons (b) The inner electrons of all the atoms

ANSWERS (MCQ) 1. (c) 2. (c) 3. (a) 4. (b) 5. (a)

6. (c).

CON V E N TI O NA L Q UE S TI O NS ( C Q ) 1. Explain the difference between conductors, semiconductors, and insulators.

2. Explain the phenomenon of conduction in solids.

22 Electrical Technology 3. What is the importance of a potential well? 4. How are energy bands formed? 5. Explain the difference between metallic bond, ionic bond, and covalent bond. 6. Explain the difference between conventional current flow and electron current flow.

7. Discuss the electrical characteristics of germanium and silicon. 8. How does diffusion current differ from conventional current?

3

Electric Circuit OBJECTIVES Gauges to measure difference in pressures

In this chapter you will learn about:           

The essential ingredients of an electric circuit The AND circuit and the OR circuit The comparison between an electric circuit and a water circuit The difference between a.c. and d.c. The concept of electric current and electromotive force Safety precautions while working on electric equipment Different types of insulators, semiconductors, and conductors used in electrical and electronic engineering and their properties Different types of switches and contact materials The types of electric circuit Ohm’s law and simple connected problems The water circuit and the electric circuit — an analogy

Resistance to flow Metre to measure litres per minute Pump Water circuit Metre to measure difference in pressures (volts) Electrons

Resistance to flow +

Electrons

Metre to measure amps

Cell The electric circuit

3.1 INTRODUCTION An electric circuit is a practical and commercial means of transporting electric energy from one place to another. The transfer of energy is done by charge carriers (electrons). An electric circuit can be in any one of the three states: (1) Open circuit, (2) Closed circuit, and (3) Short circuit. Open circuit is the passive state of an electric circuit, where energy is available but it is not being transported. Closed circuit is the active state of an electric circuit, where energy is available and it is being transported. Short circuit is the defective state of an electric circuit, where fuses of appropriate current rating are used to avoid this state.

3.2 ELECTRIC CIRCUIT An electric circuit consists of the following parts: 1. Source of electrical energy (a.c. or d.c.). 2. Load (this could be a single load or multiple loads or any of a wide range of electrical appliances). 3. The switch (to open and close the circuit). 4. Connectors (cables and wires which join parts) (see Figure 3.1). The circuit in Figure 3.2 contains two switches. The lamp will only light up when switch A and switch B are closed. If only one switch is closed, the lamp will not light up. In an AND circuit, the switches are connected in series.

R Load or resistance S

Conductor

Switch

V

–

+

Source of electrical energy

Figure 3.1 Electric Circuit

24 Electrical Technology In the arrangement shown in Figure 3.3, the lamp will light up when either switch is closed. In this circuit, which is known as an OR circuit, the switches are connected in parallel. A

A

Switch B

Load or resistance

B

Lamp

–

+

AND-circuit

Figure 3.2 In an AND-circuit the Switches are Connected in Series

–

+

OR-circuit

Figure 3.3 In an OR-circuit the Switches are Connected in Parallel

A relation, known as Ohm’s law, exists between the current and the voltage drop across the load resistance. V = R LI Voltage drop equals resistance times current.

3.3 CURRENT Owing to the intangible nature of the electric current, a reference to an analogy such as that provided by the hydraulic system, gives a useful idea of the character of the electric circuit. Let us consider the elementary hydraulic system as shown in plan in Figure 3.4(a), where a simple pump consisting of a piston P in a cylinder C, is connected to the two ends of a pipe, the whole pipe being filled with water. While the piston is at rest, the water remains stationary, since there is no motive force urging it to move. Although there is water present in all parts of the system, yet to produce the motion of the water, a force or pressure of some sort must be applied to it. When the pump is actuated, the piston on its forward stroke (shown by C full arrows) produces an increase of pressure at one end of the P pipe and a corresponding suction at the other end, resulting in G a displacement of the water molecules along the pipe in the direction of the full arrows. The reverse piston-stroke (dotted arrows) reverses the conditions of pressure and suction, and the water molecules are now displaced in the reverse direction (dotted arrows). The reciprocating piston produces alternating (a) (b) surges in the water, the water being always in a state of motion. Figure 3.4(b) shows a simple electrical system. An electric Figure 3.4 (a) Water Analogy (a.c.) generator, or alternator, G, has its two terminals connected to (b) Simple a.c. Circuit a conducting material, say, a copper wire. When the generator is at rest, there is no electric flow in the wire, although it abounds with electrons. When the generator is put into motion, it sets up an electrical potential difference between its terminals, resulting in a displacement of electrons in the conducting path. Compared with the water pump, the electric generator may be regarded as a device for pumping electrons around a conducting path or circuit. The generator continually reverses the direction or sense of its potential difference, the electron displacement being, therefore, alternated first in one direction and then the other, as shown by the full and dotted arrows. The electrons surge back and forth in the conducting path much in the same manner as the water molecules in Figure 3.4(a). The moving electrons carrying negative electrical charges produce surges of currents of electricity in the conductor. An electric current is the flow of electricity along any path or around any circuit. The symbol for current is I or i. An electric current which continually changes its direction in this manner is called alternating current (a.c.). Currents of this nature find considerable application in electrical practice; the natural occurrence of alternating quantities is also found in light and sound. An alternating current is an electric current which alternately reverses its direction in a circuit in a periodic manner, the frequency being independent of the constants of the circuit. The term frequency is applied to denote the number of times a change of direction is produced in a given interval of time, usually one second. In Figure 3.5(a) the hydraulic system includes two simple valves. On the forward piston-stroke (full arrow), the water pressure forces open the valve O and shuts the valve I, due to the difference in pressure, water molecules are displaced

Electric Circuit

25

X around the pipe in the direction of the full arrows. O S S The reverse piston-stroke (dotted arrows) closes valve O and opens the valve I, but does not produce I G Y any displacement of the water in the pipe. In this way, the reciprocating piston produces a series of gushes or pulsations of the water in the pipe; the water being constrained by the valves to always move in the same A Z B direction. With a suitable design and speed of pump, a (a) (b) sensibly steady unidirectional flow may be obtained. The dynamo G of Figure 3.5(b) is an electric generator fitted with a device called a commutator Figure 3.5 (a) Water Analogy (d.c.) (b) Simple d.c. Circuit whose purpose may be regarded as being similar to that of the valves in Figure 3.5(a). The dynamo maintains a steady potential between its terminals—usually with a very small degree of pulsation—so producing a continuous unidirectional drift of the free outer electrons which constitutes the electric current flow around the circuit. The direction of current flow indicated in the diagram is the practical or positive flow. An electric current of this nature, namely unidirectional, is usually known as a direct current (d.c.) or sometimes as a continuous current. A direct current is an electric current flowing in one direction only and sensibly free from pulsation. The direct current flows in one direction like the water of a stream, while the alternating current flows backwards and forwards like the tide in an estuary. In the hydraulic system, the flow of water can be stopped by shutting the stop-cock S, as can be seen in Figure 3.5(a). In the electric circuit, the opening of a switch S, graphically represented in Figure 3.5(b) inserts a non-conducting material, usually air, into the path and prevents the passage of electrons; the flow of current necessitates a complete conducting path or circuit from the generator around the external circuit and back to the generator. When the external circuit is opened by a switch, the generator is said to be on open circuit: with the circuit completed by closing the switch, a closed circuit is presented and the current flows uninterrupted. The rate of flow of water is measured by the quantity of water passing a given point in unit time. The intensity or strength of an electric current is similarly measured by the quantity or charge of electricity passing a given point in unit time, usually one second. The fundamental unit of electric charge is that carried by one electron. Consequently, the fundamental unit of current strength is a rate of flow of one electron per second. For practical purposes, a unit of charge known as the coulomb (C) equal to 6.28 × 1018 electrons is adopted. The ampere (A) is a rate of flow of one coulomb per second. It is the rate of flow which is to be understood by the use of the term current. In the hydraulic system, a higher rate of flow is obtainable either by an increased piston speed or by using a more powerful pump. At the same time, the diameter of the pipe and the amount of friction on the inner walls have a controlling effect on the rate of flow. Similarly, the rate of flow of current in the electric circuit is dependent not only upon the electrical pressure applied from the generator, but also upon the resistance due to the nature and dimensions of the conducting material. Resistance is that property of a body by virtue of which it resists the flow of electricity through it, causing a dissipation of electrical energy in the form of heat. The symbol for resistance is R or r. The strength of the electric current is the same at all parts of the circuit. If the circuit branches into two or more paths, the total current strength in these branches is equal to the current leaving and returning to the generator. + –

3.4 ELECTROMOTIVE FORCE The water of a stream rising in the hills flows down hills in its effort to reach the sea. This is due to the gravitational force acting on the mass of water and pulling it downwards until it reaches sea level. If an elevated tank of water, as has been diagrammatically represented in Figure 3.6 has an open pipe near its base the water will gush out, always attempting to reach a lower level. The stored water is said to possess potential energy, by virtue of its position. The water is said to be at a higher head or pressure compared to water at a lower level (lower head). The term head is, thus, relative, and the natural flow is from the higher to the lower level. The flow depends upon the relative heights (Figure 3.6) of the tanks quite independent of the tank which initially holds the greater quantity of water. The flow ceases when the upper

S

P

Figure 3.6 Water Analogy (p.d.)

26 Electrical Technology tank has emptied, but the flow could be made continuous by driving the pump to force the water—which is at the lower level—against the gravitational pull, back to the upper tank from which it would continue to flow due to its potential energy. The pump imparts kinetic energy—or energy due to motion—to the water so as to force it to an elevated position where it regains potential energy. As the water subsequently falls, the potential energy of the water—at successive points along the connecting pipe—gradually falls also until it is in the lower tank and at a relative zero level. The fall of pressure is uniform. Considering a direct current in the electric circuit, the cell or dynamo functions to produce a higher pressure or potential at one terminal rather than at the other, and a flow of current results from the higher to the lower potential around the external circuit. Potential in electric circuit is analogous to pressure in a hydraulic system. Electric potential is often referred to as electric pressure. Potential is also analogous to temperature, which is the determining condition for heat to flow between one body and another at different temperatures when the two are in contact. The current flow in a circuit is assumed to be from the positive to the negative terminal, and a point at positive potential is said to have a higher potential with respect to a point at the negative or a relatively lower potential. The flow of free electrons is, of course, from the point of negative potential to the point of higher potential. The production of a potential difference (p.d.) is, thus, an essential condition for the flow of current. Potential difference denotes a difference between the electrical states existing at two points, tending to cause a movement of electricity from one point to the other. The symbol for p.d. is V or v. In the external circuit, the magnitude of potential falls gradually from the positive terminal to the negative terminal— and at points along the circuit—successively diminishing values of potential occur and can be measured: there is a p.d. between any two points chosen at random in the external circuit. For example, point Y in Figure [3.5 (a) and (b)] is positive with respect to a point Z at a lower potential, whereas point Y is negative to a point X at a higher potential. A force is necessary to drive the current from low to high potential within the generator. This force is known as the electromotor force (e.m.f.). The electromotive force of a source is that force which tends to cause a movement of electricity in a circuit. The symbol for e.m.f. is E or e. E.m.f. and p.d. are both cause and effect. The electromotive force produces and maintains a potential difference between the terminals and without an e.m.f. there could be no p.d. The two bear to one another a relationship similar to that borne between kinetic and potential energy, respectively. In the hydraulic analogy, the potential energy acquired by the water in the upper tank would be equal to the kinetic energy imparted by the pump were it not for the losses due to friction incurred in the pump. Similarly, in the electric circuit, the e.m.f. of the generator will be slightly greater than the p.d. produced, owing to the inevitable losses within the generator. When the electric generator is on open circuit (stopcock closed in the hydraulic system), the p.d. becomes equal to the e.m.f. since no current is flowing within the generator. The e.m.f and the p.d. are measured in terms of the same unit, the volt (V). Electrical mains supplies are being standardized at a pressure of 240 V; telephone and telegraph systems operate at 50 V and 80 V, respectively. On the other hand, potential differences between points in telecommunication circuits are frequently measured either in millivolts (mV) or in microvolts (mV).

3.5 REFERENCE ZERO The water in a river flows until it finds the sea level, which is regarded as the reference zero level. In a similar respect, the earth, which is a good electrical conductor, is regarded as being at zero electric potential, the absolute potential of any point being referred to the zero potential of the earth. The potential at any point is the potential difference between that point and earth, being positive if the current tends to flow from that point to earth and negative if current tends to flow from earth to that point. To ensure the safety of users, it is a common practice to earth all those parts of electrical appliances, i.e., the mountings and casings, with which the human body is likely to come in contact. The body will then remain at one potential, that of earth, even when touching the electrical apparatus— consequently, no current will flow through the body since there can be no difference of potential. However, the practical achievement of this safety condition requires considerable care in the design of earthing systems.

3.6 SAFETY PRECAUTIONS WHILE HANDLING ELECTRIC CIRCUITS Electrical and electronic circuits can be dangerous. Safe practices are necessary to prevent electrical shocks, fires, explosions, mechanical damage, and injuries resulting from the improper use of tools. As our knowledge and experience grows, we will learn specific procedures for dealing with electricity and electronics. In the meantime: 1. Always follow safe procedures. 2. Use service manuals as far as possible. They often contain safety information.

Electric Circuit

27

3. We should investigate before we act. 4. When in doubt, do not act. We should ask our seniors. Safe practices will protect us and our fellow workers. Study the following rules. Discuss them with others, and we should ask our seniors about anything we do not understand. 1. We should not work when we are tired or we have taken medicines that make us drowsy. 2. We should not work in poor light. 3. We should not work in damp areas or with wet shoes or clothing. 4. We should use approved tools, equipment, and protective devices. 5. We should avoid wearing rings, bracelets, and similar metal items when working around exposed electric circuits. 6. Never assume that a circuit is off. Double check it with an instrument that we are sure is operational. 7. Some situations require a buddy system to guarantee that power will not be turned on while a technician is still working on a circuit. 8. Never tamper with or try to override safety devices such as an interlock. 9. Keep tools and test equipment clean and in good working condition. Replace insulated probes and leads at the first sign of deterioration. 10. Some devices, such as capacitors, can store a lethal charge. They may store this charge for long periods of time. We must be certain that these devices are discharged before working around them. 11. Do not remove grounds and do not use adaptors that defeat equipment ground. 12. Use only an approved fire extinguisher for electrical and electronic equipment. Water can conduct electricity and may severely damage equipment. Carbon dioxide (CO2) or halogenated-type fire extinguishers are usually the preferred choices. Foam-type fire extinguishers may also be desired in some cases. Commercial fire extinguishers are rated for the type of fires for which they are effective. Use only those rated for proper working conditions. 13. Follow directions when using solvents and other chemicals. They may be toxic, flammable, or may damage certain materials such as plastics. 14. A few materials used in electronic equipment are toxic. Examples include tantalum capacitors, and beryllium oxide transistor cases. These devices should not be crushed or abraded and we should wash our hands thoroughly after handling them. Other materials (such as heat shrink tubing) may produce irritating fumes if overheated. 15. Certain circuit components affect the safe performance of equipment and systems. Use only exact or approved replacement parts. 16. Use protective clothing and safety glasses when handling high-vacuum devices such as picture tubes and cathode-ray tubes. 17. We should not work on an equipment before we know its proper procedures and are aware of any safety hazards. 18. Many accidents have been caused by people rushing and cutting corners. We should take the time required to protect ourselves and others. Running, horseplay, and practical jokes are strictly forbidden in shops and laboratories. Circuits and equipment must be treated with respect. Always practice safety; our health and life depend on it.

3.7 INSULATORS An essential need in any electrical equipment is for the separation or insulation of one conducting path from another as illustrated in Figure 3.7. An insulating material is one which offers a relatively high resistance to the passage of an electric current: such a material is also referred to as a dielectric. All known dielectrics have a small electrical conductivity. Super purification considerably increases the resistivity (specific resistance), of some materials, indicating that impurities may be a major cause of dielectric conduction. The final choice of an insulating material must depend on mechanical as well as electrical properties appropriate to its intended use. Important mechanical and physical properties to be considered may include one or more of the following: tensile, compressive, shear and impact strength; modulus of elasticity; resistance to heat and shock; low water absorption; resistance to mineral oils, to acids, and to other chemical actions; specific gravity; coeffiecient of linear expansion; non inflammability; thermal conductivity and rigidity; dimensional stability; flexibility (for cables); adaptability to a wide range of moulding or machining processes; freedom from tracking (i.e., from the creation of a conducting carbon path following a high voltage discharge over the surface). Important electrical properties include adequate electric strength against breakdown under high applied voltage; low power-factor; permittivity over a range of frequencies and temperature, insulation resistance; surface and volume resistivity. Cost is naturally an important factor. The commoner gases are the most nearly perfect insulators but with complete absence of mechanical strength. For example, air is the insulating material for a route of overhead wires; most switching devices depend on the insertion or

28 Electrical Technology bridging of an air gap in a circuit, though in practice the conductors must be based upon some insulating support and this material enters largely into the degree of insulation obtained. The commoner non-metallic liquids have a small electronic conductivity. The ionic conductivity of liquids varies greatly; for oils it is very small, and these are the only practicable liquid insulating materials, an important example being transformer oil which, of course, also acts as a cooling agent. Pure water is practically a non conductor but is very slightly ionized. It is, however, practically impossible to keep it free from impurities, of which even the smallest percentage ionizes it and renders it a comparatively good conductor. For this reason, the presence of moisture is detrimental to good insulation, the water providing a path for the leakage of electric current. As an example, the failure of a cable is more often due to ingress of moisture than to any other cause. Water in the form of a surface film on insulators is harmful not because of its low conductivity, but because of its power to dissolve and dissociate certain constituents of the insulator, resulting in the presence of ions with consequent electrolytic current. To overcome this, impregnation is adopted to fill all interstices which might trap moisture and surface varnishing to keep moisture away from the surface of the insulator. With solid insulating materials, since there are no free electrons in pure substances (excepting metals), no electronic conduction takes place. Ionic conduction can occur, and if appreciable, the result may be some change in the composition of the materials. The textiles cotton, silk, and wool are used extensively for insulating flexible conductors as well as in internal cables and coil windings. The TwoConductors insulation resistance of textiles is altered in the presence of moisture conductors due to the development of a back e.m.f. The presence of impurities in textiles increases the conductivity and also the tendency for corrosion Insulation by the small currents flowing through it. Paper is used almost exclusively to separate the conductors in ThreeConductors conductors multiconductor underground cables (Figure 3.7), but by crimping the paper or by using a wrapping string between conductor and paper. The enclosed air plays an important part as the insulating material; such Jacket Conductor cables are known as air-spaced paper core cables (ASPC). Insulation To name but a selection from the extensive range of available insulating materials there are the natural minerals such as mica, asbestos, slate, and Multimarble: or vitrified materials like glass, ceramics, and porcelain: natural conductor and synthetic rubber: ebonite: waxes, enamels, varnishes, oils: natural resins (shellac) and synthetic resins, from the early bakelite to the more Jacket Insulated Tape recent introductions either of the thermosetting or thermoplastic types. conductors Thermosetting plastics undergo chemical change when subjected to heat and pressure: they are thus converted to an insoluble, infusible state Coaxial which cannot be further reformed even by the application of more intense heat and pressure. Thermoplastic compounds, on the other hand, can be softened and re-softened indefinitely by the application of heat, provided Shield Jacket Inner that the heat applied is insufficient to cause chemical decomposition. conductor Cable core Many new substances are formed by the molecular process known Figure 3.7 Cables: Strands of Insulated as polymerization. Polymerization is the building up of long-chain Electrical Conductors Laid molecules from relatively simple molecular structures. In the case of Together Usually Around a thermosetting compounds, cross-linking of the molecules occurs. Among Central Core and Surrounded these materials are many variants under trade names such as Perspex by a Heavy Insulation (acrylic resins), polythene (polymerized ethylene), neoprene (based upon synthetic rubber and polystyrene—cellulose base). Also extensively used are papers of various grades and laminated paper or fabric board such as the synthetic resin bonded paper (SRBP). A list of commonly employed insulating materials with their properties is given in Table 3.1.

3.8 SEMICONDUCTORS On the basis of electrical conduction, all solid materials can be broadly divided into three classes. If the conductivity s is greater than 103 mho/cm, the solid is called a conductor. Examples are: copper s = 5 × 105 and constantan s = 2 × 104 mho/cm. The conductivity of a conductor normally falls with increasing temperature apart from a few special alloys designed to have conductivity practically independent of temperature. At the other extreme is the group of solids having conductivity s lower than 10–9 mho/cm and known as insulators. Examples are glass s = 10–10, ebonite, s = 10–5; mica, s = 10–14 and paraffin wax s = 10–17 mho/cm.

29

Electric Circuit

Table 3.1 Insulating Materials Power Factor × 104 Material

Type

Surface Resistivity (M Ω/cm-square)

Volume Resistivity Relative Permittivity At (m Ω/cm-cube) at 1 kHz (εr) 1 kHz

At 60 kHz

Insulating oil

Oils and

–

109

2.3

10 pF

±10 per cent→A,

±5 per cent→B, ±2.5 per cent→K, ±5 per cent→C, ±1 per cent→D

Example 13.6 Find the capacitance of the capacitors as shown in Figure 13.24:

Red (a)

Orange

Orange

Black

Brown

(b)

Red Green

Brown

(c)

Figure 13.24 For Example 13.6 Note: Ceramic Capacitors of This Type are Known as Pin-Up Capacitors, They Find Extensive Use in Miniaturized Equipment

Capacitors and d.c. Transients

259

Solution: (a)

First band

Second band

Third band

Orange

Orange

Red

3

3

102

Capacitance = 3300 pF = 3k3 pF (b)

Brown 1

(c)

Brown 1

Black 0 Capacitance = 10 000 pF = 10 kpF Green 5 Capacitance = 1500 pF = 1k 5pF

Orange 103 Red 102

Example 13.7 On typical tubular capacitors: (1) B82, (2) 4100 and (3) 5L6 is written. Find their capacitance. Solution: 1. B82 → 82 pF ± 5 per cent 2. A100

→

100 pF ± 10 per cent

3. 5L6 →

5.6 pF ± 0.5 pF

Example 13.8 Find the capacitance of the following tubular ceramic capacitors. Violet

Yellow Violet Brown

White

Orange

Red Violet Brown

(a)

Green

Black

Red Red White

(b)

Red (c)

Figure 13.25 For Example 13.8 Solution: 2nd 3rd 1st a. Violet Yellow Violet N750 4 7 Capacitance = 470 pF ± 10 per cent b. Orange Red Violet N150 2 7 Capacitance = 270 pF ± 5 per cent c. Black Red RedA d 2 1 NPO 2 2 Capacitance = 2.2 pF ± 0.25 pF

4th Brown 101

5th White ± 10 per cent

Brown 101

Green ± 5 per cent

White 10–1

Red ± 0.25 pF

Example 13.9 What quantity of electricity will produce a difference of potential of 200 V between the plates of a capacity of 5 mF? Compare the magnitudes of two capacitors, one having two circular plates of 4 cm diameter and ½ mm apart and the other having two square plates 5 cm each side and ¾ mm apart.

260 Electrical Technology Solution: 1.

C = Q/V, Q = CV = 5 × 10–6 × 2 × 102 = 10–3 coulombs or 1000 mC 2 A1 = π4 = 4 π cm2 4

2.

A2 = 5 × 5 = 25 cm2

d1 = 0.05 cm, d2 = 0.075 cm A1d2 C1/C2 = = (4 π × 0.075)/ (25 × 0.05) A2 D 1 The capacitances are in the ratio 3:4.

= 3.77/5 = 0.75

Example 13.10 A simple capacitor, consisting of two insulated parallel plates, has a capacitance of 0.001 mF and receives a charge of 1 μC. What is the p.d. between the plates? What would be the capacitance if the areas of each plate were trebled and the spacing between the plates halved? Solution: 1. V = Q / C = 10−6 / 10−9

= 1000 V 2.

A1 = A, A2 = 3 A, d 1 = 2d , d 2 = d C1 = 10−9 F C2 =

A1d 2 A2 d 1

= ( A × d ) / (3 A × 2d )

= 1/ 6 The capacitance would be increased six fold to 0.006 μF. Example 13.11 A variable capacitor of 1000 μF is charged to a p.d. of 100 V. The plates of the capacitor are then separated by means of an insulated rod so that the capacitance is reduced to 300 μF. Would you expect the p.d. across the capacitor to have changed and, if so, by how much? Solution: C = Q/V For a constant value of Q. V∝1/C. If the capacitance is reduced by 10/3, the charge remains constant and the p.d. therefore increases by 10/3 to 100 × 10/3 = 333 V. Increment in p.d. = (333–100) = 233 V Example 13.12 If a capacitor is to have a value of 0.1 μF and the paper used for the dielectric is 4 cm wide by 0.02 mm with a relative permittivity of 2.25, what length of paper would be required? Solution: C = 0.1 µ F

(

A = l × 0.04 m 2

ε r = 2.25

)

ε 0 = 8.854 × 10 −16 d = 0.02 × 10 −3 m ε 0ε r A C = F d = ε 0 ε r wl × 106 µ F

Capacitors and d.c. Transients

261

where, I and w are the length and width of the dielectric.

l= =

Cd

( ε 0ε r w × 106 ) 0.1 × 0.2 × 10−3 2.25 × 8.854 × 10−14 × 0.04 × 106

= 200 / 79.7 = 2.51 m Example 13.13 A capacitor consisting of two air-spaced parallel plates, each of effective area 150 cm spaced 1 mm apart. A potential difference of 100 V is maintained between the plates. What is: (1) the electric field strength between the plates; (2) the charge held by the capacitance? This capacitor is connected in a circuit which gives it a steady charging current of 1.0 μA. Draw a graph showing the relation between the voltage across the capacitor and the time during which the charging current has been flowing. Show values on your axis. (μ0 = 8.554 × 10–12 F/m). Solution:

C= C=

ε 0ε r A d

, where, ε r = 1 for air

(1 × 8.854 × 10−12 × 150 × 10−4 ) 10−3

= 133 µµ F If a p.d. of V volts is maintained between the parallel plates of the capacitor d metres apart, the potential gradient = V/d V/m. When V = 100 V and d = 10–3m, the electric field strength = 100 / 10–3 = 105 V/m Q = CV = 133 × 10–12 × 100 = 0.0133 μC The charge Q coulombs for a current i flowing for t seconds is 7.5

Q = i × t coulombs The p.d is V = Q /C = i × t / C = ( i / C ) × t =

10−6 t = 7500 t 133 × 10−12

This relationship can be represented by a straight line with a gradient such that the voltage increases steadily at a rate of 7500 volts per second, as shown in Figure 13.26

V (volts)

t (milliseconds)

Figure 13.26 Solution For Example 13.13

Example 13.14 A capacitor consisting of two air-spaced parallel plates, each of effective area 1000 cm2, spaced 0.1 cm apart is connected across a constant voltage source of 500V. Calculate the charge on the capacitors (ε0 = 8.854 × 10–12 F/m). Solution: In air, C=

8.854 × 10− 12 × 1000 × 10 − 4 = 8.854 × 10− 12 F 0.1 × 10− 2

Q = CV = 8.854 × 10 −12 × 500 coulomb = 0.4427 µ C

262 Electrical Technology Example 13.15 A parallel plate capacitor has two metal plates, each of area 500 cm3. Calculate its capacitance, when the distance between the plates is adjusted to 0.5 cm A constant charging current of 2 μA is supplied to this capacitor. Calculate: (1) for how many microseconds must this charge continue in order to raise the potential to 1000 V, (2) Using the same two capacitor plates and the same charging current flowing for the same time as in (1) what alteration is necessary in the capacitor to ensure that the p.d. between the plates rises to twice the value obtained in 1. (ε0 = 8.854 × 10–12 F/m). Solution: In air ε A 8.854 × 10−12 × 500 × 10−4 C = 0 = d 5 × 10−3 = 88.54 µ F V = 1000 = Q / C = It / C

1.

2 × 10−6 × t × 10−6 88.54 × 10−12 t = 1000 × 88.54 / 2 = 44.227 msec . =

2.

2V = 2Q / C , and since the charge Q will be unchanged

when the alteration to the capacitor is made. 2V = Q / ( C12 ) i.e. C must be halved. The capacitance is inversely proportional to the distance between the plates so that the plates must be separated by twice the spacing, i.e., by 1.0 cm.

13.10. TIME CONSTANT

Current (Amperes)

The time required to charge a capacitor depends not only upon the capacitance C Farads and upon the applied V volts but also upon the value of any series resistance R ohms present in the circuit. In the uncharged state, the p.d. across the capacitor is zero, but immediately it commences to acquire a charge, its p.d. reaches some value v, which Figure 13.27 shows to be in the nature of a back e.m.f. The available V charging potential is reduced to (V–v) and if, V is constant and I= C R + – v gradually increases as the charging proceeds the charging p.d. (V–v) gradually falls, reaching zero when the capacitor is R fully charged and v = V. i The charging current commences at a maximum value v and falls to zero, varying as shown by the curve in Figure + 13.27. The value of the current at any time, by Ohm’s law, will depend upon the magnitude of the resistance R presi = V e –t/CR ent, and after any particular elapsed time tʹ seconds will be R equal to O

 

Time (Seconds)

Figure 13.27 Capacitor Charging Current

i ʹ = (V–vʹ ) / R

where, v ʹ denotes the back e.m.f. after t’ seconds charging. The solution to this equation is i = ( V / R )e −t / ( CR )

(13.7)

where, e = 2.718, when t = 0, –t / CR = 0 and e–0 = 1. So that i has its maximum value, i = V/R. Theoretically, i reaches zero when t =∞, which may be confirmed by substituting t = ∞ in the expression for i. If the capacitor is discharged by shunting it and removing the charging p.d., the p.d. across the capacitor gradually falls and so does the current strength. The instantaneous value of the current strength on discharge is given by i = −( V / R ).e −t / ( CR )

(13.8)

Capacitors and d.c. Transients

The negative sign for the current indicates the reversal of current direction consequent on discharge. This current curve is given in Figure 13.28. The expression for the discharge may be readily produced from the charging equation if the capacitor is considered to be discharged by the application of an equal and opposite p.d., –V. Because the capacitor is fully charged, the current i = 0 due to the original source +V, and is Hence the current is i = ( i1 + i2 ) = o = − ( V / R ) .e

− ( V /R ).e

(13.9)

− t / ( CR )

Time (seconds)

Current (amperes)

O

C

i = –V e –t/CR R

+ R

–

i2 = ( −V /R )e − t / ( CR ) for the added p.d., − V .

263

–

i V –v=0

I = –V R

Figure 13.28 Capacitor Discharging Current

− t / ( CR )

Although the current value diminishes with time for both charge and discharge, the magnitude of the charge is, of course, increasing on charge and decreasing on discharge. If q is the instantaneous magnitude of the charge whose final or maximum value is Q = CV, then during charge, q = CV 1 − e −t / ( CR ) (13.10)

(

and during discharge

)

(

q = CV e −t / ( CR )

)

(13.11)

The curves of variation of q with time during charge and discharge are similar to those for the growth and decay of current in an inductor. Expression for the instantaneous value of the voltage may be derived by dividing the equation for quantity by the capacitance C, for V = q / C. During charge, (13.12) v = V 1 − e −t / ( CR )

(

)

v = Ve -t / CR v = V .e

and during discharge,

(13.13)

For a particular charging period t = CR q = CV 1 − e −CR / CR = CV 1 − e1 = CV ( e − 1 ) / e

(

=

)

CV ( 2.718 − 1 ) 2.718

(

)

= 0.632 CV = 0.632 Q

(13.14)

The time for the quantity of charge to reach 0.632 of its maximum value Q is known as the time constant and is numerically equal to the product CR, provided C and R are expressed in farads and ohms, respectively. The time-constant is equal to the time which would be required to charge (or discharge) the capacitance if the initial charging (or discharging) rate were maintained. Considering the charge, the initial current I = V/R Coulomb/second. The total charge Q = CV Coulombs and since t = Q / l, the time required to charge the capacitance at a constant rate would be t = CV /( V /R ) = CR seconds

(13.15) For practical purposes, it may be taken that the final value of the current or charge is reached after a time period equal to five times the value of the time constant (see Figure 13.29).

13.11 GRAPHICAL DERIVATION OF THE TRANSIENT CHARACTERISTICS OF AN R-C CIRCUIT The transient characteristics of an R-C circuit can be derived graphically from the values of the resistance R, the capacitance C and the applied voltage V. For the circuit shown in Figure 13.30, the circuit current i0 at the instant of V , assuming that the capacitor is initially uncharged and that therefore the supply being switched on is given by i 0 = R the full supply voltage appears across the resistance. The voltage across the capacitor vC builds up from zero until it is

264 Electrical Technology Quality (Coulombs) O

eventually equal to the supply voltage V. The growth of the capacitor voltage is represented by the characteristic shown in Figure [13.30 (a) and (b)]. At any instant t seconds after the switch has been closed, let current be i and the voltage across the capacitor vC. It follows that VR = V − vC

Final steady charge

T Time (Seconds) (a)

i=

and

V − vC R

(13.16)

Quality (Coulombs)

vC

O

V T

+

R

–

Time (Seconds) (b)

C

vR

V

T

vR vC

vC

i t (a)

Figure 13.29 Quantity Variation Decreases (a) Charge (b) Discharge

t (b)

Figure 13.30 Growth of Voltage Vc Across the Capacitor in an R-C Circuit

If this current were to remain constant until the capacitor were fully charged and if the time required for such an action were T seconds, then the charge supplied in that period would be V − v  iT =  C T  R  During the same period, the voltage across the capacitor would increase uniformly from the value Vc to the supply voltage V. The characteristic would then take the form of a straight line which would be a tangent to the curve, as shown. The charge on the capacitor would rise from CvC to CV and the difference in charge:  V − vC  (13.17) C ( V − vC ) =  T  R  Hence,

T = RC

V A

O

D

BE

C t

Figure 13.31 Construction of the Capacitor Voltage Growth Characteristic in an R-C Circuit

This relation indicates that so long as the capacitance and the resistance of a circuit remain constant, the time taken to complete the charging of a capacitor from any given state of charge is the same, and T is, therefore, termed the time constant of the circuit. It is possible to construct the characteristic shown in Figure 13.31. Let the final capacitor voltage be V, represented by OA and let AB be constructed to represent the time constant. If we join O to B, OB represents the tangent to the curve at point O. From a point C quite near to O and lying on line OB, drop the perpendicular CD and measure along the distance DE equal to AB (the time constant). Join CE, which is the tangent to the curve of C. By drawing a family of such tangents, the shape of the characteristic becomes apparent. Obviously, the greater the number of constructions drawn, the greater the accuracy of producing the characteristic. With suitable adaptation, the method of construction can be applied to all transient characteristics.

Example 13.16 A capacitor of 1000 mF capacitance is charged to a p.d. of 200 V. If this capacitor is discharged in 1/1000 sec, what is the average value of current during discharge? Solution: Q = CV = 103 × 10−6 × 200 = 0.2 C Q = It , I = Q / t I = 0.2 × 1000 = 200 A

Capacitors and d.c. Transients

265

Example 13.17 A 0.02 mF capacitor receives a charge of 20 mC. What energy is stored in the capacitor? If the capacitor is made up of parallel plates, each having an area of 180 cm2 and placed 1 mm apart in air, how many plates are used? Indicate by a curve approximately drawn to scale, how the p.d. across the capacitor will diminish with time if the capacitor has an insulation resistance of 1000 MΩ? Solution: 1. ε r = 1, ε 0 = 8.854 × 10−12 , a = 1130 cm 2 = 0.113 m 2 , d = 1 mm = 10−3 m, C = 0.06 × 10−6 F, ( µ + 1) = number of plates W = CV 2 / 2 = Q 2/ 2 C

(

)(

)

= 202 × 10−12 / 2 × 0.02 × 10−6 = 0.01 J 2.

1000

C = e 0 e r A /d F = e0 er na / d F n = CD / (e0 er a )

800 −3

× 10

) ( 8.854 × 10

−12

× 1 × 0.113 )

= 20 Number of plates = 20 + 1 = 21 3.

P.D. (Volts)

= ( 0.02 × 10

−6

600 400 200

p.d. across capacitor = v = V .e −t / CR initial p.d. = V = Q / C = 20 / 0.02 = 1000 V

0

v = 1000e −t / CR

10

20 30 40 Time (Seconds)

50

60

Figure 13.32 For Example 13.17

Taking the two values of t = 10 and t = 50 sec for an approximate curve and substituting the value of R = 1000 MΩ. where, t = 10 sec 50 sec v = 606.5V 82V The curve is drawn in Figure 13.32. Example 13.18

A 2 mF capacitor charged to a p.d. of 150 V is connected with an uncharged 4 mF capacitor. To what voltage would the combination be charged? Draw a curve approximately to scale showing how the voltage would decrease with time, if the insulation resistance of each capacitor is 2 MΩ? Solution: Q = CV = 2 × 10–6 × 150 = 300 mC

On the charged capacitor,

When the capacitors are connected together, the total charge remains the same; the joint capacitance becomes 2 + 4 = 6 mF. V = Q/C = 300/6 = 50 V The joint insulation resistance = 1 MΩ = 106 Ω on discharge v = V.e –t/CR where, V = 50 V = initial p.d. v = V.e –t/CR = 50.e–t/6 from which the following values may be obtained. t (seconds)

0

1

4

6

12

24

v (volts)

50

42.3

25.7

18.4

6.8

0.9

Figure 13.33 has been plotted from these figures.

266 Electrical Technology 50

P.D. (Volts)

40 30 20 10

0

5

10 15 Time (Seconds)

20

25

Figure 13.33 For Example 13.18 Example 13.19 A 10 mF capacitor is connected in series with a 20 kΩ resistor across a 100-V d.c. supply. Calculate the time constant of the circuit and the initial charging current, assuming that the capacitor is initially uncharged. Draw the current/time characteristic of the circuit and estimate the current and the voltage across the capacitor at an instant of time at which the time constant has elapsed after closing the switch to apply the supply voltage to the circuit. Solution: T = RC = 20 × 103 × 10 × 10−6 = 0.2 s i0 = V / R =

100 = 5.0 × 10−3 A = 5.0 mA 20 × 103

The construction of the current/time characteristic is shown in Figure 13.34. OA represents the initial current (5.0 mA) and OB the time constant (0.25). The point D is taken 10 per cent along the line AB measured from A. This is equivalent to an instantaneous current 0.5 mA less than 5.0 mA. The procedure is repeated at 0.5 mA intervals and the current/ time characteristic constructed, as shown in Figure 13.34. From the characteristic at an instant 0.2 s after switch on the instantaneous value of the current is found to be 1.8 mA. It follows that vR = iR = 1.8 × 10–3 × 20 × 103 = 36 V vC = V – vR = 100 – 36 = 64 V

i (mA) 5

A 0 0.2 s

Percentage of maximum current voltage

and

E

4 3 1.8

2 1 0

B 0.1 0.2

0.3

0.4

0.5

0.6

0.7

0.8

Figure13.34 For Example 13.19

t (s)

100 90 80 70

A

B Curve of capacitor charge voltage

B

A Curve of capacitor charge current

60 50 40 30 20 10 0

0 .25 .50 .75 1

2 3 R – C time constants

4

5

Figure 13.35 Universal Time-Constant Chart for Capacitors and Resistors

13.12 UNIVERSAL TIME CONSTANT The capacitor opposes a voltage change while an inductor opposes a change of current. For the R-C circuit as for the R-L circuit, the resistor voltage will follow the circuit current. The time constant chart for the R-C circuits permits us to find voltages and currents at any instant during which the capacitor is undergoing a charge as shown in Figure 13.35. The chart is applicable to any source voltage value or any size capacitor or resistor. After five time constants, the capacitor is considered to have reached the full charge value (actually 99.3 per cent of full charge). After five time constants, the current flow in the circuit stops and the voltage across the resistor becomes zero.

Capacitors and d.c. Transients

267

Table 13.2 Time Constant Values Time Constant

(A) Percentage of capacitor discharge voltage or charge current

(B) Percentage of capacitor charge voltage

0.001

99.9

0.1

0.002

99.8

0.2

0.003

99.7

0.3

0.004

99.6

0.4

0.005

99.5

0.5

0.006

99.4

0.6

0.007

99.3

0.7

0.008

99.2

0.8

0.009

99.1

0.9

0.01

99

1

0.02

98

2

0.03

97

3

0.04

96

4

0.05

95

5

0.06

94

6

0.07

93

7

0.08

92

8

0.09

91

9

0.10

90

10

0.15

86

14

0.20

82

18

0.25

78

22

0.30

74

26

0.35

70

30

0.40

67

33

0.45

64

36

0.5

61

39

0.6

55

45

0.7

50

50

0.8

45

55

0.9

40

60

1

37

63

2

14

86

3

5

95

4

2

98

5

0.7

99.3

Example 13.20 For the circuit shown in Figure 13.36, solve for time constant. When tc = 3, what is the current? At tc = 3, what is the voltage across the capacitor?

268 Electrical Technology Solution:

100 µf

RC = 200000 × 100 × 10 = 20 sec when tc = 3, current drops to a value of 5 per cent = 0.025 ma when tc = 3 capacitor voltage rises to 95 per cent of full value. = 95 V –6

200 KΩ

+ −

100 V

Figure 13.36 For Example 13.20

Example 13.21 A circuit to be designed has two requirements: the time constant must be 10 m sec, and the capacitor must be 0.01 mF. What resistor will produce such a tc? Solution: 10−3 R = tc/C = 10 × × 10−6 = 106 Ω 0.01 = 1 MΩ Example 13.22 What capacitor must be used with a 20000 Ω resistor to obtain a tc of 2 sec? Solution: 2 C = tc / R = = 0.0001 f 20000 = 100 µ F Example 13.23 A series circuit is composed of a 0.0025 μF capacitor and a 3000 Ω resistor in series with a 100 V battery. At what fractional part of a second after the switch is closed will the voltage across the capacitor reach 63 per cent of full value? At what time in fractional seconds will the current flow cease? Solution: tc = RC = 0.0025 × 10–6 × 3000 = 7.5 μsec At 1 tc the voltage reaches 63 per cent of full value. Hence, the time 7.5 μsec. At 5 tc current flow ceases. Hence, the time is 37.5 μsec. Example 13.24 In an R-C circuit, the capacitor must charge to 22 per cent of the applied voltage in 200 μsec. If R = 10000 Ω, what must be the value of the capacitor? Solution: RC at 22 per cent = 0 .25 tc = ( 200 µ sec ) × 4 = 800 µ sec C = tc / R =

800 × 10−6 = 0.08 µ F 10000

Example 13.25 In an R-C circuit, the capacitor should charge to 39 per cent of the full value in 100 μsec. If the capacitor value is 0.005 μF, what must the resistor value be? Solution: RC at 39 per cent = 0.5 one tc = 200 µsec R = tc /C =

200 = 40000 Ω = 400 k Ω 0.005

269

Capacitors and d.c. Transients

Example 13.26 A direct potential of 200 V is suddenly applied to a circuit comprising an uncharged capacitor in series with a 1000 Ω resistor. Calculate the initial rate of rise of voltage across the capacitor. After the capacitor is completely charged, the source of e.m.f. is switched off and the terminals of the circuit are short circuited. Calculate the initial rate of decrease of voltage across the capacitor. Solution: Since the capacitor is initially uncharged, there is no initial p.d. between its plates. When the supply e.m.f. is applied, the corresponding voltage drop must appear across the 1000 Ω resistor. Let the initial current be i0. 200 V = = 0.2 A R 1000 ∆V ∆V C = = 100 × 10−6 × ∆t ∆t 0.2 ∆V = = 2000 V/ s = 2.0 kV/ s ∆t 1000 × 10−6 i0 =

When the capacitor is discharged, the full p.d. is applied to the resistor. This gives the same current initially as before, but the direction of the current flow is reversed. It follows that the initial rate of voltage decrease is – 2.0 kV/s.

13.13 CONNECTING CAPACITORS IN SERIES When two or more capacitors are connected in series, the total capacitance is less than that of the smallest capacitor in the circuit. This can be seen in Figure 13.37 that the effect of connecting two capacitors A and B in series, each with a dielectric thickness d, is the equivalent of having one capacitor whose dielectric thickness is 2 d. The two centre plates of the capacitors in series do not add to the capacitance in any way, because the charges produced on them are electrically opposite and, therefore, neutralize each other. The effect then is the same as though the inner plates were eliminated. Referring to Figure13.38, the capacitors C1, C2 and C3 are in series and are charged by the source voltage, ET. During the charging of the capacitors, the current is the same throughout the circuit and, therefore, at the end of the charging period, each capacitor must carry the same charge. In that case Capacitor Capacitor A B + − + − d

d

+ Equivalent to

−

C1

−Q

+ V1 −

2d

C2

C3

+Q − Q

+ Q −Q

+ V2 −

+ V3 −

+

−

Momentary electron flow

+Q

ET

Figure 13.37 The Effect of Two Capacitors in Series

Figure 13.38 Capacitors in Series

V1 = Q /C1 , V2 = Q / C2 , V3 = Q /C3 E T = V1 + V2 + V3 = Q /C1 + Q / C2 + Q / C3 E T = Q ( 1/C1 + 1/C 2 + 1/C 3 ) ET / Q =

1 1 1 + + ; C1 C 2 C 3

If CT is the total equivalent capacitance that will store the same charge 1 1 1 1 = + + CT C1 C2 C3 ET = and

Q C C C and V1 = E T T , V2 = E T T , V3 = E T T CT C1 C2 C3

1 1 1 1 + + = C1 C2 C3 CT

270 Electrical Technology and for N capacitors in series

C1

1 CT = 1 1 1 1 + + + ...... + C1 C2 C3 CN

C2

30 µF

(13.18)

20 µF

V1

V2 600 V

If the N series capacitors all have the same value CT =

ε

C N

Figure 13.39 For Example 13.27

Example 13.27 If a voltage of 600 V is applied across two capacitors of 30 MF/200 V and 20 MF/400 F connected in series, find the voltage across each. Can we connect them in parallel? Solution: E = V1 + V2 = 600 V V1 = E − V2 = 600 − V2 If the electric charge displaced in the circuit is denoted by q, then q = C1V1 = C2V2 ,

C1 V2 = C2 V1

30 × 10−6 3 V2 V2 ; = = −6 600 − V2 2 600 − V2 20 × 10 V1 = 240 V V2 = 360 V No

13.14 CONNECTING CAPACITORS IN PARALLEL It can be seen in Figure 13.40 that the effect of connecting two capacitors A and B in parallel, each with a plate area of ‘α’ the dielectric thickness remaining the same is equivalent to a single capacitor with a plate area of ‘2 α’. In Figure 13.41 the capacitors C1, C2, and C3 are in parallel, and each is charged by the source voltage ET. Therefore, Q1 = C1ET, Q2 = C2ET and Q3 = C3ET. The total charge QT is the sum of the individual charges stored in the capacitors QT = Q1 + Q2 +Q3 = ET (C1 + C2 + C3) If CT is the total capacitance,

QT = ETCT CT = C1 + C2 + C3

Therefore, For N capacitors in parallel,

CT = C1 + C2 + C3+.....CN

(13.19) + Q1

+ Q2

+ Q3 d

1 2

A d

B 1 2

A Equivalent to

d

C1

− Q1

C2

− Q2

C3

− Q3

B Equivalent capacitor C

Figure 13.40 Capacitors in Parallel

Momentary electron flow

+

− ET

Figure 13.41 Capacitors in Parallel

Capacitors and d.c. Transients

271

If N capacitors are in parallel, and all have the same value, then CT = NC The series connection of capacitors increases the distance between the plates and decreases the total capacitance. The parallel connection of capacitors increases the effective plate area and increases the total capacitance. Example 13.28 2 µF

Four perfect capacitors are connected as shown in Figure 13.42 to the three terminals ABC. 1 Calculate the value of capacitance that would be measured across the terminals BC. 2 Terminals B and C are now connected together. What capacitance would be measured across AB? 3 Find the energy stored in the whole circuit in the second condition when a 100 V battery is connected across AB. Solution: 1 1 1 1 61 1 = + + + = 1. 2 3 1 5 30 CBC CBC = 2. when B and C are joined

3µF C

A B 1µF

5µF

Figure 13.42 For Example 13.28

30 = 0.492 µ F 61

1 1 C AC = 1 /  +  = 1.2 µ F  2 3 1 1  C AB = 1 /  +  = 0.83 µ F 1 5  C AC  C AB = ( 1.2 + 0.83 ) = 2.03 µ F

3.

Energy stored = 1/2CV 2 = 1/2×2.03×10-6×1002 = 10.15 mJ

Example 13.29 A 10 μF capacitor is charged from 100 V battery. The battery is then removed. Calculate the energy stored in the capacitor. Two other capacitors of 5 and 3 μF joined in series are now connected across the terminals AB of the charged capacitor (Figure 13.43). Determine 1. The total capacitance across AB 2. The p.d. across each of the three capacitors. A A Assuming that no leakage of charge occurs, find the energy stored in each capacitor. Solution: 5 1 2 10 10 I 875 Energy stored = CV 2 3 1 = × 10 × 10−6 × 108 2 B B = 0.05 J (a) (b) 1. CAB = (10 + 1.875) = 11.875 μF 2. The total charge given to the 10 μF capacitor from the 100 V battery is Q = 10 × 10–6 × 100 = 10–3 C This charge remains constant but is redistributed when the three capacitors are connected.

5

V1

3

V2 (c)

Figure 13.43 For Example 13.29

272 Electrical Technology If VAB is the new p.d. across AB

VAB = Q / C = 10–3 / (11.875 × 10–6)

= 84.21 V The charge taken by the two capacitors in series must be equal, because electrons flow out of the one into the other. C1V1 = C2V2 ,

3 V1 = 5 V2

3 V1 = , V1 + V2 = VAB = 84.21 V 3+5 V1 + V2 V1 = 31.58 V V2 = 84.21 − 31.58 V2 = 52.63 V Example 13.30 A capacitor is made with nine conducting surfaces with alternate surfaces connected, as shown in Figure 13.44. The area of one side of one surface is 150 cm2, and the surfaces are separated by sheets of mica with a thickness of 0.4 mm and a relative permittivity of 5. What is the value of the capacitance? Solution: A µr N = 9, A = 150 × 104 m 2 square metres relative permittivity

−3 × 10 4 m 2 Nd = 90,.4A×=10150 m, ε r = 5 −3 d = 0ε .04ε ×r 10 ( N −m1 ), εAr = 5 C = d − 1) A ε 0ε r ( N C = The total capacitance= 8.854 ×d10−12 × 5 × 8 × 190 × 10−4 −3 −12 × 5 × 8 × 190 × 10 −4 = 80..854 × 10 4 × 10

d metres

4 × 10µ−3F = 0.0133 = 0.0133 µ F .5 µF 0 .25 µF

.1 µF

C1

C2

C3 .5 µF

Figure 13.44 For Example 13.30 Example 13.31 Find the total capacitance in Figure 13.45. Solution: C p = ( 0.5 + 0.5 ) = 1 µ F

C4 0 .25 µF

.1 µF

1 µF

C1

C2

CP

Figure 13.45 For Example 13.31

1 1 1  1 CT =  µF + +  µF = 15  0. 25 0.1 1  C T = 0.667 µ F

Example 13.32 A capacitor of 16 μF / 600 V connected in a circuit is burnt. The following capacitors are available: 8 μF / 400 V (Qty12); 16 μF / 400 V (Qty 2); 10 μF / 400 V (Qty 3). How will you connect them? Solution: If two 16 μF capacitors are connected in series, the total capacitance will be 8 μF / 800 V. To further increase the capacitance, a series, parallel combination of capacitors can be formed. This is shown in Figure 13.46.

16 µF

16 µF

8 µF 8 µF

8 µF

τ

8 µF

Figure 13.46 For Example 13.32

Capacitors and d.c. Transients

273

Example 13.33 Find the total capacitance in Figure 13.47. Solution: .003 µF C7

C2

=

.002 µF

10–6 (.002 · .003) = .005 µF

C3 0.2 µF

0.05 µF

C1

C7

.02 ⋅ .005 10–6(.02 + .005) = .004 µF

.007 µF

.003 µF

C9

C4 .005 µF

C2

.02 µF

C8

=

= 10–6 (.007 · .005) = .012 µF

C5

C1

.002 µF

0.12 µF

.007 µF C4

C6

C10

C9

.012 ⋅ .004 10–6(.012 + .004 )= .003 µF

.004 µF

.004 µF

.007 µF

C8

C6

.005 µF

0.04 µF

=

C3

=

.003 µF

C5

CT

C10

Figure 13.47 For Example 13.33

Figure 13.48 Solution For Example 13.33

Example 13.34 Illustrate the charging and discharging of a capacitor. Solution: This is shown in Figure 13.49. Charging

E

Discharging

E

R

100 98.1 94.9 86.4 % Fall of voltage

100

63.2 % Rise of voltage

R

0

1

2

3

4

5

36.8 13.6 5.1 1.9 0

1

2

Time constant

Figure 13.49 Solution For Example 13.34

3 Time constant

4

5

274 Electrical Technology

S UM M A RY 1. Capacitors store electric charge. 2. Although a capacitor is not a battery, it often behaves as one. 3. A capacitor has a capacitance of 1 farad; if the addition of a charge of 1 coulomb raises its potential by 1 volt. 4. A capacitor consists of two plates separated by a dielectric. 5. The dielectric responds to the presence of electric field between the plates. 6. The response of a medium to the presence of an electric field is called permittivity. 7. The permittivity of free space is unity. 8. Capacitance is determined by the area of the plates, the thickness of the dielectric and the type of dielectric. 9. The energy stored in the electric field of a capacitor is equal to ½ CV2. 10. Time constant = CR seconds 11. The total capacitance of N capacitors connected in series is 1 CT 1 1 1 1 + + + ................ + C1 C2 C3 CN

12. The total capacitance of N capacitors connected in parallel is CT = C1 + C2 + C3...................+ CN. 13. Surface density is defined as the quantity of charge per unit area of the surface of a conductor. 14. An electric field line is a line drawn in an electric field such that its direction at any point gives the direction of the electric field at that point. 15. The following properties can be ascribed to the electric field lines: (1) they begin and end on equal and opposite quantities of electric charge; (2) they are in a state of tension which causes them to tend to shorten; and (3) they repel one another sideways.

M U LT IP LE C H O I C E Q UE S TI O NS (M C Q ) 1. In a charged capacitor, the energy is stored in (a) (b) (c) (d)

The field between the plates The positive charges The negative charges None of the above

2. The capacitance of a parallel plate capacitor does not depend on (a) (b) (c) (d)

The area of the plates The medium between the plates The distance between the plates The metal of the plates

3. The law that governs the electric force between electric charges is called (a) (b) (c) (d)

Ampere’s law Coulomb’s law Faraday’s law Ohm’s law

4. Which one of the following is the unit of electric field intensity? (a) Volt metre (c) Volt Joule

(b) Volt/Joule (d) Volt/metre

5. The force between two electrons separated by a distance r varies as (a) r2 (c) r

(b) 1/r (d) 1/r2

6. Which of the following relations is correct? (a) V = q/C (c) V = q×C

(b) C = qV (d) q = V/C

7. A capacitor works in (a) (b) (c) (d)

a.c. circuits d.c. circuits both a.c. and d.c. neither in a.c. circuits nor in d.c. circuits

8. The energy stored in a capacitor of capacitance C and potential V is given by (a) C2V/2 (c) CV/2

(b) CV2/2 (d) 2CV

9. A dielectric is introduced between the plates of a capacitor kept at a constant potential difference. The charge on the capacitor (a) (b) (c) (d)

Decreases Increases Remains the same None of the above

10. A capacitor of capacity 50 μF is charged to 10 volts. The energy stored in the capacitor is equal to (a) (b) (c) (d)

2.3 × 10–3 Joule 1.2 × 10–6 Joule 25 × 10–4 Joule 5 × 10–3 Joule

Capacitors and d.c. Transients

ANSWERS (MCQ) 1. (a) 2. (d) 3. (b)

4. (d)

5. (d)

6. (a)

7. (c)

8. (b)

9. (b)

275

10. (c).

CON V E N TI O NA L Q UE S TI O NS (C Q ) 1. Explain in brief how a capacitor is formed. 2. Under what conditions are capacitors formed that are undesirable in circuiting? 3. Explain briefly in what manner the dielectric contributes to the total unit value of a capacitor. 4. Compare the dielectric constant of three materials with that of air. 5. In what manner do capacitors in series differ from inductors or resistors in series? 6. In what manner do the curves for the universal timeconstant chart of the capacitors differ from those of inductors? 7. After one time constant, what voltage and current relationships exist in an R-C circuit? 8. After five time constants, what voltage and current relationships exist in an R-C circuit? 9. Briefly explain the differences in the rise and fall of currents and voltages in a charging R-C circuit and a discharging R-C circuit. 10. In a pulse-forming circuit, a capacitor holds 9 × 10–7 coulombs when 300 V are applied to it. What is the value of the capacitance in microfarads? 11. An 8 μF capacitor in a power supply circuit holds 1.8 × 10–2 coulombs of electric charge. What is the applied voltage?

12. In a nuclear research laboratory, a 200 μF capacitor is in a circuit which applies 5000 V. What is the capacitor charge in coulombs? 13. A special purpose capacitor has no unit value marked on it. The two plates measure 3 cm by 3 cm and are separated by air for a distance of 3 mm. What is the capacity? 14. In the high-voltage power supply system of a television receiver, 15000 V are impressed across a 0.003 μF capacitor. What is the stored energy? 15. Four capacitors each of capacity 4 mμF are connected as shown in Figure 13.50. If VP – VQ = 15V, what is the energy stored in the system? 16. The equivalent capacity between the points A and B in Figure 13.51 is 17. In Figure 13.53, what will be the effective capacitance between A and B? 18. In Figure 13.54, C2 = 10 μF and all the other capacitors are 4.00 μF each. Find the capacitance between X and Y. 19. In designing a signal-filtering network circuit, it was found that a time constant of 4 milliseconds was necessary with a resistance value of 2000 Ω. What is the value of the capacitor necessary for this time constant? 20. A parallel plate capacitor with plate area A and separation d, as seen in Figure 13.56, is filled with dielectrics as shown. The dielectric constants are K1 and K2 respectively. What will the total capacitance be?

4mµF 3µF

4mµF P 4mµF

Q

4mµF

3µF B

Figure 13.51 For CQ 16

Figure 13.50 For CQ 15

3 µF

4µF

3 µF A

3µF

A

B

4µF

A

B

3 µF 2µF

Figure 13.52 Hint for CQ 16

2µF

Figure 13.53 For CQ 17

276 Electrical Technology C1

C4 X

Y C1

C2

P

X C4

C5

d/2 d/2

Q

C3

Figure 13.55 Hint For CQ 18

K1

d

K2

Figure 13.56 For CQ 20 ANSWERS (CQ) 10. 0.003 μF 11. 200 V 12. 1 coulomb 13. 26.6 pF 14. 0.3375 joules 15. 1.8 ergs

Y

C2

C3

Figure 13.54 For CQ 18

C5

16. 9 μF 17. 3 μF 18. 4 μF 19. 2 μF 20. C =

2ε 0 A d

 K1 K 2 K + K  1 2

  .

14

Dielectric Materials OBJECTIVES In this chapter you will learn about:    Electric fields, capacitance, and permittivity    Dielectric materials most commonly used    Properties of dielectric materials    Shortcomings of dielectric materials    Series and parallel representation of dielectric losses    Loss angle and its significance

Twin shielded cable

l

14.1 INTRODUCTION A capacitor consists basically of two metal plates separated by a dielectric, such as air, paper, ceramic, aluminium oxide, tantalum oxide, Teflon, and other insulating materials that have controlled impurities. The size of a capacitor is dependent on the area of the metal plates, the distance between them and the dielectric used. For better understanding of the electrical properties of capacitors, some knowledge of dielectric materials is required. In addition, other highly significant properties must be understood, especially when capacitors are used in ac applications.

14.2 DIELECTRIC MATERIALS Dielectric materials used for capacitors can be grouped into the following five main classes: Mica, Low-loss Ceramic, Glass, etc: These are used for capacitors from a few pF to a few thousand pF. (pF = 10−12 F). High-permittivity Ceramic: These ceramics can be used for capacitors from a few hundred pF to some mF. Paper and Metallized Paper: These varieties are used for capacitors from a few thousand pF to some mF. Electrolytic (Oxide Film): Such materials are used for capacitors from just under 1 mF to many thousands of mF (aluminium electrolytic and tantalum dielectric). 5. Dielectrics such as Polystyrene, Polyethylene, Terephthalate (Polyester), Polycarbonate, Polypropylene, etc: These dielectrics range from a few hundred pF to a few hundred µF.

1. 2. 3. 4.

Many factors affect the dielectric properties of a material when it is used in a capacitor. Among these are permittivity, dissipation or power factor, insulation resistance, dielectric absorption, dielectric strength, operating temperature, and temperature coefficient of capacitance.

14.3 PERMITTIVITY (DIELECTRIC CONSTANT) The permittivity, dielectric constant or specific inductive capacity of any material used as a dielectric is equal to the ratio of the capacitance of a capacitor using the material as a dielectric to the capacitance of the same capacitor using vacuum as a dielectric. The permittivity of dry air is approximately equal to one. A capacitor with solid or liquid dielectric of higher permittivity (e) than air or vacuum can, therefore, store e times as much energy for equal voltage applied across the capacitor plates. If the space between the capacitor plates is filled by a dielectric with a relative permittivity, er:

C=

ε 0ε r A d

farads

(14.1)

278 Electrical Technology and

electric flux density, D = ε 0 εr = ε electric field intensity, ε

(14.2)

where, e is the absolute permittivity. Note: Compare with magnetic flux density, B = µ 0 µr = µ magnetic field intensity, H

14.4 POWER FACTOR Dielectrics can be classified into two main groups: polar and non polar. Polar materials have a permanent imbalance in electric charges within the molecular structure. The dipoles within the structure consist of molecules whose ends are oppositely charged. These dipoles, therefore, tend to align themselves in the presence of an alternating electric field (if the frequency is not too high). The resultant oscillation causes a large loss at certain frequencies and at certain temperatures. In non-polar materials, the electric charges within the molecular structure are in balance and the dipoles do not rotate under applied fields, although they may distort. Therefore, no sharp loss peaks with frequency and temperature exist. Consequently, dielectrics which are non-polar in R structure make the best a.c. working capacitors, particularly at higher frequencies. Series R Losses occur due to current leakage, dielectric absorptions, etc., depending on the Shunt frequency of operation. The variation of permittivity with frequency is negligible as long as C the loss is low, increased losses occur when the process of alignment cannot be completed, owing to molecular collisions, and in these regions, there is a fall in permittivity. Equivalent circuits showing series and parallel loss resistance can be given, but they Figure 14.1 Losses in a are greatly dependent on the system of measurement at any particular frequency. The Capacitor important criterion is the ratio power wasted per cycle power stored per cycle This is the power factor of the material.

14.5 INSULATION RESISTANCE (OR INSULANCE) An ideal dielectric would allow the electrons to flow on, making their way from one electrode to the other through the dielectric. There is no ideal, practical dielectric available, so a current of electrons does flow always from one plate to the other, resulting in leakage current for the capacitor. Hence, the insulation resistance for a capacitor is related to leakage current by Ohm’s law (V = IR). It is usual to use leakage current for electrolytic capacitors, all other types being expressed in insulation resistance as the leakage current is low. The insulation resistance of a dielectric material may be measured in terms of surface resistivity in ohms or megohms, or as volume resistivity in ohm, centimetres. The insulation resistance of any capacitor will be lowered with the presence of high humidity (unless it is sealed) and will be reduced when operated in high ambient temperatures (whether sealed or not). For perfectly sealed capacitors used under conditions of high humidity, there should be less deterioration but for imperfectly sealed capacitors, the drop in insulation resistance will be roughly inversely proportional to the effectiveness of the sealing. Unsealed capacitors will show a large and rapid drop in insulation resistance under these conditions.

14.6 DIELECTRIC ABSORPTION If a capacitor were completely free from dielectric absorption, the initial charging or polarization current, when connected to d.c. supply, would be V −t (14.3) I = e CR R where, I = current flowing after a time t V = applied voltage R = capacitor series resistance C = capacitance and e = base of Napierian logs (2.718)

279

Dielectric Materials 100% Discharge time Charge

and the polarization current would die off asymptotically to zero. If R is small, this takes place in a very short time and the capacitor is fully charged. In all solid dielectric capacitors, it is found that after a fully charged capacitor is momentarily discharged and left open circuited for some time, a new charge accumulates within the capacitor because some of the original charge has been absorbed by the dielectric. This is shown in Figure 14.2. This produces the effect known as dielectric absorption. A time lag is, thus, introduced in the rate of charging and of discharging the capacitor which reducesthe capacitance as the frequency is increased and also causes unwanted time delays in pulse circuits.

0

Residual charge

Time

Figure 14.2 Discharge Curve With Residual Charge

14.7 DIELECTRIC STRENGTH The maximum amount of field strength that a dielectric can withstand is called the dielectric strength of the material. When an electric field is established across the faces of a material, molecular alignment and distortion of the electron orbits around the atoms of the dielectric occur. This produces a mechanical stress which, in turn, generates heat. The production of heat represents a dissipation of power, such as a loss being present in all dielectrics, especially when used in high frequency systems where the field polarity is continually and rapidly changing. A dielectric whose conductivity is not zero between the plates of a capacitor provides a conducting path along which l charges can flow and, thus, discharge the capacitor. The resistance R of the dielectric is given by R = p , l being the a thickness of the dielectric film (which may be as small as 0.001 mm) and ‘a’ being the area of the capacitor plates. The resistance R of the dielectric may be represented as a leakage resistance across an ideal capacitor. The required lower limit for acceptable resistance between the plates varies with the use to which the capacitor is put. High-quality capacitors have high shunt resistance values. A measure of dielectric quality is the time taken for a capacitor to discharge a given amount through the resistance of the dielectric. This is related to the product CR. Capacitance C µ

area 1 area and µ thickness R thickness

(14.4)

This CR is a characteristic of a given dielectric. In practice, circuit design is considerably simplified if the shunt conductance of a capacitor can be ignored (i.e., R →∝ ) and the capacitor, therefore, regarded as an open circuit for direct current. Since the capacitance of a parallel plate capacitor as represented in Eq. 14.2 is given by C = ε oε rA/d , reducing the thickness of a dielectric film increases the capacitance but decreases the resistance. It also reduces the voltage the capacitor can withstand without breakdown (since V = Q/C). Any material will break down, usually destructively, when subjected to a sufficiently large electric field. A spark may occur at breakdown which produces a hole through the film. The metal film forming the metal plates maybe welded together at the point of breakdown. Breakdown depends on electric field strength (where E = V/d), so the inner films will break down with smaller voltages across them. This is the main reason for limiting the voltage that may be applied to a capacitor. All practical capacitors have a safe working voltage stated on them generally at a particular maximum temperature. Figure [14.3 (a) and (b)] illustrate the typical shapes of graphs expected for electric field strength E plotted against thickness and for breakdown voltage plotted against thickness. The shape of the curves depends on a number of factors that include: 1. 2. 3. 4. 5.

The type of dielectric material. The shape and size of conductors associated with it. The atmospheric pressure. The humidity/moisture content of the material. The operating temperature.

Dielectric strength is an important factor in the design of capacitors, transformers, high-voltage insulators as well as in motors and generators. Dielectrics vary in their ability to withstand large fields. The ceramics

Figure 14.3 Plot of (a) Electric Field Strength Against Thickness and (b) Breakdown Voltage Plotted Against Thickness

280 Electrical Technology have very high relative permittivities and they tend to be ferroelectric, i.e., they do not lose their polarities when the electric field is removed. When ferroelectric effects are present, the charge on a capacitor is given by: Q = CV + (remanent polarization) (14.5) These dielectrics often possess an appreciable negative temperature coefficient of capacitance. Despite this, a high permittivity is often highly desirable and ceramic dielectrics are widely used.

14.8 THERMAL EFFECTS As the temperature of most dielectrics is increased, the insulation resistance falls rapidly. This causes the leakage current to increase, which generates further heat. Eventually a condition known as thermal runaway may develop, when the heat is generated faster than it can be dissipated to the surrounding environment. The dielectric will then burn and fail. Thermal effects may often seriously influence the choice and application of insulating materials. There are some important factors to be considered here: (1) the melting point; (2) ageing due to heat; (3) the maximum temperature that a material can withstand without deterioration of the essential properties; (4) flash-point or ignitability; (5) resistance to electric arcs; (6) the specific heat capacity of the material; (7) thermal resistivity; (8) the coefficient of expansion; and (9) the freezing-point of the material.

14.9 LOSS ANGLE In capacitors with solid dielectrics, losses can be attributed to two causes: 1. Dielectric hysteresis, whereby a phenomenon by which energy is expended and heat produced as a result of the reversal of electrostatic stress when dielectric is subjected to alternating electric stress. This loss is analogous to hysteresis loss in magnetic materials. 2. Leakage currents that may flow through the dielectric and along the surface paths between the terminals. The total dielectric loss may be represented as the loss in an additional resistance connected between the plates. This may be represented as either a small resistance in series with an ideal capacitor or as a large resistance in parallel with an ideal capacitor.

14.9.1 Series Representation The circuit and phasor diagrams for the series representation are given in Figure [14.4 (a) and (b)]. The circuit phase angle is shown as angle f. If resistance Rs is zero, then the current I would lead the voltage v by exactly 90°, this being the case of a perfect capacitor. The difference between 90° and the circuit phase angle f (the complement of f) is shown as d, known as the loss angle of the capacitor. From Figure 14.4 (b) Loss angle, δ = (90° − φ) tan d =

= Since, Figure 14.4 (a) Circuit Diagram for Series Representation and (b) Phasor Diagram Power factor of the capacitor,

cos f =

VRS V

=

IR S IZ S

=

VCS

=

RS 1 / ω Cs

IR S IX CS

= RS ω CS

(14.7)

Since Q = 1/ CR , then tan δ = R S ω CS = 1 / Q

(14.8)

RS R ≈ S ZS X CS

Since, X CS » ZS when d is small. Hence, Power factor = cosφ ≈ RsωCs and,

VR S

(14.6)

cos φ » tan δ

(14.9) (14.10)

Dielectric Materials

281

Dissipation factor D is defined as the reciprocal of Q-factor and is an indication of the quality of the dielectric, i.e. (14.11)

D = 1/ Q = tan d

14.9.2 Parallel Representation The circuit and phasor diagrams for the parallel representation are given in Figure [14.5 (a) and (b)]. From the phasor diagram tan d =

I RP I CP

=

V / RP V / X CP

=

X CP

(14.12)

RP

tan δ = 1 / ( R P ω CP ) Power factor of the capacitor From Figure 14.5 (b)

cos f »

(14.13) cos f =

I RP

=

I

X CP

V / RP V / ZP (14.14)

RP

since, X CP » Z p , when δ is small cos φ ≈ 1 / ( R P ω CP )

(14.15)

i.e., cos φ ≈ tan δ

(14.16)

For equivalence between RS and RP

Figure 14.5 (a) Circuit Diagram for Parallel Representation, and (b) Phasor Diagram

t from which Rs ≈ 1/( R pω 2C 2 ) Power loss in the dielectric = VI cos f. From Figure 14.5 IC p V X Cp Vω C Vω C cos δ = = = or I = cos δ I I I  Vω C  Hence, power loss = VI cos φ = V  cos φ  cos δ  However, cos φ = sin δ (complementary angles), thus

Hence,

(14.17)

(14.18)

 Vω C  power loss = V  sin δ = V 2ω C tan δ  cos δ 

(14.19)

dielectric power loss = V 2ω C tan δ

(14.20)

The relative amount of energy lost in a capacitor can be indicated by any of the three terms. The terms are: dissipation factor (DF); power factor (PF) and quality (Q). All of these terms give about the same information. Power factor is the ratio of power (P) to apparent power (Papp). In a capacitor, power represents the energy per second lost and apparent power represents energy per second stored. If a capacitor has very little energy loss its power factor is low (less than 0.01 or 1 per cent). If a capacitor has very little energy loss, R is low and Q is high (more than 100). Dissipation factor is the reciprocal of quality. In other words, dissipation factor is the ratio of resistance to reactance. Obviously, if Q is typically a large number, dissipation factor will be a small number or even a percentage. For all but the lowest quality capacitors, power factor and dissipation factor are essentially equal to each other.

14.10 DIELECTRIC MATERIALS (GENERAL) An essential need in any electrical equipment is for the separation or insulation of one conducting path from another. An insulating material is one which offers a relatively high resistance to the passage of an electric current; such a material is also referred to as a dielectric. It follows from this definition that all known dielectric have a small electrical conductivity. Super purification increases the resistivity of some materials considerably, indicating that impurities may be a major cause of dielectric conduction. The final choice of an insulating material must depend upon mechanical as well as electrical properties appropriate to its intended use. Important mechanical and physical properties to be considered may include one or more of the following: tensile, compression, shear, and impact strength; modulus of elasticity; resistance to heat and shock, low water absorption,

282 Electrical Technology resistance to mineral oils, to acids and to other chemical action, specific gravity, coefficient of linear expansion noninflammability, thermal conductivity and rigidity, dimensional stability, flexibility (for cables) adaptability to a wide range of moulding or machining processes, freedom from tracking (i.e., from the creation of a conducting carbon path following a high voltage discharge over the surface). Important electrical properties include adequate electric strength against breakdown under high applied voltage; low power factor, permittivity over a range of frequencies and temperature, insulation resistance, surface and volume resistivity. Cost is naturally an important factor.

14.10.1 Gases The commoner gases are the most nearly perfect insulators, but with complete absence of mechanical strength. For example, air is the insulating material for a route of overhead lines. Most switching devices depend upon the insertion or bridging of the air-gap in a circuit, though in practice the conductors must be mounted upon some insulating material for rigid support, and this material enters largely into the degree of insulation obtained.

14.10.2 Non-metallic Liquids The commoner non-metallic liquids have a small electronic conductivity. The ionic conductivity of liquids varies greatly: For oils it is very small, and these are the only practicable liquid insulating materials, an important example being transformer oil which, of course, also acts as a cooling medium.

14.10.3 Pure Water Pure water is practically a non-conductor but is very slightly ionized. It is, however; practically impossible to stay free from impurities of which even the smallest percentage highly ionizes it and renders it a comparatively good conductor. For this reason, the presence of moisture is detrimental to good insulation, the water providing a path for leakage of the current. As an example, the failure of a cable is more often due to the ingress of moisture than to any other cause. Water as a surface film on insulation is harmful not because of its own (low) conductivity, but because of its power to dissolve and dissociate the presence of ions with consequent electrolytic current. In order to overcome this, impregnation is adopted to fill all interstices which might trap moisture, and surface varnishing done to keep moisture away from the surface of the insulator itself.

14.10.4 Solid Insulating Materials With solid insulating materials—since there are no free electrons in pure substances (excepting metals)—no electronic conduction takes place. Ionic conduction can occur and, if appreciable, there results some change in the composition of the material.

14.10.5 Textiles Cotton, silk, and wool are used extensively for insulating flexible conductors and in internal cables and coil windings. The insulation resistance of textiles is altered in the presence of moisture due to the development of a back e.m.f. The presence of impurities in textiles increases the conductivity and also the tendency for corrosion by the small currents flowing.

14.10.6 Paper Paper is used almost exclusively to separate the conductors in multi conductor underground cables but by crimping the paper or by using a wrapping of string between conductor and paper, the enclosed air plays an important part as the insulating material: such cables are known as air-spaced paper-core cables (ASPC).

14.10.7 Natural Minerals To name but a selection from the extensive range of available insulating materials, there are the natural minerals: mica, asbestos slate and marble; or vitrified materials: glass, ceramics and porcelain; natural and synthetic rubber; ebonite, waxes, enamels, varnishes, oils; natural resins (shellac) and synthetic resins, from the early bakelite to the thermosetting or thermoplastic types. Thermosetting plastics undergo plastic change when subjected to heat and pressure; they are thus converted to an insoluble infusible state which cannot be further reformed even by the application of more intense heat and pressure. Thermoplastic compounds, on the other hand, can be softened and resoftened indefinitely by the application of heat, provided the heat applied is insufficient to cause chemical decomposition. Many new substances are produced by a molecular process known as polymerization. Polymerization is the building up of long chain molecules from relatively simple molecular structures. In the case of thermosetting compounds,

Dielectric Materials

283

cross-linking of molecules occurs. Among these materials are many variants under trade names, such as perspex, polyethylene, neoprenes and polystyrene. Also extensively used are papers of various grades and laminated paper and fabric boards such as synthetic resin bonded paper (SRBP). A list of most commonly employed insulating materials with their dielectric constants is given in Table 14.1. Table 14.1 Dielectric Constants of Materials Material

Air Amber Asbestos fiber Bakelite (asbestos base) Bakelite (mica filled) Barium titanate Beeswax Cambric (varnished) Carbon tetrachloride Celluloid Cellulose acetate Durite Ebonite Epoxy resin Ethyl alcohol (absolute) Fiber Formica Glass (electrical) Glass (photographic) Glass (Pyrex) Glass (window) Gutta percha Isolantite Lucite Mica (electrical) Mica (clear India) Mica (filled phenolic) Micaglass (titanium dioxide) Micarto Mycolex Neoprene Nylon Paper (dry) Paper (paraffin coated) Paraffin (solid) Plexiglas Polycarbonate Polyethylene Polyimide

Dielectric Constant (Approximately)

1.0 2.6–2.7 3.1–4.8 5.0–22 4.5–4.8 100–1250 2.4–2.8 4.0 2.17 4.0 2.9–4.5 4.7–5.1 2.7 3.4–3.7 6.5–25 5.0 3.6–6.0 3.8–14.5 7.5 4.6–5.0 7.6 2.4–2.6 6.1 2.5 4.0–9.0 7.5 4.2–5.2 9.0–9.3 3.2–5.5 7.3–9.3 4.0–6.7 3.4–22.4 1.5–3.0 2.5–4.0 2.0–3.0 2.6–3.5 2.9–3.2 2.5 3.4–3.5 Contd.

284 Electrical Technology Table 14.1 Contd. Polystyrene Porcelain (dry process) Porcelain (wet process) Quartz Quartz (fused) Rubber (hard) Ruby mica Selenium (amorphous) Shellac (natural) Silicone (glass) (molding) Silicone (glass) (laminate) Slate Soil (dry) Steatite (ceramic) Steatite (low loss) Styrofoam Teflon Titanium dioxide Vaseline Vinylite Water (distilled) Waxes, mineral Wood (dry)

2.4–3.0 5.0–6.5 5.8–6.5 5.0 3.78 2.0–4.0 5.4 6.0 2.9–3.9 3.2–4.7 3.7–4.3 7.0 2.4–2.9 5.2–6.3 4.4 1.03 2.1 100 2.16 2.7–7.5 34–78 2.2–2.3 1.4–2.9

14.11 THE DIELECTRIC PHENOMENON The high electrical conductivity of metals is explained by the fact that the metals have only one, or two, valence electrons, and these are so far removed from the nucleus that they are effectively screened from it by the inner completed electron shells. In consequence, they can easily detach themselves and are able to wander at random about the interspaces between the more stable portions of the atoms, which thereby, by virtue of losing their valence electrons have become positive ions. If a p.d. is applied to the ends of the conductor, then an axial drift of electrons towards the end at a higher potential is superposed on the random motions, and it is this axial component of electron motion which constitutes the electric current. All liquid or solid insulators are compounds and, therefore, the ultimate particles of which they are composed of are the molecules and, in most of the cases, the molecules are heavy and complex. From the electrical point of view, we can regard the molecules as collections of positive and negative charges. Outside the molecules, the positive charges will act as though there were only one resultant positive charge situated at some particular point. Similar is the case with all the negative charges. There are two cases to be considered: The positive and negative resultant charges occupy different E E positions, the molecules, therefore, being a dipole, as shown in + Figure 14.6 (a). In an unstressed dielectric of this nature, the + – – dipoles have random orientations and the total strain can be Dipole Induced dipole regarded as zero. If a stress in the form of an electric intensity E (a) (b) is applied, then in general, there will be a turning moment acting on each dipole, and all the dipoles will try to orient themselves Figure 14.6 The Result of the Application of along the direction of the electric force. Work has to be done an Electric Field is a Displacement against the frictional forces between neighbouring dipoles. of All the Positive Charges in the The positive and negative resultant charges act at the same Direction of the Field and of All the point. In such a case, the molecule in its unstressed state is neuNegative Charges in the Opposite tral. If a stress in the form of an electric intensity E is applied, the Direction

Dielectric Materials

285

charges are attracted in opposite directions. As a consequence, they suffer displacement, work being done against the internal binding forces which act somewhat after the manner of a spring joining the charges together. In this stressed condition, the molecule becomes an induced pole, as shown in Figure 14.6 (b). In either case, the result of the application of an electric field is a displacement of all the positive charges in the direction of the field and a displacement of all the negative charges in the opposite direction. The magnitude of the displacement will obviously increase as the field intensity is increased by: 1. 2.

Causing a greater angular movement in the case of permanent dipoles. By increasing the separation in the case of induced dipoles.

The arrangement of the molecules on two kinds of dielectric under stressed conditions is illustrated in Figure 14.7. The dotted planes indicate that in both cases, there is a boundary of negative charge adjacent to the positive electrode, and one of positive charge adjacent to the negative electrode. If we imagine an isolated positive charge placed inside the dielectric, the total force acting on it due to the whole of the molecular charges within these boundaries in zero, but the effects of the two charged boundaries are to produce an attraction in the opposite direction to that of the applied field. Thus, there is a phenomenon somewhat analogous to the demagnetizing effect of a magnet.

Figure 14.7 Dielectric Under Stress

14.12 DIELECTRIC BREAKDOWN The breakdown of the molecule is not a purely electrical phenomenon. In the case of air and other gaseous dielectrics, breakdown is due to the ionization resulting from collisions of fast moving electrons with neutral particles. This phenomenon may take place in a solid or liquid dielectric after the molecules have been split up by other causes, but the conduction current is so small that there can be, relatively, very few electrons. Furthermore, their mean free paths are so small that the velocities attained may not be high enough to cause ionization. When a dielectric is stressed by the application of an alternating field, losses analogous to the hysteresis losses in ferromagnetic materials are produced by virtue of the frictional resistance which has to be overcome in causing the molecular deformation or molecular reorientation. These losses result in the production of heat. Consequently, the resistance decreases, causing an increase in current and, consequently, in heat production; so the process becomes cumulative. A stable state will be reached if there are provisions for heat dissipation, but if there are no such facilities, then the temperature will eventually reach a sufficiently high value to cause chemical breakdown. Breakdown never occurs simultaneously through the whole mass of the dielectric and naturally, the reasons for this localization have a distinct bearing on the phenomena. There is no solid or liquid dielectric which is absolutely homogeneous since there are invariably impurities or inclusions in the material. These are of two general kinds, voids and minute particles. The voids are tiny volumes of air which can be easily trapped where the dielectric is in the form of a wrapping such as the paper of a paper-insulated cable. For a potential gradient much greater than air can withstand an electric discharge takes place with consequent disintegration of the walls of the void. This results in a concentration of energy, with consequent local heating. The inclusions in the form of minute particles can be of two forms—insulating and conducting. If we regard these as spherical, there is a local concentration of potential gradient with conducting particles such as metal, carbon or water, and a smaller concentration with insulating particles of greater permittivity than the surrounding dielectric. Particles of both – ++ – kinds are, therefore, undesirable in liquid dielectrics (such as + + –– + –– + transformer oil or high-voltage switches). The concentration – + of potential gradient causes local concentration of energy loss which may further result in local disintegration of the oil with consequent formation of solid residue which will adhere to the original inclusion. The inclusion will, therefore, tend to grow Figure 14.8 The Drops Elongate Themselves in the Direction of the Field along the direction of lines of force, as shown in Figure 14.8.

286 Electrical Technology If this phenomenon is taking place throughout the whole of the mass of the dielectric, then if we trace any particular line of force from one electrode to the other, it is clear that the total distance through the dielectric in its original form becomes progressively less and the potential gradient, therefore, becomes progressively greater. This ultimately results in breakdown. In the case of drops of water, the effect must be extremely complex for the following reasons 1. Water is a relatively good conductor and, thus, phenomena such as the above take place. 2. Water has a relative permittivity of about 80, and this also leads to concentration of potential gradients. 3. The inclusion of water lowers the electrical resistance considerably and, therefore, increases the conduction current. This increases the number of free electrons and so increases the tendency to ionization. 4. If a dielectric occupies an unsymmetrical position with respect to the electrodes, it is attracted in such a way as to increase the capacitance. This can be explained in the following manner: if the dielectric or the plates of a capacitor are free to move, then their movements will be such as to increase the capacitance. The relevance of this to the drop of water is that since E for water is so high, the drops will move into the regions of maximum electric potential gradient. The drops will then elongate themselves in the direction of the electric force, thus behaving in a manner similar to the gradually growing solid inclusions. Bubbles of air, for which ε = 1, move away from the regions of high potential gradient. Table 14.2 Properties of Dielectric Materials Material

Dielectric Strength in kV/mm at 50 Hz

Specific Resistance when dry at 25 °C, MΩ/ cm cube

Permittivity

Affected by Moisture

Safe Temperature in °C

Uses

Asbestos

3–4.5

1.6 × 105

—

Absorbent

500 or more

Bakelite

20–25

—

5–6

No

200

Bitumen (vulcanized)

14

—

4.5

No

About 60

Cotton

3–4

—

Absorbent

90

Ebonite

10–40

1000 upwards according to dryness 2–100 × 108

Covering of wires in very highly rated machines Bakelized paper made up in form of boards; there are many uses Low-voltage mining cables; cable-box filling compound Covering for wires

2–3

Slightly

40

Empire cloth

10–20

As cotton

2

90

Fibre

5

As cotton

4–6

Absorbent if varnish layer is cracked Ditto

Glass

5–12

5 × 106 to infinity

3–8

No, except on surface

90

Covers for resistance boxes, etc. Wrapping for groups of wires, e.g., armature coils

In sheet form, slot linings Room temNot used in electriperatures only cal practice except or may crack to a small extent for transmission-line insulators

Dielectric Materials

287

Table 14.2 Contd. Gutta percha

10–20

5 to 25 × 106

3–5

No

India rubber Marble

10–25 6

2 to 10 × 106 400

2–3 8

5 to 100 × 106

3–8

Slightly Somewhat absorbent No

Mica

40–150

40 40 Room temperatures 500 or more

Micanite

30

10 to 6000 × 106

6–8

No

Paper Paraffin wax.

4–10 8

As cotton 3 × 1010

2 2

Absorbent No

Porcelain

9–20

1 to 1000 × 104

4–7

Room temperatures

Shellac

5–20

9 × 109

2.5–3.5

Not when vitreous and glazed No

3

40

7

Rather absorbent

Room temperatures

Slate

130 when under pressure 90 Under 50

}

Under 60

Covering for submarine cables Cable insulation Formerly used for switchboards Not generally used for machines in its pure form Commutator segments. Slot linings for high-voltage machines; bushes Cable insulation when oil impregnated; covering for transformer conductors Insulators for overhead lines Insulating varnishes; cement for manufacture of micanite For face-plates of starters, etc., not used for switchboards anymore

S UM M A RY 1. Many factors affect the dielectric properties of a material when used in a capacitor. 2. The dielectric constant of a material is the ratio of the capacitance of a capacitor using the material as a dielectric to the capacitance of the same capacitor using vacuum as dielectric. 3. The permittivity of dry air is approximately equal to one. 4. Dielectrics can be classified into two main groups: polar and non-polar. 5. Dielectrics which are non-polar in structure make the best a.c. working capacitors, particularly at higher frequencies. 6. Power factor is the ratio of power wasted per cycle to power stored per cycle. 7. The insulation resistance for a capacitor is related to leakage current. 8. The insulation resistance is lowered in the presence of high humidity, unless it is sealed.

9. Dielectric absorption reduces the capacitance as the frequency of operation is increased. 10. The maximum amount of field strength that a dielectric can withstand is called its dielectric strength. 11. The product CR is a characteristic of a given dielectric. 12. Dielectric hysteresis is analogous to hysteresis in magnetic materials. 13. The losses in a capacitor can be represented by a series resistance Rs or by a shunt resistance Rp. 14. Dissipation factor D is defined as the reciprocal of Q factor and is an indication of the quality of the dielectric. 16. D = 1/Q = tan d 17. In unstressed dielectrics the dipoles have random orientations. 18. The breakdown of a dielectric is not a purely electrical phenomenon. 19. If the plates of a capacitor are free to move, their movements will be such as to increase the capacitance.

288 Electrical Technology

M U LT IP LE C H O I C E Q UE S TI O NS (M C Q ) 1. Capacitance varies

5. A capacitor is regarded as

(a) Directly as A

(b) Inversely as A

(c) Directly as A2

(d) Inversely as A2

(a) An open circuit for d.c. (b) A short circuit for d.c.

6. (a) Q = CV + remanent polarization

2. Power factor is given by (a)

(b) Q = CV – remanent polarization

power stored per cycle

7. Which of the following relations holds

power wasted per cycle

(b) power wasted per cycle power stored per cycle (a) Lowered in the presence of high humidity (b) Enhanced in the presence of high humidity

(b) Dissipation factor (d) power factor

(a) Building up of long chain molecules (b) Building up of short chain molecules

10. Voids and inclusions in dielectric materials are

(a) Provide a conducting path (b) Do not provide a conducting path

5. (a)

(a) Polarization (c) Quality

9. Polymerization is the

4. Even the best of dielectrics

4. (a)

(b) tan d = 1/Q (d) Q = Rs or Cs

8. Alternate terms for dielectric power loss are

3. The insulation resistance of a capacitor is

ANSWERS (MCQ) 1. (a) 2. (b) 3. (a)

(a) tan d = Q (c) tan d = Rs or Cs

(a) Desirable

6. (a)

7. (a) and (c)

8. (b) and (c)

(b) Not desirable

9. (a)

10. (b)

CON V E N TI O NA L Q UE S TI O NS (C Q ) 1. Compare the following, in terms of their suitability for use as dielectrics. (a) Paper (b) Mica (c) Ceramics (d) Porcelain. 2. Which one of these dielectrics is more suitable and where? (a) Mica (b) Tantalum (c) Electrolytes.

3. What are the requirements of transformer oil? 4. What is meant by self healing? 5. Compare magnetic hysteresis and dielectric hysteresis.

15

Field Theory OBJECTIVES

0.2747 rad

In this chapter you will learn:  Field plotting by curvilinear squares    To calculate capacitance C between concentric cylinders  To calculate dielectric stress    To show that the capacitance of an isolated twin line, pe 0 e r C= and calculate C given values of a and d log e D/ a  To calculate energy stored in an electric field    To show that the inductance of a concentric cylinder, µ0µ r  1   + log e b a  and to calculate L given values L=   2π  4 of a and b    To show that the inductance of an isolated twin line µ0µ r  1   + log e D a  and calculate L given values of L=  π  4 a and D    To calculate energy stored in an electromagnetic field

r5 r4

r1

r2 r3

Field plot for a cable

15.1 INTRODUCTION The quantity of electricity is electric charge, a concept useful in explaining physical phenomenon. Charge is said to be conservative in the sense that it can be neither created nor destroyed. Charge is said to be quantized because the charge on 1 electron (1.602 × 10−19 C) is the smallest amount of charge that can exist. Electric field effects are due to the presence of charges; magnetic field effects are a result of the motion of charges. Electric charges are defined by the force that they exert on one another. Experimentally, the forces are found to depend on the magnitudes of the charges, (charge is a scalar quantity), their relative positions and their velocities. Forces due to the position of the charges are called electric forces, and those due to the velocity of the charges are termed magnetic forces. All electrical and magnetic phenomena can be explained in terms of the forces between the charges. The field is a convenient concept in calculating electric and magnetic forces. Around a charge we are able to visualize a region of influence called an electric field. Electric and magnetic fields can be represented diagrammatically, as shown in Figure [15.1 (a) and (b)].

+

–

(a) Magnetic flux

Current direction

(b)

Figure 15.1 Simple Electric and Magnetic Fields: (a) Electric and (b) Magnetic

290 Electrical Technology A

B

I

I

r

The ampere is now regarded in the mks system as the fourth fundamental unit. Its function as a unit is to link the electrical and mechanical systems. The ampere is that current which, when flowing in each of two very long parallel conductors of negligible cross-section, and separated by a distance of one metre from one another in vacuum produces between these conductors a force of 2 × 10−7 newtons per metre of their length, as represented in Figure 15.2. Electric fields, magnetic fields, and conduction fields (i.e., a region in which electric current flows) are analogous, i.e., all exhibit similar characteristics. Thus, they may all be analyzed by similar procedures.

15.2 THE ELECTRIC FIELD +

+

The temperature at every point in a room has a definite value. You can measure the temperature at any given point or combination of points by putting a thermometer there. We call the resulting distribution of temperatures a temperature field. In much the same way, you can imagine a pressure field in the atmosphere; it consists of Figure 15.2 Mechanical Force the distribution of air pressure values, one for each point in the atmosphere. These two examples are of scalar fields, because temperature and air pressure are scalar Between Parallel quantities. Conductors The electric field is a vector field; it consists of a distribution of vectors, one for each point in the region about a charged object, such as a charged rod. In principle, we can define the electric field at some point near the charged object, such as point P in Figure 15.3 (a) as follows: We place a positive charge q0 called E a test charge at the point. We then measure the electrostatic  force F that acts F  on the test charge. Finally, we define the P electric field E at the point P due to the charged object as + +   + + Electric field + + E = F / q0 (15.1) Test charge q0 + + at point P + +  at point P + + + + Thus, the magnitude of the electric field E at point P   + + + + is E = F/q0, and the direction of E is that of the force F + + + + Charged that acts on the positive test charge. As shown in object Figure 15.3 (b), we represent the electric field at P with a (b) (a) vector whose tail is P. In order to define the electric field within some region, we must similarly define it at all points Figure 15.3 (a) A Positive Test Charge q0 Placed at Point P Near a Charged Object. An in the region.  Electrostatic Force F Acts on the The SI unit for the electric field is newton per coulomb  Test Charge (b) The Electric Field E at (N/C). Table 15.1 shows the electric fields that occur in a Point P Produced by the Charged few physical situations. Object Force Force

Table 15.1 Some Electric Fields Field Location or Situation

Value N/C

At the surface of a uranium nucleus Within a hydrogen atom, at a radius of 5.29 × 10 Electric breakdown occurs in air

3 × 1021 −11

m

5 × 1011 3 × 106

Near the charged drum of a photocopier

105

Near a charged comb

103

In the lower atmosphere

102

Inside the copper wire of household circuits

10−2

Field Theory 291 Although we use a positive test charge to define the electric field of a charged object, that field exists independently of the test charge. In our defining procedure, we assume that the presence of the test charge does not affect the charge distribution on the charged object and, thus, does not alter the electric field we are defining.

15.3 VECTORS A vector quantity is one whose full description includes direction (or angular displacement) as well as magnitude, in contrast to a scalar quantity which is completely defined by stating its magnitude and, of course, the units in which they are expressed. In addition to the magnitude and direction of a vector quantity, it may be necessary to also know the point of application and the sense. Mass, length, and volume are examples of scalar quantities. Forces, distances, and velocities are vectors since they are not completely described without reference to the direction in which they act. Alternating voltages and currents, reactances and impedances are also vectors on account of their phase displacements. For a full statement of these quantities, it is necessary to state the phase angle as well as the magnitude. Not all physical quantities are vectors. The simplest vector quantity is displacement or change of position. A vector that represents displacement is called a displacement vector. A vector can be shifted without changing its value if its magnitude (length) and direction remain unchanged. Displacement vectors represent only the overall effect of the motion and not the motion itself.

15.3.1 Components of Vectors Adding vectors geometrically can be tedious. A neater and easier technique involves algebra but requires that the vectors be placed on a rectangular coordinate system. The x and y axes are usually drawn in the plane of the page, as in Figure 15.4 (a). The z-axis comes directly out of the page at the origin. A component of a vector is the projection of the vector on an axis.  In Figure 15.4 (a), for example, ax is the component of vector a on (or along the x-axis) and ay is the component along the y-axis. The projection of a vector on the x-axis is its x-component and, similarly, the projection on the y-axis is its y-component. The process of finding the components of a vector is called resolving the vector. Note: The small arrowheads on the components indicate their direction. In Figure [15.4 (a)–(c)]: a x = a cos q and a y = a sin q  where, θ is the angle that the vector a makes with the positive direc tion of x-axis and a is the magnitude of a. Once a vector has been resolved into its components along a set of axes, the components themselves can be used in place of the vector. For example in Figure 15.5 a=

ax2 + a 2y

and

tan q = a y ax

Magnitude-angle notation

15.3.2 Multiplying Vectors

(15.2)

y

y

a

ay

a

x

ax

0

ay

q

q

ax

(a)

0

x

(b) a

ay

q ax (c)

Figure 15.4 (a) The components ax and ay of the vector a (b) The components are unchanged if the vector is shifted, as long as the magnitude and orientation are maintained (c) The components form the legs of a right triangle whose hypotenuse is the magnitude of the vector

by = – 5 m

There are three ways in which vectors can be multiplied. If we multiply  a vector a by a scalar s, we get a new vector. Its magnitude is the product  of the magnitude of a and the absolute value of s. Its direction is the direction   of a if s is positive, but the opposite direction if s is negative. To divide a by s  y (m) we multiply a by 1/s. bx = 7 m There are two ways to multiply a vector by a vector: one way produces a x (m) O q scalar (called the scalar product), and the other produces a vector (called the vector product)

15.3.3 Scalar Product

b

Figure 15.5 Resolving a Vector

  The scalar product of vectors a and b as seen in Figure [15.6 (a) and (b)], is   written as a and b defined to be  ` a.b = ab cos f (15.3)

292 Electrical Technology   where, a is the magnitude of a, b is the magnitude of b  and f is the angle between a and  b (or more properly be tween the directions of a and b. There are actually two such angles f and 360° − f. Either can be used because their cosines are the same.

Component of b along direction of a is b cos

a

Note: 1. There are only scalars on the right side of Eq. 15.3 (including the value of cos f).  b  2. a and b on the left side represent a scalar quantity.   (a) (b) 3. a . b is also known as the dot product and is spoken as ‘a dot b’.   Figure 15.6 (a)Two Vectors a and b, with an Angle f 4. A dot product can be regarded as the product of two Between Them. (b) Each Vector has a quantities: (1) the magnitude of one of the vectors and Component along the Direction of the (2) the scalar component of the second vector along the Other Vector direction of the first vector. 5. If the angle between the two vectors is 0°, the component of one vector along the other is maximum and so also is the dot product of the vectors. If, instead, f is 90°, the component of one vector along the other is zero, and so also the dot product. b Component of a along direction of b is a cos

a

Equation 15.3 can be rewritten as follows to emphasize the components.   a ⋅ b = (a cos f)(b) = (a )(b cos f).

(15.4)

The commutative law applies to a scalar product, so we can write     a⋅b = b ⋅a When two vectors are in unit-vector notation, we write their dot product as r r a ⋅ b = a x $i + a y $j + a z k$ ⋅ b x $i + b y $j + b z k$

(

)(

(15.5)

)

(15.6)

which we can expand according to distributive law. Each vector component of the first vector is to be dotted with each vector component of the second vector. By doing so   a ⋅ b = a xbx + a yb y + a z bz (15.7)

15.3.4 Vector Product

     The vector product of a and b, written as a × b, produces a third vector c whose magnitude is c = ab sin f (15.8)   Where, f is the smaller of the two angles between a and b (we use the smaller of the two angles between the two vectors  because sin f and sin [360° - f] differ in algebraic sign). Because of the notation a × b is also known as the cross product, and spoken as ‘a cross b’.   a into vector b with the fingers of one’s right hand. The outstretched thumb shows the direction of vector 1. Sweep vector    c = a × b.     2. Showing that a × b is the reverse of b × a. b c=a×b a b

c¢ = b × a

a (a)

(b)

Figure 15.7 Illustration of the Right-hand Rule for Vector Products

    If a and b are parallel or anti parallel, a × b = 0 The magnitude   of     a × b, which can be written as |a × b |, is maximum when a and b are perpendicular to each other.   The  direction of c is perpendicular to the plane that contains a and b Figure 15.7 (a) illustrates how to determine the direction    of c = a × b with  what is known as the right hand rule. Place the  vectors a and b tail to tail without altering their orientations and imagine a line that is perpendicular to the plane where they meet. Let us pretend to place the right hand around that line in such a way   that the fingers would sweep a into b through the smaller angle between them. In such a case, the outstretched thumb points in the  direction of c.

Field Theory 293  The order of vector multiplication is important. In Figure 15.7 (b), one can determine the direction of c, so that the fin  gers are placed to sweep b into a through the smaller angle between them. The outstretched thumb points in the direction    of c and so it must be that c = −c that is,     b × a = −(a × b ) (15.9) In other words, the commutative law does not apply to a vector product. In unit vector notation, it is written as r r a × b = a $i + a $j + a k$ × b $i + b $j + b k$

(

x

y

z

) (

x

y

z

)

(15.10)

which can be expanded according to distributive law, i.e., each component of the first vector is to be crossed with each component of the second vector. For example, in the, expansion of Eq 15.10, we have a x i × b x × i = a x b x i × i = 0 (15.11)

(

)

because the two unit vectors i and j are parallel and, thus, have a zero cross product. Similarly, we have a i × b j = a b i × j = a b k x

y

x y

(

)

x y

(15.12)

In the last step Eq 15.8 was used to evaluate the magnitude of i × j as unity (The vectors i and j have a magnitude of unity, and the angle between them is 90°) Also, the right-hand rule is used to get the direction of i and j as being in the positive direction of z-axis (thus, in the direction of k ). Continuing to expand Eq 15.10, it can be shown that r r a × b = ( a y b z − b y a z ) $i + ( a z b x − bz ax ) $j + ( a x b y − b x a y ) k$ (15.13) To check whether any xyz coordinate system is a right handed coordinate system, one should use the right-hand rule for the cross product i × j = k with that system. If the fingers sweep i (positive direction of x) into j (positive direction of y) with the outstretched thumb pointing in the positive direction of z, then the system is right handed. Example 15.1 r r    If a = 3$i − 4 $j and b = − 2$i + 3k$ , what is c = a × b ? Solution: When two vectors are in unit-vector notation, we can find their cross product by using the distributive law. Thus,  c = 3i − 4j × −2i + 3 k

(

) ( ) = 3i × ( −2i ) + 3i × 3 k k + ( −4j ) × ( −2i ) + ( −4j ) × 3

Next, each term is evaluated with Eq 15.8 to determine the direction with the right-hand rule. For the first term here, the angle f between the two vectors being crossed is 0. For the other term, f is 90°. r c = 6(0) + 9(− j ) + 8(−k ) − 12$i = − 12$i − 9 $j − 8k$         The vector c is perpendicular to both a and b (c . a = 0, and c . b = 0). There is no component of c along the direction of  either a or b.

15.4 ELECTRIC FIELD LINES Michael Faraday, who introduced the idea of electric fields in the nineteenth century, thought of the space around a charged body as filled with lines of force. Now usually called electric field lines, they provide a nice way to visualize patterns in electric fields. The relation between the field lines and electric field vectors is this: (1) At any point, the direction of a straight field line or the direction of the tangent to a curved field line gives the direction of E at that point. (2) The field lines are drawn in such a way that the number of lines per unit area measured in a plane that is perpendicular to the lines, is proportional to the magnitude of E. This second relation means that where the field lines are close together, E is large; and where they are far apart, E is small.

294 Electrical Technology Figure 15.8 (a) shows a sphere of uniform negative charge. If we place a positive test charge anywhere near the sphere, an electrostatic force pointing toward the centre of the sphere will act on the test charge. In other words, the electric field vectors at all points near the sphere are directed radially toward the sphere. This pattern of vectors is neatly displayed by the field lines in Figure 15.8(b), which point in the directions as the force and field vectors. The spreading of the field lines with distance from the sphere tells us that the magnitude of the electric field decreases with distance from the sphere. Electric field lines extend away from positive charge (where they originate) and toward negative charge (where they terminate). Figure 15.9 shows the field lines for two equal positive charges, while Figure 15.10 shows the pattern for two charges that are equal in magnitude but of opposite sign, a configuration we call an electric dipole. Although we do not often use field lines quantitatively, they are very useful to visualize what is going on. The charges are pushed apart in Figure 15.9 and pulled together in Figure 15.10.

––– – – – – –

––– – – – – –

E

F Positive test charge

Electric field lines

(b)

(a)

 Figure 15.8 (a) The Electrostatic Force F Acting on a Positive Test Charge Near a Sphere of Uniform Negative Charge (b) The Electric Field Vector E at the Location of the Test Charge, and the Electric Field Lines in the Space Near the Sphere. The Field Lines Extend Toward the Negatively Charged Sphere. They Originate on Distant Positive Charges

E

E

Figure 15.10 Field Lines for a Positive and a Nearby Negative Point Charge that are Equal in Magnitude. The Charges are Pulled Together

Figure 15.9 Field Lines for Equal Positive Point Charges. The Charges are being Pushed Apart

A + + + + + + + + + + + + + +

B – – – – – – – – – – – – – –

+V

–V

2 3V 1 3V

15.5 FIELD PLOTTING BY CURVILINEAR SQUARES 0

–1V 3 –2V 3

Figure 15.11 Lines of Force Intersecting Equipotential Lines in an Electric Field

Two Parallel Plates A and B are shown in Figure 15.11. Let the potential on plate A be +V volts and that on plate B be –V volts. The force acting on a point charge of 1 coulomb placed between the charges is the electric field strength E. It is measured in the direction of the field and its magnitude depends on the p.d. between the plates as also the distance between the plates.

Field Theory 295 Moving along a line of force from plate B to plate A means moving from –V to +V volts. The p.d. between the plates is therefore, 2V volts and the potential changes linearly when moving from one plate to the other. Hence, a potential gradient is followed which changes by equal amounts for each unit of distance moved. Lines may be drawn connecting together all points within the field having equal potentials. These lines are called equipotential lines and these have been drawn in Figure 15.11 for 2/3 V, 1/3 V, 0, −1/3V and −2/3 V. The zero equipotential line represents earth potential and the potential on plates A and B are, respectively, above and below earth potential. Equipotential lines form a part of an equipotential surface. Such surfaces are parallel to the plates, as illustrated in Figure 15.11 and the plates themselves are equipotential surfaces. There can be no current flow between any given points on such a surface since all points on an equipotential surface have the same potential. Thus, a line of force (or flux) must intersect an equipotential surface at right angles. A line of force in an electrostatic field is often termed a streamline. Electric field distribution for a concentric cylinder capacitor is shown in Figure 15.12. An electric field is set up in the insulating medium between two Streamlines good conductors. Any volt drop within the conductors can usually be neglected Equipotential compared with the p.d.s across the insulation since the conductors have a high lines conductivity. All points on the conductors are, thus, at the same potential so that the conductors form the boundary equipotentials for the electrostatic field streamlines (or lines of force) which must cut all equipotentials at right angles, leave one boundary at right angles pass across the field, and enter the other Figure 15.12 Electric Field boundary at right angles. Distribution for a In a magnetic field, a stream line is a line so drawn that its direction is Concentric Cylinder everywhere parallel to the direction of the magnetic flux. An equipotential surCapacitor face in a magnetic field is the surface over which a magnetic pole may be moved without the expenditure of Equipotential work or energy. lines In a conduction field, a stream line is a line drawn with a direction which is everywhere parallel to the direction of the current flow. A method of solving certain field problems by a form of graphical estimation is available which may only be applied, however, to plane linear fields; examples include the field existing between parallel plates or between two long parallel conductors. Streamlines In general, the plane of a field may be divided into a number of (or lines of force) squares formed between the line of force (i.e., streamline) and the equipotential. Figure 15.13 shows a typical pattern. In most cases, true squares will not exist, since the streamlines and equipotentials are Figure 15.13 Curvilinear Square curved. However, since all the streamlines and the equipotentials intersect at right angles, square-like figures are formed, and these are usually called ‘curvilinear squares’. The square-like figure shown in Figure 15.13 is a curvilinear square since, on successive subdivision by equal numbers of intermediate streamlines and equipotentials, the smaller figures are seen to approach a true square form. Consider the electric field established between two parallel metal plates, as shown in Figure 15.14. The stream lines and the equipotential lines are shown sketched and are seen to form curvilinear squares. Let us consider a true square abcd lying between equipotentials AB and CD. Let this square be the end of x metres depth of the field, forming a flux tube between adjacent equipotential surfaces abfe and cdhg, as shown in Figure 15.14. Let l be the length of side of the squares. Then the capacitance C1 of the flux tube is given by C1 =

e 0e r ( area of plate ) e 0e r ( Px ) = l plate separation

(15.14)

C1 = ε 0ε r x Thus, the capacitance of the flux tube whose end is a true square is independent of the size of the square. Let the distance between the plates of the capacitor be divided into an exact number of parts, say n (in Figure 15.14, n = 4). Using the same scale, the breadth of the plate is divided into a number of parts (which is not always an integer value), say m (in Figure 15.14, m = 10, neglecting fringing). Thus, between equipotentials AB and CD in Figure 15.14 there are m capacitors in parallel.

(15.15) A C

a

b

d

c

B

Figure 15.14 (a) Electric Field Between Two Parallel Metal Plates

D

296 Electrical Technology Flux tube

f

e

g

h b

a

A l C

B

x

D

c

d I

Figure 15.14 (b) Electric Field between Two Parallel Metal Plates

For m capacitors in parallel, the equivalent capacitance CT is given by CT = C1 + C2 + C3 + … + Cm. If the capacitors have the same value i.e., C1 = C2 = C3 = Cm then CT = mCt Similarly, there are n squares in series. For n capacitors connected in series, the equivalent capacitance CT is given by 1/CT = 1/C1 + 1/C2 + 1/C3 + … + 1/ Cn. If C1 = C2 = Cn then 1/CT = n/Ct, from which CT = Ct/n Thus, if m is the number of parallel squares measured along each equipotential line and n is the number of squares measured along each stream line (line of force), then the total capacitance C of the field is given by: m C = ε 0ε r x farads (15.16) n

Example 15.2 In the parallel plate capacitor shown in Figure 15.15, the dielectric has a relative permittivity of 3.5. The distance between the plates is divided into four equipotential lines, each 1 mm apart (n = 4, m = 8). Neglecting any fringing, let us find the capacitance. Solution:

(

)

C = 8.85 × 10−12 (3.5)(0.005)(8/ 4)

P

= 0.31 pF

4 mm

Using the normal equation for capacitance of a parallel-plate capacitor, C =

ε 0ε r A d

=

Q 5 mm 8 mm

( 8.85 × 10−12 ) (3.5)(0.008 × 0.005) = 0.31 pF

Figure 15.15 For Example 15.2

0.004

The capacitance found by each method gives the same value. This is expected since the field is uniform between the plates, giving a field plot of true squares.

15.6 EFFECT OF FRINGING The effect of fringing may also be considered by estimating the capacitance by field plotting, as described below. In the side view of the plates shown in Figure 15.16, RS is the medial line of force or medial streamline, by symmetry. Also, XY is the medial equipotential. The field may be divided into four separate symmetrical parts. Considering just the top left part of the field, the field plot is estimated in the following manner: 1. Estimate the position of the equipotential EF (Figure 15.16), which has the mean potential between that of the plate and that of the equipotential XO. F is not taken too far since it is difficult to estimate. Point E will be slightly closer to point Z than point O. 2. Estimate the positions of intermediate equipotentials GH and IJ. 3. All the equipotential lines plotted are 2/4 or 0.5 mm apart. Thus, a series of streamlines—cutting the equipotential at right angles—are drawn, the streamlines being spaced 0.5 mm apart, with the object of forming, as far as possible, curvilinear squares. m total capacitance, C = e 0e r x farads. (15.17) n The number of parallel squares measured along each equipotential is about 13 in this case and the number of series squares measured along each line of force is 4. Thus, for the plates shown in Figure 15.16, m = 2 × 13 = 26 and n = 2 × 4 = 8. Since x is 5 mm, m − 12 total capacitance = ε0 ε r x farads = 8 . 85 ×10 (3.5)(0.005) 266 8 n = 0.50 pF.

(

)

Field Theory 297 R 8 mm

4 mm

P X

Y

0 Q S

H R F

Medial line of force Z G

J

E

2 mm

I X 4 mm

Y

0

S

Medial equipotential

Figure 15.16 Field Plotting Taking Fringing Into Consideration Example 15.3

Plate

A field plot between two metal plates is shown in Figure 15.17. The relative permeability of the dielectric is 2.8. Determine the capacitance per metre length of the system. Solution: C = ε 0ε r x ( m / n) From Figure 15.17, m = 16 (the number of parallel squares measured along each equipotential) n = 6 (the number of series squares measured along each line of force) Hence,

(

Plate

Figure 15.17 For Example 15.3

)

C = 8.85 × 10−12 (2.8)(1)(16 / 6) = 66.08 pF

Example 15.4 A field plot for a cross-section of a concentric cable is shown in Figure 15.18. If the relative permittivity of the dielectric is 3.4, determine the capacitance for 100 m length of the cable. Solution: C = ε 0ε r x ( m / n) Equipotential From Figure 15.18, m = 13 and n = 4. Also x = 100 m

(

)

C = 8.85 × 10−12 (3.4)(100)(13/4) = 9780 pF = 9.78 nF

lines Lines of force

Figure 15.18 For Example 15.4

298 Electrical Technology

15.7 CAPACITANCE OF A PARALLEL PLATE CAPACITATOR In a parallel plate capacitor, the plates are so close together that the field between them can be regarded as uniform. If the charge density is D, the surface area of one side of the plates A square metres, the distance between the plates d metres, and the permittivity e, the total charge Q may be calculated from D = Q/A and is Q = DA coulombs. The p.d. between the plates may be calculated from E = V/d hence V= Ed From D/E = e, E = D/e, hence V = Ed = dD/e Capacitance,

C = Q / V = DA Dd / ε = ε A/ d = ε 0 ε1 , A/ d farads

(15.18)

This shows that the capacitance of a capacitor is directly proportional to the plate area and to the permittivity and inversely proportional to the thickness of the dielectric. Note: 1. A and d are in metres, er is the relative permittivity and e0 = 8.85 × 10−12. 2. The plates of the capacitor are so large and so close together that we can neglect the fringing of the electric field at the edges of the plates, so that E is uniform throughout the region between the plates. Example 15.5 What quantity of electricity will produce a difference of potential of 200 V between the plates of a capacitor of 5 mF? Compare the magnitudes of two capacitors, one having two circular plates 4 cm diameter and ½ mm apart, and the other having two square plates 5 cm each side and ¾ mm apart. Solution: 1.

V = 200 V, C = 5 ´ 10−6 F Q = CV = 5 ´ 10−6 ´ 2 ´ 102 = 10−3 coulombs or1000 m C

2.

A1 = p 42/ 4 , d1 = 0.05 cm A2 = (5 × 5) cm 2, d 2 = 0.075 cm C1 / C2 = ( A1d 2 ) / ( A2 d1 ) = (4p × 0.075) / (25 × 0.05) = 3.77 / 5 = 0.75 C1 : C2 = 3 : 4

Example 15.6 A simple capacitor, consisting of two simple insulated plates, has a capacitance of 0.001 μF and receives a charge of 1 μC. What is the p.d. between the plates? What would be the capacitance if the areas of each plate were trebled and the spacing between the plates halved? Solution: 1.

C = Q /V V = Q / C = 10−6 /10−9 = 1000 V

2.

A1 = A,

d1 = 2d ,

A2 = 3 A,

d2 = d ,

C1 = 10−9 F C2 = ?

C1/ C2 = A1d 2 / A2 d1 = ( A × d ) / (3 A × 2d ) = 1.6 i.e., the capacitance would be increased six-fold to 0.006 F

Field Theory 299 Example 15.7 A variable capacitor of 1000 μF is charged to a p.d. of 100 V. The plates of the capacitor are then separated by means of an insulated rod so that the capacitance is reduced to 300 μF. Would you expect the p.d. across the capacitor to have changed, and if so, then by how much? Solution: C = Q/V. For a constant value of Q, V µ 1/C. If the capacitance is reduced by 10/3, the charge remains constant and the potential difference therefore increases by 10/3 to (100 × 10/3) = 33 V, i.e., increases by (333 – 100) = 233 V. Example 15.8 If a capacitor is to have a capacitance of 0.1 μF and the paper utilized for the dielectric is 4 cm wide and 0.02 mm thick with a relative permittivity of 2.25, what would be the length of paper required? Solution: C = ε 0ε x , A/ d F = ε 0ε r w × l × 106 / d µ F Where, w and l are the width and length of the dielectric.

(

) = ( 0.1 × 0.2 × 10−3 ) / ( 2.25 × 8.85 × 10−12 × 0.04 × 106 )

l = C × d / ε 0ε r w × 106

= 200 / 79.7 = 2.51 m Example 15.9 A capacitor consisting of two air-spaced parallel plates, each of effective area 1000 cm2 and spaced 0.1 cm apart, is connected across a constant-voltage source of 500 V. Calculate the charge on the capacitor. Solution: In air, C=e0A/d farads A = 1000 × 10−4 m 2 , C =

d = 0.1 × 10−2 m,

−12

8.85 × 10 × 1000 × 10 0.1 × 10−2

−4

ε 0 = 8.85 × 10−12 F/m,

V = 500 V

= 8.854 × 10−12 F

Q = CV = 8.854 × 10−12 × 500 = 0.4427 µ C

15.8 CAPACITANCE OF A MULTIPLATE CAPACITATOR In calculating the capacitance of a multiplate capacitor, as has been represented in Figure 15.20, the effective area of one set of plates has to be considered in a two-plate capacitor, as shown in Figure 15.19, the charge and, hence, the area of one side of one plate only enters into the expression for calculating capacitance. For the multi-plate pattern, the area of one side

Wire

{

Plate a, area A

+Q

–Q Potential difference = Vab

d Wire (a)

Plate b, area A

E +++ + + + + + + ––– – – – – – –

(b)

Figure 15.19 (a) A Charged Parallel-plate Capacitor (b) When the Separation of the Plates is Small Compared to their Size, the Fringing of the Electric Field at the Edges is Slight

300 Electrical Technology of each pair of plates has to be considered: the number of plates connected to one terminal is normally one greater than the number of plates connected to the other, as shown in Figure [15.20 (a) and (b)]. Tinfoil

Number of dielectrics = N Number of plates = (N+1)

Waxed paper (a)

(b)

Figure 15.20 (a) Multiplate Capacitor (b) Waxed Paper Capacitor For purposes of calculation, the total plate area A is equal to the number of dielectric layers (n) multiplied by the surface area (a) of one side of one plate, or A = na. The total number of plates will be (n+1). Example 15.10 A parallel-plate capacitor has two metal plates, each of effective area 500 cm 2. (1) Calculate its capacitance, when the distance between the plates is adjusted to 0.5 cm. A constant charging current of 2 μA is supplied to this capacitor, Calculate (1) for how many microseconds this charge must continue in order to raise the potential difference between the plates to 1000 V. (3) Using the same two capacitor plates and the same charging current flowing for the same time as in (2) what alteration is necessary in the capacitor to ensure that the potential difference between the plates rises to twice the value obtained in (1)? Solution: 1.

C = ε 0 A/ d =

8.85 × 10−12 × 500 × 10−4 5 × 10−3

= 88.54 pF 2.

V = 1000 = Q / C = It / C =

2 × 10−6 × t × 10−6 = 1000 × 88.54 / 2 88.54 × 10−12

= 44.27 m sec

3. 2V = 2Q/C, and since the charge Q will be unchanged when the alteration to the capacitor is made, 2V = Q/C/2 i.e., C must be halved. The capacitance is inversely proportional to the distance between the plates so that the capacitor plates must be separated by twice the spacing, i.e.. by 1.0 cm. Example 15.11 A 0.02 μF capacitor receives a charge of 20 μC. What energy is stored in the capacitor? If the capacitor is made up of parallel plates each having an area of 1130 cm2 and placed 1 mm apart in air, how many plates are used?

Field Theory 301 Solution:

ε r = 1, ε 0 = 8.85 × 10−12 , a = 0.113 m2, d = 10−3 m, C = 0.02 × 10−6 F, (n + 1) = number of plates 1.

W = CV 2 / 2 = Q 2 / 2C = ( 202 × 10−12 ) ( 2 × 0.02 × 10−6 ) = 0.01 J

2.

C = ε 0ε r A/ d F = ε 0ε r na / d F n = C d / ( ε 0ε r a )

(

= 0.02 × 10−6 × 10−3

) ( 8.85 × 10−12 × 1 × 0.113 )

= 20 Number of plates = 20 + 1 = 21

15.9 CAPACITANCE BETWEEN CONCENTRIC CYLINDERS A concentric cable is one which contains two or more separate conductors Equipotential Streamlines arranged concentrically (i.e., having a common centre), with insulation between them. In a co-axial cable, the central conductor—which may be either solid or Inner hollow—is surrounded by an outer taubular conductor, the space in between conductor being occupied by a dielectric. If air is the dielectric, then concentric insulating r dr discs are used to prevent the conductors from touching each other. The two a kinds of cable serve different purposes. The main feature they have in common Outer b conductor is a complete absence of external flux and, therefore, a complete absence of interference with and from other circuits. The electric field between two concentric cylinders (i.e., a coaxial cable) Figure 15.21 Electric Field has been represented in the cross-section in Figure 15.21. The conductors Between Two Conform the boundary equipotentials for the field, the boundary equipotentials in centric Cylinders Figure 15.21 being concentric cylinders of radii a and b. The stream lines, or lines of force, are radial lines that cut the equipotentials at right angles. Let Q be the charge per unit length of the inner conductor. Then the total flux across the dielectric per unit length is Q coulombs/metre. This total flux will pass through the elemental cylinder of width dr at radius r (shown in Figure 15.21) and a distance of 1 m into the plane of the paper. The surface area of a cylinder of length 1 m within the dielectric with radius r is (2pr × 1) m2. Hence, the electric flux density at radius r is, D = Q / A = Q/(2π r )

(15.19)

The electric field strength or electric stress E, at radius r is given by: E = D / ε 0ε r = Q/( 2π rε 0ε r ) Volts/metre

(15.20)

Let the p.d. across the element be dV volts (since E = voltage/thickness), voltage Ex thickness. Q δ r. The total p.d. between the boundaries, Therefore, δ V = Eδ r = 2π rε 0ε r V = =

i.e.,

b

∫a

Q Q δr = 2π rε 0ε r 2π ε 0ε r

b

∫a

1 δr r

b Q Q  loog e b − log e a   log e r  = a 2πε 0ε r 2πε 0ε r 

V =

Q b log e volts 2πε 0ε r a

(15.21)

302 Electrical Technology The capacitance per unit length, C =

C =

charge per unit length = Q /V p.d.

(2πε 0ε r ) Q = farads/metre ( Q / (2πε 0ε r ) ) log e (b/a) log e (b/a)

(15.22)

Example 15.12 A coaxial cable has an inner core radius of 0.5 mm and an outer conductor of internal radius 6.0 mm. Determine the capacitance per metre length of the cable provided the dielectric has a relative permittivity of 2.7. Solution: (2πε 0ε r ) C = log e (b / a ) =

2π (8.85 × 10−11 )(2.7) = 60.4 pF F log e (6.0 /5.0)

15.10 DIELECTRIC STRESS Rearranging Eq (15.21) gives

Q V = log e (b / a ) 2πε 0ε r

However, from Eq (15.20) E = E =

Thus, dielectric stress,

Q 2π rε 0ε r

V volts/metre r log e (b / a )

(15.23)

The dielectric stress at any point is inversely proportional to r, i.e., E α 1/ r . The dielectric stress E will have a maximum when r is at its minimum, i.e., when r = a, Thus, V E max = (15.24) a log e (b / a ) E min =

It follows that r1

Equipotential lines

V b log e (b / a )

(15.25)

15.11 CONCENTRIC CABLE FIELD PLOTTING

q rad

Figure 15.22 illustrates the cross-section of a concentric cable having a core radius r1and a sheath radius r4. The capacitance of a true capacitor is given by: Line of r4 C e 0e r farads/metre force (15.26) A curvilinear square is shown shaded in Figure15.22. Such squares can be made Figure 15.22 Concentric Cable Havto have the same capacitance as a true square by the correct choice of spacing ing a Core Radius r1 between the lines of force and the equipotential surfaces in the field plot. and Sheath Radius r4 r3

r2

From Eq 15.22

C =

For a sector of q radians,

C =

2πε 0ε r log e (rb / ra )

θ 2π

farads/metre

 2πε 0ε r  θε 0ε r farads/metree   = r r log ( / ) log e b a  e ( rb / ra ) 

(15.27)

Field Theory 303 If θ = log e ( rb / ra ) then C = ε 0ε r , the same as for a true square. If θ = log e ( rb / ra ) then. eθ = ε 0ε r. Thus, if θ two equipotential surfaces are chosen within the dielectric as shown in Figure 15.22, then eθ = r2 / r1 , e = r3 / r2 , and eθ = r4 / r3 . Hence,

( eθ )3 =

r2 r1

×

r3 r2

×

r4 r3

,

i.e.,

e3θ =

r4 r1

(15.28)

e 2q = r3 / r1 (15.29) It follows that Equation 15.28 is used to determine the value of θ and, hence, the number of sectors. Hence, for a concentric cable having a core radius of 8 mm and a inner sheath radius of 32 mm, if two equipotential surfaces within the dielectric are chosen (and, therefore, form three capacitors in series in each sector) r4 32 e3θ = = =4 (15.30) r1 8 1 log e 4 = 0.462 rad = 26.47°. Thus, there will be 2p/0.462 = 13.6 sectors in the field plot. 3 (Alternatively, 360°/26.47° = 13.6). From above, Hence, 3θ = loge and θ =

e 2θ = r3 / r1 , i.e., r3 = r1e 2θ = 8e 2(0.462) = 20.15 mm eθ = r2 / r1 , from which r2 = r1eθ = 8e0.462 = 12.70 mm The field plot is shown in Figure 15.23. The number of parallel squares measured along each equipotential is 13.6 and the number of series squares measured along each line of force is 3. Hence, in Eq. 15.16 where, C = ε 0ε r x(m / n), m = 13.6 and n = 3. If the dielectric has a permittivity of, say, 2.5, then the capacitance per metre length,  13.6  C = 8.85 × 10−12 (2.5)(1)   = 100 pF  3 

(

=

r2 = 12.70 mm r4 = 32 mm

)

From Eq 15.22 C =

r1 = 8 mm

2πε 0ε r log e (r4 / r1 )

F/m

(

Figure 15.23 Field Plot of a Concentric Cable

)

2π 8.85 × 10−12 (2.5) log e (32 /8)

r3 = 20.15 mm

= 100 pF// m

Thus, field plotting using curvilinear squares provides an alternative method of determining the capacitance between concentric cylinders. Example 15.13 A concentric cable has a core diameter of 20 mm and a sheath inside diameter of 60 mm. The permittivity of the dielectric is 3.2. Using three equipotential surfaces within the dielectric, determine the capacitance of the cable per metre length by the method of curvilinear squares. Draw the field plot of the cable. Solution: The field plot consists of radial lines of force dividing the cable cross-section into a number of sectors, the lines of force cutting the equipotential surfaces at right angles. Since three equipotential surfaces are required in the dielectric, four capacitors in series are found in each sector of θ radians.

304 Electrical Technology

0.2747 rad

r5 r4

r1

r2 r3

In Figure 15.24, r1 = 20/2 = 10 mm and r5 = 60/2 = 30 mm. It follows from Eq. 15.28 that e4q = r5/r1 = 30/10 = 3, from which 4θ = loge 3 and θ = 1/4 loge 3 = 0.2747 rad Thus, the number of sectors is 2p/0.2747 = 22.9. The three equipotential lines are shown in Figure 15.24 at radii of r2, r3, and r4 From Eq. 15.28 e3θ = r4 / r1 ;

r4 = r1e3θ = 10 e3(0.2747 ) = 22.80 mm

e 2θ = r3 / r1 ;

r3 = r1e 2θ = 10 e 2(0.2747 ) = 17.32 mm

eθ = r2 / r1 ;

r2 = r1eθ = 10 e0.2747 = 13.16 mm

Thus, the field plot for the cable is as shown in Figure 15.24. Figure 15.24 Field Plot For Example 15.12

 22.9  C = ε 0ε r x(m / n) = 8.55 × 10−12 (3.2)(1)   = 162 pF  4 

(

)

15.12 CAPACITANCE OF AN ISOLATED TWIN LINE The field distribution with two oppositely charged, long conductors A and B, each of radius a is shown in Figure 15.25. The distance D between the centers of the two conductors is such that D is much greater than a. Figure 15.26 shows the field of each conductor separately. Initially, let conductor A carry a charge of +Q coulombs per metre, while conductor B is uncharged. We can then consider a cylindrical element of radius r about conductor A having a depth of 1 metre and a thickness dr as shown in Figure 15.26. The electric flux density D, at radius, r, is given by D = charge/area = Q/(2p × 1) coulomb/metre2. The electric field strength at the element, Q / 2π r Q Volts/metre E = D (ε 0ε r ) = = (15.31) 2π r ε 0ε r ε 0ε r

D

Lines of forces

B

A

Radius a

Radius a

Equipotential lines

Figure 15.25 Field Distribution with Two Oppositely Charged Long Conductors

Field Theory 305

dr

r

A

B

D

Figure 15.26 The Field of Each Conductor Shown Separately

Since E = V/d, potential difference V = Ed. Thus, p.d. at the element = E δ r =

Qδr volts. 2π r ε 0ε r

The potential may be considered as zero at a large distance from the conductor. Let this be at radius R. Then the potential of conductor A above zero, VA1, is given by R R 1 Qδr Q Q = VA1 = ∫ [log e r ] aR dr = ∫ 0 2π r ε ε 0 r 2 π ε ε 2 π ε ε 0 r 0 r 0 r (15.32) Q Q = [llog e R − log e a ] i.e., VA1 = log e ( R / a ) 2π ε 0ε r 2π ε 0ε r Since conductor B lies in the field of conductor A, by similar reasoning, the potential at conductor B above zero, VB1, is given by R Q dr Q Q VB1 = ∫ [log e x]RD = log e ( R /D) = (15.33) D 2π r ε ε 2 π ε ε 2 π ε 0ε r 0 r 0 r Repeating the above procedure, this time assuming that the conductor B carries a charge −Q coulombs per metre, while −Q log e ( R / a ) and the potential of conductor A is uncharged, gives the potential of conductor B below zero, VB 2 = 2πε 0ε r conductor A below zero, due to the charge on conductor B. −Q VA2 = log e ( R /D) (15.34) 2πε 0ε r When both conductors carry equal and opposite charges, the total potential of A above zero is  Q   −Q  VA1 + VA2 =  log e ( R / a )  +  log e ( R / D)   2πε 0ε r   2πε 0ε r      =

Q log e ( D /a ) 2πε 0ε r

(15.35)

Furthermore, by the same token, the total potential of B below zero is Q VB1 + VB 2 = ( log e ( R /D) − log e ( R /a) ) 2πε 0ε r =

−Q log e ( D /a ) 2πε 0ε r

(15.36)

306 Electrical Technology Hence, the p.d. between A and B is  Q  2 log e ( D / a )  volts/metre  2πε 0ε r    The capacitance between A and B per metre length is Charge per metre Q C = = pd 2 ( Q /(2πε 0ε r ) ) log e ( D / a ) C =

C =

1 2πε 0ε r farads/metre 2 log e ( D / a )

πε 0ε r

(15.37)

log e ( D / a )

Example 15.14 Two parallel wires, each of diameter 20 mm, are uniformly spaced in air at a distance of 50 mm between centres. Determine the capacitance of the line if the total length is 200 m. Solution: πε 0ε r π (8.85 × 10−12 )(1) π (8.85 × 10−12 ) C = = = = 9.28 × 10−12 F log e ( D /a ) log e (50 / (512)) log e 20 Capacitance of a 200 m length = 9.28 × 10−12 × 200 F = 1860 pF or 1.86 nF

15.13 ENERGY STORED IN AN ELECTRIC FIELD (ALTERNATE METHOD) Let us consider the p.d. across a parallel plate capacitor of capacitance C farads being increased by dv volts in dt seconds. If the corresponding increase in charge is dq coulombs, then dq = Cdv. If the charging current at that instant is i amperes, then dq = idt. Thus, idt = Cdv, i.e.; i = Cdv/dt i.e., instantaneous current = capacitance × rate of change of p.d. The instantaneous value of power to the capacitor p = vi watts = v(C dv/dt) watts The energy supplied to the capacitor during time dt = power × time = (vCdv/dt)(dt) = Cv dv Joules. Thus, the total energy supplied to the capacitor when the p.d. is increasing from 0 to V volts is Wf =

Wf =

i.e., energy stored in the electric field,

v

∫0

1 CV 2 Joules 2

C = Q /V , energy stored = Since,

=

1

2

QV

The electric flux density D = Q/A from which Q = DA Hence, energy stored = 1/2 (DA)V joules The electric field strength E = V/d from which V = Ed Hence, energy stored = 1/2 (DA) (Ed) joules But Ad is the volume of the field.

v

 v2  Cv dv = C    2 0

1 (Q / V ) V 2 2

(15.38)

Field Theory 307 Therefore, energy stored per unit volume, wf = 1/2 DE joules/cubic metre Since,

D/E = e0er, then D = Ee0er

Therefore,

wf = 1/2 (e0er E) (E) wf = 1/2 e0er E2 joules/cubic metre

Also, since, D/E = e0er, then E = De0er and w f = D 2 /2ε 0ε r Joules/cubic metre

(15.39)

Example 15.15 Determine the energy stored in a 10 nF capacitor when charged to 1 kV, and the average power developed if this energy is dissipated in 10 μs. Solution:

W f = 1/ 2 CV 2 = Average power developed

1

2

(10 × 10−9 )(103 ) 2 = 5 mJ

= (energy dissipated, W)/ (time, t) =

5 × 10−3 = 500 W 10 × 10−6

Example 15.16 A 400 pF capacitor is charged to a p.d. of 100 V. The dielectric has a cross-sectional area of 200 cm2 and a relative permittivity of 2.3. Calculate the energy stored per cubic metre of the dielectric. Solution: w f = D 2 / 2ε 0ε r Electric flux density D = Q / A = CV /A = Hence, energy stored,

(400 × 10−12 )(100) = 2 × 10−6 C / m 2 200 × 10−4

w f = D 2 / 2ε 0ε r =

(2 × 10−6 ) 2 2(8.85 × 10−12 )(23)

= 0.0983 J/ m3 = 98.3 mJ / m3

15.14 INDUCED E.M.F. AND INDUCTANCE A current flowing in a coil of wire is accompanied by a magnetic flux. If the current changes, so does the flux linkage (i.e., the product of flux and the number of turns) and an e.m.f. is induced in the coil. This is shown in Figure. 15.27. If the turn is connected to an electric circuit, then electric current will flow in it. It will flow in the same direction as long as the magnetic flux increases in the same direction or decreases in the opposite direction. C = N d φ / dt volts Where, df/dt is the rate of change of flux. Inductance is the name given to the property of a circuit, where, by there is an e.m.f. induced into the circuit by the change of flux linkages produced by a current change. e = L di/dt volts where,

di/dt is the rate of change of current.

If a current changing uniformly from o to I amperes produces a uniform flux change from o to f webers in t seconds, then average induced e.m.f., Eav = Nf/t = L I/t, from which inductance of coil, L = Nf/I henry. Flux linkage means the product of flux, in webers, and the number of turns, N, with which the flux is linked. Hence,

Flux linkage = N φ

and

L = N φ /I

inductance = Flux linkage per ampere.

(15.40)

308 Electrical Technology N

N

N

N

×

S Maximum flux linking the turn, zero rate of change of flux, thus zero voltage

N

×

S Zero flux linking the turn, maximum rate of change of flux, positive maximum voltage

S Maximum flux, zero rate of change, zero voltage

S S Zero flulx, Maximum flux, maximum rate zero rate of of change change, negative zero voltage maximum voltage

e

+e

π 2

F 0°

90°

π 180°

3π 2

2π 270°

360°

–e One period

Figure 15.27 Induced e.m.f.

15.14.1 Skin Effect When a direct current flows in a uniform conductor, the current will tend to distribute itself uniformly over the crosssection of the conductor. However, with alternating current, particularly if the frequency is high, the current carried by the conductor is not uniformly distributed over the available cross-section, but tends to be concentrated at the conductor surface. This is called skin effect. When current is flowing through a conductor, the magnetic flux that results is in the form of concentric circles. Some of this flux exists within the conductor and links with the current more strongly near the centre. The result is that the inductance of the central part of the conductor is greater than the inductance of the conductor near the surface. This is because of the greater number of flux linkages existing in the central region. At high frequencies, the reactance of the extra inductance is sufficiently large to seriously affect the flow of current, most of which flows along the surface (skin) of the conductor where the impedance is low rather than near the centre where the impedance is high.

15.15 INDUCTANCE DUE TO INTERNAL LINKAGES AT LOW FREQUENCY When a conductor is used at a high frequency, the depth of penetration of the current is small compared with the conductor cross-section. Thus, the internal linkages may be considered as negligible and the circuit inductance is due to the fields in the surrounding space. However, at very low frequency, the current distribution is considered uniform over the conductor cross-section and the inductance due to flux linkages has its maximum value. Let us consider a conductor of radius R, as has been shown in Figure 15.28, carrying a current I amperes uniformly distributed over the cross-section. At all points on the conductor cross-section current density, J = current/area = 1/pR2 amperes/metre2. dr Let us now consider a thin elemental ring at radius r and width dr contained, within the conductor, as shown in Figure 15.28. The r R current enclosed by the ring: dr i = current density × area enclosed by the ring = ( l /pR2)(pr2) = l r2/R2 amperes Magnet field strength, H = Ni/l amperes/metre At radius r, the mean length of the flux path, l = 2pr and N = 1 turn. 1m Hence, at radius r. Hr =

Ni (1) ( Ir 2 / R 2 ) = = Ir / 2π R 2 ampere/metre l 2π r

Figure 15.28 Inductance at Low Frequency

Field Theory 309 and the flux density,

Br = µ 0 µ r H r = µ 0 µ r ( Ir / 2π R 2 ) tesla

Flux f = B × A webers. For a 1 m length of the conductor, the cross-section are A of the element is (dr × 1) m2. Thus, the flux within the element of thickness dr.  µ 0 µ r Ir  φ = (δ r ) webers 2   2π R  The flux in the element links the portion π r 2 /π r 2 , i.e., r 2 /R 2 of the total conductor. Hence, linkages due to the flux within radius r

µ 0 µ r I r3  µ 0 µ r Ir  r2 = = δ r δ r weber turns  2 2 2π R 4  2π R R Total linkages per metre due to the flux in the conductor =

R

∫0

µ 0 µ r I r3 2π R

4

dr =

µ0µ r I 2π R

4

R

∫0

r 3 dr

R

µ0µ r I  r4  µ 0 µ r I  R4  = =     weber turns 2π R 4  4  2π R 4  4  0 =

1  µ0µ r    4  2π 

µ henry/metre 8π

or

(15.41)

It is seen that the inductance is independent of the conductor radius R.

15.16 INDUCTANCE OF A PAIR OF CONCENTRIC CYLINDERS dr

The cross-section of a concentric cable (coaxial cable) is shown in Figure 15.29. Let a current I amperes flow in one direction in the core and current I amperes flow in the opposite direction in the outer sheath conductor. Consider an element of width dr at radius r, and let the radii of the inner and outer conductor be a and b, respectively as shown. The magnetic field strength at radius r Ni (1)( I ) I Hr = = = 2π r 2π r I The flux density at radius r

Br = µ 0 µ r H r =

r a b

Figure 15.29 Cross Section of a Concentric Cable

µ0µ r I 2π r

For a 1 m length of the cable, the flux f within the element width dr is given by

µ0µ r I  µ0µ r I  φ = Br A =  dr webers  (δ r × 1) = 2π r  2π r  This flux links the loop of the cable formed by the core and the outer sheath. Thus, the flux linkage per metre length of the cable is

( ( µ0 µr I ) / 2π r ) δ r weber turns

and total flux linkages per metre = =

b

∫a

µ0µ r I 2π r

µ0µ r I 2π

dr =

µ0µ r I 2π b

 log e r  = a

b

∫a

1 dr r

µ0µ r I 2π

log e b / a weber turns

310 Electrical Technology Inductance per metre length = Flux linkages per ampere µ0µ r I = log e b / a henry/metre 2π At low frequencies, the inductance due to the internal linkages is added to this result. Hence, the total inductance per metre at low frequency is given by 1  µ0 µ  µ0 µ + L=  log ε b / a henry/metre 4  2π  2π (15.42) µ 1  L=  + log ε /  henry/metre 2π  4 Example 15.17 A co-axial cable has an inner core of radius 1.0 mm and an outer sheath of internal radius 4.0 mm. Determine the inductance of the cable per metre length. Solution: µ 1  L=  + log e b / a  H/m 2π  4  =

( 4π × 10−7 ) (1) 2π

( 0.25 + log e 4 )

= 3.27 × 10−7 H/m or 0.327 µ H/m

15.16.1 Inductance of an Isolated Twin Line Consider two isolated, long, parallel, straight conductor A and B, each of radius a metres, spaced D metres apart. Let the current in each be I amperes, but flowing in opposite directions. Distance D is assumed to be much greater than radius a. The magnetic field associated with the conductors is as shown in Figure 15.30. There is a force of repulsion between conductors A and B. Equipotential lines

Lines of force (or streamlines)

B

A

Radius a Radius a

D

Figure 15.30 Magnetic Field Associated with Two Isolated, Long, Parallel Straight Conductors It is easier to analyze the field by considering each conductor separately. At any radius r from conductor A, as seen in Figure 15.31, magnetic field strength, Hr = Ni/l = I/2pr ampere/metre.

Field Theory 311 Flux density, Br = µ 0 µ r H r = ( µ 0 µ r I ) / 2π r telsa The total flux in 1 m of the conductor,

dr

µ0µ r I  µ0µ r I  φ = Br A =  δ r webers  (δ r × 1) = π r 2 2π r  

A

r

B

Since this flux links conductor A once, the flux linkages with conductor A due to this flux, µ0µ r I = δ r weber turns 2π r D There is, in fact, no limit to the distance from conductor A at which Figure 15.31 Magnetic Field of Each a magnetic field may be experienced. However, let R be a very large Conductor radius at which the magnetic field strength may be regarded as zero. Then the total linkages with conductor A due to the current in conductor A is given by µ0µ r I R µ0µ r I R µ0µ r I R ∫a 2π r dr = 2π ∫a dr /r = 2π [ loge r ]a µ0µ r I = log e ( R a ) 2π Similarly, the total linkages with conductor B due to the current in A =

R

∫D

µ0µ r I 2π r

dr =

µ0µ r I 2π

log e

R

D

Now let us consider conductor B alone, carrying a current of −I amperes. By similar reasoning, total −µ 0 µ r I log e R a , and total linkages with conductor A due to the linkages with conductor B, due to the current in B = 2π −µ 0 µ r I µ µ I log e R D . Hence, total linkages with conductor A = 0 r log e D a weber-turn/metre and, total current in B = 2π 2π −µ 0 µ r I log e D /a weber-turns/metre. linkages with conductor B = 2π For a 1 m length of the two conductors Total inductance = flux linkages per ampere  µ0µ r  = 2 log e D / a  henry/metre  2π  =

µ0µ r π

log e D / a henry/metre

(15.43)

The above equation does not take into consideration the internal linkages of each line. From Section 15.14, inductance per metre due to internal linkages = μ/8p henry/metre. Hence, inductance per metre due to internal linkages of two conductors = μ/4p henry/metre. Therefore, at low frequency, total inductance per metre of the two conductors, µ0µ r  1  (15.44) L=  + log e D / a  henry/metre π 4  This is often referred to as the loop inductance. Example 15.18 A single-phase power line comprises two conductors each with a radius of 8.0 mm and spaced 1.2 m apart in air. Determine the inductance of the line per metre length, ignoring inter linkages. Assume the relative permeability μr = 1. Solution:

L= =

µ0µ r π

log e D / a henry/metre

( 4π × 10−7 ) (1) log π

= 4π × 10−7

1.2    −3  × 8 0 10 .   −7 × 20 . 0 10 H/m or 2.0 µ H/m log e 150 = e

312 Electrical Technology

15.17 ENERGY STORED IN AN ELECTROMAGNETIC FIELD For a non-magnetic medium, the relative permeability μr = 1 and B = µ 0 µ r H = µ 0 H , the magnetic field strength H is proportional to the flux density B and the graph of B against H is straight line, as shown in Figure 15.32. When the flux density is increased by an amount dB due to an increase dH in the magnetic field strength, then the energy supplied to the magnetic circuit = area of the shaded strip (in Joules per cubic metre). For a maximum flux density OY in Figure 15.32 the total energy stored in the magnetic field = area of triangle OYX = 1/2 base × height = 1/2 OZ × OY If OY = B teslas and OZ = H ampere/metre, then the total energy stored in a nonmagnetic medium (15.45) w f = 1 2 HB Joules/metre3 Since B = μ0 H, the energy stored w f = 1 2 µ0 H 2 Joules/metre3

(15.46)

Alternatively, H = B/μ0, thus, the energy stored is w f = 1 2 B 2 /µ0 Joules/metre3

(15.47)

X

Y Flux density B

dB

dH 0

Magnetic field strength, H

Z

Figure 15.32 Magnetic Energy in a Non-magnetic Medium

15.18 MAGNETIC ENERGY STORED IN AN INDUCTOR i

vR

R

V vL

L

Establishing a magnetic field requires energy to be expended. However, once the field is established, the only energy expended is that supplied in order to maintain the flow of current in opposition to the circuit resistance i.e. the I 2R loss, which is dissipated as heat. For an inductive circuit containing resistance R and inductance L, as represented in Figure 15.33, the applied voltage V at any instant is given by V = vR + vL V = iR + L di / dt

Miltiplying by i

Vi = i R + Li di / dt Miltiplying by dt Figure 15.33 Magnetic Energy Vi dt = i 2 R dt + Li di (15.48) Stored in an Inductor Vidt is the energy supplied by the source in time dt, i2Rdt is the energy dissipated in the resistance and Lidi is the energy supplied in establishing the magnetic field or the energy absorbed by the magnetic field in time dt seconds. Hence, the total energy stored in the field when the current increases from O to I amperes is given by 2

the energy stored,

Wf =

I

∫0

Wf =

I

 i2  Li dt = L    2 0 1 2 LI Joules 2

(15.49)

L = N f /I , Hence, total energy stored

Wf =

1 N φ I Joules 2

Also H = NI/l, from which N = Hl/I and f = BA. Thus Wf =

or

1  Hl  1   ( BA) I = HBlA Joules 2 I  2

Wf =

1 HB Joules/metre3 2

(15.50)

Field Theory 313 Example 15.19 Calculate the value of the energy stored when a current of 50 mA is flowing in a coil of inductance 200 mH. What value of current will double the stored energy? Solution:

W f = 1/ 2 LI 2 =

1 200 × 10−13 2

(

) ( 50 × 10−13 )

2

Joules

= 2.5 × 10−4 J 2 , or 0.25 mJ, or 250 µ J If the energy stored is doubled then

( 2.5 × 10−4 ) = 12 ( 200 × 10−3 ) I 2 I =

(

 (4) 2.5 × 10−4   200 × 10−3 

(

)

)   

= 70.71 mA

S UM M A RY 1. Electric field effects are due to the presence of charges. 2. Magnetic field effects are due to the motion of charges. 3. The field is a convenient concept in calculating electric and magnetic forces. 4. The ampere is now regarded in the mks system as the fourth fundamental unit. 5. The ampere is the link between the electrical and mechanical systems. 6. Electric and magnetic fields are analogous and can be analyzed by similar procedures. 7. The electric field is a vector field. 8. A vector quantity has magnitude as well as direction. 9. A scalar quantity has magnitude only. 10. Where the field lines are close together E is large, where they are far apart E is small. 11. Electric field lines extend away from a positive charge where they originate and towards a negative charge where they terminate. 12. There can be no current flow between any two points on an equipotential surface. 13. A line of force in an electrostatic field is often termed a stream line. m 14. C = ε 0ε r x farads. n

A farads. d 16. Capacitance between concentric cylinders 2πε 0ε r = farads/metre. log e b / a 15. C = ε 0ε r

17. Field plotting using curvilinear squares provides an alternative method of determining the capacitance between concentric cylinders. 18. Energy stored in an electric field, 1 W f = CV 2 Joules. 2 19. Energy stored per unit volume 1 W f = DE Joules/cubic metre. 2 20. Energy stored in a non-magnetic medium 1 1 1 W f = HB = µ H 2 = B 2 / µ0 Joules/meter 3 . 2 2 2 2 21. In the expression, Vidt = i Rdt + Lidi. Vidt is the energy supplied by the source in time dt, Lidi is the energy supplied in establishing the magnetic field, and i2Rdt is the energy dissipated in the resistance. 1 22. Energy stored in an inductor, W f = LI 2 Joules. 2

M U LT IP LE C H O I C E Q UE S TI O NS ( M C Q ) 1. Electric field effects are due to the (a) Presence of charges

(b) Motion of charges

2. Magnetic field effects are due to the (a) Presence of charges

(b) Motion of charges

314 Electrical Technology 3. The function of ampere as the fourth fundamental unit is to (a) Link electrical and mechanical systems (b) Delink electrical and mechanical systems

4. A line of force in an electrostatic field is often called (a) A stream line

ANSWERS (MCQ) 1. (a) 2. (b) 3. (a)

(b) An equipotential line

4. (a)

5. An equipotential surface in a magnetic field is the surface over which a magnetic pole may be moved (a) Without the expenditure of work or energy (b) With the expenditure of very little work or energy (c) With the expenditure of considerable work or energy

5. (a).

CON V E N TI O NA L Q UE S TI O NS ( C Q ) 1. Explain the meaning of the terms (a) Steamline (b) Equipotential with reference to an electric field. 2. A field plot between two metal plates is shown in Figure 15.34. If the relative permittivity of the dielectric is 2.4, determine the capacitance of a 50 cm length of the system. 3. A field plot for a concentric cable is shown in Figure 15.35. The relative permittivity of the dielectric is 5. Determine the capacitance of a 10 m length of the cable. 4. A coaxial cable has a capacitance of 100 pf per metre length. The relative permittivity at the dielectric is 3.2 and the core diameter is 1.0 mm. Determine the required inside diameter of the sheath. 5. A concentric cable has a core radius of 20 mm and a sheath inner radius of 40 mm. The permittivity of the dielectric is 2.5. Using two equipotential surfaces within the dielectric, determine the capacitance of the cable per metre length by the method of curvilinear squares. Draw the field plot of the cable. 6. Two parallel wires, each of diameter 5.0 mm, are uniformly spaced in air at a distance of 40 mm between centres. Determine the capacitance of a 500 m run of the line. 7. The capacitance of a 300 m length of an isolated twin line is 1522 pF. The line comprises two air conductors that are spaced 1200 mm between centres. Determine the diameter of each conductor. 8. Determine the energy stored in a 5000 pF capacitor when charged to 800 V and the average power developed if this energy is dissipated in 20 μs. 9. A 0.25 μF capacitance is required to store 2 J of energy. Determine the p.d. to which the capacitor must be charged. 10. A capacitor is charged with 6 mC. If the energy stored is 1.5 J, determine (a) the voltage across the plates, and (b) the capacitance of the capacitor.

11. After a capacitor is connected across a 250 V d.c. supply, the charge is 5 μC. Determine (a) the capacitance, and (b) the energy stored. 12. A 500 pF capacitor is charged to a p.d. of 100 V. The dielectric has a cross-sectional area of 200 cm2 and a relative permittivity of 2.4. Determine the energy stored per cubic metre in the dielectric. 13. All of the energy stored in a 0.02 F capacitor with 440 V across it is transferred to a 1.2 H inductor with negligible losses. What is the current in the inductor? 14. An air capacitor of capacitance 0.005 μF is connected to a direct voltage of 500 V, disconnected, and then immersed in oil of relative permittivity 2.5. Find the energy stored in the capacitor before and after immersion, and account for the difference. 15. A capacitor is charged to 40 μC at 100 V. It is then connected to a similar capacitor B of 4 times the plate area of A. Find the charge on each capacitor and the loss in energy. 16. A coaxial cable has an inner core of radius 0.8 mm and an outer sheath of internal radius 4.8 mm. Determine the inductance of 2.5 m length of the cable. Assume that the relative permeability of the material used is 1. 17. Determine (a) the loop inductance, and (b) the capacitance of a 500 m length of single-phase twin line having conductors of diameter 8 mm and spaced 60 mm apart in air. 18. An isolated twin line has conductors of 7.5 mm radius. Determine the distance between centres if the total loop inductance of 1 km of the line is 1.95 mH. 19. A single-phase power line comprises two conductors spaced 2 m apart in air. The loop inductance of 2 km of the line is measured as 3.65 mH. Determine the diameter of the conductors. 20. Determine the value of the energy stored when a current of 120 mA flows in a coil of 500 mH. What value of current is required to double the energy stored?

Field Theory 315

Plate

Plate

Figure 15.35 For CQ 3

Figure 15.34 For CQ 2

ANSWERS (CQ) 2. 23.4 pF 3. 1.66 nF 4. 5.93 mm 5. 200.6 pF 6. 5.014 nF 7. 10 mm 8. 1.6 mJ; 80 W 9. 4 kV 10. 500 V; 12 μF 11. 20 nF; 0.625 mJ

12. 13. 14. 15. 16. 17. 18. 19. 20.

0.147 J/m3 56.8 A Before, 6.25 × 10−8 J; After, 2.5 × 10−8 J A, 8 μC-B, 32 μC-0.0016 J 10.2 μH a: 0.592 mH; b: 5.133 nF 765 mm 53.6 mm 13.6 mJ; 169.7 mA

Single Phase Alternating Voltage and Current

16

OBJECTIVES Direct voltage Instantaneous values

In this chapter you will learn about:   The basic a.c. generator      A.c. values    Phasor diagrams    Mathematical operations on a.c. waveforms    Phase relation between voltage and current in purely resistive circuits   Phase relation between voltage and current in   purely inductive circuits   Phase relation between voltage and current in   purely capacitive circuits    Simple numerical problems on the above topics

v

v2

v1 t1

v

v3 t3

t2

v

t

Alternating voltage Positive mean value Instantaneous value v T

t

Negative mean value

D.c. and A.c. Values

16.1 INTRODUCTION Direct current sources drive current in one direction only. Pure d.c. in addition to maintaining its direction of current flow, also maintains a constant amplitude, as shown in Figure 16.1. Pure d.c. can be either positive going or negative going. If a direct current changes its amplitude while maintaining its direction of flow, it is called fluctuating d.c., which can also be positive going or negative going, as illustrated in Figure 16.2. If a load is connected across a d.c. source, current will flow through it in a pre-determined direction. If it is desired to change the direction of current through the load, the only way to do this is to invert the connections of the source of e.m.f. as shown in Figure 16.3. 0 time

0

Pulsating d.c. +ve

Pure d.c. + ve 0

time (a)

V/I

0 (b)

V/I time (a)

Figure 16.1 Pure d.c. (a) Positive-going (b) Negative-going

Source

Pulsating d.c. –ve

V/I

Pure d.c. – ve

time

(b)

Figure 16.2 Fluctuating d.c. (a) Positive-going (b) Negative-going

Load

Source

Figure 16.3 D.c. Source at Work

Load

Single Phase Alternating Voltage and Current

317

If the source of e.m.f. drives current in alternate directions and, in addition, brings about a change in the instantaneous value of the current, it is called alternating current, as can be seen in Figure 16.4. If a load is connected across such a source of e.m.f. the current through the load will first flow in one direction and then in the opposite direction, as shown in Figure 16.5. 90°

0

0°

360°

180°

Source

Load

Source

Load

270°

Figure 16.4 Alternating Current

Figure 16.5 A.c. Source at Work

Over 90 per cent of the power generated for commercial use is a.c. power. d.c. power has its own specific fields of application in the industry. An additional advantage of a.c. power is that it can be easily transformed and transported with the help of transformers. Compared with the d.c. circuit, the a.c. circuit is complicated by the fact that the continuously changing current magnitude induces back e.m.f.s in any inductance and capacitance present. As a result, opposition additional to conductor resistance is presented to the flow of current, whose rise and fall does not necessarily occur in step with the voltage, i.e., the question of phase has to be considered.

16.2 COMPARISON OF A.C. AND D.C. With a.c. the current reverses direction periodically. The current can assume a variety of waveforms. In each case, a single complete waveform is called a cycle and the number of cycles occurring in one second is called the frequency. This is the rate at which the waveform repeats itself and is measured in cycles per second (c/s) or hertz (Hz). The time taken by a complete cycle is the period T, which is equal to the reciprocal of the frequency. Thus t (seconds) = 1/f (Hz), and f = 1/T (Hz) (16.1) If the average value of the waveform is not zero, it is regarded as a combination of a d.c. component and an a.c. component. The commercial a.c. line voltage has a sine waveform and a frequency of 50 Hz. The sine function (sin θ ) is the result of plotting the sine of the angle θ, measured in either degrees or radians, against time. The radian is the angle subtended at the centre of a circle by an arc equal to the length of the radius. 360° 180° θ (degrees) = × θ (radians) = × θ (radians) (16.2) π 2π 180 degrees = 57.296 ′ = 57° 17 ′ 45 ′′ (16.3) p Sine wave a.c. is of particular importance, since Fourier analysis allows waveforms to be broken down into a series of sine waves that consist of a fundamental component and its harmonics (Figure 16.6). The fundamental frequency is the same as the + waveform’s repetition rate. The second harmonic has a frequency 2f, the third harmonic 3f, and so on. 0 Angular velocity of rotation is measured in radians per second and is denoted by the Greek letter ω (omega). If the angle of rotation is θ radians, then – 5f1 3f1 f1 θ = ωt (16.4) where, t is the time in seconds. During one cycle, a total rotation of 2π radians occurs in time t seconds, so that Figure 16.6 Complex Waveform. Fundamental Plus Third and Fifth Harmonics 2π θ ω= = = 2π f radians per second (16.5) t T 1 radian =

318 Electrical Technology

16.3 THE SINE WAVE With a.c. the cycle starts from zero amplitude, as has been represented in Figure 16.7 and builds to a positive maximum amplitude. The waveforms then declines again to zero, and thus one-half of a cycle, or one alternation, has been produced. For this time interval, the current flow is in one direction. This resembles d.c., except that the alternation of a.c. started at zero built up to a maximum and declined to zero again. After the first alternation, the current flow starts again, but in a direction opposite to that which prevailed for the first alternation. The waveform representation in Figure 16.7 could be for either voltage or current. When the waveform is of one frequency only, each alternation has the same duration and the same amplitude as the other, and each alternation has a gradual rise and fall in amplitude. Any deviations from these three basic characteristics would indicate the presence of frequencies in addition to the basic or fundamental frequency. When the circuit associated with the a.c. supply is purely resistive, the current will coincide in its amplitude changes with that of the voltage. The voltage and current are said to be in phase. One cycle

+ G –

Peak positive amplitude

Positive Zero Negative

One alternation Time

– G +

Peak negative amplitude

Figure 16.7 The Sine Wave

16.4 BASIC A.C. GENERATOR Alternating current can be produced by several devices. One such device, the oscillator, uses resonant circuits. For the generation of high-powered a.c. for distribution to consumers, the a.c. generator is employed. The basic a.c. generator is shown in its basic form in Figure 16.8. Its operations are based on the principles of electromagnetic induction. When a conductor cuts the lines of force, a potential difference is established across the conductor and if the latter consists of a closed circuit, current will flow. Permanent magnet or electromagnet

Magnetic lines of force A S

N B

Slip rings

Wire loop + 0 Carbon contact “brushes”

–

Figure 16.8 Basic a.c. Generator The basic a.c. generator has a magnetic framework which concentrates the magnetic lines of force between an air gap. A standard commercial-type a.c. generator uses an electromagnet which has a coil wrapped around the centre portion of the pole pieces. Figure 16.8 shows only a single loop which forms a coil situated between the magnetic lines of force. The top of this conductor is marked A and the bottom is marked as B. In a commercial generator, there would be many turns of this inductor around a central core known as the armature. The coil terminals end up in two slip rings. The armature is turned by an external motor and, as the coil rotates within the magnetic lines of force, an alternating current is generated. This

Single Phase Alternating Voltage and Current

319

power is picked up by two carbon brushes. These brushes are conductors of electricity and provide a continuous pickup as the slip rings and the armature rotate. The production of a.c. by the generator is illustrated in Figure 16.9. When the armature coil is in the vertical position, as shown in the upper left-hand drawing, the least number of the magnetic lines of force are intercepted. With the coil being stationary, there would be no induced voltage and, hence, none would be present across the slip rings and the external resistor.

Figure 16.9 Generation of One a.c. Cycle Let us assume that the coil is now rotated in a counter clockwise direction. As the coil turns, the A and B horizontal sections intercept more and more magnetic lines of force, since the lines are highly concentrated at the centre area between the north and south poles. Thus, the voltage induced across the inductor will start to rise and will reach its peak when the coil is in a horizontal plane, as shown in the upper right side of Figure 16.9. As the coil continues to turn, it will start moving into the vertical position, but as it does so, less and less magnetic lines of force arc cut and, hence, the amplitude of the voltage declines. When the coil has reached the vertical position again, it has made a 180º turn and is now in the position shown in the drawings at the lower left-hand side. In this position, the output voltage is zero. As the coil continues to turn, it will again produce the gradual rise of voltage, but now in a direction opposite to that for the initial turn. This is so because the A section of the coil is now rotating from the bottom in a counter clockwise position, as shown in the lower left-hand drawing, as opposed to the polarity, which prevailed when the coil was also in a vertical position as shown in the upper left-hand drawing. When the coil now assumes the horizontal position (see the lower right-hand drawing) a maximum voltage is reached in the negative direction, as shown for the 270º rotation point. As the coil continues to turn, it will assume the original vertical position and the voltage will again drop to zero. The coil has now made a 360º turn and the generator has produced one cycle of a.c. Furthermore, when the coils turns at a higher rate of speed within the magnetic lines of force, successive alternations are produced at a frequency which corresponds to the coil rotation.

In a.c., the values of voltage as well as of current are constantly undergoing a complete amplitude change from zero to a maximum at a periodic rate. Thus, a peak value of 100 V of a.c. will not do the same work as would 100 V of d.c. For one cycle of a.c. current, the average value would be zero, because the positive peak value is equal to the negative peak value and because of opposite polarities, the average value would cancel and be equal to zero. The term average value is used by the industry to refer to 0.636 of the peak value of one alternation. The average value is occasionally used for circuits involving rectification of the sine wave. Average value is the mean value taken over one half-cycle, as has been diagrammatically represented in Figure 16.10. The

Sin

16.4.1 A.c. Values

90°

Figure 16.10 Sine-wave Graph

320 Electrical Technology symbols used to denote average values are Eav and Vav for voltages and Iav for currents. For a sine waveform, the average value is equal to 0.637 (=2 / π) of the maximum (peak) value or Eav = 0.637 Emax

(16.6)

Vav = 0.637 Vmax I av = 0.637 I max

For example, if the maximum value of a sine wave current is 10 A, then the average value over one half- cycle is 6.37 A. In other words, the same effect would be produced by a current rising sinusoidally from 0 to 10 A, and falling to zero as by a steady current of 6.37 A. Example 16.1 If the average value of a sine wave voltage is 159 V, what is its peak value? Solution: Vav = 0.637 Vmax Vmax = Vav / 0.637 = 159 / 0.637 = 250 V Root mean square (r.m.s.) or effective value is defined as the square root of the mean value of the squares of the instantaneous values taken over one complete cycle; r.m.s. or effective values of voltage and current are denoted simply by the capital letters E, V, and I, respectively. The r.m.s value is indicated by most forms of a.c. measuring instruments, and it is most generally used for calculations; the appearance in a.c. work of the symbols E, V, and I may always be taken to indicate the effective values.

(

The r.m.s. value of a sine curve is equal to 0.707 = 1 /

)

2 of the peak value or

E = 0.707 Emax V = 0.707 Vmax I = 0.707 I max

(16.7)

This means, for example, that a steady current of 7.07 A will produce or consume the same amount of power (e.g., the same heating effect) as a sine wave alternating current of maximum value 10 A when maintained for an equal time period. The r.m.s. value is indicated in Figure 16.10. The mean value of a sine wave curve taken over a complete cycle is zero, because the sum of all the instantaneous values during one half-cycle is nullified by an equal but opposite sum in the subsequent half-cycle. A simple mean value cannot be used for the complete cycle. All square quantities are positive, so that if all instantaneous values over one complete cycle are squared, then a mean value of these squares can be found. The square root of this mean value will then give an effective value—the root mean square (r.m.s.) value, for the current or voltage. In Figure 16.11, the full line graph represents one cycle of a sine wave current i = Imax sin ωt. By taking a number of instantaneous values, squaring them and plotting the squared values, the dotted graph is obtained: this is the graph of 2

i 2 = ( I max sin ω t )

and it is always positive. The mean value of the dotted curve occurs at the dotted line AB about which the curve will be seen to be symmetrical. This value is still the square of a current. Taking the square root of the magnitude OA gives the full line CD (magnitude OC ) which is the root of the mean of the squares—the r.m.s value.

Figure 16.11 r.m.s. Value

Single Phase Alternating Voltage and Current

321

2

In Figure 16.11, a simple value Imax = 10 A has been taken. The curve for i 2 (10 sin ω t ) reaches a peak value of 100; the mean value for this is by symmetry equal to one half of the peak value of 50. Taking the square root of this gives an r.m.s. of 7.07, i.e., 0.707 times the peak value. The values obtained above for Iav = 0.637 Imax and I = 0.707 Imax are correct only for pure sine waveform. The ratio of r.m.s. to average value is termed the form factor. For a sine waveform, this is Form factor =

r.m.s. 0.707 = = 1.11 average 0.637

(16.8)

16.5 PHASOR DIAGRAMS A clear representation of an alternating quantity can be given on a phasor diagram. This representation depends on the fact that a sinusoidal function can be derived from the circular motion of a rotating vector known as a phasor.

Figure 16.12 Rotating Vectors The alternating voltage is represented by a phasor having a length which corresponds to the peak value Vm. Modern convention allows the phasor to have a length equal to its r.m.s. value and it is understood that in order to obtain the peak value, the former must be multiplied by 2 . The phasor rotates with angular velocity ω about its origin in a counter-clockwise direction. If a phasor is imagined to be stationary, in order to examine its angular position in relationship to other phasors, the diagram is then called a phasor diagram. The projection of the phasor on the vertical axis, for each angle of rotation, gives the corresponding instantaneous value, e.g., v1 at ω t1, v2 at ω t2, v3 at ω t3, and so on. The phasor representing an alternating quantity is sometimes written with an underlined symbol, e.g., V Example 16.2 The peak value of a sine wave is 200 V. What is the effective (r.m.s.) value? Solution: E = 200 × 0.707 = 141.4 V Example 16.3 The root mean square value of an a.c. waveform is 300 V. What is the peak value? Solution: Ep = 1.414 × 300 = 424 V Example 16.4 Illustrate the radian versus angular velocity and explain the process.

322 Electrical Technology Solution: Figure 16.13 is a vector representation of the characteristics of an alternating current. The radius is called the vector arm. For a complete revolution, the radius line would draw a sine wave. The vector arm indicates the constant changing of the degree, as well as the instantaneous amplitude. Thus, it is an indication of the velocity of the sine wave. l = 1 radius Radius

+

Amplitude

57.3°

90° 0°

360°

–

180°

270°

Velocity

Figure 16.13 Radians Versus Angular Velocity Circumference = 2π times arc of 1 radian 1º = 0.01745 radian

(16.9)

Example 16.5 The maximum value of I is 20 ma. What is the instantaneous value of current at 60º and 150º? Solution: i = I sin 60° = 20 × 0.866 × 10−3 = 17.32 ma i = I sin 150° = I cos (150° − 90° ) = I cos 60° = 20 × 0.5 × 10−3 = 10 ma Example 16.6 The maximum value of E is 50 V. What is the instantaneous value of voltage at 285º? Solution: cos 285° = − cos ( 285 − 270 ) ° e = 50 × ( − cos 15 )° = 50 × ( − 0.966 ) = − 48.3 V Example 16.7 What is the value of the angular frequency of an alternating voltage with a frequency of 50 Hz? Solution: ω = 2p F , F = 50 Hz

ω = 2p × 50 rad/s = 314 rad/s A phasor diagram, as has been shown in Figure 16.14, gives the same information as a waveform diagram represented in Figure 16.15. The length of the phasor represents the magnitude of the quantity being represented. The indication ω denotes the direction of rotation of the phasor at an angular frequency ω = 2πf.

VM

v 0

π

2π ωt

Positive horizontal reference direction

Figure 16.14 Phasor Diagram

Figure 16.15 Waveform Diagram

Single Phase Alternating Voltage and Current

323

The instantaneous value is given by v = Vm sin ωt with reference to ωt = 0. The zero value of the voltage may occur at a different instant of time, e.g., at an angle φ, which is shown prior to ωt = 0 (Figure 16.16). In this case, v = Vm sin (ωt + φv)

(16.10)

v = VM ·sin (ωt + ϕv)

ω

i = IM ·sin (ωt + ϕi)

vi 0

π

π

2

V

i

3 π 2

+ ϕi

2π

+ϕv

ωt = 0

ωt

ϕv,ϕi

Figure 16.16 (a) In Phase Voltage and Current Waveform (b) φ i = φ v. Voltage and Current in Phase At time t = 0, the phase angle is (ω t + ϕr ) = (0 + ϕr ) = ϕr . In Figure 16.16, the voltage phasor lies at an angle ϕ r to the positive horizontal reference axis. The same relation applies for alternating current also; i = I m sin(ω t + ϕ i ) . When the current and voltage have simultaneous zero values, i.e., φi = φv, they are said to be in phase. This is illustrated in Figure 16.16. In the phasor diagram, the current and voltage phasors are in the same direction or in phase. The alternating voltage and current may not be in phase v = Vm sin (ωt + φv) i = Im sin (ωt + φi)

(16.11)

In such a case, as shown in Figure 16.17, if φv > φi, the voltage has its corresponding zero value prior to the current. The voltage phasor V rotates in advance of the current phasor I. The two phasors are displaced by an angle φ to one another, such that

v i

v ,i

ω

V I

0

ϕ ϕi

π 2

π

3 π 2

2π

ϕ

ϕv ϕi

ωt = 0

ωt

ϕv

Figure 16.17 (a) Out of Phase Voltage and Current Wave Form (b) Voltage Leads Current or Current Lags Voltage

φ = φv – φi

(16.12)

φ being the phase angle or phase difference between current and voltage. The voltage leads the current by an angle φ or the current lags the voltage by an angle φ. Alternating voltage may not always lead the alternating current; it can be the other way also. In this case, it has its corresponding zero value at an angle φ later than that of the current. This is illustrated in Figure 16.18.

324 Electrical Technology

v

v ,i

i

i π

0

ϕ

2

π

3 π 2

ϕi

ϕ

V

ϕi

2π

ωt = 0

ωt

ϕi ϕv

ω

(a)

(b)

Figure 16.18 (a) Out of Phase Voltage and Current Waveforms (b) Current Leads Voltage or Voltage Lags Current

φ = φv – φi (16.13) In this case, the voltage phasor V lags the current phasor I by an angle φ. which can be derived from the difference between the two corresponding zero value points. Since, in this case φi > φv, with the voltage lagging the current, the phase angle is negative. Thus, Φ positive The voltage leads the current or the current lags the voltage Φ negative The voltage lags the current or the current leads the voltage Example 16.8 Find the peak value of an alternating voltage with V = 220 V. What is the peak to peak value? Solution: Vm =

2V =

2 × 220 V

= 316 V Vp–p = +316 V to –316 V

Figure 16.19 For Example 16.8

Example 16.9 50 per cent of max

30° (a)

70 per cent of max

45° (b)

Figure 16.20 For Example 16.9

How is the instantaneous value related to the maximum value when θ = 30° and when θ = 45°. Illustrate the same. Solution: When θ = 30° e = 50 per cent of Em When θ = 45° e = 70 per cent of Em Example 16.10 What is the instantaneous voltage if the frequency is 300 Hz, Em 75 V, and the time 2 m sec from the start of the cycle?

Solution: e = E sin ( t × f × 360° ) = 75 sin ( 0.002 × 300 × 360 ° ) = 75 sin 216 ° = 75  − sin ( 216°° − 180° ) = − 75 sin ( 36° ) = − 75 × ( 0.5878 ) = − 44.085 V

Single Phase Alternating Voltage and Current

325

225 v

125 v 100 v

175 v 150 v 100 v

Figure 16.21 Voltages with Different Amplitudes but in Phase

16.6 ADDITION OF SINUSOIDAL WAVEFORMS On many occasions, two or more waveforms of current or voltage are present simultaneously in circuits. For instance, in audio signal amplifiers, there are a number of signals present, varying in amplitude, frequency and wave shape. Also, such signals may be out-of-phase with each other. The net result in a circuit is usually the formation of a composite signal formed by the point-by-point addition of the instantaneous values of the waveform. In some other circuits, such as the polyphase type, several waveforms may have the same amplitude and frequency but differ in phase relationships. However, there are instances where several waveforms may have the same frequency and phase, and differ only in amplitude. Their combination in a circuit results in a waveform of the same frequency, but with an amplitude which is equal to the sum of the amplitudes of the original waveforms. This is illustrated in Figure 16.22. The values shown are peak values, though the r.m.s. values can also be used. Sine waves of the same phase and frequency occur across resistors in a.c. circuits and such resistors have voltage drops and current relationship, which conform to Ohm’s law. In a pure resistance: vR = iR, v – vR = 0, v = vR, v = vR = Vm sin ωt, (16.14) i = vR /R, i = Vm sin ωt /R, i = Vm /R sin ωt. (16.15) The voltage v, and the current i, pass through their zero values simultaneously. They are in phase, as can be seen in Figure 16.22.

Figure 16.22 Voltage, Current and Phase Relations in a Purely Resistive Circuit When several waveforms have the same frequency but are out of phase with each other, we can no longer employ simple addition of the peak values to find the total value of the composite wave produced. With out-of-phase wave forms, the peaks of voltage (or current) do not occur at the same time. The peak value of the composite waveform will be decreasingly less as the phase difference becomes greater, for example, two equal voltages which are exactly 180° out of phase will produce a zero resultant.

16.6.1 Out-of-phase Waveforms A current I in the coil, as shown in Figure 16.23, produces a magnetic flux, φ, which is proportional to the current. A sinusoidal current brings about the production of a sinusoidal magnetic flux as represented in Figure 16.23. A variation

326 Electrical Technology

Figure 16.23  The Induced Voltage Lags Behind the Current by π/2 Radians or 90° of the current i and with it a variation in the flux φ induces in the coil an e.m.f. E which, according to Lenz’s law, acts in opposition to the changes in current or flux. If the rate of change in current is positive, then the e.m.f. E, is negative and vice versa. At the peak value of current, the rate of change of current is zero. At the zero value of current, the rate of change of current is at its greatest. The e.m.f. varies sinusoidally, but passes through its zero value π/2 radians or 90° later than the current. It lags behind the current by 90°. This is shown in Figure 16.23. In a pure inductance, v + e = B, v = −e, i = I m sin ω t  p v = Vm sin  ω t +   2

(16.16)

For a pure inductance, the applied voltage across it leads the current through it by 90° or the current through it lags the applied voltage across it by 90°. All of these relations are shown in Figure 16.24. In a pure capacitance, When a capacitor is connected to a d.c. source, an electric charge is supplied to it, which produces an electric field between the plates, as can be seen in Figure 16.25(a). The movement of charge, the charging current, becomes zero when the capacitor attains the voltage of the source. If the polarity of the source is reversed, then the direction of current is also reversed and existing charge is then removed. The capacitor discharges to a zero voltage, and, with the direction of current remaining the same, the capacitor becomes charged to a voltage of reversed polarity. Figure 16.24 Voltage and Current Relations When an alternating voltage is applied to a capacitor, it undergoes in a Purely Inductive Circuit periodic charging and discharging, as has been represented in Figure 16.25(b). Movement of charge in alternating directions means that in the conductors, there is a current of varying direction, an alternating current.

(a)

(b)

Figure 16.25 (a) Capacitor Connected to a Direct Current Sources (b) Charge and Discharge with Alternating Current

Single Phase Alternating Voltage and Current

The voltage across the capacitor has its zero value The voltage phasor is displaced by φ = –

327

radians later than the current, the voltage lagging the current by 90°.

radian with respect to the current phasor.

The three situations are given in the form of a summary and illustrated in Figure 16.26.

(a)

(b)

(c)

Figure 16.26 Phase Relations. (a) In Phase (b) Current Lags the Applied Voltage by 90° (c) Current Leads the Applied Voltage by 90° Example 16.11 A coil of 1000 turns is rotated 3600 r.p.m. in a magnetic field having a uniform flux of 0.06 tesla. The average area of each turn is 50 cm2, and the axis of rotation is at right angles to the direction of flux. Calculate: (1) the frequency; (2) its period; (3) the angular velocity, and (4) the maximum value of the generated voltage. Write down a trigonometric expression for the instantaneous voltage and calculate its value when the coil has rotated 7 radians from the position of zero voltage. Solution: 3600 1. frequency, f = = 60 Hz 60 2. period, t =

1 1 = = 16.67 milliseconds f 60

3. angular velocity, ω = 2p × 60 = 377 radians per second 4. maximum voltage, Emax = BAn ω = 0.06 × 50 × 10−4 × 103 × 377 = 113 V Therefore, the instantaneous voltage e = Emax sin ω t = 113 sin 377t = 113 sin ϕ if

ϕ = 7 radians e = 113 sin(7 radians) = 74 V

Example 16.12 A sine wave voltage of peak-to-peak value 24 V is applied across an 80 Ω resistor. Calculate (1) the peak, average (over an alternation), and effective values of the sinewavevoltage, and (2) the values of the peak power and the average power dissipated over the cycle.

328 Electrical Technology Solution: 1.

Peak value = E p − p 2 = 24 2 = 12 V Average value =

2 × 12 = 0.637 × 12 = 7.664 V p

Effective value = 12

2 = 12 × 0.707 = 8.49 V

( 12 V ) 144 E 2W = = = 1.8 W R 80 Ω 80 2

2.

alternatively,

peak power = average power =

peak power 1.8 = = 0.9 W 2 2

average power =

Er2.m.s. = 0.9 W 80

Example 16.13 A sine wave voltage has an effective value of 160 V. Calculate its peak-to-peak value and its average value over an aternation. Solution: peak-to-peak value =

2 E r .m.s. = 2.828 × 110

= 311 V average value (over an alternation) =

2 2 × Er .m.s. p

= 0.9 × 110 = 99 V Example 16.14

π  Two alternating voltages are represented by e1 = 10 sin ωt and e2 = 15 sin  ωt +  . What is the phase relation between 6  e1 and e2? What are the trigonometrical expressions for e1 + e2. Verify these results by drawing to scale the sine waves for e1, e2, e1 + e2, e1 – e2 and e – e1. Solution: Because

p radians = 30°, e1 lags e2 by 30° or e2 leads e1 by 30° 6 e1 + e2 = 102 + 152 + 2 × 10 × 15 cos 30° sin (ω t + ϕ1 )

where,

 15 sin 30°   = 18.1° j1 = tan −1  10 + 15 cos 30° 

Therefore, e1 + e2 = 24.18 sin (ωt + 18.1°), which leads e1 by 18.1° but lags e2 by 30° - 18.1° = 11.9° e1 − e2 = 102 + 152 − 2 × 1 ×015 cos 30° sin (ωt + ϕ 2 ) where,

 15 sin 30°   = −111.7° j2 = tan −1  15 cos 30° −10 

Therefore, e1 - e2 = 8.07 sin (ωt - 117°), which leads e1 by 68.3° and e2 by 38.3° Note that e2 - e1 = 8.07 sin(ωt + 68.3°), which is 180° out of phase with e1 - e2, The waveforms for e1, e2, e1 + e2, and e2 - e1 are shown in Figure 16.27.

329

Single Phase Alternating Voltage and Current 24.18 v

e1 + e2 15 v

e2

10 v e1

0

π 2ω

8.07 v

π ω

2π ω

2ω

t

e1 – e2

Figure 16.27 For Example 16.14 Example 16.15 e2 – e1

Two alternating voltages are represented by e1 = 10 sin ωt  π and e2 = 15 sin  ω t +   6 Draw to scale a phasor diagram containing e1 and e2 and construct the phasors representing e1 + e2 and e1 – e2. Compare the results with the answers in the previous example. Solution: e1 + e2 = 24.2 sin(ω t + 18° ) e1 − e2 = 8.1 sin(ω t − 112° ) °

e2 − e1 = 8.1 sin(ω t + 68 )

8.1 30° –112°

e2 2

24.

15 18° 10

e1 + e2

e 1 – e2 e1

Horizontal reference line

8.1 e1 – e2

Figure 16.28 For Example 16.15

300 V

50 Ω

within the limits of measurement. The relevant phasor diagram is given in Figure 16.28. Example 16.16

Figure 16.29 For Example 16.16

In a resistive a.c. circuit, a source of 300 V supplies current to a resistance of 50 Ω (Figure 16.29). Calculate the different values of current.

330 Electrical Technology Solution: Applied voltage = 300 V 300 = 6A 50 I p = 1.414 × 6 = 8.484 A

I r . m. s . =

I av = 8.484 × 0.636 A = 5.38 A Example 16.17 An electrically heated oven has, in the heated condition, a resistance of 20 Ω. What is the maximum instantaneous value of the current if the oven is connected to an alternating voltage of 220 V? Solution: I = V R = 220 20 A = 11 A I m = 2 × 11 = 15.6 A Example 16.18 How will you add two voltages in quadrature? Explain with the help of phasor diagrams, with numerical values? Solution: The graphical and vector representations of two out-of-phase 140 V E3 voltages are given in Figure 16.30. Each waveform has a peak 100 V E1 E3 E2 value of 100 V and voltage E2 lags the voltage E1 by 90° (the two 45° voltages are in quadrature). In the vector diagram, the lengths of E2 45° 0 0 lines for E1 and E2 are equal, indicating the respective voltages E1 are also equal. The diagonal gives the resultant. The resultant line is displaced exactly 45º from E1 and from E2. The resultant –100 V can be considered to be the sum of the E sin q values of each –140 V waveform. The composite wave has shifted in phase. In Figure 16.31, we have two waveforms with different Figure 16.30 Adding for Equal Voltages with amplitudes but the same phase difference (in quadrature). Now, a Phase Difference of 90º instead of a square, we form a parallelogram because of unequal lengths. The resultant is nearer to the vector with the greater amplitude (34°) and further away from the vector with the lesser 175 V amplitude (56º). This has been shown in Figure 16.31. 150 V

E3

16.7 ALTERNATE TREATMENT: A.C. VALUES 1.

I half-cycle

1 = 1 T 2

+T /4

∫

−T /4

100 V

E1 E2

2pt I m cos dt T

E2

E3 34° 56° E1

 +T /4  T    si n 2pt   2p  T  −T /4  −p   p I   = m  sin   − sin   2    2  p   2 = I m = 0.637 I m p  2I = m  T

Figure 16.31 Adding Two Unequal Voltages in Quadrature

In many problems, we are interested in the energy-transfer capability of an electric current. By definition, the average value of a varying power p(t) is the steady value of power Pav that in period T would transfer the same energy W = If t +T

Pav T = W =

∫ t

T

p (t ) dt =

∫ 0

pdt then

(16.17)

Single Phase Alternating Voltage and Current

1 Pav = P = T

T

∫

(16.18)

pdt

1 T

T

∫ 0

i

Half-cycle average

0

By convention, P always means average power and no subscript is necessary. If electrical power is converted into heat in a resistance R, then P=

Im

331

pdt =

1 T

T

2

0

T 4

T 2

3T 4

T t

(16.19)

∫ i 2 Rdt = Ieff R 0

where, Ieff is defined as the steady state value of current that is equally effective in converting power. Solving Eq. 16.19, 1 T

I eff =

– T 4

T

∫ i 2 dt = I r.m.s.

(16.20)

0

cos2 ωt=

1.0

Figure 16.32 The Half-cycle Average. The Average Value Over a Cycle is Zero. In Certain Practical Problems, We are Interested in the Half-cycle Average

1 2

(1 + cos 2 ωt)

cos ωt

Mean value of cos2 ωt

0.5

0

2π ωt

π

Figure 16.33 Calculating the Root-mean-square Value of a Sinusoid and Ieff is seen to be the square root of the mean squared value or the root-mean-square current Ir.m.s.. The effective or r.m.s. value of a sinusoidal current can be found from Eq. (16.20), where, i = Im cos (2p/T) t. I r2.m.s.

1 = T

T

∫ 0

I m2

2pt 1 I m2 cos dt = T T 2

T

∫ 0

2   1 + cos 4pt dt = I m  T  2

Therefore, the effective value of a sinusoidal current is I r.m.s. =

Im 2

= 0.707 I m

(16.21)

S UM M A RY 1. An electrical voltage (or current) having an instantaneous value which is constant in magnitude and direction is called direct voltage (or current). 2. An electrical voltage (or current) which periodically varies in magnitude and direction is called an alternating voltage (or current). 3. The maximum instantaneous value of an alternating voltage (or current) is called its peak value or amplitude.

4. The angle which an alternating voltage traverses in a second is called its angular frequency. 5. The ratio of peak value to the r.m.s value is called the peak or crest factor. 6. The full sinusoidal oscillation of an alternating voltage (or current) is called its period. 7. The angle through which an alternating voltage (or current) is displaced, in turn, represents its phase.

332 Electrical Technology 8. A phasor diagram is a representation of an alternating quantity by a rotating vector called a phasor, whose length is equal to the maximum value and which rotates at the angular frequency of the alternating quantity. 9. The r.m.s or effective value for an alternating voltage (or current) produces the same heating effect as the equivalent d.c. value.

10. The ratio of the r.m.s value to the mean (average) value of an alternating voltage (or current) is called form factor. 11. The number of periods of an alternating quantity which occur in each second is called its frequency. 12. A waveform diagram is the representation of the instantaneous value of an alternating quantity in relation to angle or to time.

M U LT IP LE C H O I C E Q UE S TI O NS (M C Q ) 1. Which of the following statements regarding the capacitance of a parallel plate capacitor is wrong? (a) The capacitance of a parallel-plate capacitor varies linearly with the area of the plates (b) The capacitance of a parallel-plate capacitor varies inversely with the separation of the plates (c) The capacitance of a parallel-plate capacitor varies with the material between the plates (d) The capacitance of a parallel-plate capacitor varies with the metal of the plates

2. The capacitance of a parallel-plate capacitor does not depend upon (a) (b) (c) (d)

The area of the plates The medium between the plates The distance between the plates The metal of the plates

3. Alternating current is one which changes in (a) Magnitude (b) Direction (c) Magnitude as well as direction (d) None of the above

4. Alternating voltage is (a) (b) (c) (d)

Independent of time Varies inversely as time Varies directly with time Varies sinusoidally with time

5. One complete set of positive and negative alternating quantities is called (a) Time period (c) Amplitude

(b) (d)

Frequency Cycle

6. The frequency of a.c. mains in India is (a) 30 Hz (c) 60 Hz

(b) (d)

50 Hz 100 Hz

7. Alternating current/e.m.f. measuring instruments measure it in (a) r.m.s. value (c) Average value

(b) Peak value (d) Peak-to-peak value

ANSWERS (MCQ) 1. (d) 2. (d) 3. (c) 4. (d) 5. (d) 6. (b) 7. (a) 8. (b)

8. Alternating current is converted to direct current by (a) Transformer (c) Dynamo

(b) (d)

Rectifier Motor

9. Which of the following do not represent an alternating quantity? (a) Sine wave (b) Triangular wave (c) Rectangular wave (d) None of these

10. When two alternating waves attain their peak values simultaneously, they are said to be (a) In phase (c) Out-of-phase

(b) (d)

In quadrature None of these

11. The average value of a half-wave-rectified sine wave is given by (a) Eav / Er.m.s. (c) Eav / EP

(b) Er.m.s. / Eav (d) EP / Eav

(a) 0.637 (c) 1.11

(b) (d)

12. The form factor of a sine wave is equal to 0.7.7 1.0

13. Phasor quantities can be represented in (a) Polar form (b) Rectangular form (c) Trigonometric form (d) All of these

14. When a sinusoidal voltage is applied to a purely inductive circuit, the current (a) (b) (c) (d)

Is in phase with the applied voltage Leads the voltage by 90° Lags the voltage by 90° None of these

15. When a sinusoidal voltage is applied to a purely capacitive circuit, the current (a) (b) (c) (d)

9. (d)

Is in phase with the applied voltage Leads the voltage by 90° Lags the voltage by 90° None of these

10. (a)

11. (b)

12. (c)

13. (d)

14. (c)

15. (b)

Single Phase Alternating Voltage and Current

333

CON V E N TI O NA L Q UE S TI O NS (C Q ) 1. What is alternating current? How is it generated? 2. For an alternating waveform, define (a) Peak value (b) Period (c) Frequency (d) Angular frequency 3. What is meant by r.m.s value? What is its significance? 4. What is meant by phase angle? What is its significance? 5. How is average value calculated? 6. How is r.m.s. value calculated? 7. What is the relation between voltage and current in a (a) Resistor (b) Inductor (c) Capacitor 8. A voltage is given by the equation e = 169 sin (377 t + 0.785). Calculate (a) Maximum voltage, (b) r.m.s voltage, (c) Voltage when t = 100 ms.

ANSWERS (CQ) 8. (a) 169 V (b) 120 V (c) 169.6 V 9. 1.6 V 10. (a) 70.7 V (b) 63.7 V

9. A coil has 400 turns. If the flux in the coil changes from 0.2 mWb to 1.0 mWb in 0.2 second, find the value of the induced e.m.f. 10. The maximum value of a voltage in an a.c. circuit is 100 V. Find its (a) r.m.s value (b) Average value. 11. What are the advantages of alternating current? 12. The equation for an alternating current is i = 42.42 sin 625 t. Determine (a) Its maximum value, (b) Its frequency (c) Its r.m.s. value (d) Its average value, (e) Its form factor.

12. (a) 42.42 A (b) 100 Hz (c) 30 A (d) 11. A (e) 1.16.

Three-phase Circuits and Systems

17

OBJECTIVES In this chapter you will learn about:  Single-phase supply    Three-phase supply    Star connection and IL = IP and VL = √3 VP    Delta connection and VL = VP and IL = √3 IP    Comparing star and delta connections    The Y–D change-over switch    Balanced and unbalanced loads    Advantages of three-phase systems

Three-phase Voltage

17.1 INTRODUCTION When a piece of electrical equipment is plugged into the socket of a normal alternating current supply, it is connected between the terminal of one phase and the neutral wire, as shown in Figure 17.1. Thus, a normal alternating current circuit may also be described as a single-phase circuit. Similarly, a three-phase power consumer is supplied with the terminals of three phases, as shown in Figure 17.2.

Terminal of one phase

Alternating current load

Neutral terminal

Figure 17.1 Single-phase Supply

Terminal of phase 1 Terminal of phase 2 Terminal of phase 3 Neutral terminal

Three-phase electricity consumer

Figure 17.2 Three-phase Supply

The advantage of the three-phase a.c. supply is that it can produce a rotating magnetic field when a set of stationary three-phase coils is energized from the supply. This is the basic operating principle for modern rotating machines and, in particular, the three-phase induction motor. A three-phase electric system may be considered as three single-phase systems with a phase displacement of 2π/3 radians (120°) between them. Almost all power transmission uses the three-phase system. In the three-phase system, electrical energy originates from an alternator, which has three main windings placed 120° apart. A minimum of three wires is used to transmit the energy generated.

17.2 WHY THREE PHASE? The earliest applications for a.c. were for resistive loads, such as electric lamps and electric heaters. For these applications, the electric load is resistive and a single-phase system is more than satisfactory. However, when a.c. motors were developed, it was found that single-phase a.c. supply did not work satisfactorily, because it was unable to provide the starting torque. To make a single-phase a.c. motor self starting, it is fitted with an auxiliary winding.

Three-phase Circuits and Systems 335 The power delivered to a load from a single-phase a.c. source is pulsating, irrespective of the power factor. There is always an a.c. component of double the supply frequency. For small motors and electric lamps, this is not a serious problem. In large motors, it may cause problems such as vibration and noisy operation. The speed of the prime mover that drives a single-phase synchronous generator tends to fluctuate in response to the cyclic variation of the electrical output power of the alternator. Magneto motor force set up by the armature currents (armature reaction) is pulsating. This causes eddy currents in the field winding and field structure, thereby resulting in heating. Three-phase motors have a non-zero starting torque. The output of a three-phase machine is greater than that of a singlephase machine. Though the power in an individual phase is pulsating, the sum of powers delivered by all the phases is constant at every instant of time.

17.3 GENERATING THREE-PHASE VOLTAGE The basic principle used in generating an alternating voltage is that of rotating a wire loop at constant angular speed in a uniform magnetic field, as shown in Figure 17.3. As a result of the rotation, the area of the wire loop through which the magnetic field passes alters continuously, and hence, the magnitude of the magnetic flux through the loop varies in the same way. To generate three-phase voltages, a similar method is used, but with the difference that three wire loops L1, L2, L3, rotate at a constant angular speed about the uniform magnetic field. L1, L2, and L3 are displaced in position with respect to each other, permanently. This is illustrated in Figure 17.4.

Figure 17.3 Generating a Single-phase Alternating Voltage

An alternating voltage is induced in each wire loop. However, since the wire loops are displaced 120° from each other, and a complete revolution (360°) takes one period, the three induced voltages are delayed in time by a third of a period with respect to each other. 1. The three-phase voltages have the same frequency. 2. The three-phase voltages have the same peak value. 3. The three-phase voltages are displaced by onethird of a period in time with respect to each other. 4. At every instant of time, the sum of those voltages v A1 + vB 1 + vC 1 = 0

(17.1)

Figure 17.4 Generating a Three-phase Voltage

Voltage +vA1

v Phase A

1/3T

t1 T

Phase B

1/3T

Phase C

vB1=0

t Time

1/3T

–vC1

Figure 17.5 In a Three-phase Symmetrical Supply System v A 1 + v B1 + vC 1 = 0

This is a characteristic feature of a three-phase symmetrical supply system.

336 Electrical Technology

Rota tion

A rotating magnetic field is easily produced in a three-phase induction motor. In their order of usage, the most common types of polyphase systems are: 1. Three-phase (used for power transmission). 2. Six-phase (used for power rectification). 3. Two-phase (used for power rectification). Figure 17.6 shows the arrangement of windings in a simple a.c. generator. The coils are spaced 120 electrical degrees apart. The voltage diagram shows the relationship of the instantaneous voltages as the rotating field poles turn in the direction indicated. Internal winding connections, as represented in Figure 17.6(c) for a three-phase generator, are arranged so that any of the three or four wires may be brought out. In other words, three-phase windings may be connected either in the star pattern or delta pattern. A1 B2

C2 B1 C1

A

120°

0

B

C

A 90°

180° 270° 360°

B

90° 180° C

A2 120°

0° 120° 240° Start Start Start phase phase phase A B C

Figure 17.6 (a) Generation of Three-phase Electrical Energy, (b) Voltage Diagram, (c) Internal Winding Connections The advantages of three-phase systems apply to both the generation and transmission of electrical energy. A threephase generator may be compared to a gasoline engine. An eight-cylinder engine develops eight small pulses of power as compared to one large surge of power per cycle for a one-cylinder engine. Similarly, a three-phase generator generates energy in three windings per turn, rather than in just the single winding in a single-phase generator. In addition, the generator is actually smaller in physical dimensions rather than a single-phase generator of the same rating. Three-phase generators produce energy more smoothly than single-phase generators and provide for more economical use of space within the frame of the machine. Three-phase transmission saves materials, installation time and maintenance costs.

17.4 PHASE AND LINE VOLTAGES A three-phase network consists of three lines or phases indicated by the capital letters A, B, and C. The return lead of the individual phases consists of a common neutral conductor N. In Figure 17.7, voltmeters are connected between each of the lines A, B, and C and the neutral line N. They indicate the r.m.s. (effective) values between each of the three phases and neutral. These voltages are designated as phase voltages, VAN, VBN, and VCN. The individual phase voltages all have the same magnitude. They are simply displaced from each other by one third of a period in time.

Figure 17.7 Phase Voltages

Three-phase Circuits and Systems 337 In Figure 17.8, vAB has a sinusoidal waveform and the same frequency as the phase voltages. However, vAB has higher peak amplitude since it is compounded from the phase voltages vAN and vBN. The varying positive and negative instantaneous values of vAN and vBN at a particular time produce the instantaneous value of vAB. vAB is the phasor sum of the two phase voltages vAN and vBN. The combination of phasor displaced alternating voltages is called phasor addition. The voltage across phase to phase is called the line voltage. This is illustrated in Figure 17.8. vAB (=vAN +vNB) A

v

B

vAB

vAN

vNB (=–vBN)

vBN

C N

t VAB

Figure 17.8 The Line Voltage There are three combinations of phase voltages to give line voltages in a three-phase system. They are designated VAB, VBC, and VCA. The possibility of combining pairs of phases in a generator is a basic property of three-phase electricity. In Figure 17.9, the phase voltages are shown by the phasors VAN and VBN. The phase difference is expressed as an angle of 120° degrees between the two phasors. The phasor sum of the two phase voltages VAN and VNB can be obtained geometrically and the resultant phasor so obtained is the line voltage VAB. VAB = VAN + VNB (17.2) To obtain the line voltage VAB, the measurement is made from the A terminal through the common point N to the B terminal for a star connection. This has been illustrated in Figure 17.9. VL = 3 ×VP

17.5 STAR CONNECTION

(17.3)

v

vAN vBN

120°

t VAB

VNB

N vBN

30°

vAN 120°

N VBN

30°

120°

VAN

Figure 17.9 In a Three-phase Generating System, the Line Voltage is Always √3 Times the Phase-to-neutral Voltage

If a three-phase load is connected to a three-phase network, there are two basic possible configurations. One is the so-called star connection ( symbol Υ  ). This name comes from the arrangement of the load phase impedances, as has been illustrated in Figure 17.10. The three-phase load is shown as three equal magnitude resistances. From each phase, at any given time, there is a path to the terminal points a, b, c of the equipment, and then through the individual elements of the load resistance. All the elements are connected to one point N; the star point. This star point is connected to the neutral conductor N. The phase currents iA, iB, and iC flow through the individual elements and the same current flows through the supply lines. In a star connected system, the supply line current is equal to the phase current. The voltage across each of the impedances of the star connection is the phase voltage. Thus, for a star connection, the voltage across one of the terminals and the star point is equal to the phase voltage. The voltage and current relationships in a star connection are illustrated in the phasor diagrams in Figure 17.11. The phase voltages are displaced by 120° in phase with respect to each other, as shown in Figure 17.11(a). The line voltages are likewise displaced 120° in phase with respect to each other. Since the loads are purely resistive, the phase currents IP = (I A, I B, I C) are in phase with the phase voltages VP = (VAN, V BN, and V CN), as has been diagrammatically represented in Figure 17.11 (c). In a star-connected system, the supply line currents are also determined by the ratio of the phase voltages to the load resistances.

338 Electrical Technology

VCA

VCN

VCN

VCA

VAB

VCN

120°

120°

N

VAN

N

IB

VAN

VBN

VBN

VAB IC N IA VAN

VBN VBC

(a)

VBC

(b)

(c)

Figure 17.11 Voltage and Current Relationships in a Star-connected Network

Figure 17.10 Star Connection

Supply line current = phase current The star or wye connection is particularly suited for the A2 A distribution of power and lighting, where one three-phase A 2 B1 6 Phase B transmission line supplies the energy. All three transformers in the star bank share the single-phase load as well as the B2 B B three-phase load. The star-system also provides a ground neuC Two-phase connection tral with equal voltage between each phase and the neutral. 3 5 Three-phase systems are named after the shape of the transformer secondary winding connections. The star or 4 wye system, as shown in Figure 17.13 is shaped as the letter Figure 17.12 Single-phase Connection (Double Star) Y. Star connections are made by tying together the ends of the three transformer windings labelled X2, and bringing this termination out as the neutral line. The remaining three unidentified conductors of the four-wire, three-phase systems labelled in Figure 17.13 as A, B, and C are tied to the three X1 ends, respectively. Alternators are connected in the same manner. 60°

1

A1

Phase A

EL = 3E p

(17.4)

IL = I p

(17.5)

A four-wire transmission line usually originates from a transformer bank, or a generator connected in star. Both lighting and power circuits are connected to the four wire system. These four circuits are served by four wires, and illustrated in Figure 17.14. ILine

x1 A 1Phase

x1

x2 x2

A

A

A Transformer #1 Transformer x2 #3 C

ELine

ELine v

B

v B

x1

EB

Transformer #2

v

EA (phase) v

ELine

From power source

v C

A

EC v

Neutral Ground

Figure 17.13 Star Connection in an Alternator, or Three-phase Transformer Bank

1

N

N A B C

B 2

1

C 2

1

2

Three-phase, four-wire line

Figure 17.14 Three-phase Four Wire System

Three-phase Circuits and Systems 339 A-B-C Power A- Neutral B- Neutral Three-lighting circuits C- Neutral A bank of three transformers can be connected in star, delta, or other three-six-twelve-or eighteen-phase arrangements. Figure 17.14 illustrates a conventional method of connecting three-phase transformers in a three-phase star arrangement. Lines Lines

17.6 DELTA CONNECTION The delta connection—just like the star connection—is used to connect alternators, motors and transformers. The delta connection takes its name from the Greek letter ∆, because of its triangular appearance. The load impedances form the sides of a triangle, as shown in Figure 17.15. The terminals a, b, and c are connected to the supply lines of the A, B, and C phases. The line voltage VC is across the terminals a, b, and c, and, therefore, across each of the load resistors R. In contrast to a star connection, in a delta connection the line voltage appears across each of the load phases. The voltages with symbols VAB, VBC, and VCA are, therefore, the line voltages. The phase currents through the elements in a delta arrangement are composed of IAB, IBC, and ICA. The currents from the supply lines are IA, IB, Figure 17.15 Delta Connection and IC, and one line current divides at the connection to the ∆ to produce two phase currents. The currents in a delta connection are connected via the relationship line current = 3 phase current. The line voltages VAB, VBC, and VCA are directly across the load resistors, and in this case, the phase voltage is the same as the line voltage. The phasors VAB, VBC, and VCA, as seen in Figure 17.16 (a), are the line voltages. Because of the purely resistive load, the corresponding phase currents are in-phase with the line voltages, as illustrated in Figure 17.16 (b). Their magnitudes are determined by the ratio of the line voltage to the resistance R. The line currents IA, IB, and IC are now compounded from the phase currents. A line current is always given by the phasor sum of the appropriate phase currents, as shown in Figure 17.16 (c) The line current IA is the phasor sum of the currents IAB, and IAC, for example. Thus, in case of a balanced delta connection, the ratio of the line current to the phasor current is √3. (17.6)

line current = 3 phase current. VCA

VAB 120°

VCA

VAB

ICA

IAB

VCA

IC IAB

ICA

VAB

IA IBC VBC (a)

VBC (b)

IBC

IB

VBC (c)

Figure 17.16 Voltage and Current Relationship in a Delta–connected Network Example 17.1 A star-connected load, consisting of resistors, each of 10 Ω, is connected to a three-phase network with line voltage VL = 380 V, as shown in Figure 17.17. What is the magnitude of the supply line currents?

340 Electrical Technology Solution: The magnitude of each supply line current is: I A = I B = IC =

VP 220 = = 22 A R 10

Example 17.2 A delta-connected balanced load (i.e., the resistance of each load is the same) is given in Figure 17.18. What are the values of the line currents? Solution: The delta line currents are: I A = I B = I C = 3 × 38 = 1.73 × 38 = 66 A

Figure 17.17 For Example 17.1

Figure 17.18 For Example 17.2

To make a delta connection, connect the beginning of one phase to the end of the next phase until the last and final connection is to be closed. The delta connection should not be completed until the voltage across the last two ends, C2–A1 (Figure 17.19), is measured. If the voltmeter reads zero across C2–A1, the circuit may be closed. If the voltmeter reads twice the voltage of the phase winding, one should reverse any phase and retest. If a potential still remains across C2–A1, one should then reverse a second phase and make a final voltage test before completing the delta connection. The phase windings must have potentials 120 electrical degrees apart. 1

Line

2Phase

A

A

V C2 A1

C C1 B2

B

A

A

V

V ELine

1Phase

C2

B

A1

The delta connection may be closed when the voltmeter reads zero across C2 –A1. Be sure there is potential across each winding

A2 B1

V ELine

V ELine

C1

A2

C

Figure 17.19 Delta Connection in an Alternator (or Three Single-phase Transformers)

Figure 17.20 Test for Completion of the Delta Connection

The delta connection may be used as the source of a three-wire transmission line or even a distribution system. The three-wire delta system is used when the three phase power on three conductors is required. This is shown in Figure 17.21.

Three-phase Circuits and Systems 341 H1

H2 A x1

H1 x2

H2 B x1

H1 x2

H2 C Primary x1

A x2 x1

x2 Secondary

A B C

Three-phase line

x1 B

x2

x1

x2 C

Figure 17.21 Delta Connection for a Transformer Bank

17.7  Y– ∆ CHANGE OVER SWITCH In the star connection in Example 17.1, the line current is 22 A per line. If the same load is switched into a delta connecting (Example 17.2), the load current is 66 A, provided the magnitude of the line voltage is not changed. Thus, for a particular three-phase load, the line current in a delta connection is three times as great as for a star connection for a given line voltage. There are two √3 factors in a delta connection, giving a total multiplying factor of 3 × 3 = 3. ∆-line current = 3 (Y-line current) (17.7) This fact is used in the star-delta changeover switch shown Figure 17.22   (a) Y-∆ Changeover Switch (b)  in Figure 17.22 of the load. Let us take the example of a threeThe Windings of the Three-phase phase motor, which is first of all started up in the Y (star), conMachine with their Terminations nection (lower current, switch to the left) and then switched Separated over (full running current, switch to the right). This is one possible way to start up an asynchronous motor. A special starter is not necessary. The windings of the three-phase motor, with their terminations separated, are also shown in Figure 17.22.

17.8 SUPPLY OF THREE-PHASE ELECTRICAL ENERGY The supply of electrical energy to commercial and domestic consumers is an important application of three-phase electricity. For low-voltage distribution (in the simplest case supply of light and power to buildings), there are two requirements: 1. It is desirable to use conductors operating at the highest possible voltage but with low current in order to save on expensive conductor material. 2. For safety reasons, the voltage between conductor and earth must not exceed 250 V. A voltage distribution system in ∆ connection is, according to the second requirement, only possible with a low line voltage less than 250 V. However, this is contrary to the first criterion. On the other hand, with a star connection, a line voltage of 380 V is available. In this case, there is only 220 V between the supA ply line and the neutral conductor. In this case, the B first criterion is satisfied and, to comply with the sec- C ond requirement, the neutral conductor is earthed. N Although in this configuration a fourth conducFuses tor is necessary, a low voltage distribution using star connection is always cheaper because of the saving Switch in the cross-sectional area of the supply lines. Hence, the most convenient distribution system for electriLight cal energy supply is the 380/220 V four-wire, threeLight phase a.c. system. This offers the possibility of simultaneously supplying three-phase as well as single-phase current Power Light to users. Supply to buildings can be arranged as in Figure 17.23. The individual houses utilize one of the Figure 17.23 Three-phase Four-wire System

342 Electrical Technology phase voltages. A, B, and C to N are distributed in a sequence (light current). However, higher performance loads (for example, three-phase a.c. motors) may be fed with the line voltage (heavy current).

17.9 BALANCED AND UNBALANCED LOADS A supply to single-phase users is also possible using any one phase of a four-wire, three-phase system. However, the equipments having different performances can be connected to the individual phases so that the phases will be differently loaded. This means that there will be unbalanced loading of the phase of the four-wire three-phase network. In balanced loading, the line currents IA, IB and IC have the same magnitude. The phasors for these currents for balanced loading are shown in [Figure 17.24 (a) and (b)]. They form a closed triangle, i.e., their phasor sum is zero. However, if the line currents have different magnitudes, because of unequal magnitudes, their phasor sum is not zero. A current then flows back through the neutral conductor. These are undesirable, as can be seen in Figure 17.24 (b). IC IC IA

N N

IA

IN

IC N

IB

IA

N

IB

IB (a)

IA IC IB

(b)

Figure 17.24 (a) Balanced Current (IN = 0) (b) Unbalanced Currents (IN ≠ 0) The energy supply authorities, therefore, attempt to plan—as nearly as possible for balance in the network—so as to reduce the neutral current to a minimum. In heavy current electrical engineering, other polyphase systems are used as well as three-phase systems. However, only polyphase systems with up to 17 phases are used and, in practice, it is limited to only those systems with multiples of three phases. Terms, Symbols, UUnits, and Formulae are given in Table 17.1. Example 17.3 A three-phase system has a peak voltage per phase of 200. What are the instantaneous values of voltages when E1 is at the 45º angle? Solution: E1 e1 = 141.4 V

E2

E3

sin 45° = 0.707 +200 V

e1 = 200 × 0.707 = 141 .4 V e2 = 200 × sin(45 ° − 120 °)

e1

= 200 × si n(−75 °)

e3 = 52 V

e3

= 200 × (−0.97) = −194 .00 V e3 = 200 × sin(45° − 240°) = 200 × sin(− 195°) = 200 × sin(195° − 180°)

e2 = –194 V

e2

Figure 17.25 For Example 17.3

– 200 V

= 200 × sin(15°) = 200 × 0.26 = 52 V

Example 17.4 The peak current in a three-phase system is 30 ma. What is the instantaneous current for each waveform at the 60º phase angle?

Three-phase Circuits and Systems 343 Table 17.1 Terms, Symbols, Units, and Formulae Term

Symbol

Unit

Instantaneous alternating voltage

v

V

Effective or r.m.s. value of alternating voltage

V

V

Peak value of an alternating voltage

Vm

V

Vm =

Instantaneous value of an alternating current

i

A

I =

Effective or r.m.s. value of alternating current

I

A

T =

1 f

Time

t

s

Period

T

s

f =

1 T

Frequency

f

Hz

Resistance

R

Ω

Phase voltages for star load

VAN V  BN VCN

V V V

V VAN  1  AB VBN  = × VBC  3 V VCN   CA

Line voltages for star load

VAB V  BC VCA

V V V

VAN VAB   VBC  = 3 × VBN   VCA  VCN

Line currents for delta load

 I A  I  B  I C  I AB   I BC   I CA

A A A

 I AB I A   I B  = 3 ×  I BC   I C   I CA

A A A

IN

A

IA I AB   1   I BC  = ×  IB  3   I CA    IC

Phase currents for delta load

Neutral conductor current

Formulae

V =

Vm 2

2V Im 2

Solution: i1 = 0.03 × sin 60° = 0.03 × 0.87 = 0.0261 amp = 26.1 ma i2 = 0.03 × sin(60° − 120°) = 0.03 × sin(− 60°) = − 0.0261 amp = − 26.1 ma i3 = 0 .03 sin(60 ° − 2 40°) = 0. 03 × sin(− 180°) = 0.03 sin 0° = 0 amp Example 17.5 In a three-phase four-wire system, the line voltage is 400 V. Non-inductive loads of 17 kW, 10 kW, and 8 kW are connected between the three line conductors and the neutral point as shown in Figure 17.26. Calculate the current in each line.

344 Electrical Technology IA IA

+ VA ∠0° N VC ∠–240°

–

+

12 kW

IN = IA + IB + IC –

–

10 kW

8 kW

+ VB ∠–120° IC

A

IB

IB

N B

IC C

Figure 17.26 For Example 17.5 Solution:

The line currents are

400

VA

=

VB

=

= 230 . 94 V ∠0 ° 3 230.94 ∠ − 120 °

VC

=

230.94 ∠ − 240 °

IA =

12 × 103 = 51.96∠0°A 230.94∠0°

IB =

10 × 103 = 43.3∠120°A 230.94∠120°

IC =

8 × 103 = 34.64∠240 °A 230.94∠240 °

S UM M A RY 1. A normal alternating current circuit may also be described as a single-phase circuit. 2. A three-phase power consumer is supplied with the terminals of the three phases. 3. Three phase a.c. supply is such that it can produce a rotating magnetic field when a set of stationary three phase coils is energized. 4. Alternators have no commutators. 5. Alternators are capable of generating very high voltages. 6. When either an electromotive force or an alternating current passes through one cycle, it passes through 360 electrical degrees. 7. In an alternator, the field magnets rotate and the armature is stationary.

8. In the three-phase system, electrical energy originates from an alternator, which has three main windings spaced 120 degrees apart. 9. A three-phase generator generates energy in three windings per turn. 10. The star or wye connection is particularly suited for distribution of power and lighting where one threephase transmission line supplies the energy. 11. Three phase systems are named after the shape of the transformer secondary winding connections. 12. The delta connection is used to connect alternators, motors, and transformers.

M U LT IP LE C H O I C E Q UE S TI O NS (M C Q ) 1. What is the number of phases in a normal industrial supply system? (a) One

(b) Two

(c) Three

2. What is three-phase electricity? (a) Three-phases electricity is the combination of three phases of alternating current with the same peak value and the same frequency and which are displaced 120° in phase, with respect to each other

(b) Three-phase supply is the triple parallel connection of a single-phase supply

3. Which of the following characteristics does a threephase a.c. supply have? (a) The individual phases have equal instantaneous values of voltage at the same instant in time (b) The individual phases have the same peak values of voltage

Three-phase Circuits and Systems 345 (c) The individual phases have the same effective values of voltage (d) The individual phases have the same frequency (e) The individual phases are 120° displaced relative to each other

4. Which letters are used for the three-phase supply lines of a three-phase system? (a) A, B, C

(b) a, b, c

(c) d, e, f

5. What is meant by phase voltage? (a) It is the voltage across the phases A and B, B and C, and C and A (b) It is the voltage between the phases A or B or C and the neutral conductor N

6. What is meant by the expression line voltage? (a) It is the voltage across phases A and B, B and C, and C and A (b) It is the voltage across phases A, B, or C and the neutral conductor N

7. Which of the following ratios has the value √3? (a) The ratio of the line voltage to the phase voltage in a starconnected load (b) The ratio of the peak voltage to the r.m.s. voltage

8. What is the characteristic of a delta connection? (a) The individual phases of a three-phase load are connected in the form of a star (b) The individual phases of a three-phase load are connected in a triangular or mesh arrangement

10. What is meant by a neutral conductor? (a) It is the conductor connection to the star point of a star connection (b) It is the central lead of a delta connection

11. In a star connection (a) The phase voltage equals the line voltage (b) The phase voltage equals the voltage across an element (c) The phase current equals the supply (line) current

12. In a delta connection (a) The voltage across an element equals the line voltage (b) The voltage across an element equals the phase voltage (c) The phase current equals the supply (line) current

13. What is meant by a balanced load? (a) The load in each element has the same impedance (b) The load is arranged symmetrically

14. How many conductors are required to connect a load as a star-connection to a three-phase system? (a) Two conductors (c) Four conductors

(b) Three conductors

15. How many conductors are required to connect a load as a delta connection to a three-phase system? (a) Two conductors (c) Four conductors

(b) Three conductors

9. What is meant by supply line current? (a) It is the current in a conductor connected to the A, B, or C terminals of a three-phase load (b) It is the current that flows through the elements of a threephase load

ANSWERS (MCQ) 1. (c) 2. (a) 3. (b), (c), (d), (e) 4. (a) 5. (b) 6. (a) 7. (a) 8. (b)

9. (a) 10. (a) 11. (b), (c) 12. (a), (b) 13. (a) 14. (b), (c) 15. (b)

CON V E N TI O NA L Q UE S TI O NS (C Q ) 1. What are the advantages of polyphase over single phase? 2. Illustrate with drawings the delta and star windings of a three-phase generator. 3. What is meant by a balanced load in a polyphase operation?

4. Explain with the help of an illustration what is meant by a four-wire, three-phase system. 5. Explain with the help of a schematic, a delta-connected load across a star-connected supply. 6. Illustrate double subscript notation.

ANSWERS (CQ) 2. Figure 17.27

4. Figure 17.28

5. Figure 17.29

6. Figure 17.30.

346 Electrical Technology Stationary armature and coils

(a)

(b)

Figure 17.27 Three-phase a.c. (a) Delta and (b) Star Connection (CQ2)

Single-phase motors

3-phase gen. (or transformer secondary)

Lamps and appliances

3-phase motor

Figure 17.28 Three-phase Four-wire System (CQ4)

∠ ∠

∠ ∠

∠ ∠

Figure 17.29 A Delta-connected Load Across a Star-connected Supply (CQ5)

Figure 17.30 Double Subscript Notation (CQ6)

18

Complex Algebra OBJECTIVES

Y axis (imaginary values)

In this chapter you will learn about:    To define a complex number    To represent complex numbers    To perform mathematical operations on complex numbers (addition, subtraction, multiplication and division) both in Cartesian and polar coordinates    To represent inductive reactance, capacitive reactance, and impedance in both rectangular and polar notation    To convert from rectangular to polar and vice versa: admittance, conductance, and susceptance    Impedance and admittance triangles    Solution of problems involving complex numbers

1st quadrant

2nd quadrant (+j)

(+)

(–)

X axis (real values)

(–j ) 3rd quadrant

4th quadrant

Real and Imaginary Numbers

18.1 INTRODUCTION Alternating voltage or current is generally represented by a sine waveform, a phasor or a trigonometric expression. Although these representations are adequate for simple series or parallel arrangements, they are too cumbersome for the analysis of more complicated circuits that require the use of network theorems. What is required is a form of algebra that can be applied directly to the solution of a.c. circuits. Such a form of algebra must be capable of taking into account the circuits’ phase relationships by distinguishing between the resistive and reactive elements. This is achieved in complex algebra by the introduction of the j-operator.

18.2 THE J-OPERATOR The trigonometric and geometric aspects of vector solutions can be eliminated and mathematic operations confined solely to the algebraic in order to simplify circuit solutions. The type of algebra that expedites a.c. circuit calculations is known as complex algebra, which refers to a system wherein both real and imaginary numbers are employed. If we multiply any number by –1, we will reverse the sign of the number. If Y we once again multiply that number by –1, we will cause that number to revert A to its original plus sign because multiplication of two numbers having negative signs produces a positive product. Thus, –1 is an operator which reverses the sign of a number without changing its amplitude. We can apply the sign-reversing characteristic of the operator –1 to indicate 180° a counter-clockwise rotation of 180 electrical degrees. This has been illustrated X 180° in Figure 18.1. Thus, by multiplying the vector arm by –1, we have caused it to rotate in a counter-clockwise direction, as shown by the broken circle in Figure 18.1. This change constitutes a complete polarity reversal. Now, if we B once again multiply the same by –1, we will rotate the vector arm by another 180º and bring it back to its original position A, with a positive sign, as shown. Thus, when we multiply the vector arm amplitude twice by the –1 operator, we Figure 18.1 Vector Rotations When get a total rotation of 360º (180º + 180º). Multiplied by –1

M18_AUTH_ISBN_C18.indd 347

8/3/2012 4:07:01 PM

348 Electrical Technology For a.c. calculations, we need an operator that produces a 90º rotation. In that case, when multiplying the vector arm by such a new operator, the vector would rotate counter clockwise only 90º and if we multiply the vector magnitude again by the new operator, we would Y axis get yet another 90º shift. Thus, by two succes(imaginary values) +j sive multiplications by the new operator, we will be able to get two 90º shifts or (90º + 90º) = 180º total shift, the same as when we use the –1 ope2nd quadrant 1st quadrant rator once. Hence, the new operator must be one 4 that produces –1 when multiplied by itself. This j or 1 j2 means that the square of the new operator = –1. (+) X axis (–) The exact number of the operator is found by (real values) taking the square root of –1. But no negative 3rd quadrant 4th quadrant number has a real square root; thus, the number representing −1 is called imaginary. In mathematics, the imaginary number has the sign i, but because this letter also indicates instan3 or –j j taneous current in electricity (and electronics); the Figure 18.2 Quadrants and Signs of Values letter j is used to represent −1.

18.3 INDUCTIVE REACTANCE When a changing current is caused to flow through a coil by applying a voltage, the magnetic lines of force so produced establish an induced voltage which opposes the changing current which produced it. In d.c. this opposition is overcome after five time constants. In a.c., there is a continuous rise and fall at a periodic rate. Thus, the back e.m.f. and opposition is maintained. Moreover, a fixed unit value of inductive opposition is created for a.c. which is known as inductive reactance, XL. Inductive reactance is proportional to the voltage drop across the inductance and the current through it. Hence, XL follows the same basic Ohm’s law and also has the ohm for its unit value. Thus, XL = E I   I = E XL  (18.1) E = IX L  XL depends on the angular velocity (2π f ) as well as the inductance in henrys. X L = 2πf L = ω L

(18.2)

V

XL = 3 Ω

10

where, XL is the inductive reactance in Ω, f is the frequency in cycles per second (Hz), and L is the inductance in henrys. In using complex notation, we assign j values to the reactances, and represent resistive values by numbers +j +j alone. Inductance, with its leading voltage, has a 90º phase difference with respect to resistance and is plotted along the Y-axis. We designate an inductive reactance value by prefixing it with +j. Vector diagrams with +j 8V values are shown in Figure 18.3. Ω 5 127° = Plotting resistance values along the horizontal axis 53° Z 36.8° constitutes representation of real numbers. Plotting + – – R=4Ω + 6V reactance values along the vertical axis is representative of imaginary numbers. –j –j Resistance is the only power consumer and pure (a) (b) reactances have no real power consuming ability. Figure 18.3 Vector Diagrams with +j Values In Figure 18.3 (a) Z = 4 + 3j where, Z is called the impedance Z =

(4) 2 + (3) 2 = 5

θ = tan −1 3 = 36.8° 4 −1 3 Z = tan = 36.8 ° 4

M18_AUTH_ISBN_C18.indd 348

8/3/2012 4:07:04 PM

349

Complex Algebra

18.4 CAPACITIVE REACTANCE A capacitor opposes a change of voltage in contrast to the inductor which opposes a change of current. With the inductor, the opposition created by the back e.m.f. is overcome by the applied voltage, and the opposition declines from an infinitely high value to a zero value in five time constants. With a capacitor, the opposition is initially zero, but opposition builds up rapidly as a greater number of electrons are forced on one plate and an increasing quantity removed from the other. Thus, after five time constants, the applied battery or generator voltage is incapable of forcing any more electrons to the negatively charged-plate, or removing any more from the positively-charged plate. Now the resistance is infinite (assuming a perfect dielectric) and no more current flows. With a.c. applied to a capacitor, however, there is a continual rise and fall of voltage and current at a periodic rate. Hence, a fixed unit value of capacitor opposition is created, which is known as capacitive reactance, XC. The capacitive reactance is proportional to the voltage drop across the capacitor and the current flow to and away from it. The capacitive reactance also follows the same basic Ohm’s law, and also has the ohm for its unit value. Thus, XC = E / I   I = E/ X C  E = IX C 

(18.3)

XC depends the angular velocity of the a.c., as well as the value of the capacitor in farads. X C = 1 (2π f C ) where, XC is the capacitive reactance in Ω, f is the frequency in cycles per second (Hz), and C is the capacitance in farads. The characteristics of time constant curves for inductors are opposite to those for capacitors, and hence the formula for capacitive reactance is the reciprocal of the formula for inductive reactance. 1 1 XC = = ωC (2π f C )

(18.4)

235°

.3

23

61

Z=

µa

59°

(a)

Capacitive reactance is plotted on the Y-axis. We designate a capacitive reactance value by pre-fixing it with –j. Vector diagrams with –j values have been illustrated in Figure 18.4.

(b)

Figure 18.4 Vector Diagrams with –j value

18.5 RECTANGULAR AND POLAR NOTATION Figure 18.5 represents a phasor diagram in which OP = phasor r, OM = phasor x and OQ = phasor z. Then, ON = phasor jx and because phasor OQ = phasor ON + phasor OP, z = r + j x. This is known as rectangular notation because the phasors Imaginary axis r and jx are 90º apart. Since phasor z is specified in terms of N Q z jx two phasors whose order is important (z = 2 + j3 is not the same as 3 + j2), this representation of a phasor is referred to as a complex quantity. The polar method of denoting a phasor is in terms of both Z its magnitude and its direction. The magnitude of z is represented by the length of the line OQ, and the direction is measured by the angle, f, between OQ and the horizontal Horizontal reference reference line in polar notation. r line 0 z = Z ∠φ X M P Real axis The value of f will lie between 0º and 180º. If the phasor lies in the lower two quadrants, the angle is negative, and then z = Z∠−f. Conversions between rectangular and polar notations are: Z =

R

Figure 18.5 Rectangular and Polar Notations

R 2 + X 2 and φ = tan −1 X R

(18.5)

where, R and X are the magnitudes of the phasors r and x. Z is always considered to be positive.

M18_AUTH_ISBN_C18.indd 349

8/3/2012 4:07:09 PM

350 Electrical Technology Also R = Z cos φ ,

(18.6)

X = Z sin φ

18.6 RULES OF COMPLEX ALGEBRA 1. Equating real and imaginary parts Let Z1 = R1 + jX1 and

Z2 = R2 + jX2

If

Z 1 = Z 2,

R1 + jX1 = R2 + jX2

then

R1 = R2

and

X1 = X2

2. Addition and subtraction of phasors If Z1 = R1 + jX1 and

Z2 = R2 + jX2,

Z1 + Z2 = (R1 + jX1) + (R2 + jX2) = (R1 + R2) + j (X1 + X2)

then

and Z1 – Z2 = (R1 – R2) + j (X1 – X2) Rectangular rather than polar notation is used when adding or subtracting phasors. 3. Multiplication of phasors If Z1 = R1 + jX1 and Z2 = R2 + jX2, then Z1 Z2 = (R1 + jX1) + (R2 + jX2) = R1R2 + jR1X2 + jX1R2 + j2X1X2 = (R1R2 – X1X2) +j (R1X2 + R2X1) Using polar notation, Z1 = Z1 ∠ f1 and Z2 = Z2 ∠ f2, then Z1Z 2 = = = =

because j2 = –1

Z1 (cos φ1 + j sin φ1 ) × Z 2 (cos φ 2 + j sin φ 2 ) Z1Z 2 [(cos φ1cos φ 2 − sin φ1sin φ 2 ) + j (sinφ1cos φ 2 + cosφ1sinφ 2 )] Z1Z 2 [cos (φ1 + φ 2 ) + j sin (φ1 + φ 2 )] Z1Z 2 ∠(φ1 + φ 2 )

(18.7)

When multiplying phasors, the magnitudes are multiplied, but the angles are added. It is preferable to use polar notation when multiplying or dividing phasors. If

ET

z =Z

1

2∠ φ

2

Z1 R + jX1 = 1 R2 + jX 2 Z2

1

=8

0

4. Division of phasors. If Z1 = R1+jX1, Z2 = R2 + jX2, then

E

E2

z = Z ∠φ , Z 2 = Z 2 ∠2φ and

=

50

Figure 18.6 For Example 18.1

In order to separate the real and imaginary parts, it is necessary to eliminate j from the denominator. This is done by rationalization, which means multiplying both the numerator and the denominator by the conjugate of the denominator. The conjugate of a phasor is that particular phasor that has the same magnitude but whose angle is opposite in sign but equal in magnitude. Therefore, the conjugate of Z∠f is Z∠–f, and that of R + jX is R – jX. Then,

Z1 ( R + jX 1 )( R2 − jX 2 ) ( R R + X 1 X 2 ) j ( X1 R2 − X 2 R1 ) = 1 = 1 22 + Z2 ( R2 + jX 2 )( R2 − jX 2 ) R2 + X 22 R22 + X 22 In polar form, Z1 = Z1 ∠f1 and Z2 = Z2 ∠f2 so that

M18_AUTH_ISBN_C18.indd 350

Z1 Z (cos φ1 + j sin φ 1 ) = 1 Z2 Z 2 (cos φ2 + j sin φ2)  (cos φ1 + j sin φ1)(cos φ2 − j sin φ2)  Z = 1 ×  Z 2  (cos φ2 + j sin φ2)(cos φ2 − j sin φ2)   (cos φ1 cos φ2 + sin φ1 sin φ2) + j (sin φ1 cos φ2 − cos φ1 sin φ2)  Z = 1 ×  Z2  cos φ22 + sin φ22  Z1  cos(φ1 − φ2) + j sin(φ1 − φ 2)  = ×  1 Z 2  

8/3/2012 4:07:13 PM

Z (cos φ sin φ Z11 = Z Z11 (cos φ11 + φ 11 )) + jj sin = Z Z (cos φ sin φ Z 22 Z 22 (cos φ22 + φ22)) + jj sin Z (cos φ + jj sin sin φ )(ccos os φ φ2 − − jj sin sin φ φ2))  Z1  (cos φ1 + φ1)( = Complex Algebra 351 = Z1 × ×  (cos φ 1 + j sin φ1 )(cos φ2 − j sin φ2 )  Z 22  (cos φ22 + j sin φ22)(cos φ22 − j sin φ22)   (cos φ1 cos φ2 + sin φ1 sin φ2) + j (sin φ1 cos φ2 − cos φ1 sin φ2)  Z Z1  (cos φ1 cos φ2 + sin φ1 sin φ2) + j (sin φ1 cos φ2 − cos φ1 sin φ2)  = 2 2 = Z1 × ×  cos φ + sin sin φ Z 22  cos φ222 + φ222  Z φ11 − φ22)) + φ11 − φ 22))  cos(φ sin(φ Z11 ×  cos( −φ + jj sin( −φ =  = Z ×  1  1 Z2  2

=

Z1 ∠(φ1 − φ2) Z2

(18.8)

When dividing phasors, the magnitudes are divided, but the angles are subtracted. 1 1∠0° 1 = = ∠ − φ° Z Z ∠φ ° Z

(18.9)

Example 18.1 Combine the impedances:

E1 = 80 cos 60° + j80 sin 60° E2 = 50 cos 25° + j 50 sin 25° Solution: 80 cos 60° + j80 sin 60° = 40 + j 69.3 50 cos 25° + j 50 sin 25° = 45 + j 20.0 85 + j889.3 89.3 ,  = 46° (approx.) 85 85 85 ET = = = 121 V cos 46° 0.7

tan  =

Example 18.2 The values for the circuit in Figure 18.7 are 10 + j4. The circuit must be altered by removing C1 and R2, shown within the dotted outlines, and closing the circuit at these places. What are the resulting rectangular coordinates and the circuit impedances? Solution: = 10 + j4 = 3 − j2 = 7 + j6 = 6 7 = 0.8571 40.6 Z = 7 cos = 7 0.7593 = 9.2

Original circuit Removed circuit Resultant tann

Example 18.3

C1

6Ω

R1

L1 R2 3 Ω

2Ω Line a–c

Figure 18.7 For Example 18.2 Example 18.4

Multiply 10∠23º by 4∠11º.

Multiply 36∠45º by 20∠−15º.

Solution:

Solution:

10 × 4 = 40 and 23º + 11º = 34º Product = 40∠34º

M18_AUTH_ISBN_C18.indd 351

7Ω

36 × 20 = 720 and 45° + (−15°) = 30° Product = 720∠30°

8/3/2012 4:07:17 PM

352 Electrical Technology Example 18.5

Example 18.6

Multiply 10 + j2 by 5 + j3.

Multiply 5 + j2 by 2 – j3.

Solution:

Solution:

10 + j 2 5 + j3

5 + j2 2 − j3 10 + j 4 − j15 − j 2 6

50 + j10 j 30 + j 2 6

10 − j11 − j 2 6 = 16 − j11

50 + j 40 + j 2 6 = 44 + j 40

Example 18.8

Example 18.7 Divide 320∠50º by 8∠25º.

Divide 5∠37º by 10∠−30º

Solution:

Solution: 320 8 = 40

5 10 = 0.5

50° − 25° = 25°

37° − (−30°) = 67°

Quotient = 40∠25°

Quotient = 0.5 ∠67°

Example 18.9 Divide 12 + j4 by 3 – j2. Solution:

Numerator = 12 + j 4 Denominator = 3 − j 2 Conjugate of denominator = 3 + j 2 12 + j 4 3 + j 2 36 + j 36 + j 2 8 36 + j 36 − 8 Multiplying by conjugate: × = = 9+4 3 − j2 3 + j2 9 − j2 4 28 + j 36 = 2.2 + j 2.8 13

18.7 ADMITTANCE, CONDUCTANCE, AND SUSCEPTANCE Admittance, conductance and susceptance are the reciprocal functions of impedance, resistance and reactance, used for solving parallel circuits containing several impedances. Conductance is the ability of resistance to pass current. It is the reciprocal of R. Denoted by G, its unit is Siemen. If we have several resistors in parallel, the total circuit conductance becomes the sum of the individual conductances. 1 G= R (18.10) G T = G1 + G 2 + ....... + Gn Example 18.10 Two resistors, have conductances of 0.05 S and 0.02 S. What is the total conductance? What is the total resistance? Solution:

GT = G1 + G2 = 0. 05 + 0. 2 = 0.25 S

Note: The unit of conductance is Siemen RT =

1 1 = GT 0.25

= 4Ω

M18_AUTH_ISBN_C18.indd 352

8/3/2012 4:07:21 PM

353

Complex Algebra

We can also add individual reciprocals of impedances in a parallel circuit in a similar fashion. The reciprocal of impedance is admittance, which is the ability of a circuit to pass a.c. It is denoted by Y.

1 Z

Y =

(18.11)

In admittance, we are involved with both real and imaginary numbers. To add admittances, we split up each admittance into real and imaginary parts. Z = R + jX R − jX 1 1 = × Y = 1/ Z = R + jX R + jX R − jX R − jX R J = 2 = 2 − 2 2 2 2 R − j X R +X R + X2 Z 2 = R2 + X 2

Since,

Y =

(18.12)

R jX − 2 2 Z Z

(18.13)

R 1 and is equivalent to the conductance (a.c. = 2 R R conductance). The second term − jX / Z 2 is the reactive (imaginary) part of admittance. This is the reciprocal form of reactance, and is known as susceptance. It is the ability of an inductance or capacitance to pass alternating current. Its symbol is B. 1 1  BL = and BC = and  (18.14) XL XC  Y = G − jBL Y = G + jBC  and If the Z value of the first term represents pure resistance, it becomes

The j value is negative for inductors and positive for capacitors. Thus, XC Z2

(18.15)

18.8 IMPEDANCE AND ADMITTANCE TRIANGLES

IX I

E

EG

IR EB EY =I

Vector diagrams for impedance and admittance are shown in Figure 18.8. These are drawn to scale to illustrate the essential reciprocal function between the two. For the impedance triangle, as has been illustrated in Figure 18.8 (a), current is our reference, since this is a constant value for the series circuit. For the admittance triangle, as shown in Figure 18.8 (b), voltage is our reference, since this is a constant value for the parallel circuit.

E

BC =

and

=

XL Z2

IZ

BL =

(a)

(b)

Figure 18.8 Impedance and Admittance Triangles

Example 18.11 For the circuit shown in Figure 18.9, let us solve for admittance, impedance, cosine and angle. Solution: G = From inspection

1 1 = = 0.05 S R 20

A.c. input

R = 20 Ω

XL = 50 Ω

XC = 100 Ω

100 − 50 = 50 Ω = X T ∴ BC =

1 1 = = 0.025 XC 50

Figure 18.9 For Example 18.11

Y = 0.05 + j 0.02 (since, the reciiprocal of − j 50 equals a plus j 0. 02) Polar form: Y = M18_AUTH_ISBN_C18.indd 353

Z =

1

=

1

G2 + B2 = = 18.5 Ω

0.0029 = 0.054 S 8/3/2012 4:07:27 PM

100 − 50 = 50 Ω = X T 1 1 ∴ BC = = = 0.025 354 Electrical Technology XC 50 Y = 0.05 + j 0.02 (since, the reciiprocal of − j 50 equals a plus j 0.02) Polar form: Y =

G2 + B2 =

0.0029 = 0.054 S

1 1 Z = = = 18.5 Ω Y 0.054 B 0.02 tan θ = = = 0.4 G 0.05 θ = 22° cos θ = 0.9272 250 Ω

150 Ω Z1

1450 Ω

300 Ω

1600 Ω Z2

400 Ω

200 Ω Z3 265.5 v a–c

Figure 18.10 For Example 18.12

Example 18.12 For the circuit represented in Figure 18.10, solve for admittance, impedance, total current, and tangent of the angle. Solution: 150 R = 2 2 150 + 2502 R +X = 0.002 S

G1 =

2

−X2 −250 = 2 2 150 + 2502 R +X = −0 .0033 S

B1 =

2

300 = 0.04 S 311 + 1502 150 B2 = = 0.08 S 2 150 + 3002 200 G3 = = 0.001 S 2 200 + 4002

G2 =

2

400 = 0.002 S 200 + 4002 Y = G + jBC − jBL = 0.00566 + j 0.00012 S

B3 = Polar form:

2

Y = 0.00565 S 1 = 177 Ω Y tan θ = 0.0212 Z =

I =

E = 1.5 amp Z

S UM M A RY 1. 2. 3. 4. 5.

Numbers have both quantity and phase angle. The angle of rotation is the operator for the number. The j is usually written before the number. Numbers on the horizontal axis are real numbers. Numbers on the j-axis are called imaginary numbers.

M18_AUTH_ISBN_C18.indd 354

6. 7. 8. 9.

The j-operator indicates 90º rotation from the real axis. 0º = 1, 90º = j, 180º = j2 = –1, 270º = –j, 360º = same as 0º. Complex numbers have a real part and an imaginary part. Conjugate complex numbers have equal terms but opposite signs for the j terms.

8/3/2012 4:07:29 PM

Complex Algebra

355

CON V E N TI O NA L Q UE S TI O NS (C Q ) 1. 2. 3. 4.

What are the rules of complex algebra? Differentiate between real and imaginary numbers. What does the conjugate of a complex number express? Express polar notation.

M18_AUTH_ISBN_C18.indd 355

5. How will you carry out multiplication on numbers in polar notation? 6. How will you carry out division on numbers in polar notation?

8/3/2012 4:07:30 PM

19

Work, Power and Energy OBJECTIVES In this chapter you will learn about:  The definitions of work, power, and energy    Work done by an electric current    Electric heating    Methods of heating rooms    Types of electric heaters    Power in a resistance – (d.c.)    Power in a.c. circuits    Three-phase power    Active and reactive power    Kinetic and potential energy   

Crucible Out

High frequency a.c.

Copper coil

Shield

Cooling water in

Energy conversion

19.1 INTRODUCTION Electrical machines and appliances save a great deal of manual labour. They do the laundry, wash the dishes, transport people, provide light, supply information on call and entertain us. Time is money. The time that machines save us can be used for more pleasant, interesting and creative activities. Our lives would be very dull without these electrical devices which perform our work for us. The heating element in an ordinary radiant electric fire is a length of ordinary resistance wire which becomes raised to a temperature of about 900 °C when current is passed through it. The wire is supported on a fireclay rod or bar or coiled inside a fused mica tube. The wire is made of an alloy of nickel and chromium (nichrome) which resists oxidation in air when red hot. Many domestic electric appliances contain heating elements. The kettle, laundry iron, toaster and electric blankets are some of the appliances which do a lot of our work for us. Whenever electricity is used for raising the temperature of water or other liquids, the element is well insulated and enclosed in a metal tube or sheath. The elements of laundry iron are made of strip instead of wire, so that they are as flat as possible and also present a large surface to conducting away heat into the pole of the iron. Elements of this type are wound on mica and sandwiched between two thin sheets of the same material. When a p.d. is applied to the ends of a conductor, some of the electrons inside it are set in motion by the electric forces. Work is, therefore, done and the electrons acquire energy. The moving electrons form an electric current, and the energy of this current appears in various forms according to the type of circuit of which the conductor forms a part. An electric motor, for example, is designed to transfer as much of the electric energy as possible to mechanical energy of rotation. In the element of an electric fire, the energy of electric current is transferred to internal energy which is then given out in the form of heat. Most of our work is done in the following ways: 1. 2. 3. 4.

In lifting things, as can be seen in Figure 19.1 (overcoming gravity). In getting things to move, as has been illustrated in Figure 19.2 (overcoming inertia). In stopping things (overcoming inertia). In moving things over rough surfaces (overcoming friction).

M19_AUTH_ISBN_C19.indd 356

8/3/2012 4:23:14 PM

Work, Power and Energy

Figure 19.1 A Heavy Load is Lifted with Block and Tackle

357

Figure 19.2 The Wheel Barrow

The unit of work is the joule. One joule of work is done when a force of one newton is exerted through one metre. Work = Force × distance W=F×S Joules = newtons × metres (F stands for directional force)

19.2 WORK DONE BY AN ELECTRIC CURRENT If a p.d. of 1 volt is applied to the ends of a conductor and 1 coulomb of electricity passes through it, the work done is 1 Joule (assuming that the force acts in the direction in which the object is moving). Q = It coulombs W = VQ Joules = VIt V 2t (19.1) Joules R The work done becomes transferred to internal molecular energy in the conductor accompanied by a rise in temperature. Subsequently, this energy may be given out in the form of heat. = I 2 Rt =

19.2.1 Electric Heating There are two qualities of heat which can be measured, namely, intensity and quantity. Temperature is a measure of the intensity of heat, and is recorded in the lower ranges by thermometers. Pyrometers are used for recording higher temperatures. The absolute scale of temperature is in Kelvin (symbol K) where, temperature in kelvin = temperature in °C + 273.15. Heat is a form of energy and, therefore, the same unit is used for quantity of heat as for other forms of energy, namely Joule. Therefore, the quantity of heat required to raise the temperature is given by Q = 4187 mt where, Q is the quantity of heat in Joules m is the mass in kilograms t is the temperature rise in degrees Celsius. Example 19.1 Find the quantity of heat required to raise the temperature of 20 kg of water from 30 °C to 330 °C.

M19_AUTH_ISBN_C19.indd 357

8/3/2012 4:23:20 PM

358 Electrical Technology Solution: Q = 4187 mt = 4187 × 20 × (330 – 30) = 4187 × 20 × 300 = 25 100 000 J The heat required to raise the temperature of 1 kilogram of a material through 1 kelvin is called the Specific Heat Capacity (SHC). For water, the SHC is 4187 J/kg K. Thus, if C is the SHC, then Q = cmt A number of values of SHC for various substances are given in Table 19.1. SHC for copper = 390 J/kg K. Table 19.1 Specific Heat Capacity Substance

Specific Heat Capacity (J/kg K)

Density (kg/m3)

4187 1010 946 390 452 126 1394 444 234 225 2140

1000 1.292 2720 8930 7870 11 340 13 570 8860 10 490 7300 898

Water Air Aluminium Copper Iron Lead Mercury Nickel Silver Tin Oil (transformer)

Example 19.2 Calculate the heat energy required to raise the temperature of: 1. 4.5 litres of water from 15 °C to 100 °C (the mass of 1 litre of water is 1 kg). 2. 0.028 m3 of copper 0 °C to 60 °C. 3. 23 kg of iron from 10 °C to 80 °C. 4. 34 m3 of air from 10 °C to 18 °C. 5. 7 kg of lead from 0 °C to 90 °C. Solution: 1. Joules required = 4187 × 4.5 (100 – 15) = 1 602 000 J 2. = 390 × (0.028 × 8930) × 60 = 5 850 000 J 3. = 45 × 23 × (80 – 10) = 728 000 J 4. = 1010 × (34 × 1.292) × (18–10) = 355 000 J 5. = 126 × 7 × 90 = 79 380 J

19.2.2 Transfer of Heat The different ways by which heat is transferred are conduction, convection and radiation. Conduction is the transfer of heat through a substance, from one part to another, or between two substances in contact.

M19_AUTH_ISBN_C19.indd 358

8/3/2012 4:23:20 PM

Work, Power and Energy

359

In convection, when the air comes in contact with a heated radiator or element in a room, it receives heat from contact with the element. The heated air expands and rises, with cold air flowing in to take its place. Thus, there is a constant upwards flow of air across the heated element. These convection currents give up some of their heat to the colder parts of the room. The room and its contents are gradually heated by this means. A similar action takes place in an electric water heater. All hot objects emit heat by radiation. Radiation, like light, is mainly reflected from a bright polished surface, but almost wholly absorbed by a dark matt surface.

19.3 METHODS OF HEATING ROOMS The amount of heat required for personal comfort in an enclosed space such as a room or office varies considerably with the following criteria: 1. Number of changes of air per hour. 2. Area of windows. 3. The situation of the walls. 4. The exposure of the ceiling. 5. Material with which walls, floors, and ceiling are composed. 6. The outside air temperature. The quantity of heat developed by the heater may be reduced when the desired temperature is reached.

19.3.1 Types of Electric Heater The types of electric heaters include: 1. Open-type radiators or electric fires. 2. Tubular heaters. 3. Convector heaters. 4. Panel heaters. 5. Thermal storage systems. 6. Floor-warming arrangements. 7. Night-storage heaters. Electric Fires or Radiators: Such heaters consist of spiral resistances wound on fireclay formers and working at a luminous temperature of from 1400 °C to 1600 °C. About 50 to 60 per cent of the heat radiated is in the form of radiant heat, while the remainder is spent in warming the air convection currents. The usual sizes are from 600 W to 3 kW. They may be either portable or fixed. Radiant fires are controlled by ordinary switches and are not suitable for thermostatic control. Tubular Heaters: Tubular heaters are steel tubes about 50 mm diameter containing resistance elements arranged on mica, fireclay, or porcelain formers. The normal loading is 200 W per metre length. The tube temperature is about 200 °C, which is attained in about 20 minutes. Tubular heaters in Figure 19.3 may be fixed either as a single tube up to about 6 m in length or in vertical banks. The usual position for fixing the single tube is in the angle of wall and floor. These tubes may also be considered as convection heaters, because very little of the heat is given up as radiation. Tubular heaters are especially suitable for thermostatic control. Convector Heaters: These consist of wound resistance elements contained within a sheet-metal core with inlet and outlet openings or louvres at the bottom and top, respectively. The front and top may be of moulded plastic material for the sake of pleasing appearance. These heaters may be either portable or fixed, and may be handled without fear of burns. Figure 19.4 shows cold air entering the bottom of the heater, passing over the heater elements, and leaving the heater by the top opening as warm air. Panel Heaters: These heaters are made in the form of flat panels in which the resistance wire is embedded. Low-temperature panels operating at temperatures ranging from 30 °C to 65 °C Figure 19.3 Tubular Heater

M19_AUTH_ISBN_C19.indd 359

8/3/2012 4:23:21 PM

360 Electrical Technology E

D

F

C

A

D E

B

Figure 19.4 Cross-section of Convector Heater

may be fixed in the walls or ceiling of a room and plastered over. High temperature panels up to 340 °C are used on walls and ceilings but are not fixed on the wall surface. They are sometimes suspended from the ceiling. The greater part of the heat entering the room from these panels is of radiant form. Loadings are of the order of 430 W per m2 for the low-temperature panels and 6400 W per m2 for the high-temperature panels. Thermal Storage System: Roughly, this system is similar to the coal, coke or oil fired central heating system, except that the heat is supplied to the storage cylinder electrically. There are two types of heater: the immersion heater used on voltages up to about 650, and the electrode type used at higher voltages. The electrode type is only suitable for use on a.c. The electrodes are in direct contact with water, and the current passes through the water between the electrodes. The energy expended in passing current through water is turned into heat and raises the temperature of the water. Electrode heaters may be used either for water heating or steam raising.

Floor Warming: A floor-warming system is, in essence, a number of heating cables embedded in the floor of a room or otherwise contained in the floor space. Their function is to provide slow heat to the concrete or other floor material. The heat from the floor is employed in warming the air in the room, as well as in warming the walls. The electrical power intake is controlled by a thermostat fixed on the room wall in a suitable position. The heater cables are constructed of alloys of nickel, chromium, iron or other resistance materials, and are insulated with heat-resisting insulating materials. Mechanical protection can be by lead, copper, or p.v.c. sheathing. To obtain an even diffusion of the heat, the heating wires should be arranged to reach almost from wall to wall at regular intervals of say 100 mm. Loadings of from 110 to 160 W per m2 of floor are generally satisfactory. Night Storage Heaters: The use of this type of heater for domestic purposes is increasing rapidly, particularly since it is linked with off-peak supplies and, consequently, cheaper running costs. The heater consists of resistance elements encased in insulating tubes, situated between blocks of heat storage material. The whole structure is encased in a metal casing. The charging arrangements are similar to those of the floor-warming systems. The special off-peak circuit includes a timer which controls the automatic charging of the heater, giving a long night charge and a short midday boosting charge.

19.4 HEATING WATER Immersion heaters, as shown in Figure 19.5, consist of a copper container insulated on the outside with a thick layer of heatinsulating material, e.g., granulated cork. The whole thing is then enclosed in a thin steel casing. The cold water inlet feeds the bottom of the tank, and the hot water is drawn from the top. The water is heated by means an immersed heating element. A thermostat is connected in series with the element, to break the circuit when the water reaches the required temperature. As the water heats up, the tube expands, drawing the rod away from the micro switch mechanism and this opens the switch contacts. The microswitch contacts are sometimes provided with a mechanical or a magnetic snap action in order to prevent flutter at the critical temperature. For direct current work, a small capacitor may be connected across the switch contacts to prevent arcing. There are three systems of electric heating of water: instantaneous heaters or electric geysers, local storage, and central storage. Figure 19.5 Modern Form of Immersion In instantaneous heaters, cold water passes through the heater Heater with Rod Type Thermostat and is heated to the required temperature by contact with the sheath of the electric element. Local storage heaters have low electric loading, which is sufficient to raise the temperature of the water contents from cold to, say, 80 °C in a period of about 1 hour. These heaters are very efficient. They are fixed as near as possible to the spot at which the water is required. They are controlled by an automatic thermostat incorporated in the apparatus. Such appliances are usually of the free outlet type. Central storage is the electrical counterpart of the normal domestic coal or cokefired water installation. The water is heated by means of an immersion heater fixed in the hot-water cylinder. The heater may be controlled by a thermostat which will switch the heater into circuit when water is drawn off and the water temperature falls and switch off again when the maximum water temperature is reached.

M19_AUTH_ISBN_C19.indd 360

8/3/2012 4:23:23 PM

361

Work, Power and Energy

19.5 POWER Electrical power that is converted into another form in an electrical device is called active power. An electric circuit invariably contains ohm’s resistances. Electrical power is converted into thermal power in all these resistances. If we consider, as an example, the circuit in Figure 19.6, we find that only part of the electrical power from the source is converted with a motor to mechanical power; the remainder is dissipated in the form of heat. If the source has to supply a given amount of energy per unit of time, in other words, produce a given electrical power, and all of this power is completely converted to another form, as happens in a direct current circuit, we call this power the active power. The amount of this energy is given by the product Pt of the power P and the time t. The symbol of electrical energy is W; the unit is the wattsecond with letter symbol Ws. Electrical energy is, therefore, the quantity that an electricity supplier measures with an electricity meter at the consumer’s premises and for which a bill is submitted. The unit generally used for this purpose is a multiple of the watt-second, namely the kilowatt-hour. (To understand it better, please see Figure 19.7).

Figure 19.6 A Simple Electric Circuit. Only a Part of the Power is Converted to Mechanical Power 1 kWh = 3.6 × 106 Ws The SI unit of energy is the Joule.

P

P

w = Pt

1 J = 1 Ws (19.2) In contrast to a direct-current circuit, in an alternating current circuit, the current and voltage normally vary according to a sine function—their magnitudes change at every instant of time and their direction changes periodically.

0

t

t

Figure 19.7 Watt-second (Ws)

19.6 POWER IN A RESISTANCE If the alternating current circuit contains only a resistance, the current and voltage will be in phase, i.e., they will pass through zero at the same instant and will reach their peak values v and i simultaneously. This is shown in Figure 19.8. We obtain the instantaneous power by multiplying at every instant in time the values at that instant of v and i. If T is the length of the period of voltage (or current) then from zero upto ½ T both v and i are positive and Pt = vi is also positive. From ½ T to T, both v and i are negative, but their product is still positive.

v

Pt

Pt·v,i

i

+Pt max

vˆ

R 0

ˆˆ Pt max = v·i

Pt Pt, v

P = vˆ ˆi

v

iˆ T 2

Tt

2

0

T

T 2

t

Tp

Figure 19.8 Power in a Resistance (P = Pt max 2

M19_AUTH_ISBN_C19.indd 361

or

=

2)

8/3/2012 4:23:27 PM

362 Electrical Technology Example 19.3 By how much does the power absorbed by a resistance rise if the current in it is doubled by raising the voltage across it? Solution: If the current in a resistor is doubled by raising the voltage across it, the power dissipated in the resistance increases four folds. Example 19.4 How much current can a 10 kΩ, 1 watt resistance safely carry? Solution: P = I 2 R watts I = =

P amperes R 1 1 = = 10 milliamperes 10000 100

Example 19.5 A radio consumes 60 watt of electric power. If it is run for five hours daily, find the cost of running it for 30 days. A unit of electricity costs Rs 2. Solution: Power consumed = 50 × 5 × 30 Wh = 9 kWh (units) Cost of 9 units = 2 × 9 = Rs 18 Example 19.6 An electric iron works on 200 volts and consumes 350 watts of electric power. How much current will it draw? Find its hot resistance. If a unit of electricity costs Rs 2, find the cost of running it one hour daily for 30 days. Solution: P = E × I watts I = P/E = 350/200 = 1.75 amp Hot resistance = 200/1.75 = 80/7 Ω = 11.43 Ω Monthly consumption of power = 350 × 30 = 10.5 kWts Cost of electric power = 2 × 10.5 = Rs 21 Example 19.7 A heater draws 2 amperes of current at 220 volts. Calculate the time required to raise the temperature of 880 grams of water from 16 °C to 100 °C. The efficiency of the heater is 90 per cent. Solution: Heat required for boiling water = 880 (100 − 16) calories = (880 × 84) calories 220 × 2 × t calories 4.2 440 × t 90 × 880 × 84 = 100 4.2 t = 784 seconds = 13 min 4 sec

Heat generated =

M19_AUTH_ISBN_C19.indd 362

8/3/2012 4:23:28 PM

Work, Power and Energy

363

19.7 POWER IN A.C. CIRCUITS E When effective values of a.c. pass through a resistor, the amount I of electric energy consumed in watts is the same as it would be E I for d.c. It would seem, therefore, that the standard power formula P =EI would apply. This, however, is not the case, for with a.c. there are many occasions in electricity and electronics where voltage and current are not in phase. Sometimes, the current will lag the voltage, as shown in Figure 19.9 (a), where the positive current build up lags that of the voltage. The degree of such lag may be from a fractional degree to 180°. The voltage may also lag the current, as can be seen in Figure 19.9 (b), again from a fraction of a degree to 180°. Relative amplitudes of voltage and (a) (b) current are not a factor here. Inductors and capacitors affect the phase between voltage and current. When voltage and current are Figure 19.9 Leading (a), and Lagging (b) Voltages not in phase, true power cannot be determined from the simple product EI. This product, EI, is known as the apparent power and it is expressed not in watts, but in volt amperes (VA) or in kilovolt amps (kVA); it is not a true measure of the power consumed in the circuit. Figure 19.10 illustrates the voltage and current graphs for a circuit + containing pure inductance. If the instantaneous values of voltage and + V current are multiplied together—with due respect to sign—the power curve shown shaded is obtained. This curve has alternative positive and A B C D negative values and it is symmetrical about the zero line. The positive 0 I power consumed in building up the magnetic field during one-quarter cycle of the current is restored to the circuit during the following quarter– – cycle (negative power) when the current is falling and the magnetic Power field is decaying. The power consumed by an inductor of negligible Figure 19.10 Power in a Pure Inductance resistance is zero. The field is fully established between 0 and A and completely decayed between A and B. Similar graphs for the circuit with pure capacitance are shown in + + Figure 19.11. The power consumed in charging a capacitor during one quarter-cycle is restored to the circuit during the next quarter-cycle and I the total power absorbed during a cycle is zero. A B C D 0 In the graphs given above, the capacitor is fully charged at the instants denoted by A and C, and fully discharged at the instants v B and D. – – In the a.c. circuit, all the power consumed is due to the current flowPower ing through the resistance and it is equal to the product of the current I and the p.d. VR across the resistance. In Figure 19.12(a), one can Figure 19.11 Power in a Pure Capacitance see the illustrated version of the general case of an a.c. circuit, where power is fed into a load impedance ZÐφ . In the vector diagram that can be seen in Figure 19.12(b), the applied voltage V has two components VR = V cos f (in phase with the current) and VX = V sin f (in quadrature). The p.d. across the resistor is VR = V cos f and the power absorbed in the resistance, i.e., the true power, is

VR I = VI cos f

(19.3)

The graph for the case of a lagging current (f ≠ 90°) is shown in Figure 19.12(c). The power graph (shown shaded) is obtained from the product of the instantaneous values of v and i: this has negative values which are not equal to the positive values. The average power is equal to the magnitude OA, the power graph being symmetrical about the line AB.

(a)

(b)

(c)

Figure 19.12 Power in a.c. Circuit

M19_AUTH_ISBN_C19.indd 363

8/3/2012 4:23:32 PM

364 Electrical Technology The ratio true power/apparent power = watts/volt amps = VI cos f / VI = cos f (for a sine wave) is known as the power factor (p.f.), its value must always lie between ±1 and it is of the utmost significance in the economies of a.c. power supply systems. If the load power factor is low, on account of a preponderance of capacitive or inductive reactance, the supply generator and the transmission system carry heavy currents without a corresponding useful power in the load. Power-factor correction, to bring the phase angle as nearly as equal to zero, is applied to a.c. loads so as to reduce the supply current. For example, in an installation employing induction motors drawing a lagging current, capacitors are connected across the load terminals to reduce the power factor. In capacitors whose resistance is not negligible, the power loss can be expressed in terms of the power factor which is a direct indication of the ratio of resistance to impedance at a given frequency. Example 19.8 In a circuit where the current lags the voltage by 30°, what power is consumed by the load if the voltage is 200 V and the current is 50 ma? Solution: P = VI cos f = 200 × 0.05 × 0.866 = 8.66 W Example 19.9 In a circuit, the voltage lags the current by 54°. The total voltage is 400 V and the current is 0.2 amp. What is the true power consumed? Solution: P = VI cos f = 400 × 0.2 × 0.5878 = 47 W Example 19.10 A coil has an inductance of 0.025 H. When a 100 V 50 Hz potential is impressed across the coil the current flowing is 10 A. What are the resistance and the power factor of the coil? Solution: Z = V/I = 100/10 = 10 Ω Z2 = R2 + ω2L2 R2 = Z2 – ω2L2 2 10 – (100π) 2 × 0.0252 = 100 – 61.5 =38.5 Ω R = 6.2 Ω Power factor = cos f = R/Z = 6.2/10 = 0.62 Example 19.11 An alternating current of 1 A at a frequency of 800 Hz flows through a coil the inductance of which is 2.5 mH and the resistance of which is 5 Ω. What is the p.d. across the coil, the power absorbed in the coil, and the power factor? Solution: Z 2 = R 2 + ω 2 L2 = 52 + 50002 + 2.52 × 10−6 Z = 13.52 Ω V = IZ = 1 × 13 .52 = 13 .52 V P = I 2 R = 12 × 5 = 5 W Power factor = cos φ = R

Z

= 5

13.52

= 0.37

M19_AUTH_ISBN_C19.indd 364

8/3/2012 4:23:33 PM

365

Work, Power and Energy

Example 19.12 A magnetic bell and a capacitor are connected in series across 75 V a.c. terminals, the frequency being 17 Hz. At this frequency, the impedance of the bell is 2000/45° and the capacitor may be regarded as a pure capacitor of 2 µF capacitance. Calculate the current passing through the bell and capacitor and the power factor of the combination. Solution: Z RL = 2000∠45° R = 2000 cos 45° = 1414 Ω X L = 1414 Ω ; X C = 1

ωC

= 10

6

(2π × 17 × 2)

X c = 4680 Ω

Figure 19.13 For Example 19.12

2 = 14142 + (1414 − 4680) 2 Z RLC

Z RLC = 3556 Ω I =V Z = 75 3556 = 0.021 A Power factor = V Z = 1414 3556 = 0.398

19.8 THREE-PHASE POWER

Figure 19.14 Impedance Triangle

The load circuits which contain both resistance and inductance, or both resistance and capacitance, take both active and reactive power because of the phase difference existing between the voltage and current in them. If these two components of power are added geometrically, we can obtain the apparent power. Precisely, the same happens in each phase of threephase systems. Here we have to consider the phase difference f between the voltage and current in each phase. The components of power in three-phase systems follow from the formulae for single-phase a.c. circuits. Single Phase

Three Phase

Units

S = VI

S = 3 VL I L

VA

Active power

P = VI cos f

P = 3 VL I L cos 

W

Reactive power

Q = VI sin f

Q = 3 VL I L sin 

Var

Apparent power

Reactive power is often expressed as the product of active power and tan f. Q = VI sin φ =

P sin φ = P tanφ cos φ

Geometrical addition of active and reactive power gives the apparent power. S=

P2 + Q2

The relationships found in single-phase a.c. circuits apply also to three phase a.c. circuits.  active power P  = = Power factor  apparent power S  reacti ve power Q  sin f = = = Reactive power factor   apparent power S cos f =

(19.4)

In Figure 19.15, three loads are shown as three equal resistances, one in each phase. Three ammeters are connected in series with the resistances to measure the phase currents in them and three voltmeters are connected in parallel with the resistances to measure the phase voltages across them. For a single-phase load PP = VP IP (19.5)

M19_AUTH_ISBN_C19.indd 365

8/3/2012 4:23:39 PM

366 Electrical Technology For a combined load of three resistances in a star connection, the power must be three times as great. P = 3VP I P (19.6) If the quantities VP and IP in the individual phases are replaced by the corresponding line quantities VL and IL respectively, we obtain V P = 3 L I L = 3VL I L (19.7) 3

u

A B C

v

IP 2 A

A IP 1 yx z

V VP2

A IP 3

V VP1 V VP 3

W

Figure 19.15 Three-phase Power with The way in which the load is a Star-connected Load connected has no affect on the formula to be used, assuming that the load is balanced. Three-phase power with a delta connected load has been illustrated in Figure 19.16.

Example 19.13 In a three-phase Y-connected system, the line voltage EL = 208-V and a balanced load draws 20 kW with a power factor of 0.454. What is the line current? Solution: IL =

Figure 19.16 Three-phase Power with a Delta-connected Load

=

P 3EL cos φ 20000 = 12 amp 1.73 × 208 × 0.454

Example 19.14 In a three-phase Y-connected system, the line voltage is 208 V and a balanced load draws 1.5 A of current. What is the power factor if the power consumed is 630 W? Solution: 430 Power factor = 1.73 × 208 × 1.5 = 0.8 Example 19.15 A three-phase system has a line voltage of 440 V and a balanced load of 6000 W. Current is 20 A. What is the load power factor? What is the current for each phase and the power delivered by each phase? Solution: 6000 Power factor = 1.73 × 440 × 20 = 0.394 I NA =

20 = 11.6 amp (approx.) 1.73

P NA = 440 × 11.6 × 0.394 = 2000 W

or

= 6000 3 = 2000 W

19.9 ENERGY In an electric circuit, work is performed whenever a quantity of electricity flows between two points. If the quantity is Q coulombs and the p.d. between the two points is V volts, the work done is equal to the product Q×V

M19_AUTH_ISBN_C19.indd 366

8/3/2012 4:23:44 PM

Work, Power and Energy

and it results in the expenditure of electrical energy equal to QV Joules. Since Q = It, W = VQ = V × I × t Joules   Using the three forms of Ohm’s law, since V = IR  2 W = I × R × t Joules or since I = V R   W = V 2 × t R Joules. 

367

(19.8)

These expressions are all true, provided that V is in volts, I is in amperes and R is in ohms, and t is in seconds. The first two forms are found to be most useful in solving energy problems: the energy may be calculated for the whole or for a portion of the circuit by proper selection of the values V, I, and R. The electron-volt is the energy required to move one electron between two points having a p.d. of one volt: the electron volt = 1.6 × 10–19 Joules. The joule is sometimes known as the watt second so that it is convenient to remember that the energy. W = VIt watt-seconds (19.9) Energy is the capacity for performing work. If a machine or device possesses energy it can, by releasing the whole or part of its energy, perform work conversely; work must be done on a device to impart energy to it. Accordingly, work and energy are interchangeable and are measured in the same units. Energy exists in many forms, the most common being mechanical, electrical and chemical or in the form of heat, light, or sound. Energy may be either kinetic or potential, the former being energy due to motion and the latter due to position or state. An electric motor, when revolving, possesses kinetic energy which is converted to work when the motor is coupled up to drive a machine. Examples of both types of energy are illustrated in Figure 19.17. The unit in which energy is measured is the Joule: it is the energy expended when a force of 1 newton is exerted through a distance of 1 metre. The newton (N), a unit of force is that force which, acting on a mass of 1 kilogram, gives to it an acceleration of 1 metre per second.

19.9.1 Energy Conversion According to the law of Conservation of Energy, the amount of energy in the universe is constant. Energy can neither be created nor destroyed it may, however, change its form. Different forms of energy are interchangeable, and when energy disappears in one form, it reappears in another. The present theory postulates that the mass of a body is due to its energy and that mass and energy are interchangeable. Electrical energy in small quantities is obtained by the release of chemical energy from a voltaic cell. In large amounts, it is produced from the mechanical energy applied to a generator, the mechanical energy, in turn, deriving from the chemical energy of combustion of coal, oil, or petrol, or from the heat energy released by fission of the atomic nucleus. In favourable circumstances, the kinetic energy of wind, as represented in Figure 19.17(a), or potential energy of falling water, as shown in Figure 19.17(b) is used.

(a)

(b)

Figure 19.17 (a) Kinetic Energy and (b) Potential Energy In converting from one form of energy to another, it is not possible to ensure that all the energy appears in the desired form. For instance, in utilizing an electric motor, most of the applied electrical energy appears in the desired form of mechanical energy; some electrical energy, however, is unavoidably used up in producing heat in the machine, light in the sparking which occurs at the brushes, and sound in the characteristic hum of the rotating machine. Electrical energy is used in reversing the magnetic field, and of the total mechanical energy produced some is lost in overcoming friction in the bearings and brushes and air resistance before the useful energy output can be made available. In the design of such converting machinery, the restriction of energy losses to a minimum is the main consideration.

M19_AUTH_ISBN_C19.indd 367

8/3/2012 4:23:45 PM

368 Electrical Technology Example 19.16 An electric lamp is rated at 60 W for a 240 V supply. What is its resistance? If used with a voltage of 210 V, what power would it take, assuming that its resistance remains constant? Solution:

1.

1.

W = V2 R

W = V2 R

= V=259600 W = 240 R = V 2 W = 2402R 60 60 2= 60 960=Ω59600 60 = 960 Ω 2.

2.

W =V

2

2

R = 46 W

2 960 = 44100 960 = V= 44100 = 210 = 2102W960 960 R = 46 W

Example 19.17 A dynamo giving an output of 10 kW is sending a current of 50 A through a resistance. What is (1) the value of this resistance in ohms and (2) the voltage of the dynamo? Solution:

1.

, I 2/ W W = I 2 RW, = I 2RR= R = I 2/ W

1.

2 R = 10000 = 4 50 R =50 10000 Ω2 = 4 Ω

2.

W = VI W = VI

2.

V =W V I ==10000 W I =/ 50 10000/ 50 = 200 V= 200 V Example 19.18 A resistor A is connected in series with parallel resistors B and C, and the combination is joined to the terminals of a 40 V battery. If the watts dissipated in A, B, and C are 8, 6 and 2, respectively, what are the individual values of the resistances? Solution: Since the power in A = 8 W = the power taken jointly by B and C, the joint resistance of B and C must equal that of A, and p.d. across A = p.d. across (B and C) = 20 V for A, Hence, forHence, A,

W = V 2 RW= =20V2 2 RR==S202 R = 8 Ω R = 400 8R= =50400

for B,

for B,

= 50 Ω

for C ,

B 6w C 2w

W = V 2 RW= =20V2 2 6R = 202 6 R = 400 6R= =66400 .67 6Ω = 66.67 Ω

for C ,

A 8w

W = V 2 RW= =20V2 2 2R = 202 2

40 v

Figure 19.18 For Example 19.18

R = 400 2R= =200 400Ω2 = 200 Ω Example 19.19 A circuit consisting of three resistors of 50 Ω, 100 Ω, and 300 Ω, respectively, joined in parallel, is connected in series with a fourth resistor across a 20 V d.c. supply. What is the value of the fourth resistor if the power dissipated in heating the 50 Ω resistor is 2 W? In which of the three parallel resistors is the most amount of heat generated? Solution: Since W = V2/R, for a given p.d. the power is inversely proportional to the resistance; most power is dissipated in the 50 Ω resistor. Example 19.20 A house is equipped with ten 250 V, 50 W lamps, and twenty 250 V, 25 W lamps which are wired in parallel. What is the resistance presented to the supply mains when all lamps are switched on?

M19_AUTH_ISBN_C19.indd 368

8/3/2012 4:23:48 PM

Work, Power and Energy

369

Solution: Resistance of one 50 W lamp at 250 V R = V 2 W = 2502 W = 62500 50 = 12250 Ω Resistance of ten 50 W lamps in parallel = 1250 /10 = 125 Ω Resistance of one 25 W lamp = 62500 / 25 = 2500 Ω Resistance of twenty 25 W lamps in parallel = 2500 / 20 = 125 Ω Joint resistance of 125 Ω and 125 Ω in parallel = 62.5 Ω

Example 19.21 If in the circuit shown in Figure 19.20, the total current drawn from the cell (which has negligible internal resistance) is 1.5 A, what is the resistance of x, and what power is dissipated in each resistor? 1

50

R

3

2w 100

1.5

300

x 1.5 A

2v

20 v

Figure 19.19 For Example 19.19

Figure 19.20 For Example 19.21

Solution: Joint resistance of 3 Ω and 1.5 Ω in parallel = 1 Ω Total value of known resistors = 1 + 1 = 2 Ω Current taken by this path = V R = Z Z = 1 A Current taken by x = 1 .5 − 1 = 0 .5 A Resistance of x = V I = 2 0 .55 = 4 Ω Power in 4 Ω resistor = V 2 R = 22 4 = 1 W Power in 1 Ω resistor = I 2 R = 12 × 1 = 1 W P.D. across 3 and 1.5 Ω resistors = P.D. across 1 Ω resistor = 1 V Power in 3 Ω resistor = V 2 R = 12 3 = 1 3 W Power in 1.55 Ω resistor = V 2 R = 12 1.5 = 2 3 W

S UM M A RY 1. Newton is that force which acting upon a mass of one kilogram gives to it an acceleration of one metre per second. 2. Energy is the capability of doing work. 3. A Joule is the energy expended when a force of 1 newton is exerted through a distance of one metre.

M19_AUTH_ISBN_C19.indd 369

4. 5. 6. 7. 8. 9.

1 Joule = 1 watt-second = V 2 t/R Joules. Power is the rate of doing work or of expending energy. Electrical energy = VIt = I 2Rt = V 2/R Joules. Electric power = VI = I 2R = V 2/R watts. 1 calorie = 4.2 J = 1 Joule = 0.24 cal. 1 Horse-power = 746 W.

8/3/2012 4:23:51 PM

370 Electrical Technology

M U LT IP LE C H O I C E Q UE S TI O NS (M C Q ) 1. For a given time voltage, four heating coils will produce maximum heat when connected. (a) All in parallel (b) All in series (c) With two parallel positions in series (d) With two series positions in parallel 2. The heater elements in an electric iron is made of (a) Constantan (b) Iron (c) Nichrome (d) Tungsten 3. One kwh of electric energy is equal to (a)100 W (b) 860 kcal (c) 735.5 W (d) 4186 J 4. Two electric bulbs have tungsten filament of the same length. If one of them gives 60 W and the other 100 W, then (a) 60 W bulb has thicker filament (b) 100 W bulb has thicker element (c) both the filaments are of the same thickness 5. Two heaters rated at 1000 W, 250 volts each are connected in series across a 250 volts, 50 Hz a.c. mains. The total power drawn from the supply will be (a) 500 W (b) 1000 W (c) 2000 W 6. A 100 W light bulb burns on an average of 10 hours daily for 1 week, The weekly consumption of electricity will be (a) 0.7 units (b) 0.07 units (c) 7 units 7. A heater coil rated at 1000 W 220 V is connected to 110 V line. The power consumed is (a) 500 W (b) 2000 W (c) 290 W 8. A constant voltage is applied between the ends of an metallic wire of uniform area of cross section. The heat is doubled if (a) both length and radius are halved (b) both length and radius are doubled (c) only the length wire is doubled (d) only the radius of the wire is doubled ANSWERS (MCQ) 1. (a) 2. (c) 3. (b) 4. (b) 5. (a) 6. (c) 7. (c) 8. (b) 9. (c) 10. (d) 11. (a) 12. (a) 13. (c) 14. (a) 15. (a)

9. Electric supply is rated at 220 V. In a house, 11 bulbs of 100 W power rating are used. The rating of the fuse should be (a) 0.5 A (b) 1 A (c) 5 A 10. Electric lamps of 60 W each are connected in parallel. The power consumed by the combination will be (a) 30 W (b) 120 W (c) 20 W (d) 180 W 11. The rating of four electric appliances is given here. Washing machine 3000 W, 230 V, Electric lamp 100 W 230 V Room heaters: 1500 W 230 V; Electric fan: 80 W 230 V. Which appliance would require the thickest lead wire? (a) Working machine (b) Electric lamp (c) Room heater (d) Electrical fan 12. An electric heater takes 6 A of current as soon as it is connected to a.c. mains. After a few seconds, the current will become (a) less than 6 A (b) more than 6 A (c) will neither increase nor decrease 13. The unit to measure electric energy is (a) kVA (b) kW (c) kWh 14. Tubular heaters are (a) suitable for thermostatic control (b) not suitable for thermostatic control 15. Night storage heaters (a) include a timer (b) do not include a timer 16. The exchange of energy stored and released in reactive components is on a (a) cyclic basis (b) half cycle basis (c) quarter cycle basis 17. Work and energy are (a) interchangeable (b) not interchangeable 18. Work and energy are (a) measured in different units (b) measured in the same units 16. (c)

17. (a)

18. (b).

CON V E N TI O NA L Q UE S TI O NS (C Q ) 1. Name some types of work performed by electric machinery. 2. How do we determine the force required for moving a certain load? 3. Name the unit of force.

M19_AUTH_ISBN_C19.indd 370

4. 5. 6. 7.

Name the unit of work or energy. Define energy. Name some kinds of energy. What kind of energy drives a hydroelectric scheme?

8/3/2012 4:23:51 PM

Work, Power and Energy

8. What is the connection between heat and energy of movement? 9. What is the unit for thermal energy? 10. How are calories converted into joules? 11. How is power determined? 12. What is the unit of power? 13. Where is heat loss desirable? 14. Where is resistance heat loss undesirable? 15. What is the ratio between used and supplied electrical energy called? 16. What is the object of matching? 17. Where are resistance losses useful? 18. Where are resistance losses undesirable? 19. Where is friction useful? 20. Where is friction harmful? 21. A 220 V electric circuit is protected by a 15 A fuse. What is the greatest power in watts which can be connected without blowing the fuse?

371

22. If 1 MJ is necessary to raise the temperature of 3 litres (very nearly 3 kg of water) from 20 °C to boiling, how long should a 2 kW electric kettle take to achieve the same result? 23. For what period of time must a 50 h.p motor operate to achieve an energy output of 10 MJ, i.e., to raise 1000 kg (1 tonne) to a height above 1000 metre? 24. A weight lifter raises 200 kg to a height of 2 m in 2 s. What is the power output in watts? 25. A resistor is designed to dissipate 1 kW when connected across a 200 V d.c. supply. Two such resistors can be connected either (1) in series or (2) in parallel across the 200 V supply. What will be the power dissipated in each case? 26. Draw a circuit with switches that will give either series or parallel connection of these two resistors as might be used in an electric heater having provision for low- or high-power rating.

ANSWERS (CQ) 21. 3.3 kW 22. 8 min. 20 sec 23. 4

1 2

min. 24. 2 kW 25. 500 W (in series); 2000 W (in parallel). 40

A 40

B 200-v Supply

Figure 19.21 Solution for CQ26 Two Inter Connected Switches are Required to Operate Together

M19_AUTH_ISBN_C19.indd 371

8/3/2012 4:23:53 PM

Power Factor Correction

20

OBJECTIVES In this chapter you will learn about:    The need for power factor correction    Power factor correction    Active and reactive power    Types of compensation    Individual p.f. correction  Group p.f. correction Bank of capacitors for p.f. correction

20.1 INTRODUCTION If the load in an alternating-current circuit has only resistance, it will dissipate an active power P = VIa. If the load is a pure reactance, it will take a reactive power Q = VIr. Only rarely in practice are electrical devices pure resistances or pure reactances. Usually, they are a combination of both, and most often they consist of a resistance and an inductive reactance. This is the case for all electrical devices that need a magnetic field to perform these functions, motors and transformers, for example. Reactive energy is a numerical quantity, being the product of the reactive power Q and time t. Wr = Qt (20.1) The apparent power S of an electrical device is determined from its active power P and reactive power Q as S =

(20.2)

P2 + Q2

The apparent power embraces the entire power spectrum from active power to reactive power, depending on the phase angle between current and voltage. Apparent power is a sinusoidal function with amplitude S = VI

φ = 0°

S = VI = P

0 < φ < 90°

S contains both active power and reactive power

φ = 90°

S equals the reactive power

active power

20.2 THE NEED FOR CORRECTION An a.c. circuit may contain—in place of a resistance—an inductive or capacitive reactance, as shown in Figure 20.1, wherein the current and voltage are in quadrature. With a capacitive reactance, as has been represented in Figure 20.1(a), the current 1 1 3 i leads the voltage by 90º or by t = T/4 on the time scale. The power (vi), therefore, has zeros at T , T , T and T . 4 2 4 Between these zeros, the power alternates between positive and negative peaks, each having the same arithmetic value. The instantaneous power to a capacitor in an a.c. circuit, therefore, varies sinusoidally with twice the frequency of the alternating voltage and does this symmetrically about the time axis, that is, the zero line. From t = 0 to t = 1/4, v and i are positive, Pt = vi is also positive. At t = 1/4, i reverses its direction, although v still keeps the same direction. Pt = –vi, and the capacitor now delivers power to the source. This reversal in the flow of power occurs at every quarter period, at every 5 ms at 50 Hz. A capacitor takes no active power. This is in the case of an ideal (loss free) capacitor only.

M20_AUTH_ISBN_C20.indd 372

8/3/2012 4:31:44 PM

Power Factor Correction

373

Figure 20.1 An Ideal Capacitor and an Ideal Inductor Take No Active Power In the case of a pure inductance, as has been illustrated in Figure. 20.1 (b), the current i lags behind the voltage v by 1 1 1 = t 0= to t , and i are negative. At t = , i reverses its direction, although v still keeps the same 90º or T. From 4 4 4 direction. This reversal in the flow of power occurs at every quarter period. An ideal inductor takes no active power. This has been further elaborated in Figure 20.2. If an a.c. circuit contains both an inductive component and a capacitive component, then the two reactive powers to these components are mutually displaced in phase by 180º. In theory, no electrical energy is converted into another form of energy during the alternating exchange between a source and a reactive element so that no electrical energy is lost. However, the wires between the source Figure 20.2 Absorption and Return of Power in an and the reactive element carry this energy and, hence, Ideal Capacitor and Inductors there is a reactive current flowing between the source and the reactive element. These wires possess ohmic resistance and a fraction of the transmitted power is lost in the ohmic resistance of the lines. This I2R power loss must be drawn from the source. If in an industrial supply system a substantial reactive power is demanded by a consumer, then he will be charged for the continuous exchange of energy in the form of reactive energy because of the associated line losses. Reactive energy is not a physical quantity, but only a mathematical expression, the product of reactive power and time, denoted by Wr (var-second) Since reactive power puts a strain which grows the greater the phase shift is, on the electric supply companies, the latter require their customers to observe a given minimum power factor, usually cos φ = 0.9. The electricity supply companies install reactive energy meters for customers who need a particularly large amount of reactive power in order to register the reactive energy used.

20.3 POWER FACTOR CORRECTION The angle φ, as represented in Figure 20.3, in the power diagram equals the phase shifts between current and voltage. It is usual in power engineering to express a phase shift by means of a power factor. Cos φ is directly linked to the phase shift, as can be seen in Figure 20.3. cos φ = 1 (active power only, φ = 0°) and cos φ = 0 (reactive power on nly, φ = 90°) are possible The easiest way of reducing the inductive reactive power demands is to connect retroactive capacitances parallel to the already existing inductive loads. Depending

M20_AUTH_ISBN_C20.indd 373

Figure 20.3 Power Factor, cos φ = P/S

8/3/2012 4:31:48 PM

374 Electrical Technology on the value of the capacitors’ capacitive reactive power QC, the inductive reactive power QL taken from the main supply is, thus, wholly or partially compensated. This process, called reactive power compensation or power factor correction, has been shown in Figure 20.4. IC =

V = V⋅ω ⋅ I Xc

Q = V ⋅ I C = V ⋅ V ⋅ ωC Figure 20.4 Reactive Power Compensation

= V 2ω C (capacitor reactive power)

(20.3)

Subsequent to compensation, the electricity mains mostly supply active power. The current in the line is reduced and, thus, smaller line losses occur. The expenses charged by the electricity supply companies for the reactive energy consumed can be economized upon. By means of compensation, the reactive power and the current consumed at a constant active power are reduced and the power factor is improved. From the relation Q = V·IC, the capacitor current IC can also be calculated.

20.4 TYPES OF COMPENSATION By connecting capacitances in parallel, the inductances are compensated. This is, therefore, known as parallel p.f. correction. This kind of compensation is the one that is most commonly used, especially in three phase current installations as shown in Figure 20.5. Three kinds of compensation are usual: 1. Individual p.f. correction. 2. Group p.f. correction. 3. Central p.f. correction. In individual p.f. correction, each individual inductive load is directly allocated the necessary caW pacitor. Individual p.f. correction is generally used W in larger loads which have long operating times. In group p.f. correction, several inductive loads with powers and operating times which are as similar as possible are gathered into a group, and this group’s reactive power is compensated by a joint capacitor. Group p.f. correction is generally used for compensating fluorescent tubes. With central p.f. correction, the reactive power of a greater number of inductive loads of varying power and with varying operating times is compensated by a joint bank of capacitors. The leading (capacitor) reactive power required to cover the Figure 20.5 Compensation respective reactive power consumption is automatically switched on or off by a control system.

S UM M A RY 1. If the load is purely resistive, it will dissipate active power. 2. If the load is a pure reactance, it will take reactive power. 3. Reactive energy is a numerical quantity. 4. The apparent power of an electrical device is determined from its active power P and reactive power Q.

M20_AUTH_ISBN_C20.indd 374

5. φ = 0° S = VI = P S contains only active power. 6. 0° < φ < 90° S contains both. 7. φ = 90° S equals reactive power. 8. A pure reactance takes no active power. 9. An impedance is made up of an active and a reactive component.

8/3/2012 4:31:51 PM

Power Factor Correction

375

M U LT IP LE C H O I C E Q UE S TI O NS (M C Q ) 1. Which type of power is formed by the product of current and voltage if the current and voltage are in phase quadrature? (a) Active power (c) Apparent power

(b)

Reactive power

2. In the case of which of the three named loads does a phase shift not occur when connected to an alternating voltage? (a) (c)

Motors Light bulbs

(b)

Capacitors

3. In which unit of measurement is reactive power usually given? (a) W

(b) VA

(c) VA

4. In which lines does most of reactive current flow in compensated loads? ANSWERS (MCQ) 1. (b) 2. (c) 3. (b)

4. (c)

5. (b)

(a) In the lines between the load and the electricity supply company (b) In the lines between the capacitor and the electricity supply company (c) In the lines between the capacitor and the load 5. Which two powers are equal at a power factor cos φ = 1? (a) Active power and reactive power (b) Active power and apparent power (c) Reactive power and apparent power 6. Which requirement is decisive for dimensioning a

capacitor intended for the individual power factor correction of a load? (a) The reactive power required at nominal power (b) The reactive power required in an open circuit

6. (b).

CON V E N TI O NA L Q UE S TI O NS (C Q ) 1. What does power factor signify? 2. Why is power factor correction required?

M20_AUTH_ISBN_C20.indd 375

3. What does power factor correction perform? 4. What are the methods of power factor correction?

8/3/2012 4:31:51 PM

LCR Circuits

21

OBJECTIVES In this chapter you will learn about:           

I nductive and capacitive reactance and impedance Characteristics of electrical parameters  Series-connected impedances  Series a.c. circuits  Polar notation Parallel-connected impedances  Parallel a.c. circuits Problems on series and parallel a.c. (single phase) circuits  Components of current, leading and lagging power factor  LCR circuits  The basis of filtering using reactive components

LCR Circuits

21.1 INTRODUCTION Inductors and capacitors, when used either singly or in combination in a.c. circuits, affect circuit characteristics in a manner far different from that of resistors. The latter offer opposition to the flow of current—whether d.c. or a.c.—and consumes electric energy in proportion to the current flow through them, as also the voltage drop across them. When d.c. is applied to an inductor, the opposition that occurs is of very short duration, and prevails for approximately five time constants. Once the back e.m.f has been overcome, a steady-state current flows which is limited only by the resistance of the coil. When d.c. is applied to a capacitor, the opposition builds up from zero to an infinite value within approximately five time constants. After that, there is no current flow unless the dielectric material is defective and has some leakage resistance. When pure inductors and capacitors are used with a.c., however, they offer opposition to the current flow in a fashion similar to the resistor, but do not consume power. Since inductor, capacitor, and resistor combinations are frequently used in all branches of electricity and electronics, their circuit behaviour must be clearly understood so that their characteristics can be fully and properly utilized.

21.2 INDUCTIVE REACTANCE When a changing current is caused to flow through a coil by applying a voltage, the magnetic lines of force produced by the changing current establish an induced voltage which opposes the changing current that produced it. Counter e.m.f is produced during any voltage build up or decline. With a.c. there is a continual rise and decline at a periodic rate. While the opposition is overcome in d.c. after five time constants, in a.c. the back e.m.f and opposition are maintained because of the unbroken chain of positive and negative voltage changes. Thus, a fixed unit value of inductor opposition is created for a.c. which is known as inductive reactance. The symbol for inductive reactance is XL; (see Figure 21.1) XL is proportional to the current through and the voltage drop across the inductance. Hence, XL follows the same basic Ohm’s law and also has ohm for its unit value. The inductive reactance depends on the amplitude of the induced voltage

LCR Circuits

377

which, however, is limited by the inductance value in henrys, and the rate at which the a.c. and, hence, the magnetic field changes. The standard formula for XL is XL = 2πf L = 6.28 f L = ω L where,

XL is the inductive reactance in Ω f is the frequency in hertz (cycles per second) and L is the inductance in henrys.

V

L

f HZ

– Current or voltage +

V V

E di

dt

90°

I

0

I E

(a)

(b)

(c)

Figure 21.1 Circuit with Pure Inductance Example 21.1 A 10 H coil is in a 60 Hz a.c. circuit. What is its inductive reactance? Solution: X L = 2π f L = 6.28 × 60 × 10 = 3768 Ω Example 21.2 A 2 mH inductance has 1 MHz signal applied to it. What is its XL? Solution:

X L = 2π fL = 6.28 × 106 × 0.002 = 12560 Ω Example 21.3 The reactance of an inductor is 94200 Ω for a frequency of 5 kHz. What is its inductance? Solution: X L = 2π f L L = X L 6.28 f =

94200 = 3H 6.28 × 5000

Example 21.4 The reactance of a coil is 7536 Ω. If its inductance is 20 H, what is the frequency of the a.c. in the circuit? Solution: X L = 2π f L f =

7536 XL = 6.28 L 6 .28 × 20

= 60 Hz

378 Electrical Technology The expression for XL may also be derived as

t = 1/f There are four changes per cycle. The time for each change is ¼ f seconds Average rate of change of current is 1 f = di dt I= 4 fI max amp /sec max 4

(21.1)

Vav = L di dt = 4 fI max L × 2 π I max = 2Vmax 4π fL = Vmax 2π fL Vmax I max = 2π fL = X L

(21.2)

21.3 CAPACITIVE REACTANCE A capacitor opposes a change of voltage, in contrast to the inductor which opposes a change of current. With a capacitor, the opposition is initially zero (as electrons flow freely to one plate and away from the other), but opposition builds up rapidly as a greater number of electrons are forced on to one plate and an increasing quantity removed from the other. After five time constants, the applied battery or generator voltage is incapable of forcing any more electrons to the negatively charged plate, or removing any more from the positively charged plate. Now, the resistance is infinite (assuming a perfect dielectric) and no more current flows. When a.c. is applied to a capacitor, however, there is a continual rise and fall of voltage and current at a periodic rate. Thus, while the dq dt opposition rises to an infinite value after five time constants with d.c., I when a.c. is applied to a capacitor, the changing reversal of the polarity 0 of the applied voltage prevents the opposition from building up to an V c infinite value. Electrons not only flow to one plate at one instant, but fH Q also flow away from that plate an instant later. Hence, a fixed unit valV ue of capacitor opposition is created, known as capacitive reactance. (a) (b) Its symbol is XC (see Figure 21.2). Capacitive reactance is also in proportion to the voltage drop across I the capacitor and the current flow to and away from it (current does 90° not flow through a capacitor). Capacitive reactance also follows Ohm’s law, and its unit is also ohm. Z

V

X C = 1 2π fC

(c)

where, X C is the capacitive reactance in ohms

Figure 21.2 Circuit with Pure Capacitance

f is the frequency in hertz and C is the capacity in farads

Example 21.5 What is the reactance of a capacitor having a capacitance of 100 µF, when the frequency of the applied a.c. is 50 Hz? Solution: X C = 1 2π fC =

1 = 31.83 Ω (6.28 × 50 × 100 × 10−6 )

Example 21.6 A 600 µF capacitor is used in a circuit where the a.c. has a frequency of 500 kHz. What is the capacitive reactance? Solution: X C = 1 2π fC = 1 (6.28 × 500 × 103 × 600 × 10−12 ) = 530 Ω

LCR Circuits

379

The expression for XC may also be derived in the following manner. The charge Qmax = CVmax The total change of charge = 4 · CVmax (four changes) charge per second = 4fCVmax I av = ( 2 π ) × I max I av = ( 2 π ) × I max π I max = I av ×  2

 π  I max = I av ×    2

π  π  2 f Cmax ×   = 2 f C Vmax Vmax I max = 4 fCVmax ×I max  = 4= fCV 2 2

i.e

i.e

and (r.m.s)

and (r.m.s) I = 2π fCV

or

or

XC = V I =

I = 2π fCV 1 1 XC = V I = 2π fC 2π fC

Example 21.7 If the reactance of a capacitor is 8.4 Ω at a frequency of 30 Hz, what is the value of the capacity? Solution: X C = 1 2π fC , C = 1 6.28 f X C C = 1 (6.28 × 30 × 8.4) = 1 158.256 = 0 .0006 F = 600 µ F Example 21.8 A 5 µF capacitor has a reactance of 159 Ω in an a.c. circuit. What is the frequency of a.c.? Solution: X C = 1 2p fC f = 1 2pCX C f = 1 (6.18 × 5 × 10−6 × 159) = 1 0.0049926 = 200 Hz

21.4 FILTERING The opposing effects of the inductive and capacitive reactance can be utilized for filtering. In a low-pass filter, as has been shown in Figure 21.3—that passes low-frequency signals while attenuating high-frequency signals—the reactance of L1 must be chosen to have a high ohmic value for the signals to be attenuated and present a decreasingly lower reactance for signals having lower and lower frequencies. If a more pronounced filtering action is desired, yet another capacitor can be placed at the signal input (as shown by the dotted lines). This forms what is known as a pitype filter, because the circuit configuration resembles the Greek letter π. The extra input capacitor gives a sharper slope to the response of the filter and, hence, the higher frequency signals can be cut off more sharply. A high-pass filter with a typical response curve is shown in Figure 21.4. The action is just the reverse. It passes highfrequency signals while attenuating low-frequency signals. If a pi-type filter is formed by adding an inductance at the input (shown dotted) a more abrupt low-frequency cut-off is achieved.

C1

Low-frequency output signals

Input a-c signals

Frequency

Figure 21.3 Low-pass Filter

High-frequency output signals

Amplitude

Input a-c signals

Amplitude

L1

Frequency

Figure 21.4 High-pass Filter

380 Electrical Technology

21.5 BASIC SERIES A.C. CIRCUITS There are three possible constituent components in an a.c. circuit, R (resistance), L (inductance) and C (capacitance). Their characteristics are summarized in Table 21.1. Table 21.1 Characteristics of Electrical Parameters S. No.

1.

Characteristic

Resistor

Inductor

Capacitor

R

L

C

Symbol

2.

Unit

Ω (ohm)

H (Henry)

F (Farad)

3.

Function

Opposes the flow of electric current, whether it is d.c. or a.c. (R)

Opposes changes in the current (XL)

Opposes changes in the voltage (XC)

4.

Ohms Law

E=I×R I = E/R R = E/I

E = I × XL I = E/XL XL = E/I

E = I × XC I = E/XC XC = E/I

5.

Series connection

R1

R2

R3

RT = R1 + R2 + R3 6.

L1

L2

L3

LT = L1 + L2 + L3

Parallel connection R1 R2

C1 C2 C3

1/CT = 1/C1 + 1/C2 + 1/C3 C1 C2 C3

R3 L1 L2 L 3

7.

Reciprocal

8.

Electric power

9.

Control over value

1 1 1 1 = + + RT R1 R2 R3

1 1 1 1 = + + LT L1 L2 L3

CT = C1 + C2 + C3

The reciprocal of resistance (R) is called conductance (G) G = 1/R R = 1/G

The reciprocal of inductive reactance (XL) is called inductive admittance (bL) bL = 1/XL XL = 1/bL

The reciprocal of capacitive reactance is called capacitive admittance (bC ) bC = 1/XC XC = 1/bC

Resistors consume power Inductors store electric WR = I 2R watts power in their magnetic field This power is radiated in the 1 WL = LI 2 Joules form of heat. 2 In ideal inductors there is a 100 per cent exchange of power.

Capacitors store electric power in their electric field 1 WC = CV 2 Joules 2 In ideal capacitors, there is a 100 per cent exchange of power.

l Ω A ρ = specific resistance l = length R=ρ

A = area of cross-section A control over any one of the above factors results in a control over the ohmic value of the resistor.

L=

0 . 4π N 2 µ A H l

C=K

A F d

µ = Permeability of medium

K = Permittivity of medium

N = number of turns A = area of cross-section l = length A control over any one of the above factors results in a control over inductance.

A = area of cross-section d = distance between plates A control over any one of the above factors results in a control over capacitance.

LCR Circuits

381

Table 21.1 Continued 10.

Effect of frequency

11.

Characteristic

12.

Types

13.

Colour code

Resistance value remains constant at low frequencies. At high frequencies, the current has a tendency to flow in the skin of the resistor (skin effect) and effective resistance increases. Resistance value—Ω ohm s. Thermal tolerance—watts Tolerance (ohmic) — ± per cent Fixed resistor, variable resistor. Carbon resistor, wire wound resistor, metal film resistor, metal oxide resistor. Thermistor, varistor, VDR.

Inductors act as almost short circuits for d.c. Inductive reactance increases with frequency and at very high frequencies inductors act as open circuits.

Capacitors act as open circuit for d.c. Capacitive reactance decreases with frequency and at very high frequencies capacitors act as short circuits.

Inductance value—H Permeability—µ, core—air, iron, ferrite

Capacitance value—F Permittivity—K Working voltage —WV d.c.

Fixed inductor, variable inductor. Air core inductor, iron core inductor, ferrite core inductor.

Ohmic value and tolerance of the resistor can be found from colour code (carbon resistors).

Dots of different colours indicate circuit connections in some inductors.

Fixed capacitor, variable capacitor. Air capacitor, paper capacitor, mica capacitor, ceramic capacitor, electrolytic capacitor. NPO, N 150, N 750 Capacitance value, tolerance, working voltage and temperature co-efficient can be found from colour code.

21.6 THE CONCEPT OF IMPEDANCE The simplest series circuits consist of resistance and inductance or capacitance connected in series, as shown in Figure 21.5. In each of the combinations shown, the phase relationship between voltage and current lies between 0° and 90°. In the R-L circuit, the phase angle is a lagging one, and in R-C circuit it is a leading one. Inductance is introduced into a circuit, usually by the inclusion of a coil. The inductor forming the coil has resistance which may be negligible but is never zero. This resistance may be added to that of a series resistance to obtain the total circuit resistance, as indicated in Figure 21.6. In a series circuit, like components are associated with voltages that are separately in phase with one another. It follows that the voltages associated with like series connected components may be added arithmetically. This has been illustrated in Figure 21.7.

Figure 21.5 R-L and R-C Circuits

In the general a.c. series circuit containing resistance, inductance and capacitance, as shown in Figure 21.8, the voltage VL across the inductance and the voltage VC across the capacitance are both in phase opposition. The circuit can be effectively either inductive or capacitive, depending on whether VL or VC is predominant. For such a circuit the impedance is given by X = ( R 2 + ( X L  X C )2 ) Figure 21.6 R-L Circuit with Resistive Coil

where, ~ denotes the difference.

1

2

382 Electrical Technology

I V

VR

VL

VL

VL

R VC

V L φ

VC

I

C

VR

VC

Figure 21.7 General a.c. Series Circuit

Figure 21.8 Various Conditions of General a.c. Series Circuit

There are three possible forms of solution to this relation and these have all been illustrated in Figure 21.8.

1. If XL > XC, then the circuit is effectively inductive and the current lags the voltage. 2. If XL < XC, the circuit is effectively capacitive and the current leads the voltage. 3. If XL = XC, the circuit is effectively resistive, and it can be said to be in resonance. i. The circuit is effectively inductive. ii. The circuit is effectively capacitive. iii. The circuit is in resonance.

Example 21.9 A 500 µH inductor is connected to a 1.0. V, 5 kHz sinusoidal supply. Assuming the inductor to have negligible resistance, determine the circuit current and, hence, calculate the capacitance of a capacitor that would take the same current from the same source. Solution: X L = 2π fL = 6.28 × 5000 × 500 × 10−6 = 15.7 Ω I = V X L = 1.0 15.7 = 0. 0637 A = 63.7 mA for the same current to flow,

X C = X L = 1 2π fC C = 1 2π f X C = 1 (6.28 × 5000 × 15.7) = 2.03 × 10−6 F = 2.03 µ F

Example 21.10 A pure inductance of 5.0 mH is connected in series with a pure resistance of 69 Ω. The circuit is supplied from a 4 kHz sinusoidal source and the voltage across the 69 Ω resistor is found to be 1.5 V as shown in Figure 21.9. Calculate the supply voltage.

V 4 kHz

VL L 5.0 mH VR 1.5 V

R 69 Ω

Figure 21.9 For Example 21.10

LCR Circuits

383

Solution: VR = 1.5 V I = VR R = 1.5 69 = 0.0217 A X L = 2π fL = 6.28 × 4000 × 5 × 10−3 = 125.7 Ω

(

Z = R 2 + X L2

)

1

2

= (692 + 125.7 2 )

1

2

= 143. 4 Ω V = IZ = 0.0217 × 143.4 = 3.1 V

Example 21.11 A 240 V 50 Hz supply is applied to a coil of 60 mH inductance and 2.5 Ω effective resistance connected in series with a 68 µF capacitor as shown in Figure 21.10. Calculate the current and the phase angle of the circuit. Also, calculate the voltage across each of the circuit components. Solution: X L = 2π fL Figure 21.10 For Example 21.11

= 6.28 × 50 × 60 × 10−3 = 18.85 Ω X C = 1 2 π fC = 1 (6.28 × 50 × 68 × 10−6 ) = 46.80 Ω X = X L − X C = 18.85 − 46.80 = −27.95 Ω Z = (R2 + X 2 ) I =V Z =

1

2

= (2.52 + 27.952 )

1

2

= 28.06 Ω

240 = 8.55 A 28.06

cos φ = R Z = 2.5 28 .06 = 0 .0356

φ = 88° lead

(

2 2 Z LR = R + X L

)

1

2

= (2.52 + 18.852 )

1

2

= 19.1 Ω

VLR = IZ LR = 8.55 × 19.1 = 163 V VC = IX C = 8.55 × 46.8 = 400 V Figure 21.11 illustrates the concept of impedance. Impedance is the property of a part of a circuit relating the voltage drop V across it to the current I, i.e. Z =

V I

384 Electrical Technology

R Load Z

Z

L C

ωL

Z=

400

Figure 21.11 The Concept of Impedance The term impedance is also used to describe a part of a circuit. Thus, if we have two or more distinct parts of a circuit connected in series, as in Figure 21.11, we have series-connected impedances. The vector sum of resistance and reactance in a circuit is called impedance, which is defined as the ratio of the r.m.s. e.m.f. in a circuit to the r.m.s. current which is produced thereby. The symbol for impedance is Z and it is measured in ohms. Thus, Z = (R2 + X 2)1/2

70° R

where, X is the algebraic sum of XL, and XC, so that the full expression for impedance is 2  1    Z =  R2 +  ω L −   ω C    

Figure 21.12 For Example 21.12

1

2

1   = R + j  ωL − ω C   The impedance of any portion of a circuit may also be determined by vectorially adding the appropriate values of R, X, and (XL ~ XC). The full statement of an impedance must also include its phase angle, which may have a value of ±90° or any intermediate value as well as its magnitude. An impedance is usually written in the form Z f ; if f is positive (inductive circuit), or as Z −φ , if f is a negative angle (capacitive circuit) for the a.c. circuit the complete expression of Ohm’s law is I = V/Z, V = IZ, Z = V/I This expression is the same as for the d.c. circuit, writing Z instead of R. Z is dependent upon frequency as well as upon the values of R, L, and C. Example 21.12 A certain telephone receiver has an impedance of 400 Ω with an angle of lag of 70° when an alternating current with a frequency of 800 Hz is passing. Calculate the effective resistance and inductance at that frequency. Solution: Let us see the impedance triangle in Figure 21.12.

ω = 2π f = 6.28 × 800  5000 radn/sec ω L = Z sin φ , 5000 L = 400 sin 70° 0 . 9397 L = 400 × = 0. 075 H 5000 R = 400 cos 70° = 400 × 0.342 = 136.8 Ω Example 21.13 When an alternating voltage of 10 V at a frequency of 500 Hz is applied to an iron-cored coil having an inductance of 0.1 H, the current is found to lag behind the applied voltage by 81° as shown in Figure 21.13. With the aid of a vector diagram, determine the effective resistance of the coil and the value of the current.

LCR Circuits

385

Solution:

Z

X

81° R

Figure 21.13 For Example 21.13

21.7 SERIES CONNECTED IMPEDANCES Series circuits are more complicated when they comprise two or more impedances connected in series, as shown in Figure 21.14. As usual, in series circuit analysis, the phasor diagram is drawn with the current I as reference, and is completed by applying the following relations: Thus, the circuit impedance is found by collecting the separate resistances and reactances into groups. The complicated series circuit is now simplified into one effective impedance equivalent in every way when observed from the circuit terminals. V1 = IZ1 , V2 = IZ 2 , V = V1 + V2

(

V = ( IR1 + IR2 ) 2 + ( IX1 + IX 2 ) 2

(

)

= I ( R1 + R2 ) 2 + ( X1 + X 2 ) 2 = IZ where,

(

Z = ( R1 + R2 ) 2 + ( X1 + X 2 ) 2

)

1

)

1

1

2

2

Figure 21.14 Circuit Comprising Series Connected Impedances

2

Example 21.14 Two coils are connected in series across a 5.0 V sinusoidal a.c. supply. Coil A has a resistance of 5 Ω and a reactance of 8 Ω, and coil B has a resistance of 15 Ω and a reactance of 12 Ω as shown in Figure 21.15. Calculate the circuit current and the volt drops across each of the coils. Solution: R = RA + RB = 5 + 15 = 20 Ω X = X A + X B = 8 + 12 = 20 Ω Z = (R2 + X 2 )

1

= (202 + 202 )

2

1

I

2

V 5.0 V

= 28.3 Ω I = V Z = 5.0 28.3 = 0.18 A

(

Z A = RA2 + X A2

)

1

2

= (52 + 82 )

1

2

= 9.4 Ω

VA = IZ A = 0.18 × 9.4 = 1.69 V

(

Z B = RB2 + X B2

)

1

2

= (152 + 122 )

VB = 0 .18 × 19 .2 = 3.46 V

1

2

= 19 .2 Ω

VA

RA = 5 Ω XA = 8 Ω

VB

RB = 15 Ω XB = 12 Ω

Figure 21.15 For Example 21.14

Example 21.15 The operating coil of a relay, having an inductance of 8 mH and a resistance of 30 Ω is connected across a 5 V, 600 Hz a.c. supply. What would be the current through the coil and the phase angle of the current relative to the applied voltage?

386 Electrical Technology Solution: The relay may be regarded as being made up of a resistance and an inductance connected in series. ω = 2π f = 6.28 × 800 = 5000 rad/sec

X = ω L = 5000 × 8 × 10 −3 = 40 Ω I = V (R2 + X 2 )

1

2

= 5 (302 + 402 )

1

2

= 5 50 = 0.1 A

φ = tan −1 X L R = tan −1 40 30 = tan −1 1.3 = 53° 8′ lagging Example 21.16 The voltage across a circuit consisting of an inductance and a resistance in series is 100 V. If the resulting voltage across each of the components is the same, what is the value of this voltage? Explain with the help of a diagram. Solution: VL = VR

Since,

V = (VR2 + VL2 )

V

=

10

0

VL

VR = 100 2 φ

I

VR

2

1

2

=

2 VR = 100 V

= 70.7 V

Figure 21.16 illustrates the required diagram.

Figure 21.16 For Example 21.16 Example 21.17 The ohmic resistance of an inductor is 5 Ω and its impedance at a frequency of 50 Hz is 13 Ω. Find its impedance at 20 Hz. Solution: RL = 5 Ω Z at 50 Hz = 13 Ω Z 2 = RL2 + X L2 V

169 = 25 + X L2 X L = (169.25) 2π f L = 12 Ω,

1

2

= 12 Ω

L =

12 H 6.28 × 50

2π × 20 × 12 X L at 20 Hz = 2π × 20 × L = = 4.8 Ω 2π × 50 Z at 20 Hz = (52 + 4.82 )

V1

1

2

V1

I Z1 (R1, X1)

V2 Z2 (R2, X2)

IX1 V

IR1 IR2

V2 I

V2 IX2

Figure 21.17 Circuit Comprising Two Dissimilar Impedances

= 6.2 Ω Two impedances—one inductive and the other capacitive connected in series—are shown in Figure 21.17. When X represents a reactance, its value in ohms is positive if the reactance is inductive and negative when the reactance is capacitive. Example 21.18 Three impedances are connected in series across a 20 V, 50 Hz a.c. supply. The first impedance is a 10 Ω resistor, the second a coil of 15 Ω inductive reactance, and the third consists of a 15 Ω resistor in series with a 25 Ω capacitor as shown in Figure 21.18. Calculate: (1) the circuit current, (2) the circuit phase angle, (3) the impedance volt drops.

LCR Circuits

387

Solution: R = R1 + R2 + R3

V

= 10 + 5 + 15 = 30 Ω

20 V 50 kHz

1

2

= (302 + 102 )

1

R1 10 Ω

V2

X = X 2 + X 3 = 15 − 25 = −10 Ω Z = (R2 + X 2 )

V1

R2 5 Ω X2 15 Ω

2

= 31.6 Ω

V3

R3 15 Ω

I = V Z = 20 31.6 = 0.63 A

X3 25 Ω

cos φ = R Z = 30 31.6 = 0.95

φ = 18.5 ° lead V1 = IR1 = 0.63 × 10 = 6.3 V

V2

(

V1

V2 = IZ 2 = I R22 + X 22 = 0.63 (15 + 5 ) 2

2

1

2

(

)

1

2

V

= 10.0 V

V3 = IZ 3 = I R32 + X 32

)

1

2

= 0.63 (152 + 252 )

= 18.4 V

1

2

V3

Figure 21.18 For Example 21.18

Example 21.19 An inductor of 0.85 H and a capacitor of 15 mF are connected in series with a resistor of 60 Ω as shown in Figure 21.19. The current flowing through the circuit at a frequency of 40 Hz is 5 A. Find the applied voltage. Solution: VC = IX C =

2 6.28 × 40 × 15 × 10−6

=

106 = 530.8 V 1885

VL = IX L

Figure 21.19 For Example 21.19

= 2 × 6.28 × 40 × 0.85 = 427 V VR = 2 × 60 = 120 V

(

E APP = VR2 + (VL − VC ) 2

)

1

2

(

= 1202 + (530.8 − 427) 2 = (14400 + 10774) = (25174)

1

2

1

)

1

2

2

= 158.6 V

Example 21.20 A circuit consists of a capacitor of 2 mF and a resistance of 200 Ω connected across a supply of 1.0 V at 800 HZ. Calculate: (1) the reactance, (2) the phase angle, (3) the current flowing, (4) the p.d. across the resistance, and (5) the p.d. across the capacitor.

388 Electrical Technology Solution:

1.

ω = 2π f = 6.28 × 800  5000 rad/sec X C = 1 ω C = 106 (2 × 5000) = 100 Ω

2.

φ = tan −1 X C R = tan −1 − 100 200 = −27°

3.

I = V ( R 2 + X C2 )

1

2

= 1 (2002 + 1002 )

1

2

= 1 223 = 0.0044 A 4.

VR = IR = 0.0044 × 200 = 0.88 V (in phase)

5.

VC = IX C = 0.0044 × 100 = 0.44 V (lagging 90 °)

Example 21.21 A resistor of 10,000 Ω and a capacitor of 0.01 mF are connected in series as shown in Figure 21.20. If 100 V at a frequency of 1 kHz is applied across the circuit, what is the drop of potential across, (1) the resistor, and (2) the capacitor? Find the value of the tangent of the angle between the applied voltage and the voltage across the resistance. Solution:

Figure 21.20 For Example 21.21

21.8 POLAR NOTATION By means of polar notation, we may express the voltage—taken as a reference—across an impedance as V 0 volts, whilst the corresponding current is expressed as I f amperes. In such an instance, V and I are magnitudes, while 0 and f are the angles of rotation of the phasors relative to a datum in the diagram. The impedance is given by Z =

V0 If

(21.3)

The magnitudes are operated arithmetically. Thus, V/I becomes the magnitude of the impedance. The angles are subtracted to give (0 − f) = −f . Thus, the impedance is given by ohms. The phase angle is reversed. Hence, for a capacitive circuit, the current leads the voltage and f is positive, yet the angle of the reactance is negative and XC is taken as negative. In an inductive circuit, the current lags the voltage and f is negative and XL is taken as positive. Impedance may be expressed in the form Z -f ohms. Example 21.22 An impedance passes a current 2.5 30° A when a voltage 50 − 15° V is applied. Determine the impedance in similar form. Solution: V Z = I 50 −15° = = 20 −45° Ω 2.5 30°

LCR Circuits

Example 21.23 Two impedances 20 −45° Ω and 30 30° Ω are connected in series across a supply and the resulting current is found to be 10 A as shown in Figure 21.21. Determine the supply voltage and the supply phase angle. Solution: R1 = Z1 cos φ1 = 20 ⋅ cos(−45°) = 14.1 Ω

389

I V

V1

Z1 = 20 ∠ – 45° Ω

V2

Z2 = 30 ∠30° Ω

R2 = Z 2 cos φ2 = 30 ⋅ cos 30° = 26.0 Ω R = R1 + R2 = 14.1 + 26.0 = 40.1 Ω

Figure 21.21 For Example 21.23

X 1 = Z1 sin φ1 = 20 ⋅ sin(−45°) = −14.1 Ω (capacii tive) X 2 = Z 2 sin φ2 = 30 ⋅ sin 30° = 15.0 Ω X1 = X1 + X 2 = −14.1 + 15.0 = 0.9 Ω Z = (R2 + X 2 )

1

2

= (40.12 + 0.92 )

1

= 40.1

2

V = IZ = 10 × 40.1 = 401 V tan φ = X R = 0.9 40.1 = 0.022

φ = 1.3° The total reactance is positive and therefore the circuit is inductive. The current, therefore, lags the supply voltage and the phase angle is –1.3°.

21.9 PARALLEL CONNECTED IMPEDANCES In the series circuit, where the current is the same in all parts of the circuit, it is convenient to use the current vector as the reference vector and to draw the various voltage vectors at suitable phase angles to the current line. In the parallel circuit, the p.d. across the several branches is the same; the current usually differs in magnitude and phase in each branch. For parallel circuits it is, therefore, convenient to draw the voltage vector as the reference vector and to set off the currents in the various branches at the appropriate phase angles. A leading current (capacitance) will then appear in the first quadrant, while a lagging current (inductance) will appear in the fourth quadrant. There are three possible arrangements of simple circuits in parallel; R in parallel with L, R in parallel with C and, finally, L in parallel with C. A resistance and inductance in parallel have been illustrated in Figure 21.22.

(

)

The total current I = I R2 + I L2

1

I

2

V

= I R − jI L

( ) ( ( ) (

)  )

2  = V + V ωL  R = V − j V R ωL

( )

 =V 1  R

2

1

(

+1 1

2

2

1

ω L )  2

{ R − j 1(ω L)} =V (1 + 1 R jω L ) = tan −1

IR

{ ( V ω L ) ÷ ( V R )}

φ = t an −1 R

ωL

IR

IL

R

L IL

=V 1

From the vector diagram, φ = tan −1 I L

V φ

2

IR

I IR

Figure 21.22 L and R in Parallel

390 Electrical Technology Here, the total impedance of the circuit is

Z =V Z =

I

1 1 + 1 R jω L

(21.4)

1

Note: The relation Z = ( R 2 + X L2 ) 2 applies only to series circuits and should not be used for parallel networks. The expression for the impedance of a parallel arrangement is rather complex and is easily derived from the ratio of V to I. Example 21.24 A 20 Ω resistor is connected in parallel with an inductor of 318 μH inductance and negligible resistance to 1.0 V. 5 kHz supply. Determine: (1) the supply current, (2) the network impedance, and (3) the supply phase angle. Refer Figure 21.2. Solution: X L = 2π fL = 6.28 × 5 × 103 × 318 × 10−6 = 10 Ω 1.

IL = V IR = V

XL R

=

=

(

1.0 = 0.10 A 10

1.0 = 0.05 A 20

I = I R2 + I L2

)

1

2

= (0.052 + 0.102 )

1

2

= 0.11 A 2. 3.

Z =V

I

= 1.0 0.11 = 8.9 Ω

cos φ = 0.05 = 0.47

φ = 0.11 62° lag Example 21.25 A 50 Ω resistor is connected in parallel with a coil of negligible resistance to a 100 V, 50 Hz supply and the supply current is found to be 3.5 A as shown in Figure 21.24. By means of a phasor diagram, drawn to scale, find the inductance of the coil. Solution: The current in the 50 Ω resistor is in phase with the supply voltage and its magnitude is given by I R = 100 50 = 2.0 A

V V 1.0 V 5 kHz

IR

IL

R

L

20 Ω

318 µH

φ

IL

IR

I IR

Figure 21.23 For Example 21.24 The current in the coil lags the supply voltage by 90º. We can draw the phasor diagram shown in Figure 21.24. A circle of radius equivalent to 3.5 A is drawn, taking the supply voltage as reference. The current phasor must lie on it at some point. However, the phasor I2, when added to IR, must give the supply current I. Thus, when we project a line from IR to meet the circle, then the point of intersection is appropriate to the condition.

Figure 21.24 For Example 21.25

I L = 2. 87 A, X L = V I L = 100 2.87 = 35.8 Ω XL 35 .8 L= = = 0 .11 H 2 Ωf 6 .28 × 50

LCR Circuits

Example 21.26 A resistor of 25 Ω is connected in parallel with an inductive reactance of 50 Ω. If the applied voltage is 100 V as shown in Figure 21.25, find (1) the impedance, (2) the current in the circuit, and (3) the phase difference. Solution: 100 IR = = 4 amp 25 IL =

100 = 2 amp 50

(

)

Line current = I R2 + I L2

= (42 + 22 )

1

1

391

ILine 100 V

25 Ω

IR ILine

50 Ω

IL

Figure 21.25 For Example 21.26

2

2

= 4.47 amp Z =

100 = 22.37 Ω 4.47 tan q = R

XL

= 25

50

= 1

2

q = 26° 36 ′ The applied voltage will lead thee current. A resistance connected in parallel with a capacitance has been represented in Figure 21.26. The active current in the resistor is IR = V/R in phase with the applied voltage. The reactive current in the capacitor is IC = ω CV, leading by 90º on the applied voltage and the current IR. The total current from the supply is

(

)

I = I R2 + I C2

1

2

= I R + jI C

( ( (

) )) 1

I = I R2 + I C2 = I R + jI C 2 2 V 2 1 j CV = + ( ω CV )  = V + ω 1 2 2 2 2C = I R +2 jI  I = RIV  R R + I2C = + ( ω CV )  = V + jω CV 1 R 1 2  R 2 2 2 2 =V 1 + ( ωC ) = V= 1 V + jω+C( ω CV )2  1 = V + jω CV R R 2 R Figure 21.26 C and R in Parallel 2  R 2 =V 1 + ( ωC ) = V 1 + jω C R R 1 2 2 The phase angle φ = tan −1 I C 2 IR = V 1I + ( ωC ) = V 1 + jω C R −1 RC φ = tan I −1 −1 R V ω CV ÷ = tan = tan ω CR R φφ = tan −−11 I C IR ÷ V = tan ω CV = tan −1 ω CR R Z = V I = 1 1 + jω C R −1 tanI−1= 1ω CV 1 ÷+V jR Z = =V ω C= tan ω CR The total circuit impedance is R Z = V I = 1 1 + jω C R 1

( )

{(

( ) (( ) ) {( ) {( )

}

)

(

)

(

2

( ( ( (

)

} }

( (

) ) ) )

( (

) )) )

Example 21.27 Find the current in Figure 21.27. IL =

Eapp XL

=

100 1000

= 0.1 amp IC =

Eapp XC

100 V 60 Hz

100 = 600

= 0.166 amp

600 Ω

Figure 21.27 For Example 21.27

I X = I C − I L = 0.166 − 0.1 = 0.066 amp The current can also be calculated from m the impedance Z =

1000 Ω

(1000)(−600) XC × X L = = −15000 Ω (1000) − (600) X +X

IL =

Eapp XL

=

100 1000

= 0.1 amp IC =

Eapp

=

100 600

XC 392 Electrical Technology = 0.166 amp

I X = I C − I L = 0.166 − 0.1 = 0.066 amp The current can also be calculated from m the impedance Z =

(1000)(−600) XC × X L = = −15000 Ω (1000) − (600) XC + X L

100 = −0.066 amp −1500 The line current will lead the applied voltage by 90º. Line current =

Example 21.28 In the circuit given in Figure 21.28, find (1) line current, (2) capacitive current, (3) resistive current, (4) phase difference, and (5) power consumed. Solution: 1.

IC =

220 = 2.2 amp 100

2.

IR =

220 = 0.44 amp 500

3.

100 Ω

1 Amp

200 V

500 Ω

Figure 21.28 For Example 21.28

{

I LINE = (0.44) 2 + (2.2 − 1) 2

}

1

2

= 1.3 amp tan θ =

4.

=

IC − I L IR 1.2 = 2.75 0.44

I V

θ = 69°9′

IC C

5. Power consumed = I R 2 × R

IL L

= (0.44) 2 × 500 Figure 21.29 L and C in Parallel

= 97 watts IC

Finally, in the case of capacitance in parallel with inductance, as shown in Figure 21.29, the current in the capacitive branch is given by. IC = V X C

IC

I 90° V IL

–90°

V

I

Where IC leads V by 90º The current in the inductive branch is given by IL = V

IL

Figure 21.30 Phasor Diagrams for L and C in Parallel

XL

where, I lags V by 90 ° I = IC + I L

The phasor diagram can take one of two forms (Figure 21.30), depending on whether IC is greater or less than IL. Here, f can be either +90º or –90º, depending on whether the capacitance or the inductance predominates.

21.10 COMPONENTS OF CURRENT It is possible to resolve a current into its components that are in phase or in quadrature with the supply voltage. This has been illustrated in Figure 21.31.

Figure 21.31 Resolving a Current into its Components

LCR Circuits

The current can be resolved into two components that are at right angles to one another; I cos f, which is in phase with the voltage, is termed the active or power component, and I sin f which is in quadrature, is termed the quadrature or reactive component. The value of current I can be determined either by drawing a phasor diagram to scale or by resolving the currents into components (Refer Figure 21.32), so that

I2

I

I cos φ = I1 cos φ1 + I 2 cos φ2

I1

I1 sin φ1

I1 cos φ1 I2 cos φ2 V

I1

I2 cos φ2

V

I1 sin φ1

(21.5)

2

2

I2 sin φ2 I1 cos φ1

Figure 21.32 Addition of Vectors

I sin φ = I1 sin φ1 + I 2 sin φ2

and

I

I2 sin φ2

I2

393

2

but

I = ( I cos φ ) + ( I sin φ )

hence,

I 2 = ( I1 cos φ1 + I 2 cos φ2 ) 2 + ( I1 sin φ1 + I 2 sin φ2 ) 2

(21.6)

Finally, cos φ =

I1 cos φ1 + I 2 cos φ2 I

(21.7)

Example 21.29 A parallel circuit consists of two branches A and B, as shown in Figure 21.33, all phase angles being relative to the supply voltage. Determine the supply current and the phase angle. Solution: I = I A + IB I cos φ = I A cos φ A + I B cos φ B = 10 cos(−60°) + 12 cos 90°

Figure 21.33 For Example 21.29

= 5.0 + 0 = 5.0 A I sin φ = I A sin φ A + I B sin φ B = 10 sin(−60°) + 12 sin 90° = −8.66 + 12.0 = 3.34 A I 2 = ( I cos φ ) 2 + ( I sin φ ) 2 I = (5.02 + 3.342 ) cos φ =

1

2

= 6.0 A

I cos φ 5.0 = 0.83, φ = 33.5° = I 6.0

I = 6 33.5° A

Thus Example 21.30

An inductor of 1 mH and a capacitor of 0.1 µF are connected in parallel across a frequency of 100 kHz as shown in Figure 21.34. If the current in the inductor is 1 amp, find: (1) applied voltage, (2) capacitive current, and (3) line current Solution: 1.

Inductive current = Eapp X L 1 = Eapp 628 = 628 V

2.

Capacitive current = Eapp X C IC =

628 1 2π × 100 × 103 × 0.1 × 10−6

= 39.44 amp

Figure 21.34 For Example 21.30

1.

Inductive current = Eapp X L 1 = Eapp 628 = 628 V

394 Electrical Technology 2. Capacitive current = Eapp X C IC =

628 1 2π × 100 × 103 × 0.1 × 10−6

= 39.44 amp 3.

Line current = I C − I L = 39.44 − 1 = 38.44 amp

Example 21.31 A resistance of 1 MΩ is connected in parallel with a capacitor of 100 µF. If the frequency of the applied voltage is (1) 50 MHz and (2) 50 Hz, find the branch in which the current will predominate. Solution: XC = 1 XC =

At 50 MHz,

2π fC

1 6.28 × 5 × 10 × 100 × 10−12 6

X C = 318.5 Ω Figure 21.35 For Example 21.31

Xc = 31.8 M Ω

At 50 Hz,

At 50 MHz, all of the current will flow through the capacitive branch. At 50 Hz, the whole of the current will flow through the resistive branch, as shown in Figure 21.35.

100 Ω 40 Hz

Example 21.32 A coil having an inductance of 0.4 H and an effective resistance of 100 Ω is connected in parallel with a noninductive resistance of 141 Ω across a sinusoidal a.c. supply with a frequency of 40 Hz as shown in Figure 21.36. With the aid of a vector diagram, determine the approximate impedance (modulus and angle) of the complete circuit. Solution:

φ

141 Ω 0.4 H

(a)

Y

(b)

Figure 21.36 For Example 21.32

LCR Circuits

Example 21.33 A circuit consists of two branches in parallel. One branch consists of an inductance L in series with a resistance R, and the other branch consists of a capacitor C in series with a resistance R as shown in Figure 21.37. If L/C = R2, prove that the impedance of the circuit is independent of frequency and equal to R. Solution: Z L = R + jω L ZC = R + 1

L

R

C

R

395

Figure 21.37 For Example 21.33

jω C

If the joint impedance = Z , then 1

Z

=

1 1 + R + jω L R + 1

= 1 But

L 1

C Z

( R + jω L )

jω C

+ jω C

( jω CR + 1)

= R 2 , L = CR 2 = 1

( R + jω CR 2 )

= (1 + jω CR )

+ jω C

{ R(1 +

( jω CR + 1)

jω CR ) } = 1 R

Z = R (i.e. independent of frequency) Example 21.34 A 50 Hz supply at a pressure of 100 V is applied across a resistance of 1000 Ω, at an inductance of 0.5 H and of negligible resistance, and a capacitor of 10 µF, all of which are connected in parallel. What is the total current drawn from the supply and what is the phase angle of this current relative to the applied voltage? Illustrate your answer by a vector diagram. Solution: I R = V R = 100 1000 = 0.10 A I L = V (ω l ) = 100 (100 Ω × 0.5) = 0.637 A I C = Vω l = 100 × 100 Ω × 10−5 = 0.314 A I X = I L − I C = 0.323 A

I R = V R = 100 1000 = 0.10 A I L = V (ω l ) = 100 (100 Ω × 0.5) 396 Electrical Technology = 0.637 A I C = Vω l = 100 × 100 Ω × 10−5 = 0.314 A I X = I L − I C = 0.323 A Total current = (0.12 + 0.3232 )

1

2

= 0.3376 A

φ = tan −1 0.323 0.1 = tan −1 3.23 = 72°46 ′ (current laggging since I L > I C ) The vector diagram is given in Figure 21. 38.

(a)

(b)

Figure 21.38 For Example 21.34

Example 21.35

A resistor of 1500 Ω and an inductance of 5 H are connected in parallel across a 50 Hz a.c. supply of 1000 V r.m.s. as shown in Figure 21.39. What will be the total current taken from the mains? Solution: I R = V R = 1000 1500 = 0.66 A IL = V ω L =

1000 = 0.64 A (100 Ω × 5)

Total current I = ( I R2 + I L2 )

1

2

= (0.662 + 0.642 ) = 0.8452

(b)

(a)

1

2

1

2

= (0.4356 + 0.4096)

1

2

= 0.92 A

φ = tan −1 I L I R = tan −1 0.64 0.66

Figure 21.39 For Example 21.35

= tan −1 0.9696 = 44°7 ′ Example 21.36 In Figure 21.40, C = 35 F and R = 3.3 kΩ. If the source voltage is 50 mV, 1.8 MHz, calculate the values of the total impedance, IR, IC, IS, true power factor, and phase angle. Solution: X C = 1 2π fC = 1 / 2π × 1.8 × 106 × 35 × 10−12 ) = 2.53 kΩ Total impedance ZT =

R × XC 3.3 × 2.53 = 1 R + XC (3.32 + 2.532 ) 2

= 2.0 kΩ IR =

50 mV = 15.2 µ A 3.3 kΩ

IC =

50 mV = 19.8 µ A 2.53 kΩ

Supply current = I S = ( I R2 + I C2 )

1

2

= (15.22 + 19.82 )

1

2

= 25.0 µ A 2 True power = E

2 Reactive power = E

R

=

XC

(50 mV) 2 = 0.758 µ W 3.3 kΩ =

(50 mV) 2 = 0.988 µ VA 2.53 kΩ

Apparent power = E × I S = 50 mV × 25µ A = 1.25 µ VA

Figure 21.40 For Example 21.36

IC =

50 mV = 19.8 µ A 2.53 kΩ

Supply current = I S = ( I R2 + I C2 )

1

2

= (15.22 + 19.82 )

1

2

= 25.0 µ A 2 True power = E

2 Reactive power = E

R

=

XC

(50 mV) 2 = 0.758 µ W 3.3 kΩ =

LCR Circuits

397

(50 mV) 2 = 0.988 µ VA 2.53 kΩ

Apparent power = E × I S = 50 mV × 25µ A = 1.25 µ VA Power factor =

0.758 = 0.61, leading 1.225

Phase angle, φ = cos −1 0.61 = −53°

S UM M A RY 1. Impedance is a combination of resistance and reactance. 2. Impedance causes phase shift. 3. Impedance may cause either a leading or a lagging current. 4. The symbol for impedance is Z and its unit is ohm. 5. In RC circuits, the current leads the voltage. 6. In series RC circuits ZT > R or XC 7. In parallel RC circuits ZT < R or XC 8. The current phasor is the reference phasor in all series circuits, because there is only one current and it is the same in all series-connected loads.

9. The voltage phasor is the reference phasor in all parallel circuits, because there is the same voltage across all parallel-connected loads. 10. Resistance and reactance are in quadrature. 11. In RL circuits the current lags the voltage. 12. In series RL circuits ZT > R or XL 13. In parallel RL circuits ZT < R or XL 14. In a series LCR circuit Z may be > or < XL or XC 15. When XL = XC resonance occurs

M U LT IP LE C H O I C E Q UE S TI O NS (M C Q ) 5. Cos f in an electric circuit is called

1. Ohm’s law, expressed as E = IR, (a) Cannot be applied to a.c. circuits (b) Can be applied to a.c. in the same manner as to d.c. (c) Applied to a.c. circuits by substituting Z for R

2. In a circuit containing inductance only

(a) (b) (c) (d)

3. In a circuit containing capacitance only (a) Current lags voltage by π/2 (b) Current leads voltage by π/2 (c) Both are in phase Effective resistance due to capacity Effective wattage Effective voltage None of the above

4. (a)

5. (b)

Pure inductor Pure capacitor Pure resistor Either an inductor or a capacitor

7. In an LCR series a.c. circuit

4. The capacitive reactance in an a.c. circuit is

ANSWERS (MCQ) 1. (b) 2. (b) 3. (b)

Phase factor Power factor Frequency factor Form factor

6. Power factor is one for

(a) Current leads voltage by π/2 (b) Current lags behind voltage by π/2 (c) Current and voltage are in phase

(a) (b) (c) (d)

(a) (b) (c) (d)

6. (c)

7. (a).

(a) The current is in phase with voltage (b) The current always lags the voltage (c) The current always leads the voltage

398 Electrical Technology

CON V E N TI O NA L Q UE S TI O NS (C Q ) 1. What is the necessary reactance at 220 V a.c. to have a current flow of 0.04 A? 2. If the inductive reactance is 2000 Ω and the current is 3 ma, what is the voltage? 3. A 10 H coil is in a 60 Hz a.c. circuit. What is its inductive reactance? 4. A 2 mH inductance has a 1 MHz signal applied to it. What is the XL? 5. The reactance of an inductor is 94,200 Ω for a frequency of 5 kHz. What is its inductance? 6. The reactance of a coil is 7536 Ω. If the inductance is 20 H, what is the frequency of the a.c. in the circuit? 7. What is the capacitive reactance when the applied a.c. has a value of 300 V and the current is 1 ma? 8. What is the current in an a.c. circuit if the voltage is 500 kV and the capacitive reactance is 10 MΩ? 9. What is the reactance of a capacitor having a capacitance 100 µF when the frequency of the applied a.c. is 50 Hz? 10. A 600 p.f. capacitor is used in a circuit where the a.c. has a frequency of 500 kHz. What is its capacitive reactance? 11. If the reactance of a capacitor is 8.4 Ω at a frequency of 30 Hz, what is the value of its capacity? 12. A 5µF capacitor has a reactance of 159 Ω in an a.c. circuit. What is the frequency of the a.c.? 13. If the voltage is 20 kV, and the current is 800 ma. What is the impedance? 14. When a capacitor is in a circuit and the frequency of the a.c. is increased, in what manner does the reactance change? If the frequency is increased on an inductor, how does its reactance change? 15. In a radio a 35 mH coil has a 549.5 V a.c. signal across it. The signal has a frequency of 1000 kHz. What current flows through the coil? 16. An inductor of 10 H, capacitor of 2 µF and a resistance of 1500 Ω are connected in series across a 230 V, 50 Hz, supply. This has been illustrated in Figure 21.41. Find: (a) The current flowing in the circuit (b) Power factor (c) The phase angle between the applied voltage and current 17. In the circuit given in Figure 21.42, find: (a) Z, (b) apparent power, (c) power factor, and (d) power consumed. 18. In the circuit given in Figure 21.43, find the XL, RT, Z, power factor, I, apparent power and true power. 19. In the circuit given in Figure 21.44, find: IX, IR, IT, Z, power factor, apparent power, and true power.

20. In the circuit given in Figure 21.45, find IR1, IR2, IRT, I, XL and Z. 21. In the circuit given in Figure 21.46, find Z, phase angle, IT, apparent power, true power, ER, EX, ET.

22. A series circuit has one resistor of 45 Ω and another of 2.5 Ω. In series with these are two capacitances, one with a value of 45 Ω, and another of 50 µF. If the frequency is 50 Hz, what is the total capacitive reactance? What is the phase angle? 23. For the circuit shown in Figure 21.47, find RT, XC, IR, IX, IT, Z and power factor. 24. In the circuits of a burglar alarm system a 2 H coil is connected in series with a 600 Ω resistor. Another inductor of 3 H and a second resistor of 42 Ω are also in this series circuit. The frequency of the a.c. is 60 Hz. What is the true power if the apparent power is 50 W? 25. In an electronic tester, an inductance having a reactance of 2000 Ω is in parallel with a 3000 Ω resistor. The source voltage is 180 V a.c. What are the total current, the impedance, and the cosine of the angle? 26. In the frequency-stabilizing circuit of a remote control system, a resistor of 1200 Ω is in parallel with an inductor. The total current flow is 425 ma, and the total voltage impressed on the circuit is 300 V. What is the impedance, the current through the reactor, the inductive reactance value and the power factor? 27. Calculate the value of the total impedance phasor, Z0, in Figure 21.48. 28. Calculate the total impedance in Figure 21.49. 29. A circuit is composed of a parallel branch with three sections, in series with a series impedance as shown in Figure 21.50. Calculate the total impedance, the total current, and the true power. 30. Calculate the impedance of the circuit in Figure 21.51. 31. Find VAB, VBC and VAC in Figure 21.52. 32. In Figure 21.53, find the total impedance. 33. The equation of an alternating voltage is E = 220 sin (ωt + π/6) and the equation of the current in the circuit is I = 10 sin (ωt – π/6). What is the impedance of the circuit? 34. In the circuit given in Figure 21.54, find the impedance, the voltage across the inductance and the power factor of the circuit. 35. Calculate the values of Z and IT for the circuit in Figure 21.55. 36. An impedance passes a current 2.5 30 A when a voltage of 50 - 15 V is applied across it. Determine the impedance in a similar form. 37. In the network shown in Figure 21.56, calculate the total current, the impedance, and the power.

LCR Circuits

38. The network in Figure 21.57 consists of a 128 Ω resistance in parallel with a 40 µF capacitor and is connected a 240 V, 50 Hz supply. Calculate the branch currents and the supply current. 39. In the circuit in Figure 21.58, find the impedance and power factor. 40. In the circuit given in Figure 21.59, find the power factor. 41. In the circuit given in Figure 21.60, find XL, XC, RT, Z, power factor, current, and true power. 42. For the circuit in Figure 21.61. Find the total current, power factor and true power. 43. A certain telephone receiver has an impedance of 400 Ω with an angle of lag of 70° when an alternating current with a frequency of 800 Hz is passing. Calculate the effective resistance and inductance at this frequency. 44. The impedance of a series circuit consisting of a resistor and an inductor is 200 Ω when the frequency is 500 kHz. If the value of the resistance is 100 Ω what is the value of the inductance? 45. A circuit consists of a resistance of 100 Ω, in series with an inductance. When the frequency of an applied voltage changes from 200 Hz to 500 Hz, the impedance doubles in value. Calculate the value of the inductance. 46. A circuit consisting of a resistance of 50 Ω, an inductance of 10 H and a capacitance of 1µF in series

47.

48. 49. 50.

51. 52.

53.

is found to pass 1 A with a certain applied e.m.f. alternating at 50 Hz. What current will it pass with the same e.m.f alternating at 100 Hz? What is the impedance at a frequency of 900 kHz of a circuit consisting of a capacitor of 0.0002 µF in series with a coil of 200 µH inductance and 25 Ω resistance? Which one is predominant in the circuits shown in Figure 21.62, inductive reactance or capacitive reactance? The ohmic resistance of an inductor is 5 Ω and its impedance at a frequency of 50 Hz is 15 Ω. Find its impedance at 20 Hz. A series circuit consists of a fixed inductor and a variable resistance connected across a source of alternating current at a fixed frequency. Explain the effect—on the overall circuit—of varying the resistance from zero to its maximum value with the help of vector diagrams. In the parallel branches shown in the circuit in Figure 21.63, what is the total admittance and impedance? In a series LR circuit, L = 500 mH, R = 1 kΩ, and the supply voltage is 10 V, 400 Hz. Calculate Z, I, VL, VR, phase angle, true power, apparent power and power factor. In a series LCR circuit, L = 500 mH, C = 2.2 µF, R = 1 kΩ, E = 24 V, 200 Hz. Calculate: Z; Ii; VL, VC, VR; phase angle; true power; apparent power; power factor.

XL = 20

4Ω

R = 15 Ω

XL = 126 Ω = 8 amp

R = 168 Ω = 6 amp

Figure 21.44 For CQ 19

Figure 21.42 For CQ 17

72 V a–c

0.7 Ω

3.5 V a–c 60 Hz

75 V a–c

1008 V a–c

0.02 h

0.005 h

= 3 amp

Figure 21.41 For CQ 16

399

L1

36 Ω R1 24 Ω

Figure 21.43 For CQ 18

L2

18 Ω

R2

Figure 21.45 For CQ 20

12 Ω

L3

12 Ω

400 Electrical Technology R = 12 Ω

XC = 20 Ω

116.5 V a–c

Figure 21.46 For CQ 21

zT

Figure 21.47 For CQ 23

2Ω

3Ω

8Ω

z1

z2

z3

4Ω

5Ω

1Ω

Figure 21.48 For CQ 27

Figure 21.49 For CQ 28

200 Ω 45 Ω 50 Ω

44 Ω

75 Ω Z3

R1

90 Ω

40 Ω

60 Ω

Z1 10 Ω

357 v a– c

C 1 µF

Z2 100 Hz 40 V

Figure 21.51 For CQ 30

Figure 21.50 For CQ 29

400 Hz 60 V

Figure 21.52 For CQ 31

R 1000 Ω

C 0.02 µF

R 27 kΩ

Figure 21.53 For CQ 32

LCR Circuits L 10 mH 50 V 50 kHz

50 kHz 10 V

C 0.005 µF

L 5 mH

R 1000 Ω

R 2700 Ω

Figure 21.55 For CQ 35

Figure 21.54 For CQ 34

35 Ω

50 kHz 50 V

V 240 V 50 Hz

R 2700 Ω

L 10 mH

Figure 21.56 For CQ 37

I

IR R 120 Ω

75 Ω

IC C 40 µF

25 Ω

Figure 21.57 For CQ 38

Figure 21.58 For CQ 39

3480 Ω 0.5 µfd 1520 Ω 140 V

300 V 100 V

R1 10 h

C1

R2

L1

200 V

9615 v 60 Hz

244 V

Figure 21.59 For CQ 40

L2

232 Ω

C2

2693 Ω

Figure 21.60 For CQ 41

Figure 21.61 For CQ 42 L=2H 1 µF

EAPP 30 Hz

L=5H

C

(a)

0.01 µF

EAPP 60 Hz

C 200 Ω

(b)

L = 5 mH

50 Ω

L = 200 mH

100 Ω

500 Ω EAPP 20 µF 1000 Hz (c)

C

EAPP 200 Hz (d)

Figure 21.62 For CQ 48

0.1 µF

C

1000 Ω a– c

Figure 21.63 For CQ 51

401

402 Electrical Technology ANSWERS (CQ) 1. 5500 Ω 2. 6 V 3. 3768 Ω 4. 12560 Ω 5. 3 H 6. 60 Hz 7. 300 000 Ω 8. 0.05 A 9. 31.83 Ω 10. 530 Ω 11.600 μF 12. 200 Hz 13. 25000 Ω 15. 2.5 ma 16. (a) 2156 Ω (b) 0.6957 (c) 45º 52′ (leading) 17. (a) 25 Ω (b) 225 W (c) 0.8 (d) 180 W 18. 0.4 Ω; 4.7 Ω; 10.5 Ω; 0.4384; 0.3 A; 1.05 W; 460 mW 19. 8 A; 6 A; 10 A; 100.8 Ω; 0.6; 10.080 W; 6048 W 20. 3 A; 6 A; 9 A; 12 A; 6 Ω, 48 Ω 21. 23.3 Ω; 59º; 5 A; 582.5 W; 300 W; 60 V; 100 V; 116.5 V 22. 95.2 Ω, 64 23. 90 Ω; 270 Ω; 0.9 A; 0.3 A; 0.95 A; 852.5 Ω; 0.94 24. 22 W 25. 108 ma; 1,666.6 Ω; cos = 0.55 26. 847 Ω; 0.3 A; 1200 Ω; 0.7 27. 3.15 53°6′ Ω 28. 121 V, 46º

29. 102 Ω; 3.5 A; 1217.5 W 30. 1880 Ω = 50= 45° 31. VAB = 36.4 74°3 ; VBC = 31.6 71°6′ ; VVAC AC 32. 16.1 kΩ 33. 22 Ω 34. 4141 Ω; 37.7 V; 0.652 35. 730 Ω; 13.7 mA 36. 20 −45° Ω 37. 24.4 mA; 2049 Ω; 924 mW 38. IR = 2.0 A, IC = 3.0 A, I = 3.6 A 39. 61 Ω (approximately); 0.573 40. 0.573 41. 3768 Ω; 5307 Ω; 5000 Ω; 6410 Ω; 0.78; 11,249.5 W 42. 1 A; 0.8; 753.6 W 43. 0.075 H; 136.8 Ω 44. 55.1 µH 45. 93 µH 46. 0.0134 A 47. 248. 48. (a) Inductive reactance; (b) capacitive reactance; (c) inductive reactance; (d) capacitive reactance 49. 6.2 Ω 51. 0.0128 υ; 78 Ω 52. 1.6 kΩ; 6.23 mA; 7.82 V; 6.23 V; –51.5º; 38.8 mW; 62.3 mVA; 0.62 lagging 53. 1.04 kΩ; 23.2 mA; 14.6 V; 8.4 V; 23.2 V; 14.9º; 0.56 VA; 0.54 W; 0.97 leading.

22

Resonance

ZS

ZP Reactance

+

ωL

O

Frequency

R O

fr

–

In this chapter you will learn about:  The definition of resonance as it applies to any a.c. circuit with only one source  The properties of series resonant circuits  The quality (magnification) factor Q, and its interpretation as the voltage magnification factor  The degree of selectivity  Bandwidth and its relation to Q  The properties of parallel resonant circuits  Impedance, current and voltage relations in series and parallel resonant circuits  Power factor of resonant circuits

Impedance (ohms)

OBJECTIVES

–l ωC

Reactance Sketches

22.1 INTRODUCTION In Chapter 20 of this book, titled ‘Power Factor Correction’ we corrected for a low power factor in a circuit predominantly inductive, by introducing into the circuit a capacitor having a reactance equivalent in numerical value to the inductive reactance. Through this method, the undesired circuit effects of the inductance were cancelled, leaving only resistance as an active circuit element. As a consequence, the phase angle became 0° and the power factor increased to one. The principle of selecting capacitors and inductors to provide for equal reactance values in series or in parallel circuits is also known as resonance. A resonant circuit has the ability to discriminate against unwanted signals while at the same time it presents a highly selective network to the desired signal. Without resonance, all types of radio, television, radar and other familiar forms of transmission and reception would be impossible. The resonant circuit is one of the most important circuits ever designed. The key feature of a resonant circuit is its ability to select a signal of specific frequency from among hundreds of other signals having different frequencies. Since capacitors and inductors can be made variable with respect to their unit values, we can tune such components in circuits in order to achieve resonance for a particular signal. Thus, we are provided with a manual means of signal selection widely used in all aspects of electrical and electronic engineering. We explore the various aspects of series and parallel VR VL VL resonance so that we may be able to apply this knowledge to practical situa- V tions as the necessity arises.

22.2 SERIES RESONANCE When the inductive reactance (XL) in a general series circuit is numerically equal to the capacitive reactance (XC), VL = VC and, consequently, V and I are in phase with one another. This is an instance of a condition known as resonance—in this case series resonance—because it concerns a series circuit. At resonance, the circuit is said to resonate. The phasor diagram for a series resonant circuit is shown in Figure 22.1.

M22_AUTH_ISBN_C22.indd 403

VL

VR

VC

V

I

VC VC

Figure 22.1 Series-resonant Circuit

8/6/2012 1:15:05 PM

404 Electrical Technology Many circuits in electronic and communication networks are supplied either from variable-frequency sources or from sources supplying a number of frequencies. A general series circuit is shown in Figure 22.1. When the frequency is zero, i.e., corresponding to a d.c. supply, 1 X L = 2p f L = 0 and X C = = ∞. 2p f I As the frequency increases, XL increases in direct proportion, whilst XC decreases inversely with increase of frequency, as can be seen in Figure 22.2. At resonance XL = XC, VL = VC. X The frequency at which this condition is satisfied is known as the XL Z resonant frequency and denoted by fr. We have, for XL = XC, in a general series circuit (XL – XC) fr

f

XC

Figure 22.2 Variation of Reactance with Frequency The expression for fr is found as

Z = (R 2 + (X L − X C ) 2 )1/2 = R (22.1) To denote that this is the impedance at resonance, the subscript r is added to the symbol. Thus, Zr = R. The same condition XL = XC could have been achieved by keeping the frequency constant and varying one of the reactances—probably the capacitance—to meet the above relation. To sum up, there are two methods of achieving resonance in a series circuit: 1. Varying the frequency till resonance occurs. 2. Varying one of the components (L or C) until resonance occurs.

X L = XC 2p f r L = fr 2 = fr =

1 2p f r C 1 (2p ) 2 LC 1 2p LC

Hz

(22.2)

At resonance 1. V and I are in phase. 2. The circuit impedance is a minimum. 3. The current at resonance is Ir = V/R. This may be considerably greater than that at other frequencies. 4. The voltage drops across the reactive components may reach very high values possibly many times that of the supply voltage. z 5. The power dissipation in the circuit is a maximum. 6. The peak rates of energy storage for the reactors become equal Ir and maximum. The energy stored by the reactors is constant and oscillates between the electric and magnetic modes of storage. In other words, first it is stored in the inductor, then it is transferred to the capacitor and then it transfers back to the inductor and so on. The energy, therefore, remains with the reactors and it is this oscillation of energy between them that makes the circuit appear to be purely Z = R resistive. A circuit is said to resonate whenever this effect of energy oscillation predominates, i.e., when the peak rate of energy storage in each of the reactors—given by I 2X—is at least ten times the power fr f associated with the circuit resistance (Refer Figure 22.3). If this condition is not met, then the circuit is a power-oriented arrangement Figure 22.3 Series Resonance Characteristics

M22_AUTH_ISBN_C22.indd 404

8/6/2012 1:15:10 PM

Resonance

405

which just happens to have the current in phase with the voltage. Figure 22.4 illustrates the action of oscillation of energy between an inductor and a capacitor. A simple mechanical analogy of resonance is provided by the pendulum, which can be set swinging to a large amplitude from a series of minute forces, provided that these impulses are properly tuned in their application at the resonant frequency of the particular pendulum as shown in Figure 22.5.

Kinetic energy + + + + – – – –

Potential energy

WC = 1/2 CV 2

WL = 1/2 LI 2

Figure 22.4 Energy Transfer

Figure 22.5 A Simple Mechanical Analogy of Resonance

Example 22.1 Examine the circuit in Figure [22.6 (a) and (b)] under several signal frequencies. Gen 1200 kHz

1.76 m.m.f. 0.04 henry 75,360 W 301,440 W

Load = 56,520 W

Gen 300 kHz

(a)

75,360 W 301,440 W Load = 56,520 W

(b)

Figure 22.6 Circuit for Example 22.1 Solution: In Figure 22.6(a), we have a generator operating at 1200 kHz and a load having a resistance of 56520 Ω in series with an inductor of 0.04 H and a capacitor of 1.76 ρ F( P = µµ = Pico = 10−12 ). Under these conditions, XC = 75,360 W and XL = 301,440 W X = XL – XC = 301,440 - 75,360 = 226,080 W (Primarily inductive) In polar form, we have X 226,080 tan θ = = = 4 and θ = 76°, cos θ = 0.24 R 56520 56,520 R Z = = = 235,500 76° Ω cos θ 0.24 If we increase the frequency, XL will get still higher, and XC will get lower (decrease). If we decrease the frequency, the reverse will happen, i.e., XL will decrease and XC will increase. To simplify our analysis, let us assume that the frequency is decreased from 1200 kHz to 300 kHz. XL will now decline to one-fourth of its original value (75,360 W), XC will now increase to four-times its original value (301,440 W). We now still have the same total reactance, but X = – j226,080 instead of +j226,080 and Z = 235,500 Ω −76°. Let us assume that the voltage of the generator is 23,550 V; then I =

E 23,550 = = 0.1 amp; Z 23,550

P = EI cos θ = 23,550 × 0. 1 × 0.24 = 565.2 Ω

Eload = IR = 56,520 × 0.1 = 5652 V When, however, a frequency of 600 kHz is applied to this circuit, resonance occurs, as shown in Figure [22.7 (a) and (b)]. Both XL and XC have the same ohmic value of 150,720 W. The voltage and current are in phase, making the circuit purely resistive, and Z = 56,520 Ω∠0ο

M22_AUTH_ISBN_C22.indd 405

8/6/2012 1:15:16 PM

406 Electrical Technology Gen 600 kHz

150,720 Ω 150,720 Ω

23,550 V Load = 56,520 Ω

600 kHz

V + i = 62,000 V

(a)

i = 62,000 V – – + V

Load = 23,550 V

(b)

Figure 22.7 Resonance for the Circuit in Figure 22.6 (a) for Example 22.1 Since the net opposition of the individual reactances to current flow is zero, the current now rises. E 23,550 I = = = 0.411 amp. Z 56,520 Because of the current increase, the power will also rise; however, the power factor is now one, there will be an additional increase over that which prevailed for the other frequencies P = 23,550 × 0.411 × 1 = 9.679 W. For XL = XC

E = IX = 150,720 × 0.411 = 62,000 V ER = IR = 56,520 × 0.411 = 23,550 V X L − X C = 0,

wL −

1 1 = 0 = and LC = 2 wC w

Example 22.2 If an inductance of 0.2 mH and a capacitor of 0.8 are placed in series. What is the resonant frequency? Solution: 1 fr = 2p LC 1 = = 398 kHz 6.28 0.2 × 0.8 × 10−12 Example 22.3 If a 0.49 rF capacitor is in series with a 0.2 H coil, what is the resonant frequency? What is the LC product? Solution: 1 fr = = 509 kHz 6.28 0.2 × 0.49 × 10−12 X L = ω L = 639,304 Ω XC =

1 = 639,304 Ω ωC

Converting L to micro henrys and C to microfarads gives the following LC product: 200,000 × 0.00000049 = 0.098 Note: 1. The product of LC is the determining factor with respect to the resonant frequency. 2. There are a number of inductance and capacity combinations, which will produce the same resonant frequency as long as their product is the same. In Example 22.3, for example, we can select 0.196 mF for the capacitor and 0.5 mH for the inductor. The product is still 0.098, but the reactance values differ. The resonant frequency is still the same.

M22_AUTH_ISBN_C22.indd 406

8/6/2012 1:15:26 PM

Resonance

407

X L = ω L = 1.6 Ω X C = 1/ω C = 1.6 Ω The originally high reactance values of 639,304 W are considerably reduced to a low value of 1.6 W, while the same resonant frequency is maintained. This wide latitude in the selection of reactance values at resonance provides us with a considerable variation in the choice of circuit characteristics, to suit particular requirements. It is sometimes advisable to find the LC product so that different values of L and C can be chosen if the existing ones do not provide the circuit characteristics desired in terms of original selectivity. Example 22.4 A series resonant circuit has an inductance of 1.1 mH and a capacitor of 0.2 mF. What is the LC product? What is the resonant frequency? Do the individual reactances exceed 2 megaohms? Solution: f r = 339 kHz X L = X C = 2.341,812 Ω each (exceed 2 megohms) LC = 0.22

Example 22.5 If a capacitor of 0.05 mF in a series circuit provides a resonant frequency of 833 kHz, what is the value of the inductance? Solution: L=

1 4π 2 f r 2 C

= 1/ [39.5 × (833,000) 2 × 0.05 × 10−6 ] = 0.65 µ H Example 22.6 A coil has an inductance of 84.5 mH. With what value capacitor will it resonate at 387 kHz? Solution: C = 1 / (4π 2 f 2 L) = 1 / [39.5 × (387,000) 2 × 8.45 × 10−6 ] = 0.002 µF Example 22.7 Assuming that a magneto bell has a constant inductance of 12.5 H and a resistance of 1000 W, calculate (1) the joint impedance of the bell and a 2 mF capacitor in series at ringing frequency (100 rad/sec), and (2) the frequency at which this circuit element would be in a state of resonance. Solution: 1.

R = 1000 Ω ω L = 100 × 12.5 = 1250 Ω

1 ω C = 106 /(100 × 2) = 5000 Ω 12

Z = { R 2 + (ω L − 1/ω C ) 2 }

12

= {10002 + (1250 − 50002 ) }

= (10002 + 37502 )1 2 = 3880 Ω

M22_AUTH_ISBN_C22.indd 407

8/6/2012 1:15:30 PM

408 Electrical Technology X = tan −1 3750 1000 = tan −1 3.75 R φ = −75°6′ (negative because 1 ωC > ω L)

φ = tan −1

2.

ω L = 1/ω C ,

ω 2 = 1/( LC ),

f r = 1/{2p ( LC )1/ 2 } = =

ω = 1/( LC )1/ 2 1

6.28 × (12.5 × 2 × 10−6 )

1/ 2

103 = 200 / 6.28 = 31.6 Hz 6.28 × 251/ 2

Example 22.8 A series tuned circuit has a capacitance of 0.2 mF. What must be its inductance for resonance to occur at a frequency of 500 Hz? Solution:  L = 1/  C For resonance, L = 1/(4π 2 f 2 C ) = 1/(4 π2 ×52 ×1010 ×0.2 ×10−6 ) = 1/(0.2π 2 × 106 ) = 0.51 µH Example 22.9 A resonant circuit consists of a coil of inductance 1000 mH and a high frequency resistance of 20 W in series with a variable capacitor, which may be assumed to be free from loss. An e.m.f. of 1 V at a frequency of 100 kHz is induced into the coil from a nearby circuit. What will be the value of the capacitance to produce resonance with this e.m.f. and what will then be the p.d. across this capacitor? Solution: At resonance,

ω L = 1/ω C C = 1/(ω 2 L) = 1/(4π 2 × 1010 × 10−3 ) = 10−6 / 40π 2 = 0.00254 µ F

At resonance,

I = V / R = 1/ 20 amp VC = 1/ω C = 1/(2p × 105 × 0.00254 × 10−6 × 20) = 1/(4p × 0.00254) = 31.4 V

Example 22.10 A circuit is tuned to a frequency of 1000 kHz when its capacitance is 0.05mF. What is its inductance? Solution: At resonance,

ω L = 1/ω C L = 1/(4π 2 f 2 C ) = 106 /(4π 2 × 10−12 × 0.05) = 0.507 µH

Example 22.11 A resistor, a capacitor, and a variable inductor are connected in series across 200 V a.c. mains, the frequency being 50 Hz. The maximum current which can be obtained by varying the inductance is 314 mA, and the pressure across the capacitor, which has a negligible resistance is then 250 V. Calculate the values of the capacitance, the resistance, and the inductance of the circuit.

M22_AUTH_ISBN_C22.indd 408

8/6/2012 1:15:36 PM

Resonance

409

Solution: The circuit conditions are shown in Figure 22.8  = 314 rad/sec. Since the current has its maximum value, the circuit is at resonance and Z = R 0.314 A

R = V/I = 200/0.314 = 637 W The p.d. across the capacitance

R

VC = I /(  C ) = 250, C = I (  VC )

200 V 50 Hz

C = 0.314 /(314 × 250) = 4 × 10−6 F  L = 1/(  C ),

C

L = 1/( 2 C )

250 V

L

L = 106 /{ (100p 2 ) × 4 } = 2.5 H Figure 22.8 For Example 22.11

Example 22.12

The current flowing through a circuit consisting of R ohms L micro henrys and C micro farads in series is 0.5 A when a voltage of 30 V is applied at the resonant frequency of the circuit and is 0.3 A when the frequency of the applied voltage is doubled. What is the reactance of the circuit in the second case, and the phase angle between the applied voltage and the resulting current? Illustrate your answer by vector diagrams. I = 0 .5 Solution: (0 .3) L The current is shown in Figure 22.9 (a). At resonance, the illustration is given in Figure 22.9 (b) 30 v

R

wC

When the frequency is doubled, it is as shown in Figure 22.9 (c) Z = V/I = 30/0.3 = 100 Ω Reactance X = (Z 2 - R2)1/2 = (1002 - 602)1/2 = 80 Ω as seen in Figure 22.9 (d)

C IwL = –I

R = 30 / 0.5 = 60 Ω

fr (2fr)

(a)

(b)

IwL R = 60

= tan

1.33 = 53° 8 '

vX

1f The angle is positive because at frequencies > f r , ω L > ωC 1 > fr , ω L > , When the frequency is doubled, ωL is doubled –I ωC wC and 1/ωC is halved. The vector diagram and impedance triangle are given in Figure [22.9 (c) and (d)].

v

f = tan−1 X /R = tan−1 80 60 −1

I

vR

f

ω L = 1/(ω C ) and R = V /I

Z

=

10

0

I

vR

X = 80

(c)

(d)

Figure 22.9 For Example 22.12

Example 22.13 A circuit consisting of a non-reactive resistance R ohms, an inductance L henrys, and a capacitance C farads joined in series is connected across a source of alternating current which has a constant sinusoidal voltage output of 1 V over the range of frequencies concerned. If R = 1000 W, L = 0.2 H, and C = 0.2 mF, determine the frequency at which the r.m.s. voltage across L equals that across C. What is the value of this voltage? Also, what is the voltage across R and the power absorbed by the circuit at this frequency? Solution: Voltage across L = ω LI Voltage across C = I /ω C When these are equal  L = 1/  C and 2 = 1/LC or  = 1/( LC )1/ 2  = 1/(0.2 × 0.2 × 10−6 )1/ 2 = 103 / 0.2 = 5000 rad/sec f = 5000 / 2 p = 796 Hz

M22_AUTH_ISBN_C22.indd 409

8/6/2012 1:15:43 PM

410 Electrical Technology At resonance,

I = V /R = 1/1000 = 0.001 A VC = VL = ω LI = 5000 × 0.2 × 0.001 = 1V VR = IR = 0.001 × 1000 = 1 V power, P = I 2 R = 10−6 × 103 = 0.001 W

At resonance,

22.3 Q-QUALITY FACTOR OF A SERIES CIRCUIT The quality factor (Q) is based on the ratio of reactive power which is the peak rate of energy oscillation to the active power which is the power dissipated in the circuit. Thus, Reactive power Q Q= = Active power P =

I2XL X X = L or C 2 R R I R

=

ωr L 1 or ωr CR R

At resonance, X L = XC and Q=

ω rL 1 = R ω r CR

(22.3)

Also V and VL = IX L = I ω r L R ω L V VL = ⋅ ω r L = r ⋅ V and VL = Q ⋅ V R R I =

(22.4)

Similarly, VC = IX C = I V

I=R

I

V R VL L VC

C

Figure 22.10 Series Resonance

V 1 1 1 = ⋅ = ⋅ V = QV R ω rC ω rC ω r CR

(22.5)

Q is, therefore, a factor of magnification of the supply voltage. The voltage across the inductor as well as the voltage across the capacitor is many times the supply voltage. Q may be very large, especially if R is only the resistance of the inductor coil, and values of Q can be many hundreds of times that of the supply voltage. V VLR = IZ LR = ( R 2 + ω r 2 L2 )1/ 2 R 1/ 2  ω 2 L2   , V = (1 + Q 2 )1/ 2 = V 1 + r R   When Q >> 1, the voltages across the coil and that across the capacitor tend to be equal and of opposite polarity. (Refer Figure 22.10)

Example 22.14 A series resonant circuit has a capacitor of 50 rF in series with a 2.2 mH coil. The resistance of the circuit is 1658 W. What is the bandwidth of the circuit and the Q? What is fr? What are the frequencies at the 0.707 point of the curve?

M22_AUTH_ISBN_C22.indd 410

8/6/2012 1:15:48 PM

Resonance

411

Solution: 1 R

L 1 = × 6632 = 4 C 1658

Current

Q=

Bandwidth (Refer Figure 22.11) R 1658 = = = 12.3 kHz 6.28 L 13,816 × 10−6 f r = Q( f 2 − f1 ) = 4 × 120 = 480 kHz f2 = fr +

f1 = f r −

0.707 max

f1

f2

Frequency, kHz

Figure 22.11 Bandwidth

R 1658 = 4p L 12.56 × 2 .2 × 10−3 = 480 + 60 = 540 kHz

R = 480 − 60 = 420 kHz 4p L

VLR

Knowing the frequency, we can also find the value of Q. 480 fr Q= = =4 120 f 2 − f1

v 2.0 v

I

VLR

L 1.0 mH R 5.0 Ω

Example 22.15

VC

V

C 0.2 mF

A coil of resistance 5.0 W and inductance 1.0 mH is connected in series with a 0.2 mF capacitor. The circuit is connected to a 2.0 V variable frequency supply as shown in Figure 22.12. Calculate the frequency at which resonance occurs, the voltages across the Figure 22.12 For Example 22.15 coil and the capacitor at this frequency and the Q of the circuit. Solution: 1 1 fr = = = 11, 250 Hz 1/ 2 −3 2p ( LC ) 2p (1 × 10 × 0.2 × 10−6 )1 / 2

VC

= 11. 25 kHz I r = V /R = 2 . 0 / 5 . 0 = 0 . 4 A X L = X C = 2p f r L = 2p × 11, 250 × 1 × 10−3 = 70.7 Ω Z LR = ( R 2 + X L2 )1/ 2 = (5.02 + 70.7 2 )1/ 2 = 71.0 Ω VLR = I r Z LR = 0.4 × 71.0 = 28.4 V VC = I r X C = 0.4 × 70.7 = 28.3 V Q = VC / V = 28.3/ 2.0 = 14.15

22.4 SELECTIVITY AND BANDWIDTH For a circuit with a high value of Q (>100) VL max and VC max coincide with the maximum circuit current at fr; as seen in Figure 22.13. However, for a circuit with a low value of Q (10.   = 2p  ( LC ) L2  2p ( LC )1/ 2

M U LT IP LE C H O I C E Q UE S TI O NS (M C Q ) 1. At resonant frequency (a) X L >X C (b) XL = XC (c) X L < X C

(d) X L =

6. fr(series) = fr(parallel), provided (a) Q > 10

1 XC 2

2. Resonant frequency is given by 2p 1 (a) (b) 1 1 2 ( LC )2 2p( LC ) (c)

1 1 2pL(C ) 2

(d)

1 1 2pC ( L) 2

3. At resonance

4. The series resonant circuit, at resonance is a (a) Current multiplier (c) Voltage multiplier

(b) Power amplifier

5. The parallel resonant circuit at resonance is a (a) Current multiplier (b) Power amplifier (c) Voltage multiplier

M22_AUTH_ISBN_C22.indd 419

(c) Q = 10

7. Q is the value of (a) Reactive power to active power (b) Active power to reactive power (c) Reactive power to average power (d) Average power to reactive power

8. A series tuned circuit, above resonance is (a) Resistive

(b) Inductive

(c) Capacitive

9. A parallel tuned circuit above resonance is (a) Resistive

(a) V and I are in phase (b) V leads I by 90° (c) V lags I by 90°

(b) Q < 10

(b) Inductive

(c) Capacitive

10. A series tuned circuit below resonance is (a) Resistive

(b) Inductive

(c) Capacitive

11. A parallel tuned circuit below resonance is (a) Resistive (c) Capacitive

(b) Inductive

12. The higher the Q of a tuned circuit (a) The narrower the bandwidth (b) The broader the bandwidth (c) The same the bandwidth

8/6/2012 1:17:17 PM

420 Electrical Technology ANSWERS (MCQ) 1. (b)

2. (a)

3. (a)

4. (c)

5. (a)

6. (a)

7. (a)

8. (b)

9. (c)

10. (c)

11. (b)

12. (a).

CON V E N TI O NA L Q UE S TI O NS (C Q ) 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17.

Explain the phenomenon of resonance. What are the characteristics of a series resonant circuit? What are the characteristics of a parallel resonant circuit? Explain the difference between series and parallel resonance. Derive the expression for series resonance. From the first principles, derive the expression for parallel resonance. What is the power factor of a resonant circuit? Explain Q and selectivity. How are they related? What is bandwidth and how does it relate to Q? Which one is more selective and why, a high-Q circuit or a low-Q circuit? What is fly wheel effect? How is it related to resonance? Explain in detail the exchange of energy between an inductor and a capacitor with the help of illustrations. What effect does a shunt resistor have in a parallel resonant circuit? What effect does the inductor resistance have in a parallel resonant circuit? A tuner for v.h.f. signals has a capacitor of 900 pF and an inductance of 5 mH. What is the resonant frequency? A control circuit contains a 705 µH inductor in series with a capacitor. Resonance is at 300 kHz. What is the value of the capacitor? In a ship-to-shore communication system, a series resonant circuit had an inductive reactance of 2700 Ω. The series resistor has a value of 45 Ω. What is its Q?

ANSWERS (CQ) 16. 400 pF 17. 60 18. 5 19. 871.2 kHz, 967 kHz 20. 4 µH, 4.5 pF 21. 12.6 Hz

M22_AUTH_ISBN_C22.indd 420

18. In an industrial motor control system, a circuit contained a 500 µH inductance and a 0.0005 µF capacitor in a series resonant circuit. If the series resistor is 20 Ω. What is the circuit Q? 19. In the design of a resonant circuit for the radio broadcast band a capacitor of 0.001 µF was placed in parallel to an inductor of 270 µH. The inductor had an internal resistance of 500 Ω. What is the resonant frequency if the inductor resistance were reduced to 50 Ω value? 20. In a v.h.f. receiver being designed for 37.5 kHz, the tuning capacitor had a value of 9 pF. It was decided to double the L/C ratio for a broader band-pass. What are the new values of capacity and inductance? 21. Calculate the resonant frequency for an 8-H inductance and a 20 µF capacitance. 22. Calculate the resonant frequency for a 2 µH inductance and a 30 F capacitance. 23. What value of capacitance resonates with a 239 µH inductance at 1000 kHz? 24. What value of inductance resonates with a 10 pF capacitance at 1 MHz? 25. A series circuit resonant at 0.4 MHz develops 100 mV across a 250 µH inductance with a 2 mV input. Calculate Q. 26. What is the a.c. resistance of the coil in the preceding example? 27. An LC circuit resonant at 2000 kHz has a Q of 100. Find the total bandwidth and the edge frequencies f1 and f2.

22. 65 MHz 23. 106 pF 24. 239 µH 25. 50 26. 12.56 Ω 27. 20 kHz, 1990 kHz, 200 kHz

8/6/2012 1:17:18 PM

23

The Fourier Series OBJECTIVES In this chapter you will learn about:    Synthesis of non-sinusoidal waveforms by combining a number of sine waves and cosine waves    Fourier series and its ability to express finite, continuous and singlevalued functions in terms of sine and cosine series    Harmonic content of the Fourier expressions, which represent a number of non-sinusoidal waveforms   Calculating the effective value of a non-sinusoidal waveform      The expression for the non-sinusoidal current in a series a.c. circuit    Determining the expression for the total non-sinusoidal current in a series-parallel a.c. circuit

+

0

–

5f1 3f 1

f1

Complex waveform-fundamental plus third plus fifth harmonics

23.1 INTRODUCTION We frequently come across waveforms of complex shape. If they repeat the same pattern, complex waveforms can be analyzed into two or more components, each of which is itself of pure sine waveform. It is for this reason that a.c. theory can be developed on the assumption of a pure sine waveform. In the eighteenth century, a French mathematician by the name of Jean Baptist Fourier (1768–1850) was trying to solve a problem involving the flow of heat. He developed a means of analysis by which any finite, continuous, and periodic waveform could be analyzed into a series of sinewaves. The general series contains a fundamental sinewave whose frequency is the same as that of the periodic waveform; together with the fundamental are harmonic components, so that the derivation of the Fourier series is sometimes known as harmonic analysis.

23.2 COMPLEX WAVEFORMS The construction of complex waveforms is demonstrated more easily by graphical synthesis rather than by analysis. In Figure 23.1 is shown a sine-wave current if = Imax sin ωt, of peak value Imax and frequency

ω Hz, the second current shown, having a smaller amplitude and three 2π ′ sin 3ωt. If the times the frequency of if may be represented by i3 = I max two voltages which produce these currents are applied simultaneously to a simple circuit, the instantaneous total current may be determined by adding the ordinates of the component current waves at successive instants; this produces the complex waveform of the dotted line which may be written as: ′ sin 3 ω t ) i C = ( I max sin ω t + I max

(23.1)

iC = (IM sin ωt + I′M sin 3ωt)

+

0 –

i3 = I ′M sin 3ωt)

if = IM sinωt)

Figure 23.1 Complex Wave: Fundamental Plus Third Harmonic

422 Electrical Technology In any complex waveform, the lowest frequency is called the fundamental frequency; other sinusoidal components whose frequencies are exact multiples of the fundamental are called harmonics. In Figure 23.1, if has the fundamental frequency ω f1 = , while i is a third harmonic of frequency f3 = 3 ω (2π ) . A component of frequency f 4 = 4ω (2π ) would 3 (2π ) be a fourth harmonic; the first harmonic coincides with the fundamental frequency. In electronics, a practical example of a periodic waveform is the sawtooth voltage. Such a voltage is applied to a cathode ray tube so that the beam is deflected horizontally from left to right with a constant velocity. This non-sinusoidal waveform, illustrated in Figure 23.2, is composed of a fundamental component and a large number of harmonics. If all these components are added together (a process known as synthesis) the result, of course, is the original sawtooth voltage. A complex waveform containing, fundamental frequency f1, third harmonic frequency 3f1, and fifth harmonic frequency 5f1, is built up in Figure 23.3. This shows the general symmetrical waveform produced by the presence of odd harmonics and in the limit, in case all the odd harmonic frequencies up to infinity are taken the square topped waveform is produced (Figure 23.4).

Figure 23.2 Non-sinusoidal Waveforms (a) Negative Sawtooth Waveform (b) Positive Sawtooth Waveform (c) Triangular Waveform (d) Symmetrical Squarewave with a d.c. Average Value (e) The Symmetrical Square Waveform (f) Half Wave (HW) Rectification (g) Full Wave (FW) Rectification

The Fourier Series

–

+ 0 –

423

0

5f1 3f1

+

f1

Figure 23.3 Complex Wave; Fundamental Plus Third and Fifth Harmonics

Figure 23.4 Square-topped Waveform

Complex waveforms produced by the addition of fundamental, 2nd and 3rd harmonics are shown in Figure 23.5, and in Figure 23.6 for the addition of the fundamental and second to seventh harmonics. Harmonic components of a complex wave may differ from the fundamental in three ways: (1) in relative peak values; (2) in frequency; and (3) in phase displacement. The complete analysis of a complex wave is stated by Fourier’s theorem.

Figure 23.5 Synthesis of a Non-sinusoidal Voltage Waveform

424 Electrical Technology

Figure 23.6 Synthesis of the Sawtooth Wave

23.3 SYNTHESIS OF NON-SINUSOIDAL WAVEFORMS Consider a non-sinusoidal voltage waveform that contains fundamental, second harmonic, and third harmonic components, as illustrated in Figure 23.5. Let us assume that the amplitudes of the second and third harmonics are, respectively, one-half and one-third of the fundamental amplitude. Then, if the fundamental wave is represented by e1 = E1 sin ωt, the expressions for the second and third harmonic are as follows: E sin 2ω t 2 E e3 = sin 3ω t 3 The complete equation for the nonsinuusoidal voltage is e2 =

e

E

E E sin 2ω t sin 3ω t 2 3 1.00 sin ω t 0.500 sin 2ω t 0.333 sin 3ω t

E

1.00 sin 2 ft

E

1 .00 sin

E sin ω t

0.500 sin 4 ft

0 .500 sin 2

(23.2)

0.333 sin 6 ft

0.333 sin 3

(23.3)

The Fourier Series

425

where, e = instantaneous value of the non-sinusoidal wave (V) E = amplitude of the fundamental sine wave (V)

ω = angular frequency (rad/s) f = frequency (Hz)

θ = angle (rad)



t = time (s) The three components (fundamental and harmonics) start off together at the beginning of their positive half cycles. This accounts for the positive signs in front of all the terms in the equations for the voltage. To derive the non-sinusoidal wave, we can plot the fundamental and the two harmonics on a common graph by using a horizontal radian (or degree) scale. At a number of conveniently chosen points, we can then take the algebraic sum of the vertical distances associated with the instantaneous values of the three components. The resultant waveform has a striking similarity with the sawtooth voltage in Figure 23.2(a) with a negative slope. If we add more and more harmonics (of the correct size) we will be able to obtain a closer and closer approximation to the negative sawtooth waveform. This has been illustrated in Figure 23.6. The perfect sawtooth can be obtained only by including an infinite number of harmonics. However, the higher the order E sin 100ω t , of the harmonic, the less is its amplitude; for example, the amplitude of the hundredth harmonic is, e = 100 only 1 per cent of the fundamental amplitude. Thus only a limited number of harmonics are required to obtain a good approximation to the negative sawtooth voltage. When sufficient harmonics are included, the peak value of the sawtooth waveform is approximately 1.57 × E volts. Theoretically, the Fourier series contains an infinite number of harmonics, but in practice, only a limited number are necessary to obtain a good approximation to the required waveform.

23.4 THE FOURIER SERIES The Fourier theorem can be stated as follows: any finite, continuous, single-valued function, f (t) that has a period of 2π radians (360°) can be expressed by the following series: (23.4)

(23.5) f(t) is the mathematical expression for a function of time; any quantity whose instantaneous value is dependent on time; finite indicates that f (t) contains no infinities; single-valued means that f (t) cannot have more than one value at a particular time; continuous means that there are no discontinuities in the function. Note:

1. f (t) = 1/(t – 1) cannot be expressed by a Fourier series because as t → 1, f (t) → ∞. 2. The f (t ) =

t is double valued; such a function cannot be analyzed into a Fourier series.

3. f (t ) = (t − 1)(t − 2) has no real value between t = 1 and t = 2, because, we cannot take the square root of an imaginary number and the sum of the Fourier series never can be an imaginary number. 4. The term αo is a non-alternating component, which may or may not be present in a given complex wave. It represents the mean level of f (t).

426 Electrical Technology 5. If the positive and negative excursion of a non-sinusoidal waveform are not equal (asymmetrical waveform), the waveform can be regarded as composed of a d.c. value together with the alternating components. 6. The fundamental and harmonic components are each associated with phase angles. The existence of phase angles means that the fundamental and the harmonics do not start off together at the beginning of their positive half cycles. 7. The presence of the cosine terms allows for the phase difference between harmonic components.

23.5 ANALYZING A COMPLEX WAVEFORM To analyze a complex waveform: 1. We need to determine the coefficients a1, a2, a3,……….., and b1, b2, b3, ……., by multiplying both sides of the Fourier equation by a suitable factor; 2. We may then integrate between the limits 0 and 2π. If the multiplying factor is chosen correctly, all the terms disappear except those that give the required coefficient. a0 =

1 2π

2π

∫0

f (t ) d (ω t )

(23.6)

Where, a0 is the mean value of f (t) between the limits 0 and 2π an = where,

2π

∫0

f (t ) cos nω t d (ω t )

(23.7)

n is any positive integer. The value of an is twice the mean value of f (t) cos n ωt between the limits of 0 and 2π.

bn = where,

1 π

1 π

2π

∫0

f (t ) sin nω t d (ω t )

(23.8)

n is any positive integer. Therefore, bn is twice the mean value of f (t) sin nωt between the limits of 0 and 2π.

Example 23.1 The e waveform of Figure 23.5 has a fundamental component whose amplitude is 6 V. Calculate the amplitudes of the second and the third harmonics. Solution: Second harmonic amplitude = 6/2 = 3 V Third harmonic amplitude = 6/3 = 2 V Example 23.2 Find the values of the coefficients in the Fourier analysis of the square waveform in Figure 23.2(d). Solution: The square wave in Figure 23.2(d) is a single-valued, finite, continuous, periodic function of ωt and has a period of 2π. It can, therefore, be analyzed by Fourier’s theorem. 1. From t = 0 to t = π/ω, the equation of the function is f (t) = D. 2. From t = π/ω to t = 2π/ω, the equation of the function is f (t) = 0. Then, 1 2π 1 π 1 f (t )d (ω t ) = f (t )d (ω t ) + 2π ∫ 0 2π ∫ 0 2π 1 D = × D×π ×0 = 2π 2 D . This verifies that the mean value of f (t ) is 2 Also a0 =

an =

1  π × D cos nω t d (ω t ) + π  ∫ 0 π

2π

∫π

2π

∫π

0 cos nω t d (ω t )  

f (t )d (ω t )

Then, 1 2π 1 π 1 f (t )d (ω t ) = f (t )d (ω t ) + ∫ 2π 0 2π ∫ 0 2π 1 D = × D×π ×0 = 2π 2 D . This verifies that the mean value of f (t ) is 2 Also a0 =

an =

1  π × D cos nω t d (ω t ) + π  ∫ 0

2π

∫π

2π

∫π

f (t )d (ω t )

The Fourier Series

427

0 cos nω t d (ω t )  

π

1  D s i n nω t  ×  =0 π  π 0

=

All cosine terms are therefore zero. This is because of the symmetry of the waveform Finally 2π 1 × ∫ f (t ) sin nω t d (ω t ) 0 π π 2π 1 =  ∫ D sin nω t d (ω t ) + ∫ 0 sin nω t d (ω t )    π 0 π 

bn =

=

1  − D cos nω t  ×  π  π 

2π

=

D (1 − cos nπ ) nπ

When n is odd, (1 – cos nπ) = 2 When n is even, (1 – cos nπ) = 0 This yields b1 =

2D , π

b2 = 0,

b3 =

2D , 3π

b4 = 0, and so on.

The required equation for the square wave is D 2D  1 1 1  f (t ) = + sin ω t + sin 3ω t + sin 5ω t + sin 7ω t + ......   π  2 3 5 7  Example 23.3 Find the values of the coefficients in the Fourier analysis of the square waveform in Figure 23.2 (e). Solution: D The waveform, under consideration, is symmetrical about the time axis. Therefore, the first term, , does not appear in 2 the equation, which is 2D  1 1  f (t ) = ×  sin ω t + sin 3ω t + sin 5ω t + ......  π 3 5   = E ( +1.00 sin θ + 0.333 sin 3θ + 0.200 sin 5θ + .... ) E D The synthesis of this square wave has been illustrated in Figure 23.7; its peak value is  =π× 4  2 approximately.

  , 0.785 E 

Example 23.4 Find the values of the coefficients in the Fourier analysis of the modified sawtooth waveform in Figure 23.8. Solution: In symmetrical waveforms, there are no cosine terms. However, there are a number of cosine terms in asymmetrical waveforms. D 2D  cos 3ω t cos 5ω t  f (t ) = − 2  cos ω t + + + .....  + 4 π  32 52  D π D 4 D π

si n 2ω t sin 3ω t   + + ......   sin ω t − 2 3   2D π2

cos 3ω t cos 5ω t   + + .....  +  cos ω t + 9 25  

sin 2ω t sin 3ω t   + + ......   sin ω t − 2 3  

428 Electrical Technology

Figure 23.7 Synthesis of the Square Wave

Figure 23.8 Modified Sawtooth Waveform for Example 23.4

Example 23.5 Give the formula for the triangular waveform in Figure 23.2(c). Solution: f (t ) = =

8D π2

cos 3ω t cos 5ω t   + + .....   cos ω t + 32 52  

cos 3ω t cos 5ω t 8D   cos ω t + + + ......  2  9 25 π  

Example 23.6 Give the formula for half-wave rectified a.c. waveform. Solution: 1 1 1 1 π  cos 2ω t − cos 4ω t − cos 6ω t.....   + sin ω t − 1× 3 3×5 5×7 2 4 

=

2D π

=

2D  1 π 1 1 1  cos 4ω t − cos 6ω t.....   + sin ω t − cos 2ω t − π 2 4 3 15 35 

Example 23.7 Give the formula for the full-wave rectified a.c. waveform. Solution: 4D  1 1 1 1  f (t ) = cos 2ω t − cos 4ω t − cos 6ω t.....   − π  2 1× 3 3×5 5×7  =

4D π

1 1 1 1  cos 4ω t − cos 6ω t.....   − cos 2ω t − 2 3 15 35  

The Fourier Series

429

Example 23.8 The half-wave rectified a.c. waveform of Figure 23.2 (f) has a peak value of 100 V. Calculate the amplitudes of the fundamental component and of all the harmonies up to and including the sixth. Solution:

Example 23.9 A non-sinusoidal source has a voltage represented by the equation e = 100 sin ωt – 80 sin (2ωt + 40°) + 40 sin (3ωt – 20°) V, where, the frequency f = ω / 2p = 400 kHz. This voltage is then applied across the series circuit in Figure 23.9. Obtain the equation for the sinusoidal current and then obtain its effective value and the total power dissipated. Solution: The fundamental component X L = 2π fL

Figure 23.9 Non-sinusoidal Wave Applied to a Series LCR Circuit

= 2 × 3.14 × 400 × 103 × 150 × 10−6 = 377 Ω

XC =

1 2π fL

1 2 × 3.14 × 400 × 103 × 250 × 10 −14 = 1592 Ω =

Total impedance

Z = 30 + j377 – j1592 = 30 – j1215 = 1215 −88.6° Ω

Peak value of the fundamental current, i1 =

100 V 1215 −88.6° Ω

= 0.082 −88.6° A Effective value of the fundamental current = 0.082 × 0.707 = 0.058 A The second harmonic component X L = 2 × 377 = 754 Ω X C = 1592 / 2 = 796 Ωc Z = 30 + j 754 − j 796 Ω = 30 − j 42 Ω = 51. 6 − 54.5° Ω Peak value of the second harmonic current, i2 =

−80 40° V 51.6 −544.5° Ω

= 1.55 −85.5° A Effective value of the second harmonic current = 1.55 × 0.707 = 1.10 A

430 Electrical Technology The third harmonic component XL = 3 × 377 = 1131 W XC = 1592/3 = 531 W Z = 30 + j1131 – j 531 W = 30 + j 600 W

Total impedance

= 600.7 / – 87.1° W Peak value of the third harmonic current, i3 =

40 −20° V 600.7 −87.1° Ω

= 0.067 −107.1° A Harmonic current = 0.067 × 0.707 = 0.047A Total non-sinusoidal current, i = 0.082 sin (ωt + 88.6°) + 1.55 sin (ωt – 85.50°) + 0.067 sin (3 ωt – 107.1°) Effective sinusoid current,

I =

0.0582 + 1.102 + 0.047 2

= 1.1025 A Total power dissipated = 1.10252 × 30 = 36.5 W Example 23.10 In the circuit given in Figure 23.10, a sinusoidal voltage source e = 14.14 sin (ωt + 20°) + 3.535 sin (3ωt + 40°) – 1.414 sin (5ωt – 60°) V is impressed. The reactance values shown correspond to the fundamental frequency. Obtain the equation of the instantaneous line current and then obtain its effective value. Solution: Fundamental current component 14.14 20° V = 1.09 −47. 4° mA i1′ = 5 + j12 kΩ i1′′ =

14.14 20° V = 1.41 4 56.8° mA 8 − j 6 kΩ

Figure 23.10 Non-Sinusoidal Wave Applied to a Parallel LCR Circuit

Fundamental line current, i1 = 1.09 − 47.4° + 1.414

− 56.8°

= 0.738 − j 0.802 + 0.774 + j1.28 = 1.512 + j0.378 = 1.56 14.04° mA Effective value of,

i1 = 1.56 × 0.707 = 1.10 mA

Third harmonic component XL = 3 × 12 = 36 kW,

XC = 6/3 = 2 kW

3.535 40° × (5 − j 36) = 0.097 −42.1° mA 52 + 362 3.535 40° × (8 + j22) i3′′ = = 0.43 −54° mA 82 + 2 2 i3 = 0.097 −42.1° + 0.43 54° i3′ =

= 0.431 41.05 ° mA Effective value of i3 = 0.431 × 0.707 = 0.305 mA

The Fourier Series

431

Fifth harmonic component XC = 6/5 = 1.2 kW

XL = 5 × 12 = 60 kW,

−1.414 −60° × (5 − j 60) 52 + 602 = −0.0235 −145.2° mA

i 5′ =

A = 0 .0235 34 .8 ° mA −1.414 −60° × (8 + j1.2) 82 + 1 . 2 2 = −0.175 −51.47° = 0.175 128 .5° mA

i 5′′ =

i 5 = 0. 0235 34. 8° + 0. 175 128. 5 ° = −0. 0901 + j 0. 1504 = 0 .175 120 .1 ° mA Effective value of i 5 = 0.175 × 0.707 = 0.124 mA Instantaneous line current i = i1 + i3 + i5 = 1.56 sin (ωt + 14.04°) + 0.431 sin (3ωt + 41.05°) + 0.175 sin (5ωt +120.1°) The effective line current is, I = 1.102 + 0.3052 + 0.1242 = 1.15 mA

23.6 SUMMARY OF PROPERTIES OF FOURIER ANALYSIS Once the type of symmetry (even or odd) of a waveform has been established, the following properties appear. 1. For an even waveform, all the terms of its Fourier series are cosine terms. No sine terms are present. However, the function does have an average value. 2. For an odd waveform, the series contains only the sine terms. There is no average value and no cosine terms. In special cases, the given waveform may be odd but only after its average value is subtracted. 3. If the given waveform is of half wave symmey y y try, only odd harmonics are present in the series. The series would contain both sine and cosine terms when n is odd. The average value is zero.

23.6.1 Waveform Symmetry It is not necessary to calculate the coefficient by observing the symmetry of the non-sinusoidal waveform, it is possible for us to predict the absence of certain coefficients. The following wave symmetries are helpful in this prediction. 1. If the area of the + ve loop in a period is equal to the area of the –ve loop, then A0=0. 2. If the functions describing the non-sinusoidal wave is even, then Bx=0. The Fourier consists of only cosine terms. A function Y=f( θ) is said to be even when f(θ) = –f(– θ). The waveforms shown in Figure [23.11 (a), (b) and (c)] are described by even functions. If the function describing the non-sinusoidal wave is odd, then An=0. The Fourier series consists of only

0

0

θ

(a)

θ

0

(b)

y

(c) y

y

0

0

0

θ

(d)

θ

(e)

(f)

y

0

π

θ

2π

θ

(g)

Figure 23.11 Waterform Symmetry

θ

432 Electrical Technology sine terms. A function is said to be odd when f(θ)= –f(– θ). The waveforms shown in Figure [23.11 (d), (e), and (f)] are described by odd functions. If the function describing the non-sinusoidal waveform is having half-wave symmetry, then the series contains only odd harmonics. All even harmonics are absent, i.e. A2, A4, B2, B4, etc. will be zero. However, the function may contain both sine and cosine terms unless the function is odd or even. A function y=f(θ) is said to be having half wave symmetry, if f(θ) = –f(θ±p) or f(t) = –f(θ±T/2). The waveforms shown in Figure 23.11 (g) is described by a function having half wave symmetry.

23.6.2 Complex Waveform Considerations It is sometimes possible to predict the harmonic content of a waveform on inspection of particular waveform characteristics. 1. If the periodic waveform is such that the area above the horizontal axis is equal to the area below the horizontal axis, then its mean value is zero. (See Figure [23.12 (a–e)]). 2. An even function is symmetrical about the vertical axis and contains no sine terms. (See Figure 23.12 (b)). 3. An odd function is symmetrical about the origin and contains no cosine terms. (See Figure 23.12 (c)). 4. f (x) = f (x+p) represents a waveform which repeats after half a cycle and only even harmonics are present. (See Figure 23.12 (d)). 5. f (x) = – f (x+p) represents a waveform for which the positive and negative cycles are identical in shape and only odd harmonics are present. (See Figure 23.12 (e)).

Figure 23.12 Predicting the Harmonic Content of a Waveform by Inspection (a) a0 = 0 (b) Contains No Sine Terms (c) Contains No Cosine Terms (d) Contains only Even Harmonics (e) Contains only Odd Harmonics

S UM M A RY 1. Complex waveforms—if they repeat the same pattern—can be analysed into two or more components, each of which is itself of pure sine waveform. 2. Any finite, continuous, and periodic waveform can be analysed into a series of sine waves.

3. The higher the order of the harmonic, the less is its amplitude. 4. Fourier’s theorem states that, any finite, continuous, single-valued function that has a period of 2π (360°) radians can be expanded by the series.

M U LT IP LE C H O I C E Q UE S TI O NS (M C Q ) 1. The derivation of the Fourier series is known as (a) Fourier analysis

(b) Fourier synthesis

2. Complex waveforms are demonstrated more easily by graphical (a) Analysis

(b) Synthesis

3. In a complex waveform the lowest frequency is called (a) First harmonic (c) Second harmonic

(b) Fundamental

4. Harmonic components of a complex wave may differ in (a) Peak values (b) Frequency (c) Phase displacement

5. If the area of positive and negative loops of a non-sinusoidal waveform are not equal. The waveform can be regarded as composed of (a) A d.c. value (b) Alternating components (c) D.c. value together with alternating components

The Fourier Series

6. In a symmetrical waveform, there are (a) Sine terms

10. If the function describing the non-sinusoidal waveform is even, then

(b) Cosine terms

(a) Bn = 0

7. For an even waveform (a) No sine terms are present (b) No cosine terms are present

8. An odd waveform contains (a) Only sine terms

(b) Only cosine terms

9. If the area of the positive loop in a period is equal to the area of the negative loop, then (a) A0 ≠ 0

(b) A0 = 0

ANSWERS (MCQ) 1. (a) 2. (b) 3. (a) and (b)

4. (a), (b) and (c)

433

5. (c)

(b) Bn ≠ 0

11. A function y = f ( θ) is said to be odd when (a) f(0)= –f( θ) (b) f(0) = f(–  θ) (c) f(θ) = –f(–  θ) 12. If the function describing the non-sinusoidal waveform has half wave symmetry, then (a) The function contains only odd harmonics (b) The function contains only even harmonics

6. (b)

7. (a)

8. (a )

9. (b)

10. (a)

11. (c)

12. (a).

CON V E N TI O NA L Q UE S TI O NS (C Q ) 1. 2. 3. 4.

List four different types of periodic waveforms. What does the term Fourier analysis convey? State Fourier’s theorem. What is the difference between sine and cosine terms in Fourier series? 5. What are harmonics? How does their frequency and amplitude relate to that of the fundamental frequency? 6. Explain the difference between symmetrical and asymmetrical waveforms. ANSWERS (CQ) 7. 12, 120 HzS, 180 Hz 8. (a) 17.0 V (b) 5.4 V (c) 8.5 V (fundamental), 3.61 V (second harmonic)

7. The waveform in Figure 23.6 has an amplitude of 24 and a fundamental frequency of 60 Hz. Calculate the amplitude and frequency of the second and third harmonics. 8. A 12 V r.m.s., 440 Hz, sine wave is rectified into a half-wave rectified signal. Calculate: (a) peak value; (b) d.c. value; (c) amplitudes of the first three non-zero harmonics. 9. Using the voltages from CQ 8, approximate the effective voltage.

9. 0.72 V (fourth harmonic)

24

Networks (a.c.) OBJECTIVES In this chapter you will learn about:    Application of Kirchhoff’s laws to single phase a.c. circuits  Voltage division in series-connected impedances    Current division in parallel-connected impedances  The principle of superposition  Thevenin’s and Norton’s theorems  Constant voltage and constant current generators    Maximum power transfer theorem and its application    Millman’s theorem and its application to multiple current source circuits  Reciprocity theorem and its application  The concept of duality  Mesh-current analysis  Node voltage analysis   Problems concerning the afore-mentioned network theorems   and their systematic solutions

i2 Solder

No i1 Leads

i4

i2 Solder

Leads

Node i1

i3 i4

Kirchhoff's Currernt Law (KCL)

24.1 INTRODUCTION 100 – Ω Circuit theorems are ideas about circuits expressed in a wellResistor defined manner. Most of them have been defined formally by R b c their originators. These definitions may be applied to the solution +v of problems which could otherwise be awkward or complicated. R 12–V 2–H Battery Inductor v + Many network problems involve only a restricted analysis (see vL+ L S Figure 24.1). Such network analysis problems can be expeditiously solved by the application of certain network theorems. a d A common problem in network analysis is to determine the (a) (b) response of the circuit to an input signal. Complicated twoterminal networks can be reduced to simpler circuits that are Figure 24.1 Restricted Analyses (a) A Voltage Experiment (b) Voltage Calculation equivalent in nature. Two circuits are equivalent if they present the same v-i characteristic. Replacing a complicated network with a simpler equivalent network is advantageous in network analysis. Other useful tools in network analysis include R1 R2 R3 the voltage divider and the current divider. i

vR1

24.2 VOLTAGE DIVISION v

Figure 24.2 Voltage Division

Figure 24.2 shows three resistors connected in series with a supply voltage, v. In order to find the p.d. across R1 we proceed in the following manner.

Networks (a.c.)

435

i = v / RT amperes v i= amperes; vR1 = iR1 volts R1 + R 2 + R3 ∴

vR 1 = v ⋅

R1 R1 + R 2 + R3

volts

(24.1)

Equation 24.1 shows that the p.d. across R1 is given by the supply voltage multiplied by the ratio R1/ (R1 + R2 + R3). There is no need to calculate the current to obtain the required p.d. The method of voltage division is only valid as long as no additional connections are made to the series circuit. The method can also be applied to an a.c. circuit, but care must be taken to work in reactances and to carry out phasor addition. Figure 24.3 shows a series network where R3 (Figure 24.2) is replaced by a capacitor of reactance −jXc ohms. The p.d. across R2 is found by voltage division. R2 vR 2 = vZ × Z But

R1

–jXc vz

Figure 24.3 Voltage Division with Reactive Component

Z = R1 + R 2 − jX c

The magnitude of the impedance is Z = ∴

( R12 + R22 ) + X c2 VR 2 = vZ

ohms R2 2

( R1 + R 2 )

volts + X c2

Example 24.1 In Figure 24.3 the components have the values R1 = 200 Ω;

R 2 = 500 Ω;

Calculate the p.d. across R2 Solution:

X c = 600 Ω;

Z = 200 + 500 − j 600 Ω 1/ 2

Z = ( 7002 + 6002 )

Ω

= 922 Ω  500  vR 2 = 25  V  922  = 13.56 V Example 24.2 In Figure 24.3 find the p.d. across the capacitor Solution: vc = 25 ×

R2

− j 600 700 − j 600

600 V 922 = 16.26 V

vc = 25 ×

Note: We are not concerned with the phase angle but only the magnitude.

v z = 25 V

436

Electrical Technology

24.3 CURRENT DIVISION In Figure 24.4 we want to find the magnitude of current which flows in one of the parallel arms. The p.d. across the parallel network of resistors is R1 R 2 R1 v = i× volts i1 R1 + R 2 i

R2

By Ohm’s law

i2

i1 =

Figure 24.4 Current Division

v amperes, R1

∴ i1 = i ⋅

R2 R1 + R 2

amperes

(24.2)

Equation (24.2) is similar to Eq (24.1) except that in order to find the current which flows in one resistor, we use the other resistor as the numerator. The immediate usefulness of the formula is limited to two resistors. For more than two resistors in parallel, it is better to use step-by-step solutions. If one of the resistors illustrated in Figure 24.4 is replaced by a reactive component, no simple formula for the current in one branch can be found. If a capacitor of reactance −jXc is used, then the current in the resistive branch will be iR = iZ

X c2 − jX c R amperes R 2 + X c2

(24.3)

Equation 24.3 is complex and, when shown in the magnitude form, it becomes very clumsy. It is better to insert numerical values for each problem.

24.4 SUPERPOSITION THEOREM The circuit in Figure 24.5 consists of two generators E1 and E2 whose internal resistances are r1 and r2, respectively. These E1 superposition =0 generators are connected to two resistors R1 and R2 . To find the p.d. vR1 , |the theorem can be used. The p.d. across R1 will be the sum of the two p.d.s which would result R1 B if only E1 or E2 were in circuit. However, if the complete E1 supply is A removed, then r1 is also removed which, of course, will change the r r1 2 vR + + resistance in the circuit. In order to avoid any change of resistance, E2 R2 E1 we must make sure that when E1 is removed, we leave the internal – – resistance r1 in the circuit and replace E1 by a short circuit. When vR1 | E1 = 0 has been calculated, we may repeat the process. This time we can replace E2 with a short circuit and calculate vR 2 | E 2 = 0 . Figure 24.5 Superposition Theorem | E1 =both 0 E1 and E2 are in circuit we calculate To find the value of vR1 when 1

vR1 = vR1 | E1 = 0 + vR 2 | E2 = 0 volts

(24.4)

taking into account the polarity between the points A and B. The two separate circuits for the solution by the superposition theorem are shown in Figure 24.6 r2 R1 R1 B A A B In Figure 24.6(a), vR1 will be given by the voltage division. vR 1 | E2 = 0 = E1

R1 r1 + R1 + R 2

r2

volts

When R2 || r2 means R2 and r2 in parallel. In Figure 24.6(b), a double voltage division is required: vR 1 | E1 = 0 = E2

R1

R2

r1 + R1 R 2

( R1 + r1 ) ( R1 + r1 ) r2

volts

r1

+ –

vR1

E1

R2 (a)

vR1 r2

r1

+ E2

R2

–

(b)

Figure 24.6 Solution by Superposition Theorem

The polarities shown on the generators indicate that the e.m.f.s have a phase relationship which causes the current to flow through R1 in the opposite direction; consequently, the resultants p.d. across R1 will be the difference between the separate p.d.s. Example 24.3 Two voltage generators are connected together. One generator has an e.m.f. of E1 volts and an internal resistance of 5 Ω. The second generator has an e.m.f. of E2 volts and an internal resistance of 600 Ω. Find the p.d. at the output terminals.

Networks (a.c.)

Solution: The equivalent circuit is shown in Figure 24.7, 600 5 + E2 ⋅ volts vx = E1 600 + 5 600 + 5 = 0.992 E1 + 0.0083 E2 volts

600 Ω

5Ω E1

The e.m.f. produced by E2 generator is almost completely lost because of the 600 Ω internal resistance when compared with the 5 Ω of the other generator.

437

E2

vx

Figure 24.7 For Example 24.3

Example 24.4 5Ω E1

1 MΩ

1 MΩ

1 MΩ

600 Ω

vx

E2

Figure 24.8 With Summing Network \

vx =

How can the above drawback be overcome? Solution: If a summing network is added, an output will be obtained from E2. Figure 24.8 shows the equivalent circuit with the summing network added. The network consists of three resistors of equal value. If the summing resistors have a high resistance, the internal resistance of the generator becomes insignificant.

1 1 E1 + E2 volts 3 3

S

R vc

C

E

Example 24.5

Find the p.d. across the capacitor after a time interval t. Figure 24.9 For Example 24.5 Solution: When the switch S is closed, a constant voltage E is connected. If we want to know the p.d. across the capacitor after a time interval t, it can be found by using the superposition theorem, as shown in Figure 24.10. In Figure 24.10(a) the capacitor p.d. is zero, and rising R R exponentially towards E. In Figure 24.10(b) the capacitor p.d. is vc and falling exponentially towards zero. From the charge and vc = 0 vc E C C discharge formulas (a)

(b)

Figure 24.10 Solution of Example 24.5 by Superposition Theorem

vc = V (1 − e −t / cR ) volts vc = V ′e − t / cR

→ →

volts

charge discharge.

Using the superposition theorem for Figure 24.9 vc = V ′e −t / cR + V (1 − e −t / cR ) volts 1

24.5 THEVENIN’S THEOREM One of the most useful theorems suggested by Thevenin, it states that a complicated network of e.m.f.s and impedances can be reduced to a single e.m.f. in series with single impedance. The advantage of using this theorem is that once we have obtained the so-called Thevenin equivalent circuit, any load may be connected, and the output voltage, or output power, can be quickly found. Thevenin’s theorem has been illustrated in Figure 24.11. V = IZ = I/Y, I = V/Z = VY and Z = V/I.

1

Z I=V

+ V –

Z

Z

(a)

2

(b)

2

Figure 24.11 As Viewed from Terminals 1 and 2, V Represents the Open-Circuit Voltage and I, Example 24.6 the Short-Circuit Current C A Apply Thevenin theorem to the circuit in Figure 24.12, across CD. 20 Ω 14 Ω Explain the theorem vis-à-vis the figure below. E vTH Load 10 Ω Solution: 6V In Figure 24.12, the terminals of the supply are AB. In addition to the internal resistance, there are two extra resistors connected between AB B D and CD. Thus, C, and D become the output terminals to which any load can be connected. If the load voltage is to be found by methods Figure 24.12 For Example 24.6 other than using Thevenin’s theorem, each value of load resistor (or impedance) will require a completely new set of calculations. To avoid this, we proceed as in the following manner.

438

Electrical Technology A

1. Disconnect the load from CD and find the open-circuit voltage, vTH. When no load is connected, there will be no current through the 14 Ω resistor. So we can find the open-circuit voltage by voltage division. 10 vTH = 6 × = 2V 10 + 20

C 14 Ω

20 Ω

10 Ω

rTH

2. Replace the e.m.f. with a short circuit and find the impedance looking into the network at CD. This is called the Thevenin equivalent impedance rTH. The circuit for this is shown in Figure 24.13. Looking in at CD,

D

B

Figure 24.13 Thevenin Equivalent Resistance

rTH = 14 + 10 || 20 Ω = 20.67 Ω

20.67 Ω

The complete Thevenin equivalent generator is shown in Figure 24.14. It is now a simple matter to find the load voltage, irrespective of the value of the load.

C

rTH vTH

Load

2V

Example 24.7 The generator of Figure 24.15 has an internal impedance that consists of a 1000 Ω resistor in parallel with an inductor of reactance 1000 Ω. A resistance load of 600 Ω is to be connected to the generator. Find the Thevenin equivalent generator for the circuit. What is the p.d. at the generator terminals when the load is connected?

D

Figure 24.14 Thevenin Equivalent Generator

Solution: 1000 Ω

R

L

Z TH = R || jX L Ω

XL = 1000 Ω

=

C

E 4V

vo

Z TH =

600 Ω

=

D

RjX L Ω R + jX L j 1000 × 1000 Ω 1000 + j 1000 j 1000(1 − j′) j 1000 Ω Ω= 2 1 + j′

Z TH = 500 + j 500

Figure 24.15 For Example 24.7 When no load is connected at CD, no current will flow

ZTH

∴ vTH= E= 4 V The equivalent generator is shown in Figure 24.16. The load voltage is given by voltage division. 600 v0 = 4 ⋅ V 600 + 500 + j 500 ∴

v0 = 4 ⋅

+9V

v0 = 4 ⋅

22 kΩ

6 V 11 + j 5 6

(11

2

1/ 2

+ 52 )

vTH

4V

C

500 + j 500 Ω vo

600 Ω

D

Figure 24.16 Equivalent Generator for Example 24.7

= 1.986 V

Ib 14 kΩ

vb

Figure 24.17 For Example 24.8

Example 24.8 The circuit of Figure 24.17 shows an n.p.n transistor connected in the grounded emitter configuration. The two resistors form the base biasing network. The base current is 5 μA, and the transistor supply is 9 V. What is the base voltage?

439

Networks (a.c.)

Solution: 14 V 14 + 22 ∴ v TH = 3.5 V

22 kΩ

vTH = 9 ⋅

E 9V

22 × 14 kΩ 22 + 14 = 8.555 kΩ

r TH =

rTH VTH

Ib

14 kΩ

Figure 24.18 Thevenizing the Circuit for Example 24.8

∴ Vb = VTH − I b rTH

Vb

Vb = 3.5 − 5 × 10−6 × 8.555 × 103 V

Figure 24.19 Equivalent Generator for Example 24.8

Vb = 3.457 V

Example 24.9

0

Find the Thevenin’s circuit for the circuit in Figure 24.20 across terminals AB. Solution: 50 ∠0 ° I1 = = 0 .0481 ∠−1 .1 ° A 1040 + j 20 I2 =

VTH

The equivalent generator is shown in Figure 24.19. When the transistor is connected 5 μA will flow through VTH

8.555 kΩ

3.5 V

Ib = 0

j2

I1

Ω 40 A 10 00 Ω

50∠0° V

I2

50

Ω

B

0 00

Ω

–

00

j4

Ω

1

50 ∠0 ° = 0 .0445 ∠20 .85° A 1050 − j 400

Figure 24.20 For Example 24.9

E 0 = VAB = 50 I 2 − ( 40 + j 20 ) I1 = 50 × 0.0445 ∠20.85° − 44. 72 ∠26.57° × 0.0481 ∠−1.1°

Z0 =

1000 × ( 40 + j 20 ) 1040 + j 20

+

50 (1000 − j 400 ) 1050 − j 400

= 88.517 ∠11.553° Ω Thevenin’s equivalent circuit is shown in Figure 24.21.

= 0. 191 ∠ − 44.15° V 88.517∠11.53° Ω

Ω

A

24.6 CONSTANT VOLTAGE GENERATOR

Howsoever complicated the original circuit may be, it can always be converted into a single source of e.m.f. in series with an impedance. What we have done in applying B C Thevenin’s theorem is to Figure 24.21 Thevenin’s Equivalent Circuit separate the losses which ra µ – v occur in the circuit from the actual e.m.f. being generated. In doing so, we have g produced a perfect source such that no matter how much current is drawn, the e.m.f. will still have the same value. In this way, we have produced the concept of a constantD voltage generator. A constant-voltage generator produces a constant voltage for any load because Figure 24.22 The Primitive Triode it has zero internal load impedance. A practical example of a constant-voltage Constant-voltage generator is the primitive triode valve which produces a constant voltage µ times Generator as large as the a.c. signal applied to its input as has been illustrated in Figure 24.22 (µ being the voltage amplification factor of the triode) – µvg is the constant voltage which has been generated by the input voltage vg. The minus sign shows inversion and ra is the internal impedance of the triode, called the anode impedance. The concept of the constant-voltage generator is useful for the analysis of circuits, because if such a generator is connected to two points in a circuit, the voltage is, by definition, forced to remain constant between these points. 10 Ω 0.191∠–44.15° V

24.7 CONSTANT-CURRENT GENERATOR This is the so-called dual concept of the constant-voltage generator. Whereas in the case of the voltage generator the voltage is constant for any load, the constant-current generator produces a constant current under all load conditions.

440

Electrical Technology

Also, by direct contrast while the voltage generator has zero internal C impedance, the constant-current generator is considered to have r infinitely high internal impedance. This means that if we look into the I terminals of a constant-current generator, we see an open circuit. The D equivalent circuit of a practical arrangement shows a parallel internal impedance. The constant-current generator with its effective internal Figure 24.23 Constant-current Generator impedance is shown in Figure 24.23. An example of a constant-current generator is the primitive pentode valve or the transistor. Reference to their characteristics shows that a large change of anode or collector voltage produces only a very small change of output current. The constant current equivalent circuits of the pentode and transistor are shown in Figure 24.24.

24.8 NORTON’S THEOREM hfev1

gmvg

Thevenin’s theorem is a way of converting a complicated network into a constant-voltage generator in series with an internal impedance. In contrast, (a) (b) Norton’s theorem is a way of converting a complicated network into a constant-current generator in parallel Figure 24.24 Constant Current Circuits with an internal impedance. (a) Transistor (b) Pentode In Figure 24.25 for example, instead of the opencircuit voltage at CD, we require the short-circuit current at CD. With a short-circuit at CD, the impedance seen by the 6 V is ICD IT C 14 × 10 R = 20 + Ω = 25.833 Ω 14 Ω 20 Ω 14 + 10 hoe

RL

rot

RL

I CD = 0.232 ×

Figure 24.25 Norton Equivalent Current

10 = 0.0966 A 10 + 14

( vTH = 2 V )

vCD = I N r N volts

C IN

10 Ω

D

By current division

0.0966 A

E

6V

6 IT = = 0.232 A 25.833

= 0.0966 × 20.67 volts

vN 20.67 Ω

vCD = 2 volts

D

The Norton equivalent circuit is presented in Figure 24.26. Depending on the need, either equivalent circuit can be used. It is possible to change the Thevenin circuit into a Norton circuit and vice versa.

Figure 24.26 Norton Equivalent Generator Example 24.10

In the circuit presented in Figure 24.27 use the superposition theorem to determine the value of current iL – j4 Ω

B

C

E iL 3Ω

+ 15 60° V

+j5 Ω

+ 10 – 45° V

A

D

Figure 24.27 For Example 24.10

F

Networks (a.c.)

441

Solution: Replace the 10 −45°V source by a short circuit, as shown in Figure 24.28. The total impedance Z presented to the 15 60° V source is –j4 Ω iL 3Ω

+ 15 60° V

+j 5 Ω

i1T

Figure 24.28 Step 1 Z lT = − j 4 + = − j4 + ilT =

j15(3 − j 5) 3 × j5 = − j4 + 3 + j5 32 + 52 75 + j 45 = − j 4 + 2.21 + j1.32 = 2.21 − j 2.68 Ω 34

15 60° V 15 60° 3 45 60° and ilT = × = 2.21 − j 2.68 2.21 − j 2.68 3 + j 5 3.47 −50° × 5.83 59.04°

ilT = 2.22 51.46° A The next step is to replace the 15 60° V source by a short circuit (Figure 24.29). The total impedance ‘Z2T’ presented to the 10 −45° V source is 20 (− j) × j 5 Z 2T = 3 + =3+ j5 − j 4 j1 –j4 Ω

= 3 − j 20 Ω i2L

Then

3Ω +j 5 Ω

+ 10 45° V

and

i2T =

10 −45° V 3 − j 20

i2L =

(− j 4) 10 −45° × j 4 + j5 3 − j 20

Figure 24.29 Step 2

=

−40 −45° 3 − j 20

=

−40 −45° = −1.98 36.47° A 20.22 −81.47°

Using the principle of superposition iL = i1L + i2L = 2. 22 51.46° − 1.98 36.47 ° = 1 .38 + j1 .736 − 1 .592 − j1 .177

j10 Ω

= −0.212 + j 0.559 = 0.6 110° A Example 24.11 Obtain Norton’s and Thevenin’s equivalents for the circuit shown in Figure 24.30 at the terminals AB and, hence, find the current through the load impedance of Z L = 30 0° Ω .

j10 Ω I

100∠0° V

~

A

I –j20 Ω

Eo B

Figure 24.30 For Example 24.11

Electrical Technology A IL

E AB = E0 = I (− j 20) = 10 90° × (− j 20) = 200 V Z 0 = j10 + I SC

(a)

j10 (− j 20) = j 30 Ω j10 − j 20

B

A 30∠0° Ω

j 30 Ω

100 0° 100 = = 10 90° A j10 − j 20 − j10

200 V

I =

ISC = 6.67∠–90° A

Solution:

ZL = 30∠0° Ω

442

j 30

(b)

B

Figure 24.31 Thevenin’s and Norton’s Equivalent Circuits

E 200 = 0 = = 6.67 −90° A Z0 j 30

The Thevenin’s and Norton’s equivalent circuits are shown in Figure [24.31(a) and (b)], respectively. 200 E0 IL = = = 4.71 −45° A Z0 + Z L j 30 + 30 Example 24.12 Derive the Norton’s equivalent circuit in Figure 24.32, between the terminals X and Y, and then obtain the load current in polar form. Solution: iL

8Ω

2.83 111.87° A

zN

iN

X 5Ω

–j6 Ω

Load

Y

Figure 24.32 For Example 24.12 Remove the load and replace it by a short circuit. The total impedance then presented to the voltage source is Z T = 1 − j3 + = 1 − j7 Ω

j 4(− j 2) 8 = 1 − j3 + = 1 − j3 − j 4 j 4 + (− j 2) j2

The current drawn from the source is iT =

10 30° A 1 − j7

The short-circuit current between the terminals XY is iN =

j4 10 30° 20 30° × = = 2.83 111.87 ° A j 4 + (− j 2) 7.07 −81.87° 1 − j7

ZN = 8 − j6 Ω Remove the short circuit and replace the load. By the current division rule 8 − j6 8 − j6 iL = iN = = 2.83 111.87° × 8 − j6 + 5 13 − j 6 =

2.83 111.87° × 10 −36.87° 14.32 −24.78°

= 1.98 99.78° A

443

Networks (a.c.)

24.9 PROCEDURE FOR SOLVING A NETWORK USING THEVENIN’S THEOREM Follow the following steps in sequence. 1. Remove the load impedance through which the current is to be found. 2. Find the open-circuit voltage across these terminals. 3. Replace all the voltage sources by short-circuits on their internal impedance and all the current sources by open circuits. 4. Find the equivalent impedance Z0, looking into these terminals. 5. Reconnect the load impedance and calculate the load current.

24.10 PROCEDURE FOR SOLVING A NETWORK USING NORTON’S THEOREM Follow the following steps in sequence. 1. Remove the load impedance through which the current is to be found. 2. Short circuit these terminals. 3. Find the short circuit current. 4. Find Z0 as in Section 24.9 (3). 5. Calculate the load current.

Zo

A

A IL

Eo

~

IL

ZL

Zo

Isc

B

ZL

B

Figure 24.33 Conversion from Thevenin’s Equivalent to Norton’s Equivalent and Vice Versa

Note: Thevenin’s equivalent circuit can be converted to Norton’s equivalent circuit and vice versa (see Figure 24.33).

lL =

l 2Ω

20 V

3Ω

1.5 Ω

235 µH

E0 Z0 + ZL

I L = I SC

Z0 Z0 + ZL

Example 24.13

10 V

Figure 24.34 For Example 24.13 current, I1 =

Determine for the network shown in Figure 24.34 the value of current I. Each of the voltage sources has a frequency of 2 kHz. Solution: The impedance through which current I is flowing is initially removed from the network as shown in Figure 24.35. 20 − 10 = 2A 2+3

Hence, the open circuit e.m.f. E = 20 − I1 (2) = 20 − 4 = 16 V When the sources of e.m.f. are removed from the circuit, the impedance, Z, looking in at the break is given by Z =

2×3 = 1.2 Ω 2+3

2Ω E

l1

20 V

3Ω

10 V

Figure 24.35 Remove the Load

The Thevenin equivalent circuit is shows as Figure 24.36. X L = 2π fL = 2 × 3.14 × 2000 × 235 × 10−6 = 2.95 Ω

i E = 16 V

z = 1.2 Ω

Hence, current, 16 16 = = 40 −47.53 A or (2.70 − j 2.95) A 1.2 + 1.5 + j 2.95 4.0 47.53°

1.5 Ω

I =

j 2.95 Ω

Example 24.14

Figure 24.36 Thevenin’s Equivalent Circuit

Use Thevenin’s theorem to determine the power dissipated in the 48 Ω resistor of the network shown in Figure 24.37. Solution: The (48 + j144) Ω impedance is initially removed from the network, as shown in Figure 24.38.

444

Electrical Technology –j 400 Ω

–j 400 Ω

300 Ω

50∠0° V

48 Ω

Figure 24.37 For Example 24.14 Current, i =

300 Ω

50∠0° V

j 144 Ω

Figure 24.38  Remove (48 + j144) Ω 

50 0° = 0.1 53.13° A (300 − j 400)

I E = 30∠53.13° V

Open circuit voltage, E = i (300) = (0.1 53.13°)(300) = 30 53.13° V when the 50 0° V source is removed

48 Ω

(192 – j144) Ω

(− j 400)(300) (400 −90°)(300) Z = = 300 − j 400 500 −53.13° = 240 −36.87° Ω

E

z

or

j144 Ω

Figure 24.39 Thevenin’s Equivalent Circuit

(192 − j144) Ω

The Thevenin equivalent circuit is shown in Figure 24.39 connected to the (48 + j144) Ω load. C

Current, I =

15 Ω

40 Ω

30 53.13° 30 53.13° = = 0.125 53.13° A 240 0° (192 − j144) + (48 + j144)

Hence, the power dissipated in the 48 Ω resistor

–j25 Ω A

B

2 = I= R (0.125) 2 (48) = 0.75 W

j20 Ω

5Ω j5 Ω

Example 24.15

20 Ω D

Figure 24.40 For Example 24.15

For the a.c. bridge network shown in Figure 24.40 determine the current flowing in the capacitor and its direction by using Thevenin’s theorem. Assume the 30 0° V source to have negligible internal impedance.

Solution: The −j25 capacitor is initially removed from the network as shown in Figure 24.41. p.d. between A and C,

 Z1  15 × ( 30 0° ) VAC =  V =   Z1 + Z 4  15 + 5 + j 5 = 21.83 −14.04° V

p.d. between B and C,

 Z2  40 VBC =  V = × ( 30 0° )  40 + 20 + j 20  Z 2 + Z3  = 18.97 −18.43°

Let us assume the point A is at a higher potential than point B. Then the p.d. between A and B AB = 21.83 −14.04° − 18.97 −18.43° = (3.181 + j 0.701) V

or

3.257 12.43 V

That is, the open circuit voltage across AB is given by

C Z1 = 15 Ω A

Z2 = 40 Ω E

B

E = 3.257 12.43° V Point C is at a potential of 30 0°. Between C and A is a volt drop of 21.83 −14.04° V . Hence, the voltage at point A is: 30 0° − 21.83 −14.04° = 10.29 −30.98° V

Z3 = (20 + j20) Ω

Z4 = (5 + j 5) Ω D

Figure 24.41 Remove −j25

Networks (a.c.)

445

The network is shown in Figure 24.42(b) and simplified in Figure 24.42(c). Hence the impedance, Z looking at terminals AB is given by Z=

(15) (5 + j5) + ( 40) ( 20 + j 20) (15 + 5 + j5) ( 40 + 20 + j 20)

= 5.145 30.96° + 17.889 26.57° = ( 20.41 + j10.65) Ω C 15 Ω

15 Ω

40 Ω

(5 + j 5) Ω

(20 + j 20) Ω

40 Ω B

A

B

A

C

(5 + j 5) Ω D (20 + j 20) Ω

D (a)

(b) 15 Ω

40 Ω C

A

B

D (5 + j 5) Ω

(20 + j 20) Ω

(c)

Figure 24.42 The Various Steps Involved in Solution The Thevenin equivalent circuit is shown in Figure 24.43 where 1 is given by I=

3.257 12.43°

( 20.41 + j10.65) − j 25

=

3.257 12.43° 44.95 −35.11°

I E = 3.257∠12.43° V (20.41 + j10.65) Ω

= 0.131 47.54° A

– j25 Ω

Thus a current of 131 mA flows in the capacitor in a direction from B to A.

24.11 MAXIMUM POWER TRANSFER THEOREM

Figure 24.43 Thevenin Equivalent Circuit

A network that consists of linear impedances and one or more voltage or current sources can be reduced to a Thevenin equivalent circuit. When a load is connected to the terminals of this equivalent circuit, power is transferred from the source to the load. The maximum power transferred from the source to the load, as illustrated in Figure 24.44, depends on the following four conditions. Condition 1: Let the load consist of a pure variable resistance R (i.e., let X = 0). I E Then current I in the load is given by: I =

E and the magnitude of current, (r + R) 2 + jx I =

Z = r + jX

Load Z = R + jX

E

{ (r + R)2 + x 2 }

Figure 24.44 Maximum Power Transfer Theorem

1/ 2

2

The active power P delivered to the load R is given by P = I R =

E2R (r + R)2 + x 2

To determine the value of R for maximum power transferred to the load P is differentiated with respect to R and then equated to zero (this being the normal procedure for finding maximum or minimum values using calculus). Using the quotient rule of differentiation

446

Electrical Technology

  (r + R) 2 + x 2  (1) − ( R)(2)(r + R)  dP   2  =E   = 0 for a maximum or minimum value. 2 2 2 dR  (r + R) + x     dP to be zero, the numerator of the fraction must be zero. dR

For

(r + R) 2 + x 2 − 2 R(r + R) = 0 or r 2 + 2rR + R 2 + x 2 − 2rR − 2 R 2 = 0

Hence,

r 2 + x2 = R2

from which

(

)

R = Ö r 2 + x2 = Z

or

(24.5)

Thus, with a variable purely resistive load, the maximum power is delivered to the load, if the load resistance R is made equal to the magnitude of the source impedance. Condition 2: Let both the load and the source impedance be purely resistive (i.e., let x = X = 0). Maximum power is transferred when R = r (d.c. condition) Condition 3: Let the load Z have both variable resistance R and variable reactance X From Figure 24.44 E (r + R) + j ( x + X )

Current I =

and

The active power P delivered to the load is given by P = I P=

2

I =

E Ö  (r + R ) + ( x + X ) 2  2

R (since power can only be dissipated in a resistance), i.e.,

E2R (r + R)2 + ( x + X )2

If X is adjusted such that X = −x then the value of power is a maximum. If X = −x then P =

E2R and (r + R)2  (r + R) 2 (1) − ( R)(2)(r + R)  dP = E2   = 0 for a maximum vallue dR (r + R)4  

Hence,

(r + R)2 − 2 R (r + R) = 0

i.e.,

r 2 + R 2 + 2rR − 2 Rr + 2 R 2 = 0 r 2 − R 2 = 0 and R = r

from which

(24.6)

Thus, with the load impedance Z consisting of a variable resistance R and variable reactance X maximum power is delivered to the load when X = −x and R = r, i.e., when R + jX = r − jx. Hence, maximum power is delivered to the load when the load impedance is the complex conjugate of the source impedance. From Figure 24.44, the magnitude of current, I =

E Ö  (r + R) 2 + ( x + X ) 2 

and the power dissipated in the load P =

E2R (r + R)2 + ( x + X )2

  (r + R) 2 + ( x + X ) 2  (1) − ( R )(2)(r + R )  dP  = E2    = 0 for a maximum value 2 dR  ( ) ( x + X ) 2  + + r R   

447

Networks (a.c.) I

Hence, (r + R) 2 + ( x + X ) 2 − 2 R (r + R ) = 0 from which R 2 = r 2 + ( x + X ) 2 and R = Ö  r 2 + ( x + X ) 2 

60∠0° V

(24.7)

R 4Ω

Example 24.16 –j 7 Ω

j10 Ω

Figure 24.45 For Example 24.16

In the network shown in Figure 24.45, the load consists of a fixed capacitive reactance of 7 Ω and a variable resistance R. Determine (1) the value of R for which the power transferred to the load is a maximum and (2) the value of the maximum power. Solution: Maximum power transfer is achieved when

1.

R = Ö  r 2 + ( x + X ) 2  = Ö  42 + (10 − 7) 2  = 5 Ω

2.

Current I = =

60 0° 60 0° = (4 + j10) + (5 − j 7) (9 + 3 j )

A

60 0° = 6.324 −18.43° A 9. 48 18.43°

20 V 5Ω

Maximum power transferred = I R = (6.324) (5) = 200 W 2

2

Example 24.17 Determine the value of the load resistance R shown in Figure 24.46 that gives maximum power dissipation and calculate the value of this power. Solution: Using the procedure of Thevenin’s theorem 1. Remove R (Figure 24.47) 2. p.d. across AB, E = [15/(15 + 5)] (20) = 15 V 3. Impedance looking into AB, with 20 V source removed

5Ω

R

B

Figure 24.46 For Example 24.17

5 × 15 = 3.75 Ω 5 + 15

A

is given by r =

E

From the equivalent Thevenin circuit (Figure 24.48), for maximum power transfer R = r i.e., R = 3.75 Ω

20 V 15 Ω

15 Ω

B

Figure 24.47 Remove R

I =

E = 15/ 7.5 = 2 A R+r 2

A I

E = 15 V

2

Pmax = I R = (2) (3.75) = 15 W

R

r = 3.75 Ω

24.12 MILLMAN’S THEOREM Several voltage sources with their internal impedances connected in parallel may be replaced by a single voltage source in parallel with a single internal impedance. Let E1, E2, E3 … En with their internal impedances Z1 , Z2 , Z3 …. Zn be connected in parallel (Figure 24.49), where, E = =

E1 / Z1 + E2 / Z 2 + E3 / Z 3 +  + E n / Z n 1/ Z1 + 1/ Z 2 + 1/ Z 3 +  + 1/ Z n E1Y1 + E2Y2 + E3Y3 +  + EnYn Σ EY = ΣY Y1 + Y2 + Y3 +  + Yn

B

Figure 24.48 Equivalent Circuit

448

Electrical Technology A Z1 E1

Z3

Z2 E2

A ZL

I1

I2

Z1

I3

Z2

Z3

ZL

E3 (a)

B

B

(b)

Figure 24.49 (a) Voltage Sources (b) Voltage Sources of Figure 24.49 (a) converted to current sources I = I1 + I2 + I3 [Figure 24.49(a)] and 1/Z = 1/Z1 + 1/Z2 + 1/Z3 [Figure 24.49(b)] In Figure 24.49(a), three voltage sources along with their internal impedances are connected in parallel across the load impedance ZL. The voltages sources in Figure 24.49(a) are converted into current sources in Figure 24.49(b). Refer Figure 24.50 for more details. A A

E = IZ = ( I1 + I 2 + I 3 ) / (1/ Z ) E1 E2 E3 + + E Y + E2Y2 + E3Y3 Z Z 2 Z3 (24.8) = 1 = 11 1 1 1 Y1 + Y2 + Y3 + + Z1 Z 2 Z 3 1 1 1 1 = + + Z Z1 Z 2 Z 3

(24.9)

or Y = Y1 + Y2 + Y3

Figure 24.50 (a) The Three Current Sources in Figure 24.49 (b) Converted to a Single Current Source. Figure 24.50(b) The Current Source in Figure 24.50 (a) Converted to a Voltage Source in Figure 24.50(b).

Using Millman’s theorem, find the current flowing through the (4 + j3) Ω in the circuit in Figure 24.51. Solution: The two current sources are converted into voltage sources as illustrated in Figure 24.52 and the circuit is redrawn. In Figure 24.52 Y1 + Y2 + Y3

141.4 75° 5 30° 20 −83.13° + − 14.14 45° 5 5 −53.13° E = 1 1 1 + + 14.14 45° 5 5 −53.13° = Z =

9.65 51.02° = 25 34. 46° V 0.386 16.56° 1 1 1 = = Y Y1 + Y2 + Y3 0.36 16. 56°

= 2. 591 −16. 56° Ω 25 34.46° I = = 3.64 15.2 3° A 2.591 −16.56° + 4 + j 3

B

(b)

(24.10)

E1Y1 + E 2Y2 − E 3Y3

ZL

E

B

(a)

Example 24.18

where, E =

Z

ZL

Z

I

A

10∠30° A

10 Ω

4Ω

j 10 Ω

j3 Ω

5∠30° V

3Ω 4∠–30° A

5Ω

–j4 Ω B

Figure 24.51 For Example 24.18 14.14∠45° Ω 141.4∠75° V

4Ω

20∠–83.13° V

5∠30° V

5Ω

j3 Ω

5∠–53.13° Ω

Figure 24.52 The Two Current Sources in Figure 24.51 Converted to Voltage Sources ∠

A I

2.591 –16.56° Ω

4Ω

E 25 34.46° V

j3 Ω B

Figure 24.53 A Redrawn Version of Figure 24.52

449

Networks (a.c.) B

Z1

Z3

V volts

Z5

Z2

24.13 RECIPROCITY THEOREM

C

In any linear, bilateral network, that contains only one independent source, the ratio of excitation to response remains constant, even when their positions are interchanged. Refer to Figures 24.54 and 24.55. As a proof of the reciprocity theorem, the circuit elements in Figure 24.56(a) (given circuit) are replaced, in Figure 24.56(b), with their equivalent impedance evaluated at the frequency of the voltage source V. Using the current divider concept, the current

I amps

Z4

A

D

Figure 24.54 Circuit for Reciprocity Theorem read by the ammeter (AM) in branch 2 is I2 =

Z3 Z2 + Z3

, Z2 +

V Z 1Z 3

=

Z3 Z 1Z 2 + Z 2 Z 3 + Z 3 Z 1

V (24.11)

B

Z1

Z3

Z5

C

Z1 + Z 3 I amps V volts Z2 Z4 Now if we suppose the voltage source and the ammeter are interchanged as in Figure 24.56(c), then using the current divider concept once again, we get A D Z3 Z3 V I1 = ⋅ = V (24.12) Figure 24.55 The Positions of Excitation and Z 1Z 3 Z1 + Z 3 Z 1Z 2 + Z 2 Z 3 + Z 3 Z 1 Response Interchanged Z2 + Z1 + Z 3 The current is exactly the same illustrating reciprocity theorem, which can be seen in Figure 24.56. L1

V +

R2

R3

C3

I2

AM

Z2

Z1 V +

Z3

I2

I1

AM

Z1

Z2

Z3

AM

+ V

R1 (a)

(b)

(c)

Figure 24.56 Reciprocity in Linear Networks (a) Given Circuit (b) General Network (c) Interchanged V and AM In any passive, linear network, if a voltage V applied in branch 1 causes a current I to flow in branch 2, then voltage V applied in branch 2 will cause current I to flow in branch 1. The ratio of a voltage V1 in one part of a network to a current I in another part is called the transfer impedance Z12 = V1/I2. An important conclusion of the reciprocity theorem is: B In passive, linear networks, the transfer impedance Z12 is equal to the recipI2 rocal of transfer impedance Z21. 4Ω 3Ω In Figure 24.56, the transfer impedance is 5∠90° A C Z 1Z 2 + Z 2 Z 3 + Z 3 Z 1 V V Z 12 = 1 = Z 21 = 2 = (24.13) j5 Ω I2 I1 Z3 Vx –j3 Ω Example 24.19

A

D

In the circuit in Figure 24.57, find the voltage Vx and verify reciprocity theorem. Figure 24.57 For Example 24.19 Solution: B 4 + j5 = 4.4 125.39° A I 2 = 5 90° × 7 + j2 4Ω 3Ω

Vx

j5 Ω Ii

A

Vx = I 2 ( − j 3) = 13.2 35.39° V

C 5∠90° A

–j3 Ω D

Figure 24.58 Interchange V and I

Interchange the positions of 5 90°Aa nd Vx 2 , as shown in Figure 24.58 − j3 I1 = 5 90° × = 2.06 −15.95° A 7 + j2 Vx = (4 + j 5) I1 = 13 .2 35 .39 ° V

450

Electrical Technology

24.14 DUALITY Two circuits are said to be duals when the mesh equations that describe the behaviour of one circuit are found to be identical in form to the nodal equations that describe the other. Duality, therefore, is a property of the circuit equations. There are times when the construction of the dual of a circuit is helpful in reducing the effort needed in analyzing simple, conventional circuits. The origin of the principle of duality can be traced back to the laws of Ohm, Faraday, and Coulomb. In the case of a voltage e applied to a series resistor, R, by Ohm’s law, e =Ri. However, an alternative form of this equation states that the current i is expressed as the product of the conductance G and the voltage e appearing across G, i.e., i = Ge. A comparison of the two equations makes it plain that if the roles of voltage and current are interchanged, then the resistance R is replaced by the conductance G. Similarly, by Faraday’s laws, the voltage and current of an inductor are related by, v = L di/dt. For the capacitor, this relationship is, i = C de/dt. v = L di/dt is a mesh-oriented equation since it is an expression for voltage. On the other hand, i = C de/di is a nodeoriented equation, since it is an expression for current. One may proceed from the mesh orientation to node orientation by merely interchanging the role for voltage and current in the describing equations. The reverse procedure is also valid. One useful feature of duality is that one can study, for example, the manner in which the voltage across a capacitor varies, by investigating the variation of current in an inductor subject to a voltage source which is of the same type as the current source. Further examples of the usefulness of duality are that the current through an inductor cannot change instantaneously. So also, the voltage across a capacitor cannot change instantaneously. Here the dual words are voltage and current, through and across, inductance and capacitance. There are dual quantities which can be identified when networks are studied using network topology. All the quantities and concepts and their duals have been listed in Table 24.1. Table 24.1 Dual Quantities 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16.

Quantity or Concept

Dual Quantity or Concept

Current Branch current Mesh Loop Loop current Mesh current Number of loops Link Tie-set Short circuit Series circuit Inductance Resistance Thevenin’s circuit Kirchhoff’s current law Closing switch

Voltage Branch voltage Node Node pair Node pair voltage Node voltage Number of nodes Twig Cut-set Open circuit Parallel circuit Capacitance Conductance Norton’s circuit Kirchhoff’s voltage law Opening switch

To take the concept of duality a step further, let us consider the series RL circuit of Figure 24.59. The mesh equations for this circuit constitutes di e = Ri + L dt

i

G

C

e

Figure 24.60 Dual of the Circuit in Figure 24.59

i

(24.14)

From the interchangeability of the roles of voltage and current, resistance and conductance, inductance and capacitance, we can directly write the current (or nodal) equation that identifies the dual circuit of Figure 24.59, shown in Figure 24.60. Thus,

R

e

L

Figure 24.59 Series RL Circuit

i = Ge + C

de dt

(24.15)

When the magnitude of G is equal to that of R and C equal to L, an exact dual circuit results. The solution for e in Eq. 24.14 exactly becomes the solution for i in Eq 24.15.

Networks (a.c.)

451

A direct graphical procedure can be employed for the identification of the dual of a circuit. Thus, to draw the dual of the circuit in Figure 24.59 directly, let us begin by drawing a dashed line completely encircling the mesh circuit as illustrated in Figure 24.61(a). This closed loop represents the reference node of the nodal method. Next, place a dot at the centre of each mesh of the planar network. Each dot denotes a node of the dual circuit. For the circuit under discussion, there is just one node required. Then identify appropriate branches between the node points by using a dual element for each element that appears in the original mesh circuit. The voltage source e is replaced by the current source i, the resistance R is replaced by the conductance G, and the inductance L is replaced by the capacitance C. The resulting dual circuit is more clearly depicted in Figure 24.61(b). The configuration is identical to that represented in Figure 24.60. R

1

G=R e

i

L

1

i

C=L

0 Reference node

C=L

G=R

e

0

(a)

(b)

Figure 24.61 (a) Illustration of Graphical Procedure for Identifying the Dual Circuit (b) The Resulting Dual Circuit Drawn in Conventional Form 2

1

Example 24.20 Find the exact dual of the two-mesh circuit depicted in Figure 24.62 using the graphical procedure. Solution: We can draw a dashed line enclosing both meshes and call this the reference node O. Place two dots in each mesh and call these the nodes 1 and 2. If we replace each element of the original circuit by its dual counterpart, the final result is shown in Figure 24.63.

C=L

G2 = R2

L=C

0

Figure 24.62 Original Two-mesh Circuit for Example 24.20 Together with Graphical Procedure Solution for Finding the Dual Circuit Shown in Dashed Lines in Figure 24.63

R2

L C=L i

i

0

0

e

G1 = R1

1

R1

G1 = R1 G2= R2

2

0

C L=C

0

Figure 24.63 Dual Circuit (Dashed Lines) for the Circuit in Figure 24.63 (Full Lines)

Example 24.21 Obtain the dual of the network in Figure 24.64. Solution: The nodes are marked for the given network, as shown in Figure 24.65(a) and its dual network is shown in Figure 24.65(b).

Figure 24.64 For Example 24.21

452

Electrical Technology

8F

a 100 sin(ω t – 30°) A

1 Ω 10

b

5H

1 Ω 20

o (b)

Figure 24.65 (a) Original Network with Marking (b) Dual Circuit

24.15 A.C. CIRCUIT ANALYSIS If the various network theorems applied to d.c. circuit analysis are also to be used for sinewave a.c. circuit analysis we must recognize the added difficulties of reactance, impedance, and phase difference, all of which do not exist in d.c. circuits. These are resolved by the use of the j operator and the rules of complex algebra. We are then able to solve network problems without any reference to phasor diagrams. All the voltages and currents are assumed to have the same frequency. In networks that involve more than one voltage (or current) source, these sources are not necessarily in phase. Each source, therefore, has its own polar angle that is related to some reference sine wave.

24.15.1 Kirchhoff’s Current Law (KCL) The phasor sum of currents existing at any electrical junction point is Zero. In d.c. circuits, we adopt a convention to distinguish between positive and negative d.c. voltages as they appear in the algebraic equations. However, we cannot assign a constant direction to an alternating current or a fixed polarity to an alternating voltage. Instead, we give an instantaneous polarity (+ or −) to a source voltage and then indicate the associated direction of the current.

24.15.2 Kirchhoff’s Voltage Law (KVL) In any closed loop in a network, the phasor sum of the voltage drops (i.e., the products of current and impedance) taken around the loop is equal to the phasor sum of the e.m.f.s acting in that loop. In a.c. circuit analysis involving Kirchhoff’s laws or circuit theorems, the use of complex numbers is essential. The above laws and theorems apply to linear circuits, i.e., circuits containing impedances whose values are independent of the direction and magnitude of the current flowing in them. Example 24.22 For the network shown in Figure 24.66, use Kirchhoff’s laws to determine the magnitude of the current in the (4 + j3) impedance. Solution: Currents I1, I2 and I3 with their directions are shown in Figure 24.67. The current in the (4 + j3) Ω impedance is specified by I3 only, which means that the three equations formed need to be solved for only one unknown current. A

l1

(l1–l2)

B

12 V

15 V

4Ω

4Ω 8Ω

–j 5 Ω H

D

(l1–l2–l3)

l2 10 V

l3

C

G

F

j3 Ω E

Figure 24.67 Current I1, I2, and I3 with their Directions

10∠0° V

12∠0° V

4Ω

–j 5 Ω

15∠0° V 8Ω

4Ω j3 Ω

Figure 24.66 For Example 24.22

Three loops are chosen. From loop ABGH, and moving clockwise (in the direction of I1) 4 I1 − j 5 I 2 = 10 + 12

(1)

From loop BCFG, and moving anticlockwise (in the direction if I2) − j 5 I 2 − 8( I1 − I 2 − I 3 ) = 15 + 12

(2)

From loop CDEF, and moving clockwise (in the direction of I3) −8( I1 − I 2 − I 3 ) + 4(4 + j 3)( I 3 ) = 15

(3)

Networks (a.c.)

453

Hence, 4 I1 − j 5 I 2 − 0 I 3 − 22 = 0 −8 I1 + (8 − j 5) I 2 + 8 I 3 − 27 = 0 −8 I1 + 8 I 2 + (12 + j 3) I 3 − 15 = 0 Solving for I3 using determinants gives

−1 I3 = − j5 4 0 − j5 −22 4 −8 (8 − j 5) −8 (8 − j 5) −27 8 −8 −15 8 −8 8 (12 + j 3)

Thus, I3 (8 − j 5) −27 + j 5 −8 −27 − 22 −8 (8 − j 5) 8 −15 −8 8 −8 −15

=

−1 8 ( − j ) 8 5 8 + j 5 −8 4 (12 + j 3) 8 −8 (12 + j 3) Hence,

I3 −1 = from which, 384 + j 700 308 − j 304 I3 =

−(384 + j 700) 798.4 −118.75° = (308 − j 304) 432.76 −44.63°

I 3 = 1.85 −74.12° –j4 Ω

B

Example 24.23 In the circuit represented in Figure 24.68, find the value of the current i3. Solution: The KVL equation for loop ABCDA is

C i3

i1

E iL

3Ω

+

i2

15 60° V

+j5 Ω

+ 10 – 45° V

i1 × (− j 4) − i2 × (3) = 15 60° − 10 −45° = 7. 5 + 12. 99 − 7. 07 + j 7.07 = 0.43 + j 20.06

A

Note: The two voltage sources are opposing in accordance with their polarities as shown.

D

F

Figure 24.68 For Example 24.23

The KVL equation for the loop CEFDC is i2 × 3 + i3 × j 5 = 10 −45° = 7. 07 − j 7. 07 At the junction point, C, the KCL equation is i3 = i1 + i2

or

i2 = i3 − i1

The two preceding equations yield 3i3 − 3i1 + i3 × j 5 = 7.07 − j 7.07 −3i1 + i3 (3 + j 5) = 7.07 − j 7.07

(1)

454

Electrical Technology

From the KVL equation for the loop ABCDA i1 × (− j 4) − (i3 − i1 ) × 3 = 0.43 + j 20.06 i1 × (3 − j 4) − 3i3 = 0.43 + j 20 .06 Adding the last two equations gives i × (− j 4) + i3 × ( j 5) = 7.5 + j12.99

(2)

i1 × j12 + i3 (− j12 + 20) = − j 28.28 − 28.28

(3)

i1 × j12 − i3 × j15 = −22.5 − j 38.97

(4)

Multiplying (1) by –j4 and (2) by −3

Subtracting (4) from (3) i3 (− j12 + 20 + j15) = −5.78 + j10.69 i3 =

−5.78 + j10.69 12.15 118.4° = 20 + j 3 20.20 8.53°

i3 = 0.60 109.9°

24.16 MESH-CURRENT AND NODAL ANALYSIS Mesh-current analysis is merely an extension of the use of Kirchhoff’s laws. Figure 24.69 shows a network whose circulating currents I1, I2, and I3 have been assigned to closed loops in the circuit rather than to branches. Current I1, I2, and I3 are called meshcurrents or loop-currents. In mesh-current analysis, the loop currents are all arranged to flow in the same direction (clockwise in Figure 24.69). KVL is applied to each of the loops, in turn, which in the circuit of Figure 24.69 produces three equations in three unknowns which may be solved for three unknowns. The three equations produced from Figure 24.69 are 4Ω

10∠0° A

6Ω

E1

I3

I2

I1

E2

Z4

Z2

Z5

Z1

Figure 24.69 Mesh-current Analysis I1 ( Z 1 + Z 2 ) − I 2 Z 2 = E1

j3 Ω

10 Ω

Z3

I 2 ( Z 2 + Z 3 + Z 4 ) − I1Z 2 − I 3 Z 4 = 0 I3 (Z 4 + Z 5 ) − I 2 Z 4 = − E 2

5Ω

j5 Ω

–j5 Ω

The branch currents are determined by taking the phasor sum of the mesh currents common to the branch. Example 24.24

Figure 24.70 For Example 24.24

Find the current flowing through (4 + j3) Ω, using meshcurrent analysis, in the circuit in Figure 24.70.

Solution: The current source is first converted into a voltage source, as represented in Figure 24.71. The mesh equations are (20 + j8) I1 − 6 I 2 = 111.8 26.57° −6 I1 + (11 − j 5) I 2 = 0 Solving the above equations gives I1 = 6.022 5.19°

10 Ω

111.8 26.57° V

j5 Ω

4Ω

I1

j3 Ω

6Ω

5Ω I2

–j5 Ω

Figure 24.71 The Current Source in Figure 24.70 is Converted into a Voltage Source

Networks (a.c.)

455

Example 24.25 In the circuit drawn in Figure 24.72, use the method of mesh-current analysis to determine the value of i2. – j4 Ω

B

C

i1

+

E

3Ω

15 60° V

i2 +i5 Ω

+ 10 – 45° V

A

D

F

Figure 24.72 For Example 24.25 Solution: As far as the polarities of the voltage sources are concerned, we can observe the normal KVL convention. In the mesh ABCDA i1 (3 − j 4) − i 2 × 3 = −15 60° + 10 45 ° i1 (3 − j 4) −

In the mesh CEFDC,

1 × 3 = −7.5 − j 12.99 + 7.07 − j 7.07 2 = 0.43 − j 20.06

(1)

−3i1 + i2 × (3 + j 5) = −10 45° = −7.07 + j 7.07

(2)

Multiplying the first equation by 3 and the second equation by 3 −j4 yields i1 (9 − j12) − i2 × 9 = −1.29 − j 60.18 −i1 (9 − j12) + i2 (3 + j 5)(3 − j 4) = (−7.07 + j 7.07)(3 − j 4) −i1 (9 − j12) + i2 (29 + j 3) = 7.07 + j 49.49 Adding the two preceding equations yields i2 (20 + j 3) = 5.78 − j 10.69 Therefore, i2 =

5. 78 − j 10. 69 12.15 −61.6° = 20 + j 3 20.22 8 .35 °

i 2 = 0 .60 −69 .1 ° A Example 24.26 For the a.c. network shown in Figure 24.73 determine using mesh current: (1) the mesh currents I1 and I2 (2) the current flowing in the capacitor; and (3) the active power delivered by the 100 0° V voltage source. Solution: 1. For the first loop (5 − j 4) I1 − (− j 4) I 2 = 100 0° For the second loop (4 + j 3 − j 4) I 2 − (− j 4) I1 = 0 Rewriting (5 − j 4) I1 + j 4 I 2 − 100 = 0 j 4 I1 + (4 − j ) I 2 + 0 = 0 Using determinants, I1 −I2 I3 = = j4 j 4 −100 (5 − j 4) −100 (5 − j 4) j4 4j 0 j4 0 (4 − j )

100 0° V

5Ω

I1

I2

–j4 Ω

4Ω

j3 Ω

Figure 24.73 For Example 24.26

456

Electrical Technology

I1 −I 2 1 = = (400 − j100) (32 − j 21) j 400 I1 =

Hence,

(400 − j100) 412.31 −14.04° = (32 − j 21) 38.28 −33 .27° I1 = 10.8 −19.2°

I2 =

400 − 90° = 10.5 −56 .7° 38.28 −33.27

2. Current flowing in capacitor = I1 − I 2

ZA

4

= 10.77 19.23° − 10.45 −56.73° = 4.44 + j12.28 = 13.1 70.12°

VX

ZB

1 ZD

3. Source power P = VI cos φ = (100)(10.77) cos 19.23°

ZC

2 ZE

5 VY

3

= 1016.9 W A node of a network is defined as a point where two or more branches Figure 24.74 Node Analysis are joined. If three or more branches join at a node, then that node is called a principal node or junction. In Figure 24.74 points 1, 2, 3, 4, and 5 are nodes, and points 1, 2 and 3, all constitute principal nodes. A node voltage is the voltage of a particular node with respect to a node called the reference node. The object of nodal analysis is to determine the values of voltages at all the principal nodes with respect to the reference node. When such voltages are determined, the currents flowing in each branch can be found. The branches leading to node 1, and to node 2, are shown separately in Figures 24.75 and 24.76. The currents are all assumed to be leaving the node. ZA

VX

ZB

1

2

1

ZB

ZC

2 ZE

ZD 3

VY

3

Figure 24.75 Branches Leading to Node 1 In Figure 24.75,

In Figure 24.76,

Figure 24.76 Branches Leading to Node 2

V1 − Vx V V − V2 + 1 + 1 =0 ZA ZD ZB

(1)

V2 − V1 V2 V2 + Vr + + =0 ZB ZE ZC

(2)

Any selection of the direction of the branch currents may be made—the resulting equations will be identical. Rearranging equations (1) and (2) gives  1 1 1 + +   ZA Z B Z D

  1   1   V1 −   V2 −   Vx = 0  ZA    ZB 

 1   1 1 1 − + + V1 +    ZB   Z B ZC Z E   

  1  V2 +    ZC

  Vr = 0 

(3)

(4)

Equations (3) and (4) may be written in terms of admittances (Y = 1/Z).

( YA + YB

+ YD ) V1 − YBV2 − YAVx = 0

−Y BV1 + ( Y B + YC + Y E ) V2 + YCVr = 0

(5) (6)

Networks (a.c.)

457

Thus, V1 −Y B

( YB + YC + YE )

−YA YC

=

−V2 = ( YA + YB + YD ) −YA YC −Y B

The number of equations necessary to produce the solution for a circuit is, in fact, always one less than the number of principal nodes. Example 24.27 In the network shown in Figure 24.77 write down the node voltage equations for the network. Solution: The node voltage equations are

( YA + YB

−YB

+ YD )

1 −YB

( YB + YC + YE ) V1 3 Ω

5Ω

j4 Ω

j10 Ω 5 45° V

10 30° V

V2

6Ω

4Ω

–j8 Ω

Figure 24.77 For Example 24.27

1 1 1   1  10 30° 5 45°  1  + − −  +  V1   V2 =  V1 + j + j + j 5 10 3 4 3 4 5 j10      3 + j4   1 1 1 + +  V2 = 0  3 4 6 8 4 + − j j  Example 24.28 In Figure 24.78 determine the voltage at the point N. Solution: In Figure 24.78 we may regard the capacitive reactance of –j4Ω as the internal impedance of 15 60° V. Similarly, let us regard the resistance of 3Ω as the internal impedance of the 10 45° V source. We can then convert both voltage sources into their equivalent current generators, as illustrated in Figure 24.79. –j4 Ω

B

N

3Ω

+

+j 5 Ω

15 60° V 10 45° V

Figure 24.78 For Example 24.28 N

I4 I1

–j 4 Ω

I5

I3 +j5 Ω

3Ω

I2

Figure 24.79 A Redrawn Version of Figure 24.78 with Voltage Sources Converted to their Equivalent Current Sources

458 Electrical Technology i1 =

15 60° V 15 60° V = = 3. 75 150° A 4 −90° Ω − j4 Ω

i2 =

10 −45° V = 3.333 −45° A 3Ω

and

The current i1 and i2 are leaving the node, while the currents i3, i4 and i5 are entering the point N. Therefore the nodal equation is 3.75 150° + 3.333 −45° = Then,

VN V V + N + N 3 j5 − j 4

−3.248 + j1.875 + 2.352 − j 2.357 = VN (− j 0.2 + j 0.25 + j 0.333) VN (0.333 + j 0.05) = −0.891 − j 0.482 1.013 −151.6° = 3 −160° V (approx.) 0.337 8.5°

VN = 4Ω

i3 Ω

IY

Example 24.29

A

10 Ω

IX

For the network in Figure 24.80, determine the voltage using nodal analysis. Solution: Figure 24.80 contains two principal nodes (at 1 and B) and, thus, only one nodal equation is required. B is taken as the reference node. The equation for node 1 is obtained as follows.

B

1

16 Ω

Figure 24.80 For Example 24.29 I X + IY = I

i.e.

 1 V1 V1 1  + = 20 0°; V1  +  = 20 16 (4 + j 3) + j3  16 4 

4 − 3j   V1  0.0625 + 2  = 20; 4 + 32  

V1 (0.0625 + 0.16 − j 0.12) = 28

V1 (0.2225 − j 0.12) = 0 From which, V1 =

20 20 = (0. 2225 − j 0. 12) 0.2528 −28.34°

V1 = 79.1 28 .34 ° V The current through the (4 + j3) branch, IY = V1 /(4 + 3 j ) Hence, the voltage drop between points A and B (VAB) is denoted by VAB = ( IY ) (4) =

79.1 28.34° V1 (4) = ( 4) 5 36.87 ° (4 + j 3)

= 63 .3 −8.53 ° V Example 24.30 In the network of Figure 24.81 use nodal analysis to determine: (1) the voltage at nodes 1 and 2, (2) the current in the j4 inductance (3) the current in the 5 Ω resistance and (4) 2.5 Ω 5Ω 2Ω 1 2 the magnitude of the active power dissipated in the 2.5 Ω resistance. Solution: 1. At node 1,

–j4 Ω

j4 Ω

3

V1 − 25 0° V V − V2 + 1 + 1 =0 2 − j4 5

Figure 24.81 For Example 24.30

Networks (a.c.)

459

Rearranging gives 1 1 1 25 0° 1 +  V1 −   V2 − =0  + − 2 4 5 5 2 j     i.e., (0.7 + j 0.25) V1 − 0.2 V2 − 12.5 = 0

(1)

At node 2, V2 − 25 90° V2 V2 − V1 + + =0 2.5 j4 5 Rearranging gives  1 1 1 25 90° 1 −   V1 +  + +  V2 − =0 2.5 5  2.5 j 4 5  i.e.,

−0.2 V1 + (0.6 − j 0.25) V2 − j10 = 0

(2)

Thus, two simultaneous equations have been formed with two unknowns V1 and V2. Using determinants, if (0.7 + j 0.25) V1 − 0.2 V2 − 12.5 = 0 and −0.2 V1 + (0.6 − j 0.25) V2 − j10 = 0 then, V1 −12.5 0.2 j 0.25 ) − j10

( 0.6 −

=

−V2 = ( 0.7 + j 0.25 ) −12.5 −0.2 − j10

( 0.7 +

j 0.25 ) −0.2

1 −0.2 j 0.25 )

( 0.6 −

i.e., V1 2 −V2 = = ( j 2 + 7. 5 − j 3. 125) (− j 7 + 2. 5 − 2. 5) (0. 42 − j 0. 175 + j 0.15 + 0.0625 − 0.04 and −V2 1 V1 = = 7.584 −8.53° −7 90° 0.443 −3.23 ° Thus, the voltage,

and voltage,

V1 =

V2 =

7.584 −8.53° = 17.12 −5.30° V 0.443 −3.23° 7 90° = 15. 8 93.2° V 0. 443 −3. 23°

2. The current in the j4 inductance is given by V2 15.80 93.23° = = 3.95 3.23° j4 4 90°

(flowing away from node 2)

3. The current in the 5Ω resistance is given by V1 − V2 17.12 −5.30° − 15. 80 93.23° = 5 5 (17.05 − j1.58) − ( −0 .89 + j15.77) = 5 17.94 − j17.35 24.96 −44.04 ° = = 5 5 I 5 = 4.99 −44.04° A (from nodde 1 to node 2)

I5 =

460

Electrical Technology

4. The active power dissipated in the 2.5 Ω resistor is given by 2

(0.89 + j15.77 − j 25) 2  V − 25 90°  2 P2.5 = ( I 2.5 ) (2.5) =  2 (2.5) =    2.5 2.5 (9.273 −95.51°) 2 85.99 −191.02° = 2.5 2.5 = 344 169° W

=

by De Moivre’s Thheorem

S UM M A RY 1. Two circuits are equivalent if they present the same v_i characteristic. 2. Replacing a complicated network with a simple equivalent network is sometimes advantageous. 3. Voltage divider and current divider are useful tools in network analysis. 4. It is possible to change the Thevenin circuit into a Norton circuit and vice versa. 5. In any linear, bilateral network, containing only one independent source, the ratio of excitation to response remains constant even when their positions are interchanged.

6. Two circuits are said to be duals when the mesh equations that describe the behaviour of one circuit are found to be identical in form to the nodal equations that describe the behaviour of the other. 7. Duality is a property of the circuit equations. 8. The phasor sum of currents existing at any electrical junction is zero. 9. In a.c. circuit analysis involving Kirchhoff’s laws or circuit theorems the use of complex numbers is essential.

M U LT IP LE C H O I C E Q UE S TI O NS (M C Q ) 1. Superposition theorem requires as many circuits to be solved as there are (a) Nodes (c) Sources

(b) Meshes (d) Branches

7. For any voltage source (a) (b) (c) (d)

The terminal voltage is greater than the source voltage The terminal voltage is always less than the source voltage The terminal voltage is equal to the source voltage None of the above

2. Six resistors, each of R Ω, are connected as shown in Figure 24.82. The resistance between CB is

8. The basis of Superposition theorem is

3. To determine the sense of voltage drop, it is necessary to know the

9. Application of Millman’s theorem to a multi-source network yields

(a) R (c) R/3

(b) 2R (d) R/2

(a) Value of the resistance (b) Direction of current through the resistance (c) (a) and (b) (d) None of the above

4. Superposition theorem is valid only for

(a) Duality (c) Non-linearity

(a) (b) (c) (d)

(b) (d)

Equivalent current source Equivalent voltage source Equivalent impedance Equivalent voltage or current source

(a) Linear circuits (b) Non-linear circuits (c) Linear-bilateral circuits (d) Bilateral circuits

A

5. Thevenin’s equivalent circuit consists of (a) (b) (c) (d)

Current source in series with impedance Current source in parallel with impedance Voltage sourced in parallel with impedance Voltage source in series with impedance

R R C

6. A voltage source of internal impedance Zs is employed to feed the signal to a load impedance ZR. The maximum power that can be delivered to ZR is (a) 75 per cent (c) 60 per cent

ANSWERS (MCQ) 1. (c) 2. (d) 3. (b)

5. (d)

6. (d)

7. (b)

8. (b)

O R

R B

Figure 24.82 For MCQ 2

(b) 100 per cent (d) 50 per cent

4. (c)

R

R

9. (a).

Linearity Reciprocity

461

Networks (a.c.)

CON V E N TI O NA L Q UE S TI O NS (C Q ) 1. Develop mesh equations for the network shown in Figure 24.83 and find the power absorbed by the 3 Ω resistor. 2. Using Millman’s theorem, find the current flowing through the (4 + j3) Ω impedance in the network in Figure 24.84. 3. Find the drop across 2 Ω resistor in the network shown in Figure 24.85. 4. Find the current through the 10 Ω resistor in Figure 24.86 using Thevenin’s theorem. A 3Ω

5Ω

j2 Ω

–

1 Ω I2

I1

+

j1 Ω

1 0° AI3

90∠45° V

5Ω

80∠30° V

10 Ω 20 Ω

10 0° V

E

F

7. Give in details the procedure for finding the dual of a network. 8. Draw the dual of the network in Figure 24.89. 9. Draw the graph (dual) of the network in Figure 24.90.

100∠0° V

C

B

5. Using Thevenin’s theorem, find the current through the coil (5 + j4) Ω in the bridge circuit in Figure 24.87. 6. In the circuit shown in Figure 24.88 find the maximum power received by 12 Ω resistance.

4Ω

D

j3 Ω

Figure 24.84 For CQ 2

Figure 24.83 For CQ 1

Figure 24.85 For CQ 3

15 A

4Ω

2Ω – 30 V +

(3+j4) Ω

–j10 Ω

j10 Ω

100 45° V

24 Ω

10 Ω A

C

1H

R

R – +

C1

180 V –+ 12 Ω

C 20 V

V

6Ω

R1

2F 10 Ω

2H

10 Ω

20 Ω

RL 10 A

Figure 24.89 For CQ 8

Figure 24.90 For CQ 9

ANSWERS (CQ) 1. 13.36 W 2. 8.89 −12.84° A

3. 7.80 84.41° V

4. 7.35 28.3° A

B

Figure 24.88 For CQ 6

Figure 24.87 For CQ 5

Figure 24.86 For CQ 4

– +

+

45 A

5. 0.094 − 17.4° A

6. 1012.5 W.

Delta Wye Transformations

25 A

OBJECTIVES Z2

In this chapter you will learn about:     

A Z3

Za

∆ – Zc

Appreciate the concept behind delta and Y connections How to convert from Δ to Y  How to convert from Y to Δ  How to solve simple problem using these conversions  Simplify complicated networks by using these conversions 

C

B

Z1

Zb

C

B A

A Za S

Zc

S

Z2

– ∆ Zb

C

C

B

Z3

Z1

B

Wye Transformations

Delta

25.1 INTRODUCTION Three impedances connected nose-to-tail, as shown in Figure 25.1, are said to be delta (∆) or mesh connected, because they form a mesh. Three impedances connected together to a common point, as shown in Figure 25.2, are said to be star (or Y ) connected. If their nodes (1, 2 and 3)—to which the two sets of impedances are connected—form part of a larger network, it is possible to assign values to them so that they have exactly the same overall effect on the network. It is possible to transform from one to the other. The ultimate aim of this transformation is to simplify the network so that fewer variables and equations are required to solve the network. 1

1

ZA

Z1

ZB Z3

Z2

ZC 3

2

3

Figure 25.1 The Delta or Mesh Connection

2

Figure 25.2 The Star or Y Connection

25.2 DELTA AND STAR CONNECTIONS The network shown in Figure 25.3(a), consisting of impedances ZA, ZB and ZC, is said to be π-connected. The network can be redrawn, as shown in Figure 25.3(b). The redrawn network is referred to as a delta-connected or mesh-connected network. 1

1

2 ZB

ZA

ZB

ZC

ZA

2

3 3 (a)

ZC (b)

Figure 25.3 (a) π-connected Network (b) Delta-connected Network

M25_AUTH_ISBN_C25.indd 462

8/6/2012 3:51:04 PM

Delta

Wye Transformations

463

The network shown in Figure 25.4(a), consisting of impedances Z1, Z2 and Z3 is said to be T-connected. This network can be redrawn, as shown in Figure 25.4(b). The redrawn network is referred to as a star-connected network. Z1

1

Z2

2 1

Z1

Z3

S

Z3

3 (a)

3

Zb

2

(b)

Figure 25.4 (a) T-connected Network (b) Star-connected Network

25.3

TRANSFORMATIONS

It is possible to replace the delta connection by an equivalent star connection in such a way that the impedance measured between any pair of terminals is the same in star as in delta. The equivalent star network will consume the same power and operate at the same power factor as the original delta network. A delta-star transformation may alternatively be termed as ‘π to T transformation’. These transformations are shown in Figures 25.5 and 25.6. 1

1

1

2

1

Z1

Z2

2

ZB Z1

ZA

ZC

ZA

ZB Z3

Z3

Z2

ZC 2

3

3

2

(a) Delta

(b) Star

3

3

(a) π

(b) T

Figure 25.6 π to T Transformation

Figure 25.5 Delta to Star Conversion

Considering terminals 1 and 2 of Figure 25.5(a), the equivalent impedance is given by ZB in parallel with ZA in series with ZC. Z12 =

Z B ( Z A + ZC

)

Z B + Z A + ZC

By the same taken,

Z 23 =

and

Z 31 =

Delta = Z1 + Z 2 Star

ZC ( Z A + Z B ) ZC + Z A + Z B Z A ( Z B + ZC ) Z A + Z B + ZC

(25.1)

= Z 2 + Z3

(25.2)

= Z 3 + Z1

(25.3)

Subtracting Eq. 25.2 from Eq. 25.1 Z A Z B − Z A ZC = Z1 − Z 3 Z A + Z B + ZC Adding Eq. 25.3 to Eq. 25.4

M25_AUTH_ISBN_C25.indd 463

(25.4)

2Z A Z B = 2Z1 Z A + Z B + ZC

8/6/2012 3:51:14 PM

464 Electrical Technology and

Z2 =

Z B ZC Z A + Z B + ZC

(25.5)

Z1 =

Z AZB Z A + Z B + ZC

(25.6)

Z2 =

Z B ZC Z A + Z B + ZC

(25.7)

Z3 =

Z A ZC Z A + Z B + ZC

(25.8)

In the same way, Z1 and Z3 can be found. where,

and

Note: (1) Z1 is given by the product of the two impedances in delta joined to terminal 1 (i.e., ZA and ZB), divided by the sum of the three impedances. (2) Z2 is given by the product of the two impedances in delta joined to terminal 2 (i.e., ZB and ZC) divided by the sum of the three impedances. (3) Z3 is given by the product of the two impedances in delta joined to terminal 3 (i.e., ZA and ZC), divided by the sum of the impedances. (Refer to Figure 25.7)

1

1

Z1

ZA

ZB Z3 ZC 2

3

3

(a)

Example 25.1 What are the values of Z1, Z2, and Z3 for the delta network given in Figure 25.8?

Z2

(b)

2

Figure 25.7 Summarizing Delta to Star Conversion

Solution: The star equivalent network for the delta network in Figure 25.8 is given in Figure 25.9.

Z1 =

(2)(3) = 0.6 Ω 2+3+5

Z2 =

(3)(5) = 1.5 Ω 2+3+5

Z3 =

(2)(5) = 1.0 Ω 2+3+5

ZB = 3 Ω

ZA = 2 Ω

Z1 Z3

Z2

Zc = 5 Ω

Figure 25.8 For Example 25.1

Figure 25.9 Solution for Example 25.1

Example 25.2 Find the current through the galvanometer G of resistance 50 W in the circuit shown in Figure 25.10. Solution: The galvanometer is removed and an open circuit is created, as shown in Figure 25.11.

M25_AUTH_ISBN_C25.indd 464

8/6/2012 3:51:23 PM

Delta

Wye Transformations

465

A A G 100 Ω

90 Ω

90 Ω

B 12 Ω

12 Ω

12 Ω

2Ω

B

100 Ω

2Ω

12 Ω

4Ω

12 Ω

12 Ω 4Ω

40 V

40 V

Figure 25.11 The Galvanometer, G, is Removed, Creating an Open Circuit between Points A and B

Figure 25.10 For Example 25.2

The delta connection consisting of three 12 W resistances is converted to an equivalent star network and the circuit is redrawn as shown in Figure 25.12. No current flows through the 4 W resistance. Hence, it can be neglected. 190 × 14 = 13.04 Ω Total resistance, RT = 190 + 14 40 V = I = 3.07 A 13.04 A

I1 = 3.07 ×

14 = 0.21 A; I 2 = 3.07 − 0.21 = 2.36 A 204

− E0 + 100 I1 − 6 I 2 = 0; E0 = (100 × 0.21) − (6 × 2 .86) = 3.84 V R0 = 4 +

106 × 98 = 54.92 Ω 106 + 98

The Thevenin equivalent circuit is given in Figure 25.13. Ig =

3.84 = 0 .0366 A 54.92 + 50

A 100 Ω

90 Ω

B

I1

4Ω

12 Ω

I1 54.92 Ω

A Ig

2Ω

I2

4Ω

4Ω

I2

4Ω 3.84 V

40 V

+ –

G 50 Ω

I

Figure 25.12 The Circuit in Figure 25.11 Redrawn after Delta to Star Conversion

M25_AUTH_ISBN_C25.indd 465

B

Figure 25.13 The Thevenin Equivalent Circuit

8/6/2012 3:51:29 PM

466 Electrical Technology Example 25.3 Replace the delta-connected network shown in Figure 25.14 with an equivalent star network.

10 Ω

ZB

Z A =20 Ω

Solution: Let the equivalent star network be as shown in Figure 25.15. (20)(10 + j10) (20)(1.414 45°) Z1 = = 20 + 10 + j10 − j 20 31.62 −18.43°

j 10 Ω

Z C =–j 20 Ω

= 8.944 63.43° Ω or (4 + j8) Ω

(10 + j10)(− j 20 ) (1.414 45°)(20 −90°) = 31.62 −18.43° 31.62 −18.43°

Z2 = Z1

Figure 25.14 For Example 25.3

= 8.944 −26.57° Ω or (8 − j 4) Ω

Z3

Z3 =

Z2

= Figure 25.15 Equivalent Star Network

(20)(− j20) 31.62 −18.43°

(400 −90°) 31.62 −18.43°

= 12.650 71.57 ° Ω or (4 − j12) Ω A

Example 25.4 A star-network, in which N is the star point, is made up as follows: AN = 70 Ω, BN = 100 Ω, CN = 90 Ω. Find an equivalent delta network. If the above star-delta networks are superimposed, what will be the measured resistance between A and C? Solution: The equivalent delta network is shown in Figure 25.17. R1 = 223 Ω,

R2 = 247.78 Ω,

RA = 70 Ω

R

0Ω

RC

B

=9

N

=1

00

Ω

C

B

Figure 25.16 For Example 25.4

R3 = 318.57 Ω

The star-delta networks are superimposed in Figure. 25.18 and the equivalent network is shown in Figure 25.19. R1 = 111.5 Ω R2 = 123.89 Ω R3 = 159.28 Ω A A

3Ω 22

22 3Ω

70 Ω

Ω 22 3

=2 1

Ω

8Ω

R

8 7.7 24

7.7

Figure 25.17 The Equivalent Delta Network

Ω

24

Ω B

R1

8Ω

.78

90

R 3 = 318.57 Ω

M25_AUTH_ISBN_C25.indd 466

7.7

47

C

24

=2

23

R2

R2

Ω

A

10

0Ω

318.57 Ω B

C 318.57 Ω

Figure 25.18 The Two Networks Superimposed

C

318.57 Ω

R3

B

Figure 25.19 The Equivalent Network

8/6/2012 3:51:36 PM

Delta

Wye Transformations

467

Example 25.5 Six equal resistors, each of 4 W, are connected as shown in Figure 25.20. Find the equivalent resistance between any two corners. Solution: After conversion, the network in Figure 25.20 may be redrawn as shown in Figure 25.21. The resistance between any two 3× 6 corners is = 2 Ω. This is shown in Figure 25.22. 3+6 B B

B

3Ω

3Ω

4Ω 4Ω

Ω

4Ω

12

4Ω

A

Ω

12

4Ω

C 3Ω

12 Ω 4Ω

4Ω

Figure 25.22 The Resistance between Any Two Corners is 3 W 4Ω 4Ω Shunted by 3 W in Series with 3 W → i.e., 2 W. Figure 25.20 For Example 25.5 Figure 25.21 After Conversion C

A

A

C

Example 25.6 How can the network in Figure 25.23 be reduced to a single resistance? Illustrate. Solution: The steps involved in reduction are illustrated in Figure 25.24. 1Ω 2 4Ω

1Ω

3Ω 8

1Ω 2

3Ω 2

159 Ω 71

3Ω 13 Ω 2

19 Ω 8 2Ω

5Ω

Figure 25.23 For Example 25.6

2Ω

5Ω

Figure 25.24 The Steps Involved in Reduction of the Network Given in Figure 25.23 (2.5 –j 5) Ω

Example 25.7 For the network shown in Figure 25.25, determine (1) the current flowing through the (0 + j10) impedance, and (2) the power dissipated by the (20 + j0) W impedance. Solution: 1. The network in Figure 25.25 can be simplified by transforming the delta PQR to its equivalent star connection as shown in Figure 25.26.

(25–j 5) Ω 120∠0° V

P

(15–j 10) Ω

Q

(20–j 30) Ω R

(20+j 0) Ω (0+j 10) Ω

S

Figure 25.25 For Example 25.7

M25_AUTH_ISBN_C25.indd 467

8/6/2012 3:51:41 PM

468 Electrical Technology )(25 15 + 10)( 25 − + jj10 − jj 55)) ((15 Z1 = Z 1 = (15 + j10) + ( 25 − j 5) + ( 20 − j 30) (15 + j10) + (25 − j 5) + (20 − j 30) )(25 ((15 15 + + jj10 10)( 25 − − jj 55)) = = 60 − 25)) ((60 − jj 25 (18.03 33.69°)(25.50 −11.31°) = (18.03 33.69°)(25.50 −11.31°) = 65 − 22..62 62°° −22 65

(15+j 10) Ω

P

(25–j 5) Ω

P

Z1

Z2

Q

(20–j 30) Ω Z3

07 45 45°° Ω Z1 = = 77..07 Ω oorr ((55 + + jj 55)) Ω Ω Z 1 Z2 =

Q

(15 + j10)(20 − j 30) (18.03 33.69°)(36.06 −56.31°) = (65 −22.62°) 65 −22.62°

R

R

(a)

(b)

Figure 25.26 Delta PQR (a) and its Star Equivalent (b)

Z 2 = 10.0 0° Ω or (10 + j 0) Ω Z3 =

(25 − j 5)(20 − j 30) (25.56 − 11.31°)(36.06 −56 .31 °) = (65 − 22.62°) 65 − 22.62°

Z 3 = 14.15 −45° Ω or (10 − j10)Ω The network, after transformation, is redrawn in Figure 25.27. I1 = 8 A, I2 = 2 A,

I3 = 6 A

Z 1=(5+j 5) Ω

(2.5–j 5) Ω

Z 2=(10+j 0) Ω Q

P Z 3=(10–j 1 0) Ω

(20+j 0) Ω

R

120∠0° V

(0+j 10) Ω

S

Figure 25.27 The Network of Figure 25.25 after Transformation of Delta PQR to its Star Equivalent The current flowing in the (0 + j10) Ω impedance of Figure 25.27 is 6 A 2. The power P dissipated (Figure 25.28) with (20 + j0) Ω impedance is: 2 = P I= (2) 2 (20) = 80 W 2 ( 20)

I1

7.5 Ω

I2 I3

120∠0° V

10 Ω

30 Ω

Figure 25.28 A Further Simplified Version of Figure 25.27 for Calculation of Power

M25_AUTH_ISBN_C25.indd 468

8/6/2012 3:51:46 PM

Delta 1

1 Z1

Z3

ZA

ZB

Z2 ZC

3

2

3

2 (a)

(b)

Figure 25.29 Conversion of Star Network (a) to its Equivalent Delta Network (b)

Wye Transformations

469

It is also possible to convert the star network as shown in Figure 25.29(a) to its equivalent delta network shown in Figure 25.29(b). Note: 1. The numerator in each expression is the sum of the products of the star impedances taken in pairs. 2. The denominator of the expression for ZA (connected between terminals 1 and 3) is Z2, which is connected to terminal 2. 3. Similar reasoning applies to the expressions for ZB and ZC.

Example 25.8 Find the delta equivalent of the star circuit represented in Figure 25.30. Z 1=0.6 Ω

Solution: (Refer to Figure 25.31) Z A = 3.0 /1.5 = 2Ω;

Z B = 3.0 /1.0 = 3Ω;

Z C = 3.0 / 0.6 = 5Ω

Z 3=1.0 Ω

Z 2=1.5 Ω

Example 25.9 Determine the delta-connected equivalent network for the star-connected impedances shown in Figure 25.32.

Figure 25.30 For Example 25.8

20 Ω

10 Ω

ZA

ZB

j5 Ω

ZC

Figure 25.31 Delta Equivalent for the Star-connected Network of Figure 25.30

Figure 25.32 For Example 25.9

Solution: Refer Figure 25.33.

ZA =

(10)(20) + (20)( j 5) + ( j 5)(10) 200 + j150 = 20 20

= (10 + j 7. 5) Ω (200) + ( j150) − j 5(200 + j150) = s ZB = 25 j5 = (30 − j 40) Ω ZC

M25_AUTH_ISBN_C25.indd 469

(200) + ( j150) = = (20 + j15) Ω 10

Z 1=10 Ω Z 3=j5 Ω

ZA

ZB

Z 2=20 Ω ZC (a)

(b)

Figure 25.33 The Network of Figure 25.32 Redrawn (a) and the Equivalent Delta Connection (b)

8/6/2012 3:51:51 PM

470 Electrical Technology

S UM M A RY 1. The star-delta transformation can also be applied to a.c. networks by using impedances instead of resistances. 2. The equivalent delta impedance between any two terminals is the sum of the two star impedances connected to those terminals in addition to the product of the same star impedances divided by the third star impedance. Z Z Z1 = Z b + Z c + b c Za Z2 = Zc + Z a +

Zc Z a Zb

Z3 = Z a + Zb +

Z a Zb Zc

3. The equivalent star impedance connected to a given terminal is equal to the product of the two delta impedances connected to the same terminal divided by the sum of the delta impedances. Za =

Z 2 Z3 Z1 + Z 2 + Z 3

Zb =

Z 3 Z1 Z1 + Z 2 + Z 3

Zc =

Z1Z 2 Z1 + Z 2 + Z 3

M U LT IP LE C H O I C E Q UE S TI O NS (M C Q ) 1. The delta connection is formed when (a) The starting point of one coil is connected to the finishing point of the other coil (b) All the starting points of the coils are connected together (c) The starting and the finishing points of each coil are left unconnected (d) None of these

2. A good example for a delta-star transformation is in (a) The series-parallel circuit (b) The parallel-series circuit (c) The bridge circuit (d) None of these

ANSWERS (MCQ) 1. (d)

2. (c)

CON V E N TI O NA L Q UE S TI O NS (C Q ) 1. Determine the delta equivalent network for that shown in Figure 25.34. 2. Determine the star equivalent network for that shown in Figure 25.35.

3. Convert the delta-connected networks shown in Figure 25.36 to their equivalent star-connected networks. 4. For the network shown in Figure 25.37 determine the equivalent impedance. A

6Ω

15 Ω 20 Ω

8Ω 10 Ω

4Ω 8Ω 10 Ω

Figure 25.34 For CQ 1

M25_AUTH_ISBN_C25.indd 470

15 Ω

10 Ω

Figure 25.35 For CQ 2

8/6/2012 3:51:54 PM

Delta – j100 Ω 1 1Ω

4Ω

3

40∠0° V

5Ω 2 (a)

Figure 25.36 For CQ 3

j100 Ω (b)

B

10 Ω

–j8 Ω j15 Ω

2

471

j

A 100 Ω

Wye Transformations

2

j10 Ω 1

3

j25 Ω

Figure 25.37 For CQ 4

ANSWERS (CQ) 1. (4 + j4) Ω, (5.2 + j13.6) Ω, (17 − j12) Ω 2. (6.1 − j5.1) Ω, (6.9 + j2.0) Ω, (5.1 + j6.1) Ω

M25_AUTH_ISBN_C25.indd 471

3. (a) Z1 = 0.4 Ω, Z2 = 2 Ω, Z3 = 0.5 Ω (b) Z1 = −j100 Ω, Z2 = j100 Ω, Z3 = 100 Ω 4. Z1 = j3 Ω, Z2 = j7.5 Ω, Z3 = j5 Ω

8/6/2012 3:51:56 PM

26

Attenuators and Filters OBJECTIVES π Section

In this chapter you will learn about:    Logarithmic ratios; decibels and nepers and conversion from decibels to nepers and vice versa  The function of an attenuator  Different types of attenuators  Two-port networks   Terms like characteristic impedance, iterative im  pedance, image impedance, insertion loss and so on  The L-pad    Balanced and unbalanced and symmetrical and unsymmetrical networks  Characteristic impedance in terms of Zoc and Zsc  Different types of filters  The pass band and the attenuation band  Properties of ideal filters  Constant k and m-derived filters  Active and passive filters  Problems connected with the design of filters  Lattice and bridged T networks  Circular and hyperbolic functions  Theorem connecting a and Z0  Frequency response of filters

Z1 2

Z1 2Z2

Z1 2 Z1 2

2Z2

(O Network)

Half section

Z1 2

Z1 2

Z2

2Z2

(π Network)

2Z2

T Section

(T Network) Unbalanced

Z1 4

Z1 4

Z1 4

Z2 Z1 4

(H Network) Balanced

2Z2 (L Network)

Z1 4 Z1 4

2Z2

(C Network)

Unbalanced and balanced networks

26.1 INTRODUCTION An attenuator is a device for introducing a specified loss between a signal source and a matched load without upsetting the impedance relationship that is necessary for matching. The loss introduced is constant, irrespective of frequency. Since reactive elements (L or C) are frequency dependent, it follows that ideal attenuators are networks containing pure resistances. A fixed attenuator section is usually known as a pad. Attenuation is a reduction in the magnitude of a voltage or current due to its transmission over a line or through an attenuator. Any degree of attenuation may be achieved with an attenuator by suitable choice of resistance values but the input and output impedances of the pad must be such that the impedance conditions existing in the circuit into which it is inserted are not disturbed. Thus, an attenuator must provide the correct input and output impedances as well as the required attenuation. The loss is expressed in the form of a power ratio. Three forms of resistance networks that can conveniently be used are the T section, the π section, and the bridged T section. These networks may be designed to have any resistive value of characteristic impedance if symmetrical, or of any image impedance if asymmetrical. One of these networks may, therefore, be used in place of a transformer for matching between circuits of different resistive impedance. The attenuation introduced will be of little consequence if amplification is included in the circuit.

Attenuators and Filters

473

There are three conditions that the attenuating network must fulfill. 1. It must give the correct input impedance. 2. It must give the correct output impedance. 3. It must provide the specified attenuation.

26.2 THE DECIBEL Consider a network, as illustrated in Figure 26.1, connecting a generator to a load. Let the input power be P1 and the output power be P2. The ratio of the output power to the input power is then P2/P1. The network may introduce a loss (P2/P1 less than unity); or it may introduce a gain (P2/P1 greater than unity).

P1

Network

P2

Output

Figure 26.1 Network Connecting a Generator and a Load If a number of such networks are connected in tandem, as in Figure 26.2, and the individual power ratios are known, the overall power ratio Pn/P1 is obtained by multiplying together the individual power ratios. This follows from the fact that P1

M1

P2

M2

P3

M3

P4

Pn–2

Mn–2

Pn–1

Mn–1

Pn

Figure 26.2 Series of Networks in Tandem Pn P P P P = 2 × 3 × 4 × n P1 P1 P2 P3 Pn −1

(26.1)

= M1 × M 2 × M 3  × M n −1

(26.2)

Where, M1 M2, etc., are the individual power ratios. Example 26.1 Five networks in Figure 26.3 are inserted in tandem between A and B. The individual power ratios are: M1 = 0.215, M2 = 20.3, M3 = 0.246, M4 = 0.251 and M5 = 25.2. Find the power at B if l mW of power is a applied to A. A 1 mW

Los s

Gain

Loss

Loss

Gain

M 1= 0.215

M 2= 20.3

M 3= 0.246

M 4= 0.251

M 5= 25.2

Net wor k 1

Net w o r k 2

Network 3

Network 4

Network 5

B

Figure 26.3 For Example 26.1 Solution: Overall power ratio = 0.215 × 20.3 × 0.246 × 0.251 × 25.2 = 0.679 (a loss) Thus, if 1 mW is applied to A, the output power at B will be 0.679 mW.

26.2.1 Logarithmic Units In a complex system containing a large number of component circuits—each contributing a gain or loss—calculation of the overall power ratio may become extremely laborious. To simplify this calculation, the individual power ratios are expressed in a logarithmic unit enabling addition to be employed in place of multiplication. The logarithmic unit employed is the decibel (abbreviated to db) and power gain or loss D of a network expressed in this unit is defined as:

474 Electrical Technology D = 10 log10 P2 where,

(26.3)

P1

P2 = output power and P1 = input power.

If P2 /P1 is less than unity, then 10 log10 P2 /P1 will be negative. A negative sign, thus, indicates a power loss and a positive sign a gain. It should be noted that since 10 log10 P2

P1

= − 10 log P1

P2

the numerical answer will be the same whether P2 /P1 or P1 /P2 is considered, but to obtain the correct sign P2 /P1 must be considered. Power ratios expressed in decibels are given in Table 26.1. Table 26.1 Power Ratios Expressed in Decibels Input

Output

Ratio

2 mW

2000 mW

2000 = 1000 2

10 log10 1000 = 10 × 3 = 30 db

3 mW

600 mW

600 = 200 3

10 log10 200 = 10 × 2·301 = 23 ·01 db

5 mW

500 mW

500 = 100 5

10 log10 100 = 10 × 2 = 20 db

20 mW

2000 mW

2000 = 100 20

10 log10 100 = 10 × 2 = 20 db

2 mW

20 mW

20 = 10 2

10 log10 10 = 10 × 1 = 10 db

40 mW

200 mW

200 =5 40

10 log10 5 = 10 × 0·699 = 6.99 db

6W

6 =2 3

10 log10 2 = 10 × 0·301 = 3·01 db

12·6 mW

12 ⋅ 6 = 1⋅26 10

10 log10 1·26 = 10 × 0·10 = 1 db

5 mW

5 1 = 500 100

3W 10 mW

500 mW

Gain in db (negative sign indicates a loss)

10 log10

 1    = 10 × −log10100  100  = 10 × (−2) = −20 db

100 mW

21·6 mW

21.6 = 0.216 100

10 log10 0·216 = 10(1·335) = 10 (−1+0·335) = −6·65 db

The decibel is fundamentally a unit of power ratio and not of absolute power, but if some standard reference level of power may be assumed, then any absolute power can be expressed as so many decibel above or below the reference standard. The standard adopted is 1 milli Watt (0.001 W). Using this standard, any power P can be expressed as 10 log10 (P/1mW) db referred to 1 mW. The expression ‘db with respect to 1 mW’ is usually abbreviated to dbm. Other abbreviations sometimes used are as follows:

Attenuators and Filters

475

‘db wrt 1 mW’; ‘db ref 1 mW’ and ‘vu’ (voice unit) Thus, 20 dbm = 20 db wrt 1 mW = 20 db ref 1 mW = 20 vu = 100 mW The absolute powers expressed in dbm have been listed in Table 26.2. Table 26.2 Absolute Powers Expressed in dbm Powers expressed in decibels referred to 1 mW

μμ W − 90 dbm μμ W − 80 dbm μμ W − 70 dbm μ W − 60 dbm μ W − 50 dbm μ W − 40 dbm μ W − 30 dbm μ W − 27 dbm μ W − 24 dbm μ W − 23 dbm μ W − 21 dbm μ W − 20 dbm μ W − 17 dbm μ W − 14 dbm μ W – 13 dbm μ W − 11 dbm μ W − 10 dbm μ W − 7 dbm μ W − 4 dbm μ W − 3 dbm μ W − 1 dbm

1 10 100 0.001 0.01 0.1 1.0 2 4 5 8 10 20 40 50 80 100 200 400 500 800

1 mW 2 mW 4 mW 5 mW 8 mW 10 mW 20 mW 40 mW 50 mW 80 mW 100 mW 200 mW 400 mW 500 mW 800 mW 1000 mW 1W 2W 4W 5W 8 mW 10 W 100 W 1 kW 10 kW 100 kW

0 dbm + 3 dbm + 6 dbm + 7 dbm + 9 dbm + 10 dbm + 13 dbm + 16 dbm + 17 dbm + 19 dbm + 20 dbm + 23 dbm + 26 dbm + 27 dbm + 29 dbm + 30 dbm + 33 dbm + 36 dbm + 37 dbm + 39 dbm + 40 dbm + 50 dbm + 60 dbm + 70 dbm + 80 dbm

26.2.2 Two-port Networks Networks in which electrical energy is fed in at one pair of terminals and taken out at a second pair of terminals are called two-port networks, as has been shown in Figure 26.4, as are transmission lines, transformers, and electronic amplifiers. The network between the input port and the output port is a transmission network for which a known relationship exists between the input and output currents and voltages. ZA

ZB

ZC

( a)

ZD

ZF

ZE

(b)

Figure 26.4  Two-port Networks (a) T- Network (b) π Network

476 Electrical Technology If a network contains only passive circuit elements—such as an attenuator—the network is said to be passive. If a network contains a source of e.m.f., such as an electronic amplifier, the network is said to be active. Figure 26.5(a) shows a T-network, which is termed symmetrical if ZA = ZB in Figure 26.5(a) and ZE = ZF in Figure 26.5(b). If ZA ≠ ZB and ZE ≠ ZF, the network is termed asymmetrical. Both the networks shown have one common terminal, which may be earthed and are, therefore, said to be unbalanced. The balanced forms of T- and π-networks are shown in Figure 26.5. Z A/ 2

Z B/ 2

Z D/ 2

ZC

Z A/ 2

ZF

ZE

Z D/ 2

Z B/ 2 ( a)

(b)

Figure 26.5 (a) Balanced T-network (b) Balanced p-network

26.2.3 Power, Current, and Voltage Ratios The conversion of power ratios expressed in decibels (and of powers expressed in dbm) to actual power ratios (and actual powers) is effected by exactly the reverse process from that used for expressing power ratios in decibels (and actual powers in dbm). 10 log10 P2/P1 = D Log10 P2/P1 = D/10 P2/P1 = antilog D/10

(26.4)

Example 26.2 What power in Watts is represented by 25 dbm? Solution:

P/1mW = antilog 25/10 = antilog 2.5 = 316.2 P = 316.2 mW = 0.316 W

Example 26.3 Five networks are inserted in tandem between A and B. The decibel gains and losses of the individual networks are as follows. Network 1

..

..

..

Network 2 Network 3

.. ..

.. ..

.. ..

Network 4

..

..

..

Network 5

..

..

..

-6.68 db +13.08 db -16.09 db -6.00 db +14.01 db

Find the power at B if 0 dbm is applied to A. O dbm A

Los s

Ga i n

Loss

Loss

Gain

–6.68 db

+13.08 db

–16.09 db

–6.00 db

+14.01 db

Net wor k 1

Net w o r k 2

Network 3

Network 4

Network 5

Figure 26.6 For Example 26.3

B

Attenuators and Filters

477

Solution:

Total decibel gain on loss = −6.68 + 13.08 − 16.09 − 6.00 + 14.01 = −1.68 db Power at B = 0 − 1.68 dbm = −1.68 dbm Note: This is the decibel approach to Example 26.1. It should be noted that −1.68 dbm corresponds to a power of −679 milliwatts. When it is desired to compare the powers developed in two equal resistors, it is sufficient to measure the two voltages or the two currents; and then the power ratio in decibels is equal to twenty times the logarithm (to the base 10) of the current or voltage ratio. Let us consider two equal resistors of R Ohms, carrying currents of r.m.s. values I1 and I2 and having voltages across them of r.m.s. values E1 and E2, respectively. Then the powers developed in these two resistors are P1 = E1 I1 = R ⋅ I12 = P2 = E2 I 2 = RI 22 =

and

1 ⋅ E2 R 1

1 ⋅ E2 R 2

The ratio between these two powers, therefore, is 2  E2 2 P2  E2 I 2   I 2   =   =   =  E1  P1  E1 I1   I1 

Expressing this in decibels, P   I 2 D = 10 log10  2  = 10 log10  2   P1   I1  I  = 20 log10  2   I1 

(26.5)

E  P   E 2 D = 10 log10  2  = 10 log10  2  = 20 log10  2   E1   P1   E1 

and

(26.6)

Thus, the power ratio in decibels is equal to 20 log10 (current ratio) = 20 log10 (voltage) ratio, provided that the two resistances, through which the two currents I1 and I2 (or across which the two voltages E1 and E2) are measured are equal. When it is required to compare the powers in two impedances by measurement of the current through them, it is desirable that their resistive components be equal since, in this case, the power ratio will be equal to the (current ratio)2. This may be verified as follows. Z1 ≡ Z1 ≡ R1 + jX1 Z 2 ≡ Z 2 ≡ R2 + jX 2

and

Let the magnitudes of the currents flowing in Z1 and Z2 be I1 and I2, respectively. The power ratio is P2

P1

=

I 22 R2 I12 R1

Hence, when R1 = R2, P2 / P1 = ( I 2 / I1 ) 2

(26.7) When considering the voltages across the two impedances the (voltage ratio) will be equal to the power ratio if the conduction components are equal. This may be verified as in the following manner. Let the impedances be Z1 ≡ Z1 φ 1 ≡ R1 + jX1 2

and

Z 2 ≡ Z 2 φ 2 ≡ R2 + jX 2

478 Electrical Technology Let their admittances be Y1 ≡ Y1

φ1 ≡ G1 + jB1

and

Y2 ≡ Y2 φ 2 ≡ G2 + jB2

Power

P1 ≡ I12 R1 =

E12 E12 . cos φ = cos φ1 Z 1 1 Z1 Z12

= E12 Y1 cos φ1 = E12 G1 Similarly,

P2 = E22 G2

The power ratio is:

E2 G P2 = 22 2 P1 E1 G1

Hence, when

G1 = G2 ,

2 P2  E2  =   P1  E1 

(26.8)

26.2.4 The Decibel and the Neper If logarithms to the base 10 are used; then the ratio is said to be in bels. The bel is a large unit and the decibel is more often used, where 10 decibels = 1 bel. If logarithms to the base e (i.e., natural or Napierian logarithms) are used, then the ratio of two powers is said to be in nepers (Np), i.e., Power ratio in decibels = 10log10 P2

(26.9)

P1

Power ratio in nepers = 1 log e P2 2 P1

(26.10)

The decibel is fundamentally a unit of power ratio. It can be used to express current ratios when the resistive components of the impedances through which the current flows are equal and voltage ratios when the conductive components of these impedances are equal. The neper is fundamentally a unit of current ratio, but it can be used to express power ratios when the resistive components of the impedances are equal. The loss of power in a transmission line or an electrical network is known as attenuation. Attenuation may be measured using either the decibel or the neper notation. As a result of its derivation from the exponential e, the neper is the most convenient unit for expressing attenuation in theoretical work. The decibel, on the other hand, being defined in terms of logarithms to the base 10, is a more convenient unit in practical calculations using the decimal system. The conditions under which the two units may be used can be summarized in the following equations, the notation of which is indicated in Figure 26.7. Power in = Ps

Power out = PR IR

Is

Es Z1

Network

ER

Z2

Figure 26.7 Attenuation Measured Using Decibel and Neper Notations Attenuation db = 10 log10

PS PR

= 20 log10

IS IR

(provided R1 = R2 )

= 20 log10

ES ER

(provided G1 = G2 )

(26.11)

Attenuators and Filters

Attenuation in nepers = log e

IS IR

= log e

IS IR

=

(provided Z1 = Z 2 )

1 P log e S 2 IR

479

(26.12)

(provided R1 = R2 )

If the resistive components of the impedances at the input and output of the network are equal, then the attenuation may be readily converted from one notation to the other Attenuation in db = 20 log 10 = 20 log e

Is IR

IS × log10 e IR

(26.13)

IS IR Attenuation in db = 8.68 × (attenuation in nepers) provided R1 = R2 Attenuation in nepers = 0.115 × (attenuatio on in db) = 8.68 log e

}

Example 26.4 The ratio of output power to input power in a system is (1) 2 (2) 25 (3) 1000 and (4) 1/100. Determine the power ratio in each case in decibels and in nepers. Solution: Power ratio in dbs = 10 log10 (P2 / Pl ) 1. P2 / P1 = 2, 10 log 2 = 3 db 2. P2 / P1 = 25, 10 log 25 = 14 db 3. P2 / P1 = 1000, 10 log 1000 = 30 db 4. P2 / P1 = 1/100, 10 log 1/100 = −20db Power ratio in nepers = ½ loge (P2/P1) 1. 2. 3. 4.

P2/P1 = 2, ½ loge 2 = 0.347 Np P2/P1 = 25, ½ loge 25 = 1.609 Np P2/P1 = 1000, ½ loge 1000 = 3.454 Np P2/P1 = 1/100, ½ loge 1/100 = −2.303 Np

26.3 CHARACTERISTIC IMPEDANCE The input impedance of a network is the ratio of voltage or current (in complex form) at the input terminals. With a twoport network, the input impedance often varies according to the load impedance across the output terminals. For any passive two-port network, it is found that a particular value of load impedance can always be found which will produce an input impedance having the same value of impedance as the load impedance. This impedance is called the iterative impedance for an asymmetrical network and its value depends on which pair of terminals is taken to be the input and which to be the output. Thus, there are two values of iterative impedance, one for each direction. For a symmetrical network, there is only one value for the iterative impedance, and this is called the characteristic impedance of the passive symmetrical two-port network. Figure 26.8 illustrates a symmetrical T-network terminated in an impedance Z0. Let the characteristic impedance be denoted by Z0 and the impedance ‘looking-in’ at the input port be Z0. Then

480 Electrical Technology ZA

I1

ZA

V1

I2

Z0

V2

ZB

Z0 Input port

Output port

Figure 26.8 Symmetrical T-network Terminated in Z0 V1/I1 = Z0 = V2 / I2 Z0 = Z A +

Z B (Z A + Z0 ) , since (Z A + Z 0 ) is in parallel with Z B . Z B + Z A + Z0

Z 0 Z A + Z 0 Z B + Z 02 = Z A2 + 2 Z A Z B + Z A Z 0 + Z B Z 0 Z 02 = Z A2 + 2 Z A Z B Characteristic impedance, Z 0 = ( Z A2 + 2 Z A Z B )

(26.14)

Now, if the output terminals of Figure 26.8, are open circuited Z OC = Z A + Z B Short circuiting the output terminals, Z SC = Z A +

Z 2 + 2 Z AZB Z AZB = A Z A + ZB Z A + ZB

Z oc Z sc = Z A2 + 2 Z A Z B

(26.15)

Z 0 = ( Z oc Z sc )

Thus

Figure 26.9 illustrates a passive symmetrical π-network terminated in an impedance Z0. Z1

I1

Z2

V1

I2

Z2

V2

Z0

Z0 Input port

Output port

Figure 26.9  Symmetrical π-network Terminated in Z0 If the impedance ‘looking in’ at the input port is also Z0, then Z0 =

( Z 2 ) ( Z1Z 0 + Z1Z 2 + Z 0 Z 2 ) / ( Z 0 + Z 2 ) Z 2 + ( Z1Z 0 + Z1Z 2 + Z 0 Z 2 ) / ( Z 0 + Z 2 )

After simplification, 2 Z 2 Z 02 + Z1Z 02 = Z1Z 22

Attenuators and Filters

 Z1Z2  2  Z0 =   Z1 + 2 Z2 

Characteristic impedance,

481

(26.16)

If the output terminals of Figure 26.9 are open-circuited, then the open-circuit impedance, Z oc =

Z 2 ( Z1 + Z 2 ) Z1 + 2 Z 2

Now if the output terminals of Figure 26.9 are short circuited, then the short-circuit impedance Z sc = Z oc Z sc =

Z 2 Z1 Z1 + Z 2 Z 2 ( Z1 + Z 2 ) Z1 + 2 Z 2

×

Z1Z 22 Z 2 Z1 = Z1 + Z 2 Z1 + 2 Z 2

Z 0 = ( Z oc Z sc )

(26.17)

Thus, the characteristic impedance Z0 is given by Z 0 = ( Z oc ⋅Z sc ) whether the network is a symmetrical T or a symmetrical π.

26.4 SYMMETRICAL T-ATTENUATOR The ideal attenuator is made up of pure resistances. Asymmetrical T-attenuator is shown in Figure 26.10 with a termination R0. R0 =

( R12 + 2 R1R2 )

(26.18)

R0 = ( Roc ⋅Rsc )

(26.19)

If the resistance R0 is the termination, the input resistance of the pad will also be equal to R0. If the terminating resistance is transferred to port A then the input resistance looking into port B will again be R0. The pad is, therefore, symmetrical in impedance in both directions of connection and may thus be inserted into a network whose impedance is also R0. The value of R0 is the characteristic impedance of the section. Attenuation may be expressed as a voltage ratio V1/V2 or quoted in decibels as 20 log (V1 / V2 ) or, alternatively, as a power ratio 10 log ( P1/ P2 ) . If a T-section is symmetrical, i.e., the terminals of the section are matched to equal impedances, then: 10 log

10 log

or Let N = V1 /V2 or I1 / I 2 or ratio. From Figure 26.10.

R1

R1 I2

V1

V

R2

V2

R0

R0 Port A

Port B

Figure 26.10 Symmetrical T-pad Attenuator

P1 V I = 20 log 1 = 20 log 1 P2 V2 I2 RIN = RLOAD = R0, i.e.,

Since

From which

I1

 V 2  I 2 P1 = 10 log  1  = 10 log  1   V2   I 2  P2  V1 2  I1 2 P1 =   =    V2   I 2  P2

P1 / P2 = V1 /V2 = I1 / I 2 P12 / P2 , where N is the attenuation. For a matched network, N is, in fact, the insertion loss

482 Electrical Technology Current

I1 = V1/ R0

Voltage,

V  V = V − I1 R1 = V1 −  1  R1  R0   R  V = V1 1 − 1   R0 

 R0  Voltage V2 =   V by voltage division  R1 + R0 

 R0   R   V1 1 − 1  V2 =    R0   R1 + R0    R0   R0 − R1     = V1   R1 + R0   R0 

V1 R − R1 V R + R1 = 0 or 1 = N = 0 V2 R0 + R1 V2 R0 − R1

Hence,

(26.20)

It is possible to derive expressions for R1 and R2 in terms of N and R0, thus, enabling an attenuator to be designed to give a specified attenuation and to be matched symmetrically into the network. R0

(N –1) (N +1)

R0

(N –1) (N +1)

 N − 1  R1 = R0   N + 1 

(26.21)

 2 N  R2 = R0  2  N − 1 

(26.22)

and

R0

2N N 2–1

R0

R0

Figure 26.11 Symmetrical T-pad

Thus, if the characteristic impedance R0 and the attenuation N ( = V1 / V2 ) are known for a symmetrical T-network then the values of R1 and R2 may be calculated. Figure 26.11 shows a T-pad attenuator having input and output impedances of R0 with resistances R1 and R2 expressed in terms of R0 and N.

26.5 SYMMETRICAL Π-ATTENUATOR A symmetrical π-attenuator is shown in Figure. 26.12, to terminate in R0 characteristic impedance  R1 R 2  2  R0 =   R1 + 2 R2 

(26.23)

R0 = ( Roc ⋅Rsc )

(26.24)

R1

IB IA

Given the attenuation factor N ( = V1/ V2 = I1/ I 2 ) and the characteristic impedance R0, it is possible to derive expressions for R1 and R2 in a similar way to the T-pad attenuator, to enable a π-attenuator to be effectively designed. It can be shown that  N 2 − 1   R1 = R0 =   2 N  and

I1

R2

V1

ID

I2

IC R2

V2

R0

R0

Figure 26.12   Symmetrical π-attenuator

 N + 1  R2 = R0   N − 1 

Figure 26.13 shows a π-attenuator having input and output impedances of R0 with resistances R1 and R2 expressed in terms of R0 and N.

Attenuators and Filters

483

There is no difference in the functions of the T- and π-attenuator pads and either may be used in a particular situation. 2 R0 N –1 2N

R0 N +1 N–1

R0

R0 N +1 N–1

R0

Figure 26.13  Symmetrical π-attenuator With R1 and R2 Expressed in Terms of R0 and N Example 26.5 Determine the characteristic impedance of each of the attenuator sections shown in Figure 26.14. 8Ω

8Ω

10 Ω

10 Ω

21 Ω

200 Ω

200 Ω

15 Ω

(a)

56.25 Ω

(b)

(c)

Figure 26.14 For Example 26.5 Solution:

1.

R0 =

( R12 + 2 R1R2 )

R0 =

( 82 ) + ( 2 )( 8 )( 21)

=

400 = 20 Ω 2

R0 = (10 ) + ( 2 )(10 )(15 )

2.

=

400 = 20 Ω 2

R0 = ( 200 ) + ( 2 )( 200 )( 56.25 )

3.

= 62500 = 250 Ω Example 26.6 For each of the attenuator networks shown in Figure 26.15, determine (1) the input resistance, when the output port is opencircuited, (2) the input resistance when the output port is short-circuited, and (3) the characteristic impedance. 15 Ω

15 Ω

15 Ω

10 Ω

Input port

5Ω

Output port (a)

5Ω

Input port

Output port (b)

Figure 26.15 For Example 26.6

484 Electrical Technology Solution: For the T-network (Figure 26.15a) 1.

Roc = 15 + 10 = 25 Ω

2.

Rsc = 15 +

3.

R0 = ( Roc ⋅ Rsc ) = ( 25 )( 21) = 22.9 Ω

10 × 15 = 21 Ω 10 + 15

For the π-network (Fig. 26.15b) 5 × (15 + 5 )

1.

Roc =

2.

Rsc =

3.

R0 = ( Roc )( Rsc )

5 + (15 + 5 )

=

100 = 4Ω 25

5 × 15 75 = = 3.75 Ω 5 + 15 20

= ( 4 )( 3.75 ) = 15 = 3.87 Ω Example 26.7 Design a T-type pad to give 25 db attenuation and to have a characteristic impedance of 600 Ω. Solution: N = antilog D/20 = antilog 25/20 =17.8  N − 1  16.8 = 600 × R1 = R0   N + 1  18.8 = 536 Ω 2N 1200 × 17. 8 = 2 316 N −1 = 67.6 Ω

R2 = R0 ×

26.6 INSERTION LOSS Figure 26.16 (a) shows a generator E connected directly to a load ZL. Let the current flowing be IL and p.d. across the load VL. Z is the internal impedance of the source. I1 IL

E VL

I2

E V1

ZL

Two-port network

z

z

(a)

(b)

Figure 26.16 Insertion Loss

V2

ZL

Attenuators and Filters

485

Figure 26.16 (b) shows a two-port network connected between the generator E and the load ZL. The current through the load shown as I2, and the p.d. across the load, shown as V2, will generally be less than current IL and voltage VL of Figure 26.16 (a), as a result of the insertion of the two-port network, between generator and load. The insertion loss ratio AL, is defined as AL =

voltage across load when connected directly to the generrator voltage across load when the two-port network is insserted (26.25)

= I L /I 2 AL V= L / V2

Since VL = IL ZL and V2 = I2 Z2, both VL and V2 refer to p.ds across the same impedance ZL, the insertion loss ratio may also be expressed as V I (26.26) insertion loss ratio = 20 log L db or 20 log L db V2 I2 When the two-port network is terminated in its characteristic impedance Z0, the network is said to be matched. In such circumstances, the input impedance is also Z0 and the insertion loss ratio is simply the ratio of the input voltage to the output voltage, V1/V2. For a network terminated in its characteristic impedance AL = 20 log

V1 I db or 20 log 1 db V2 I2

(26.27)

Example 26.8 A 0.3 kΩ potentiometer is connected across the output of a signal generator of internal resistance 500 Ω. If a load of 2 kΩ is connected across the rheostat, determine, the insertion loss at a tapping of (1) 2 kΩ (2) 1kΩ. Solution: The circuit diagram is given in Figure 26.17. Without the rheostat in the circuit (Figure 26.18), the voltage VL across the 2 kΩ load is given by

E E

3 kΩ 500Ω

V2

VL

2 kΩ

2 kΩ

500Ω Input port

Output port

Figure 26.17 For Example 26.8

Figure 26.18 Solution for Example 26.8   2000 VL =  E = 0.8E  2000 + 500 

1. With the 2 kΩ tapping, the network of Figure 26.17 may be redrawn as shown in Figure 26.19.   1000 V2 =  E = 0.4 E  1000 + 1000 + 500  Insertion loss ratio =

0.8E V1 = =2 0.4 E V2

Insertion loss = 20log 2 = 6.02 db

1 kΩ

E

500 Ω

2 kΩ

V2

2 kΩ

Figure 26.19 Solution for Example 26.8

486 Electrical Technology 2.

  (1000 × 2000 ) / (1000 + 2000 )  E V2 =   (1000 × 2000 ) / (1000 + 2000 ) + 2000 + 500    666.7  E = 0.211E =   666.7 + 2000 + 500  0.8E VL = = 3.79 0.211E V2

Insertion loss ratio =

Insertion loss = 20log 3.79 = 11.97 db Note: The insertion loss is not doubled by halving the tapping.

26.7 ASYMMETRICAL T - AND Π - SECTIONS Figure 26.20 (a) shows an asymmetrical T-pad section in which R1 ≠ R3. Similarly, Figure 26.20 (b) shows an asymmetrical π-section in which R2 ≠ R3. When viewed from port A, in each of the sections, the output impedance is R0B. When viewed from port B, the input impedance is R0A. Since the sections are asymmetrical R0A ≠ R0B. R1

R3

R 0A

R1

R 0B

R2

Port A

Port B

(a)

R 0A

Port A

R3

R2

(b)

R 0B

Port B

Figure 26.20  (a) Asymmetrical T-section (b) Asymmetrical π-section Iterative impedance is the term used for the impedance measured at one port of a two-port network when the other port is terminated with an impedance of the same value. For example, the impedance looking into port 1 of Figure 26.21(a), say, 500 Ω when port 2 is terminated in 500 Ω and the impedance looking into port 2 of Figure 26.21(b) is, say, 600 Ω when port 1 is terminated in 600 Ω. In symmetric T- and π- sections, the two iterative impedances are equal, this value being the characteristic impedance of the section. An image impedance is defined as the impedance which, when connected to the terminals of a network, equals the impedance presented to it at the opposite terminals. For example, the impedance looking into port 1 of Figure 26.22(a) is, say, 400 Ω when port 2 is terminated in, say, 750 Ω, the impedance seen looking into port 2, Figure 26.22(b), is 750 Ω when port 1 is terminated in 400 Ω. An asymmetrical network is correctly terminated when it is terminated in its image impedance. If the image impedances are equal, the value is the characteristic impedance.

500 Ω

500 Ω

Port 1

600 Ω

600 Ω

Port 2 (a)

Figure 26.21 Iterative Impedance

(b)

Attenuators and Filters

750 Ω

400 Ω

Port 1

Port 2

400 Ω

487

750 Ω

Port 1

Port 2

(a)

(b)

Figure 26.22 Image Impedance Example 26.9 An asymmetrical T-section attenuator is shown in Figure 26.23. Determine for the section (1) the image impedances, and (2) the iterative impedances. 200 Ω

Solution: R0 A =

1.

( R0C ) ( RSC )

ROC = 200 + 100 = 300 Ω, and RSC

Similarly,

(100)(300) = 200 + = 275 Ω 100 + 300

R0 A =

(300)(275) = 287.2 Ω

R0 B =

( ROC )( RSC )

R0 B =

100 Ω

R 0A

Port 1

R 0B

Port 2

Figure 26.23 For Example 26.9

R0C = 300 + 100 = 400 Ω, and RSC = 300 + Hence,

300 Ω

(200)(100) = 366.7 Ω 200 + 100

(400)(366.7) = 383 Ω

Thus, the image impedances are 287.2 Ω and 383 Ω, and are shown in the circuit of Figure 26.24. 2. The iterative impedances at port 1 (Figure 26.25), R1 200 Ω

287.2 Ω

300 Ω

200 Ω

100 Ω

383 Ω R 0B

R 0A Port 1

Port 2

Port 1

Figure 26.24 Image Impedances R1 = 200 + from which and

R1

(100)(300 + R1 ) 100 + 300 + R1

R12 + 100 R1 − 110000 = 0 − 100 ±  (100) 2 − (4)(1)(− 110000)  2 670.8 R1 = − 100 ± = 285.4 Ω 2

(Neglecting the negative value)

100 Ω

R1

Port 2

Figure 26.25 Iterative Impedance at Port 1

400 R1 + R12 = 80000 + 200 R1 + 30000 + 100 R1

R1 =

300 Ω

488 Electrical Technology The iterative impedance at port 2 (Figure 26.26), R2 200 Ω

300 Ω

R2 = 300 + from which

100 Ω

R2

Port 1

(100)(200 + R2 ) 100 + 200 + R2

300 R2 + R22 = 90000 + 300 R2 + 20000 + 100 R2

R2

R22 − 100 R2 − 110000 = 0

and

Port 2

100 ±  (− 100) 2 − (4)(1)(− 110000)  2 100 ± 670.8 = = 385.4 Ω 2

R2 =

Figure 26.26 Iterative Impedance at Port 2

Thus, the iterative impedances of the section shown in Figure 26.23 are 285.4 Ω and 385.4 Ω.

26.8 THE L-SECTION ATTENUATOR An L-section attenuator pad is shown in Figure 26.27. Such a pad is used for matching purposes only, the design being such that the attenuation introduced is a minimum. In order to derive values for R1 and R2, let us consider the resistances seen from either end of the section. R0 A and

(26.28)

R0 A R2 + R0 A R0 B = R1 R2 + R1 R0 B + R2 R0 B R0 B

and

R1

R2 R0 B looking in at port 1 = R1 + R2 + R0 B R ( R + R0 A ) = 2 1 R1 + R0 A + R2

R2

R 0B

Looking in at port 2 (26.29)

R0 B R1 + R0 A R0 B + R0 B R2 = R1 R2 + R2 R0 A

R1 =

Port 1

Port 2

Figure 26.27 L-section Attenuator Pad

Adding equations (26.28) and (26.29) gives 2R0AR0B = 2 R1R2 and

R 0A

R0 A R0 B R2

(26.30)

Substituting this expression for R1 into equation (26.28) and simplifying gives  R R2  R2 =  0 A 0 B   R0 A − R0 B 

(26.31)

From equation (26.30) R1 = and

R0 A R0 B ( R0 A R02B )( R0 A

− R0 B )

R1 = [ R0 A ( R0 A − R0 B ) ]

(26.32)

Figure 26.28 shows an L-section attenuator pad with its resistances expressed in terms of the input and output resistances R0A and R0B. √ [ R 0A ( R 0A – R 0B) ]

R 0A

√

R 0A R 20B R 0A – R 0B

R 0B

Figure 26.28  L-section Attenuator Pad with Resistances Expressed in Terms of R0A and R0B

Attenuators and Filters

489

26.9 CASCADING TWO-PORT NETWORKS Often, two-port networks are connected in cascade, i.e., the output from one network becomes the input to the second network and so on, as shown in Figure 26.29. Thus, an attenuator may consist of several cascaded sections to achieve a particular desired overall performance. I1

I2 1

V1

I3 2

V2

I n–1

V3

In Vn–1

n –1

Vn

Z0

Figure 26.29 Two-port Networks Connected in Cascade If the cascade is arranged in such a way that the impedance measured at one port and the impedance with which the other port are terminated have the same value, then each section (assuming they are symmetrical) will have the same characteristic impedance Z0 and the last section will be terminated in Z0. Thus, each network will have a matched termination and, hence, the attenuation in decibels of section 1 as illustrated in Figure 26.29 is given by a1 = 20 log V1

R1

V2

Similarly, the attenuation of section 2, is given by a2 = 20 log V2

E 100 Ω

R2

V3

500 Ω

The overall attenuation is given by a = 20 log V1

Vn

V V V V  = 20 log  1 × 2 × 3 × .... × n −1   V2 V3 V4 Vn 

Figure 26.30 For Example 28.10

the overall attenuation, a = a1+a2+a3+.....an−1 Thus, the overall attenuation is the sum of the attenuations (in decibels) of the matched sections. Example 26.10 A generator having an internal resistance of 500 Ω is connected to a 100 Ω load via an impedance matching resistance pad as shown in Figure 26.30. Determine (1) the values of resistance R1 and R2, (2) the attenuation of the pad in decibels, and (3) its insertion loss. Solution: R1 = [ 500(500 − 100) ] = 447.2 Ω

1.

 (500)(100) 2   = 111.8 Ω R2 =   500 − 100  2. As the terminals of the pad are not matched to equal impedances Attenuation = 10 log P1 Current I1 (Figure 26.31) =

E 500 + 447.2 + (111.8 × 100) / (111.8 + 100)

=E and

1000

 111.8  I 2 =  I =E 1894.5  111.8 + 100  1

(

The input power, P1 = E

P2

db

2

1000 )

(500)

l1

R 1 = 447.2 Ω

E V1 500 Ω

R2 = 111.8 Ω

l2

V2

100 Ω

Figure 26.31 Solution for Example 26.8

490 Electrical Technology

(

3.

2

)

The output power, P2 = E 1894.5 Hence, attenuation = 10 log P1

(100)

P2

IL

= 12.54 db

E

E/6 0.0528 E AL = VL (Figure 26.32) = 3.157 V2

VL

Insertion loss =

AL = 20 log VL

V2

100 Ω

500 Ω

Figure 26.32 Generator Connected Directly to the Load

= 20 log 3.157 db

= 9.99 db

26.10 FILTERS

26.11 TYPES OF FILTERS There are five types of filters: low-pass, high-pass, band-pass, bandelimination (also referred to as band-stop, band reject or notch) and all-pass filters. Figure 26.35 illustrates the frequency response plots for the first four types of filters.

Attenuation

A network that is designed to attenuate certain frequencies and pass others without loss is called a filter. A filter, therefore, possesses at least one pass band and at least one attenuation band. Attenuation is zero in the pass band while it is finite in the attenuation band. The frequencies that separate the various pass and attenuation bands are called cut-off frequencies, and are denoted by F1, F2, etc., or by FC if there is only a L L 2 2 L single cut-off frequency. Typical filters are shown in Figure 26.33. Attenuators are constructed from purely resistive elements. An C C C important characteristic of all filters is that they are 2 2 constructed from purely reactive elements, for otherwise the attenuation would never become zero. (a) (b) The simplest type of filter has only one pass band, one attenuation band, and a single cut-off frequency. If L L L it passes all frequencies up to the cut-off frequency and 4 4 2 attenuates all the frequencies above, it is called a lowpass (LP) filter. If, on the other hand, if it attenuates all C C frequencies below the cut-off frequency and passes all the C 2 2 frequencies above, it is called a high-pass (HP) filter. An ideal filter would have zero attenuation in the pass L L band and infinite attenuation in the attenuation band. L 4 4 2 For example, an ideal low-pass filter might have an (c) (d) attenuation frequency curve as shown in Figure 26.34. This cannot be achieved in practice. In the first place, in a Figure 26.33   Symmetrical Low-pass T and π Sections  practical filter it is found that the attenuation outside the in Unbalanced and Balanced Forms pass band is finite: it can, however, be made as large as (a) Symmetrical Unbalanced T Section required by using a sufficient number of sections in series. (b) Symmetrical Unbalanced π Section  Secondly, if any resistance is present (it is impossible to (c) Symmetrical Balanced T Section construct an inductance that does not possess a certain (d) Symmetrical Balanced π Section amount of resistance), the attenuation in the pass band will not be zero. Usually, however, it is only one or two decibels. Finally, mismatch losses must be considered; for although the characteristic impedance of the section may vary with frequency, it will probably be terminated in a fixed resistance, or an impedance that does not vary in frequency in the same way as the characteristic impedance of the section. Pass band

Attenuation band

Cut-off frequency

Frequency

Figure 26.34 Attenuation-frequency Curve of an Ideal Low-pass Filter

Attenuators and Filters |V0|

491

|V0|

Pass band

Stop band

Frequency

fc (a) |V0|

fc

Stop band

Frequency

(b)

|V0|

Stop band

Pass band

Stop band

Pass band

Stop band

Pass band

Passband

f i f r fh

Frequency

(c)

fi f r f h (d)

Frequency

Figure 26.35   Frequency Response of Filters (a) Low-pass Filter (b) High-pass Filter  (c) Bandpass Filter (d) Band-elimination Filter A low-pass filter is a filter that has a constant output voltage from d.c. up to a cut-off frequency fc. as can be seen in Figure 26.35 (a) As the frequency increases above fc, the output voltage is attenuated. The solid line is a plot for the ideal low-pass filter, while the dashed lines indicate the curves for practical low-pass filters. The range of frequencies that are transmitted is known as the pass band. The range of frequencies that are attenuated is known as the stop band. The cut off frequency, Fc, is also called the 0.707 frequency, the half-power frequency, the –3db frequency, the corner frequency, or the break frequency. A high-pass filter is a circuit that attenuates the output voltage for all frequencies below the cut off frequency fc. Above fc, the magnitude of the output voltage is constant. Figure 26.35 (b) describes the plot for ideal and practical high-pass filters. The solid line is the ideal curve; the dashed curves show how the practical high-pass filters deviate from the ideal. A band pass filter is a circuit that passes only a band of frequencies while attenuating all frequencies outside this band, Figure 26.35 (c). Band-elimination filters reject a specified band of frequencies while passing all frequencies outside this band, Figure 26.35(d). All pass filters are designed to provide constant gain to signals at all frequencies.

26.12 ACTIVE AND PASSIVE FILTERS The simplest approach to building a filter is with passive components (resistors, capacitors and inductors). In the radio frequency range, this works quite well. However, as the frequency comes down, inductors begin to have problems. Audio frequency inductors are physically large, heavy and, therefore, expensive. To increase inductance (for low-frequency applications), more turns of wire must be used, which adds to the series resistance, degrading the inductor’s performance. Input and output impedances of passive filters (especially below r.f.) are both a problem. The input impedance is low which loads down the source. The output impedance is relatively high, which limits the load impedance that the passive filter can drive. There is no isolation between the load impedance and the passive filter. This means that the load must be considered as a component of the filter and must be taken into consideration while determining the filter response or design. Any change in load impedance may significantly alter one or more of the filter’s response characteristics. Active Filters incorporate an amplifier (normally an op-amp) with resistor resistor/capacitor network to overcome the problems of passive filters. By enclosing a capacitor in a feedback loop, the inductor (with all of its low-frequency problems) can be eliminated. Properly configured, input impedance can be increased. The load is driven from the output of the op-amp giving a very low output impedance. Not only does this improve the load-drive capability, but the load is now isolated from the frequency determining network. Variations of load will have no effect on the active filters characteristics. The high input impedance of the op-amp allows for a reduction in the size and cost of the capacitor. By selecting a quad op-amp IC, steep roll offs can be built economically in very little space. Active filters also have their limitations. High-frequency response is limited by the gain bandwidth and the slew rate of the op-amps. High-frequency op-amps are more expensive, making passive filters a more economical choice for r.f. applications. Active filters require a power supply. For op-amps, this may be two supplies. Variations in the power supplies output voltage show up, to some extent, in the signal output from the active filter. In multistage applications, the common

492 Electrical Technology power supply provides a bus for high-frequency signals. Feedback along the power supply lines can cause oscillations unless decoupling techniques are rigorously employed. Active filters are much more susceptible to radio frequency interference than are passive RLC filters. Practical considerations limit the Q of the band pass and notch filters to less than 50. For circuits requiring very narrow (selective) filtering, a crystal filter is most appropriate.

26.13 FREQUENCY RESPONSE The gain and phase shift of a filter change as the frequency changes. This is the purpose of a filter. The frequency response of a filter is shown by plotting its gain (or loss) versus frequency on a logarithmic graph paper. The sinusoidal steady-state response of a linear system as a function of the frequency of the excitation is a fundamental characteristic of that system. It is called the frequency response. The frequency response of any linear system can be obtained directly from the system function. Let us suppose that a source waveform is given by x(t ) = A cos (ω t + φ )

(26.33)

Where, A is the amplitude, w is the angular frequency and f is the phase angle. Equation (26.33) can also be written in the form x ( t ) = Re  Ae jφ e j ωt 

(26.34)

where ( Ae jφ ) is the complex amplitude of e j wt excitation. If we denote the steady-state forced response by y(t), we can write y (t ) = Re  H ( jω ) Ae jφ e jω t 

(26.35)

y ( t ) = Re

{( H

( jω ) A ) e j (φ +θ )e jωt }

(26.36)

To find the real part, the steady-state response y(t) is y ( t ) = H ( jω ) A cos  ω t + (θ + φ )

(26.37)

Angle (degrees)

where, H(jw) emphasizes the fact that when discussing frequency response, one always evaluates the system function H(S) at the particular value S = jw. Once the value of S is specified in this fashion, the system function H(jw) becomes an ordinary complex number, with a magnitude and an angle. Denoting 120 the magnitude of H(jw) by |H(jw)| and the angle of H(jw) by θ, we 80 can write the expression for y(t). 0 – 40 – 80 – 120

where, for comparison, the source waveform was

1

2

3

(26.38)

Comparing the source x (t) and the response y(t), we can observe three significant features characteristic of the sinusoidal response of linear systems. 1. The frequency of the response is the same as the frequency of the source. 2. The amplitude of the response is equal to the amplitude of the source multiplied by the magnitude of the system function | H(jw) |. 3. The phase angle of the response is equal to the phase angle of the source in addition to the phase angle of the system function. These three quantities comprise the frequency response of the linear system. The frequency response of a filter is shown by plotting its gain (or loss) versus frequency on logarithmic graph paper. The two types of logarithmic graph paper are semi-log (Figure 26.36(a)), and log-log (Figure 26.36(b)). On semi-log graph paper the divisions along one axis are spaced logarithmically, while the other axis has conventional linear spacing between divisions.

5 10 20 30 Frequency (rad/s)

50

100

50

100

(a)

1000 500

200 Magnitude

x(t ) = A cos (ω t + φ )

40

100 50

20 10

1

2

3

5 10 20 30 Frequency (rad/s) (b)

Figure 26.36 (a) Semi-log Coordinates and (b) Log-log Coordinates

Attenuators and Filters 1

2

3 4 5 6 7 891

2

3 4 5 6 7 8 91

2

493

3 4 5 6 7 8 9 1 1 100 9 8 7 6 5 4 3 2

One octave 8–16

1 10 9 8 7 6 5 4 3 2

One octave 20 –40

One octave 5,000 – 10,000

One decade 0.1–1.0

1 1.0 9 8 7 6 5 4 3 2

One octave 0.1– 0.2

One decade 10 – 100

10

One decade 1,000 – 10,000

100

1 0.1 9 8 7 6 5 4 3 2

One decade

1 0.01 9 8 7 6 5 4 3

300 – 3,000

2

1,000

1 0.001 10,000

Figure 26.37  Log-log Graph Paper

On log-log graph paper, both axis have logarithmic spacing between divisions. Logarithmic spacing results in a scale that expands the display of smaller values and compresses the display of larger values. On a logarithmic graph paper, a 2-to-1 range of frequencies is called an octave and a 10-to-1 range of frequencies is called a decade. Note:

1. Each octave corresponds to a 2-to-1 range of values. 2. Each decade corresponds to a 10-to-1 range of values. Figure 26.38 (a) illustrates a five-decade log scale, while Figure 26.38 (b) illustrates one logarithmic decade expanded. Figure 26.37 illustrates a log-log graph paper.

494 Electrical Technology 1

10

100

1000

10,000

100,000

(a)

1

2

3

4

5

6

7

8

9 10

(b)

Figure 26.38  Logarithmic Graph (a) 5 Decade (Cycle) Log Scale (b) 1 Logarithmic Decade Expanded One advantage of logarithmic spacing is that a large range of values can be shown in one plot without losing resolution in the smaller values. For example, if frequency values between 10 Hz and 10 kHz were plotted on 100 divisions of a linear graph, each division would represent approximately 1000 Hz and it would be impossible to plot values in the decade between 10 Hz and 100 Hz. On the other hand, by using logarithmic graph paper, the decade between 10 Hz and 100 Hz would occupy the same space on the graph as the decade between 10 kHz and 100 kHz. Log-log or semi-log paper is specified by the number of decades it contains. Each decade is a graph cycle. For example, a 2-cycle by 4-cycle log-log paper has two decades on one axis and four on the other. The number of cycles must be adequate for the data being plotted. A typical sheet of log-log graph paper is shown in Figure 26.37. Because there are three decades on the horizontal axis and five decades on the vertical axis, this graph paper is called 3-cycle by 5-cycle log-log paper. Several octaves and decades are also shown here. When semi-log graph paper is used to plot a frequency response, the calculated values of gain (or loss) must first be converted to decibels before plotting. On the other hand, since decibel voltage gain is a logarithmic function, the gain (or loss) values can be plotted on log-log paper without first converting to decibels.

26.14 SYMMETRICAL NETWORKS Symmetrical networks have important electrical characteristics, namely, characteristic impedance (Z0) and propagation constant (r). Two networks having the same characteristic impedance and the same propagation constant are said to be equivalent. If an infinite number of identical symmetrical networks are connected in tandem (one after the other), as in Figure 26.39, the impedance measured at the input terminals of the first network will have some definite value depending only on the composition of the networks. This impedance, an important property of the network, is called its characteristic impedance, represented by Z0. It may be calculated from a knowledge of the component values of the network. If the first network of this infinite chain is disconnected, the number of networks remaining will still be infinite and, therefore, the input impedance looking into the second network will still be Z0 Figure 26.39(b). It follows that, if the first section be connected to an impedance equal to Z0, as in Figure 26.39(c) instead of to the infinite chain of networks, its input impedance will still be Z0.

To infinity

Z0 (a)

To infinity

Z0 (b) Z0

Z0 (c)

Figure 26.39  Characteristic Impedance of a Four-terminal Network Thus, if any symmetrical network is terminated in its characteristics impedance, the input impedance will be Z0. Similarly, if its input terminals are connected to a generator of impedance Z0, then its output impedance will be equal to Z0. When both of these conditions are satisfied, the network is said to be correctly terminated.

Attenuators and Filters

495

Figure 26.40 shows a T-section terminated in its Z0. It is one of the most important networks encountered. A ladder network, for example, could be regarded as being made up of these sections. To give a total series-arm impedance of Z1 in the ladder network, the two series-arm impedances in the T section must each be Z1/2. Z1

Z1

Z2

Z1

Z2

Z2

Z2

1Z 2 1

1Z 2 1

1Z 2 1

Z2

Z2

Z2

1 Z 2 1

1 Z 2 1

(a) 1 Z 2 1

Z2

1 Z 2 1

(b)

1 Z 2 1

1 Z 2 1

1 Z 2 1

Z2

1 Z 2 1

Z2

1 Z 2 1

1 Z 2 1

Z2

1 Z 2 1

Z2

(c) Z1

Z1

2Z22Z2

2Z22Z2

Z1

Z1

2Z22Z2

2Z2 2Z2

Z1

Z1

(d) Z1

Z1

Z2

Z2

Z2

Z2

(e)

Figure 26.40   Ladder Networks (a) Unbalanced (b) Balanced (c) Composed of T-sections (d) Composed of π  Sections and (e) Composed of L-sections

26.14.1 Asymmetrical Networks Asymmetrical networks generally have different characteristic impedances on the two sides. When dealing with such networks, the terms iterative impedance and image impedance are used in place of characteristic impedance. These are illustrated in Figures 26.41 and 26.42, respectively. When the two iterative impedances are equal—as they are in symmetrical networks—their common value is the characteristic impedance of the networks. When the image impedances are equal as they are in the case of symmetrical networks, their common value is equal to the characteristic impedance of the network. An asymmetrical network is said to be correctly terminated when it is terminated in its image impedances. 3

1

600 Ω

600 Ω 2

3

1

400 Ω

400 Ω 2

4

4 (b)

(a)

To infinity

600 Ω (c) To infinity

400 Ω (d)

Figure 26.41 Illustrating Iterative Impedance of a Symmetrical Network

496 Electrical Technology 1

1

3

300 Ω

800 Ω

300 W

800 Ω 2

4

(a)

2

Z01 (300 Ω)

Z01 (300 Ω)

3 4

(b)

Z02 (800 Ω)

Z02 (800 Ω)

(c)

Figure 26.42 Illustrating Image Impedance of a Symmetrical Network

26.14.2 Recurrent Networks The type of recurrent network most commonly encountered is the ladder network. It exists in two forms—the balanced ladder network and the unbalanced ladder network. Both are illustrated in Figure 26.40. Ladder networks (both) may be considered as being built up of a number of sections which are known by reason of their shape, as T, π, and L-sections in the unbalanced form, and H-, O-, and C- sections in the balanced form. These sections, illustrated in Figures 26.43 and 26.44 are all arranged to have a total series impedance Z1 and a total shunt impedance Z2. In addition to the ladder structure, two other forms of recurrent networks are the lattice, as represented in Figure 26.45, and the bridged-T networks. The lattice section is usually a balanced symmetrical structure. The bridge-T sections (Refer Figure 26.46) may be balanced or unbalanced, symmetrical or asymmetrical, though the unbalanced symmetrical form is the most usual. 1 2 Z1

1 2 Z1

Z2

1 2 Z1

1 2 Z1

1 2 Z1

1 2 Z1

1 2 Z1

Z2

Z2

1 2 Z1

Z2

Z1

Z1

Z1

Z1

2Z2 2Z2

2Z2 2Z2

2Z2 2Z2

2Z2 2Z2

Figure 26.43 Unbalanced Ladder Network Represented as a Series of Sections 1 Z 4 1

1 Z 4 1

Z2 1 Z 4 1

1 Z 4 1

1 Z 4 1

1 Z 4 1

Z2

Z2 1 Z 4 1

1 Z 4 1

1 Z 4 1

1 Z 4 1

1 Z 4 1

1 Z 4 1

1 Z 4 1

Z2 1 Z 4 1

1 Z 4 1

1 Z 4 1

(a) 1 2 Z1

1 2 Z1

2Z2 2Z2

2Z2 2Z2

1 2 Z1

1 2 Z1

2Z2 2Z2

1 2 Z1

1 2 Z1

1 2 Z1

2Z2 2Z2 1 2 Z1

(b) 1 Z 2 1

Z2 1 Z 2 1

1 Z 2 1

1 Z 2 1

Z2

Z2

1 Z 2 1

1 Z 2 1

1 Z 2 1

Z2 1 Z 2 1

(c)

Figure 26.44 Balanced Ladder Networks Represented as a Series of Sections (a) Ladder Network Composed of H or Balanced T Sections (b) Ladder Network Composed of O or Balanced π  Sections (c) Ladder Network Composed of C or Balanced L Sections

Attenuators and Filters ZA ZB

ZA ZB

ZA

ZB

ZB

ZA

497

ZB

ZB

ZA

ZA

Figure 26.45 Lattice Network Z3

Z1 2

Z3

Z1 2

Z2

Z1 2

Z3

Z1 2

Z2

Z1 2

Z1 2

Z2

(a) Z3 2

Z1 4 Z1 4

Z3 2

Z1 4 Z1 4

Z2

Z1 4 Z1 4

Z3 2

Z3 2

Z1 4 Z1 4

Z2

Z1 4 Z1 4

Z1 4 Z1 4

Z2

Z3 2

Z3 2

(b)

Figure 26.46 Unbalanced and Balanced Forms of Bridged T Sections (a) Unbalanced (b) Balanced

26.15 EQUIVALENCE OF BALANCED AND UNBALANCED SECTIONS Both the balanced and unbalanced sections have identical transmission properties, as long as no connections are made between the input terminals external to the network. In Figure 26.47 (a) and (b) are equivalent, provided that no connection is made as in (c). Z1 2

Z1 ZG E

2Z2 2Z2

ZL Z G E

(a)

2Z2 2Z2

Z1 2

ZL ZG E

2Z2

2Z2

Z1 2

Z1 2

(b)

(c)

ZL

Figure 26.47 Equivalence of Balanced and Unbalanced Sections (a) Unbalanced (b) Balanced

26.16 MACLAURIN’S THEOREM Many functions can be expanded as a series of powers of x (e.g., (1 + x)n, ex). Maclaurin’s theorem enables an expansion to be found for a general function of x, i.e., y = f(x) Let Differentiating

f ( x) = A0 + A1 x + A2 x 2 + A3 x3 + A4 x 4 + ..... f ′ ( x)

= A1 + 2 A2 x + 3 A3 x 2 + 4 A4 x3 + ...

f ′′ ( x) =

2 A2 + 2.3 A3 x + 3.4 A4 x 2 + ...

f ′′′ ( x) =

2.3 A3 + 2.3.4 A4 x + ...

f ′′′′ ( x) =

2.3.4 A4 + ...

498 Electrical Technology This is true for all values of x; hence, when x = 0 A0 = f (0) A1 = f ′(0)

Then

A2 =

f ′′(0) f ′′(0) = 2 2

A3 =

f ′′′(0) f ′′′(0) = . 23 3

A4 =

f ′ ′′′(0) f ′′′′(0) = 2.3.4 4

f ( x ) = f ( 0) + x f ′ ( 0) +

x2 x3 x4 f ′′ (0) + f ′ ′ ′ ( 0) + f ′′′′ (0) + ... 2 3 4

(26.39)

where, f "(0) means the value of f "(x), etc, when x = 0 This is known as Maclaurin’s theorem. If no derivatives of f (x) vanish, then this expansion will involve an infinite number of terms. It holds for (1 + x)n and e n.

26.17 CIRCULAR FUNCTIONS A number of important series can be obtained from this theorem. Take, for example, f (x) = sin (x). To find the series, one must calculate the successive derivatives and their value when x = 0. f ( x) = sin x

∴ f (0) = 0

f ′( x) = cos x

f ′(0) = 1

f ′′( x) = − sin x

f ′′(0) = 0

f ′′′( x) = − cos x

f ′′′(0) = − 1

f ′′′′( x) = sin x

f ′′′′(0) = 0, etc.

Hence, the series is

sin x = x −

x3 x5 x7 + − + ... 3 5 7

Similarly cos x = 1 − since, the expansion for e jx is

x2 x4 x6 + − + ... 2 4 6

ex = 1 + x +

x2 x3 + + ... 2 3

e jx = 1 + jx +

( jx) 2 ( jx)3 + + ... 2 3

e jx = 1 + jx −

x2 x3 x4 x5 x6 −j + + j − + .... 2 3 4 5 6

e jx = (1 −

x2 x4 x3 x5 ...) + j ( x − ...) + + 2 4 3 5

Attenuators and Filters

499

Hence, e− jx = cos x − j sin x

(26.40)

e− jx = cos x − j sin x

(26.41)

Similarly,

This shows that trigonometrical ratios can be treated from an algebraic aspect as well as from a geometrical. Both cos x and sin x may be obtained as expressions involving e by adding and subtracting these two equations. cos x =

e jx + e− jx 2

sec x =

sin x =

e jx − e− jx j2

cos ec x =

2 e jx + e− jx e

jx

j2 − e− jx

Also tan x = − j

(e jx − e− jx ) (e jx + e− jx )

cot x = j

e jx + e− jx e jx − e− jx

26.18 HYPERBOLIC FUNCTIONS The value of Thus,

e x + e− x is also important and this is known as cosh x or hyperbolic cosine of x. 2 e x + e− x cosh x = 2 e x − e− x 2

Similarly,

sinh x =

Where,

cosh x + sinh x = e x

and

(26.42)

cosh x − sinh x = e

−x

(26.43) (26.44)

These hyperbolic functions bear the same relation to a hyperbola as sine and cosine function bear to a circle. Since and

x2 x3 x4 + + + ... 2 3 4 x2 x3 x4 =1− x + − + + ... 2 3 4

ex = 1 + x + e− x

It follows that the series for the hyperbolic functions are cosh x = 1 +

x2 x4 + + ... 2 4

(26.45)

sinh x = x +

x3 x5 + + ... 3 5

(26.46)

From the definitions it can be seen that the following conversion rules apply. Circular to Hyperbolic

Hyperbolic to Circular

sin x = −j sinh jx cos x = cosh jx sin jx = j sinh x cos jx = cosh x

sinh x = −j sin jx cosh x = cos jx sinh jx = j sin x cosh jx = cos x

Figure 26.48 gives the graphs of the hyperbolic functions. Note: 1. cosh x is always greater than 1. 2. tanh x lies between +1 and −1. 3. Unlike circular functions, hyperbolic functions are not periodic.

500 Electrical Technology y 6

5

cosh x

3

x sinh

cosh

x

4

2

tanh x

1

–3

–2 tanh x

–1

0

1

2

3

–1

–2

–3

sinh

x

–4

–5

–6

Figure 26.48   Graphs of the Hyperbolic Functions  sinh x, cosh x and tanh x

coth x =

e x + e− x e2 x + 1 = 2x x −x e −e e −1

sech x =

2 e + e− x

cosech x =

2 e − e− x

x

x

x

Attenuators and Filters

501

26.18.1 Hyperbolic Identities Hyperbolic identities are similar to corresponding circular identities and may be readily deduced from them. As a general rule, identities hold if (−sinh)2 is written instead of (sin)2 and (cosh)2 instead of (cos)2. Thus, cos2x + sin2x = 1 becomes cosh2x − sinh2x = 1 2 2 cos 2x = cos x − sin x becomes cosh 2x = cosh2x + sinh2x = 2 cosh2x − 1 = 1 + 2 sinh2x It can be shown that sinh (A + B) = sinh A cosh B + cosh A sinh B (26.47) sinh (A - B) = sinh A cosh B - cosh A sinh B (26.48) cosh (A + B ) = cosh A cosh B + sinh A sinh B (26.49) cosh (A − B) = cosh A cosh B − sinh A sin B (26.50) tanh (A + B) =

tanh A + tanh B 1 + tanh A tanh B

(26.51)

tanh A − tanh B 1 − tanh A tanh B sinh 2x = 2 sinh x cosh x tanh (A − B) =

tanh x =

(26.52)

cosh 2 x − 1 sinh 2 x

(26.53)

26.18.2 Differentiation of Hyperbolic Functions The differential co-efficients of hyperbolic functions can easily be obtained by using the exponential form of the functions. Thus, we can take the example of d

dx

(sinh x) = d =

 e x − e− x  1  1   = d  e x − e− x  dx  dx   2  2 2 

e x + e− x = cosh x 2

(26.54)

Some of the most important derivatives have been given in Table 26.3. Table 26.3 Differential Co-efficients of Some Hyperbolic Functions y

dy dx

sinh x cosh x tanh x coth x sech x cosech x

cosh x sinh x sech2 x − cosech2 x − sech x tanh x − cosech x coth x

sinh−1 x

cosh−1 x

tanh−1 x

1 1 + x2 1 x2 - 1 1 1 - x2

502 Electrical Technology

26.18.3 Complex Hyperbolic Functions Expressions such cosh (a + jβ) are often encountered in the transmission theory. Their values can be encountered from first principles and tables. For example, if sinh (a + jβ) = A + jβ, we can find A and B in terms of a and β. For A + jβ = sinh (a + jβ) = sinh a cosh jB + cosh a sinh jβ. = sinh a cosh β + j cosh a sinh jβ = sinh a cos β + j cosh a sin β Equating real and imaginary parts: A = sinh a cos β and B = cosh a sin β (26.55) Similarly, if cosh (a + jβ) = A + jβ A + jβ = cosh (a + jβ) = cosh a cosh jβ + sinh a sinh jβ = cosh a cosβ + j sinh a sin β Equating real and imaginary parts A = cosh a cos β and B = sinh a sin β. (26.56) Example 26.11 Evaluate cosh (3 + j Solution:

p ) 4 cosh (3 + j

p p p ) = cosh 3 cosh j + sinh 3 sinh j 4 4 4 p p = cos 3 cos + j sinh 3 sin 4 4 = 10.07 × 0.7071 + j 10.02 × 0.7071 = 7.12 + j 7.07

26.19 THEOREM CONNECTING

a AND Z0

If a filter is correctly terminated, then the following theorem applies. Over the range frequencies for which the characteristic impedance Z0 of a filter is purely resistive (real), the attenuation a is zero. Over the range of frequencies for which Z0 is purely reactive (imaginary), the attenuation is greater than zero. The case where Z0 is partly resistive and partly reactive cannot arise in purely reactive filters. If Z0 is real, the filter and its termination will absorb power from any generator connected to it; if the filter is composed entirely of reactances, it cannot itself absorb power since in a reactance the current and voltage are always 90° out of phase. Hence, all the power delivered by the generator must be passed through to the load and, therefore, there is no attenuation (a = 0). If, on the other hand, Z0 is purely reactive, the filter and its termination cannot absorb any power, and no power is therefore passed to the load.

26.19.1 Cut-off Frequency As Z0 is real in a pass band and imaginary in an attenuation band, f0 is the frequency at which Z0 changes from being real to being imaginary. Considering a T section: if X1 and X1 + X 2 have the same signs Z0 will be purely imaginary and the filter will attenu4 ate. However, if X1 and X1 + X 2 have opposite signs, then Z0 will be real and the attenuation zero. The easiest method 4 to determine cut-off frequencies is to draw reactance sketches for X1 and X1 + X 2 against frequency (Figure 26.49). 4

Attenuators and Filters

503

x x1 4

+ x2

Reactance x1

x1 4

0

f1

fE

f5

+ x2

f4

Frequency

–x Pass Pass band band Attenuation Attenuation band band

Attenuation band

Figure 26.49  Reactance-frequency Sketches for Double Band Pass Filter Frequencies for which the curves are on opposite sides of the frequency axis are in the pass band; frequencies for which the curves are on the same side of the frequency axis are in the attenuation band. The change-over points give the cutoff frequencies. For example, the filter whose reactance sketches are given in Figure 26.49 has two pass bands and three attenuation bands.

26.20 PROTOTYPE (CONSTANT K) FILTER SECTIONS A constant-k section is a T or π section in which the series and shunt impedances Z1 and Z2 are connected by the relationship Z1Z 2 = R02 , where R0 is a real constant known as the design impedance of the section. Z 0T =

The equation

Z1Z 2 Z 0p

(26.57)

connects the characteristic impedances of T and π sections composed of the same series and shunt impedances, as has been represented in Figure 26.50. If the section is a constant-k section Z 0T =

R02 Z 0p

(26.58)

then Z0T and Z0π will be real or imaginary together, and when Z0T changes from real to imaginary, so also will Z0π. Hence, the two sections will have the same pass bands and the same cut-off frequencies.

Z0T

Z1

Z1

2

2

Z2

Z1

Z0T

Z0Π

2Z2

2Z2

Z0 Π

Figure 26.50  T and π Sections Composed of the Same Series and Shunt Impedances The constant kT or π sections of any type of filter are known as the prototype.

504 Electrical Technology

26.20.1 Low-pass Filters The prototype T and π sections are shown in Figure 26.51. L 2

L 2

L

C 2

C

C 2

Figure 26.51  Prototype T and π Low-pass Filter Sections Here, Z1 = jwL and Z2 = −j/wC Z1Z2 = L/C and the sections are, therefore, constant k sections with R0 =

L /C

As both sections have the same cut-off frequency, it is sufficient to calculate this for one section only (say, T section). The reactance sketches must first be drawn. Z1 = jω L

∴ X1 = ω L

Z2 = − j

∴ X 2 = −1

ωC

ωC

X1 + X2 = ωL − 1 ωC 4 4

and

The curves are shown in Figure 26.52. It is worth noting here that all reactance frequency curves slope upwards and to the right: that is, they have positive slope. Further, 1. 2. 3. 4. 5. 6.

The curves are on opposite sides of the frequency axis as far as point A. The curves are on the same side from point A onwards. The pass band includes all frequencies up to point A. The attenuation band includes all frequencies above point A. The point A gives the cut-off frequency given by w = wc. The section is, therefore, a low-pass filter with fc =

7. wc is the point where the curve

(26.59)

X1 + X 2 crosses the frequency axis; that is, when 4

ωL 4 Hence,

ωc 2π

−

ωc =

1 4 = 0 or ω 2 = LC ωC 2 LC

or f c =

1

π LC

which gives the cut-off frequency of a low-pass T or π section. Z 0T =

Z12 + Z1Z 2 4

= R0 1 −

ω 2 LC ω2 = R0 1 − 2 4 ωc

(26.60)

Attenuators and Filters

Since Z 0p =

R02 , it follows Z 0T R0

Z 0π =

1− ω

ωc 2

+x

x 1–wL x1 4

Reactance A

0

x1 +x 2 4

wc

–x

1 x 2– w c

w-2pf

Attenuation band

Pass band

Figure 26.52  Reactance Frequency Sketch For a Prototype Low-pass Filter Example 26.12 A simple T-section low-pass filter has a design-impedance R0. Find Z0π at f = 0.9fc. 95.5 mH

95.5 mH

191.0 mH

0.2652 µ F

0.5304 µ F

0.2652 µ F

(a)

(b)

Figure 26.53 Solution for Example 26.13 Solution:

ω

ωc

= f

Z 0p =

fc

= 0.9 R0

1 − ( 0 . 9) 2

= 2.3 R0

Example 26.13 A low-pass filter section is required with R0 = 600 Ω and fc = 1 kHz. Find L and C. Solution: R L = 0 = 191.0 mH p fc C=

1 = 0.5304 µ F π R0 f c

505

506 Electrical Technology 2C

2C

26.20.2 High-pass Filters

C

Figure 26.54 shows the prototype high-pass T and π sections. 2L

L

(a)

Z1 =

Here,

2L

L C

R0 =

(b)

Figure 26.54   Prototype High-pass Filter Sections  (a) T Section (b) π Section

and

+x

x2

0

1

–x

4p LC

Z 0T = R0 1 −

ω c2 ω2

1−

Attenuation band

• Z0

0

x1 4

(26.63)

ωc2 ω2 +Π

0

B

x1 +x 2 4 w

0

R0

–Π

0

fc Frequency (a)

0

fc Frequency

Z0Π

Z0T 0

(b)

fc Frequency (c)

Figure 26.56  Reactance Frequency Sketches For Low-pass Filter Sections +Π Z0Π

Z0

0 0

fc Frequency (a)

0

R0

–Π

0

0

fc Frequency (b)

x1

Pass band

Figure 26.55 Reactance Frequency Sketch for a Prototype  High-pass Filter

(26.62)

R0

Z 0π =

(26.61)

The cut-off frequency may be determined by the reactance sketch method, as shown in Figure 26.55.

The following points are worth noting. 1. The curves are on the same side of the horizontal axes up to point B. 2. The curves are on the opposite sides of the horizontal axes from point B onwards. 3. The attenuation band includes all frequencies up to point B. 4. The pass band includes all frequencies from point B onwards. 5. The point B gives the cut off frequency. 6. The section is, therefore, a high-pass filter with fC =

−j and Z 2 = jω L and ωC

Z0T 0

fc Frequency (c)

Figure 26.57  Reactance Frequency Sketches for High-pass Filter Sections

507

Attenuators and Filters

Example 26.14 Calculate the components of a prototype high-pass filter section with a design impedance R0 = 600 Ω and a cut-off frequency fC = 10 kHz as shown in Figure 26.58. Solution: L = R0 /4π f c

0.02652 µ F

0.02652 µ F

0.01326 µ F

9.548 mH

4.774 mH

9.548 mH

= 4.774 mH C =

1

4 π R0 f c

(a)

= 0.01326 µ F

(b)

Figure 26.58 Solution for Example 26.14

26.21 M-DERIVED FILTERS In a low-pass filter, a clearly defined cut-off frequency followed by a high attenuation is needed. In a high-pass filter, high attenuation followed by a clearly defined cut-off frequency is needed. It is not practicable to obtain either of these conditions by wiring appropriate prototype constant k sections in cascade. An equivalent section, therefore, requires the following criterion. 1. The same cut-off frequency as the prototype but with a rapid rise in Z1 Z1 mZ 1 mZ 1 attenuation beyond cut-off for a low-pass type or a rapid decrease, 2 2 2 2 at cut-off from a high attenuation for the high-pass type. 2. The same value of nominal impedance R0 as the prototype at all frequencies (otherwise the two forms could not be connected Z2 Z'2 together without mismatch). The equivalent section is called an m-derived filter section as shown in Figure 26.59. Let us consider any T section and construct a new section from it having (a) (b) a series arm of the same type but of different value. For convenience, we Figure 26.59   Derivation of the  can make Z1 = mZ, where m is some constant. The new shunt arm will m-derived T-section not be Z2 but, say, Z 2¢ . It is required to find that value of Z 2¢ which will (a) Prototype (b) m-derived make the two sections possess the same value of Z0. For the prototype section Z 0T =

Z12 + Z1Z 2 4

For the new section Z 0T =

m12 Z12 + mZ1Z 2′ 4

The two impedances will be the same if m2 Z 2 Z2 + Z1Z 2 = 1 1 + m Z1Z 2 4 4

Z1 2

Z1 2

mZ 1 2

mZ 1 2

(26.64)

Z2 m

Z2

1 − m2 Z Z2 = 2 + Z1 4m m

1-m 2 Z 4m 1

′

Z2 in series with an m and both these impedances can be constructed

This means that Z 2¢ must be an impedance impedance Z1

(1 − m2 )

4m if 0 < m < 1. The complete m-derived T section is shown in Figure 26.60.

(a)

(b)

Figure 26.60 Comparison of Prototype and m-derived T-sections (a) Prototype (m=1) (b) m-derived

508 Electrical Technology In a similar manner, a new section may be derived for a π section, having the same Z0 at all frequencies. The complete m-derived π section is shown in Figure 26.61 (b). 4m Z 1-m 2 2 Z1 mZ 1 2Z 2

2Z 2

2Z 2 m

2Z 2 m

(a)

(b)

Figure 26.61 Comparison of Prototype and m-derived π Sections (a) Prototype (b) m-derived Note: If m = 1, both T and π m-derived sections reduce to the corresponding prototype sections. The characteristic impedance of the m-derived section is the same as that of the prototype.

26.21.1 Low-pass m-derived Sections Figure 26.62 shows both the T and π low-pass filter sections. In deducing these from the general case, we must keep in mind that to divide the impedance of a capacitor by m, its capacity must be multiplied by m. mL 2

1-m 2 C 4m

mL 2

mL

mC 1-m 2 L 4m

mC 2

(a)

mC 2 (b)

Figure 26.62  m-derived Low-pass Filter Sections (a) m-derived T (b) m-derived π

26.21.2 High-pass m-derived Sections Figure 26.63 shows the T and π m-derived high-pass filter sections. 2C m

4m L 1-m 2

2C m

C m

L m 4m C 1-m 2

2L m

(a)

2L m (b)

Figure 26.63  m-derived High-pass Filter Sections (a) m-derived T Section (b) m-derived π Section

S UM M A RY 1. An attenuator must provide the required input and output impedances as well as provide the required attenuation. 2. The decibel is fundamentally a unit of power ratio. 3. Networks in which electrical energy is fed in at one pair of terminals and taken out at a second pair of terminals are called two-port networks

6. An asymmetrical network is correctly terminated when it is terminated in its image impedance.

4. Z 0 = ( Z oc ⋅Z sc )

9. Attenuators are constructed of purely resistive elements.

5. An image impedance is defined as the impedance which, when connected to the terminals of a network is equal to the impedance presented to it at the opposite terminals.

7. An L-section attenuator pad is used for matching purposes only. 8. A network that is designed to attenuate certain frequencies and pass others is called a filter. 10. Filters are constructed of purely reactive elements. 11. A low-pass filter is a filter that has a constant output voltage from d.c. up to a cut-off frequency.

Attenuators and Filters

12. A high-pass filter is a circuit that attenuates the output voltage for all frequencies below the cut-off frequency. 13. A band-pass filter is a circuit that passes only a band of frequencies while attenuating frequencies outside of this band. 14. Band-elimination filters reject a specified band of frequencies while passing all the frequencies outside this band. 15. All pass filters are designed to provide constant gain to signals at all frequencies. 16. Log-log or semi-log paper is specified by the number of decades it contains. 17. Two networks having the same characteristic impedance and the same propagation constant are said to be equivalent.

509

18. A ladder network could be regarded as being made up of T-sections. 19. sin x = x −

20. cos x = 1 −

x3 x5 x7 + − + .... 3 5 7 x2 x4 x6 + − + ... 2 4 6

21. e jx = cos x + j sin x 22. e−jx = cos x − j sin x 23. f0 is the frequency at which Z0 changes from being real to being imaginary. 24. Prototype filters are also known as constant k-filters.

M U LT IP LE C H O I C E Q UE S TI O NS (M C Q ) 1. Attenuators are used to introduce.

4. For a symmetrical network

(a) A specified amount of loss (b) A specified amount of gain (c) The desired amount of phase shift

(a) There is only one value of iterative impedance (b) There are two values of iterative impedance, one for each direction

2. The decibel is basically a (a) Voltage ratio (c) Power ratio

(b) Current ratio

3. Attenuation in db is equal to

5. Filters can have (a) Only one pass band and one attenuation band (b) More than one pass band and one attenuation band

}

(a) 8.58 × (attenuation in nepers provided R1= R2 (b) 0 .115× (attenuation in nepers)

ANSWERS (MCQ) 1. (a) 2. (c) 3. (a) 4. (a) 5. (b).

CON V E N TI O NA L Q UE S TI O NS (C Q ) 1. Establish the equivalence between T- and π-network 2. With the help of suitable illustrations, explain the difference between balanced and unbalanced attenuators 3. Design a T-type attenuates to give an attenuation of 20 db and to have a characteristic impedance of 75 Ohms. 4. Find the arms of a T-section symmetrical resistive attenuator which would introduce an attenuation of 15 db when working between a source and a load, each of impedance equal to 75 00 . 5. It is desired to design a T-type attenuator pad with 40 db loss to work between a source impedance of 70 Ohms and a load impedance of 600 Ohms. Determine the resistance value of the elements of the pad. 6. Design an L-Type attenuator to operate into a resistance of 500 Ohms, and to provide an attenuation of 15 dbs. 7. Differentiate between a constant-k and an m-derived filter.

8. Calculate the values of the inductors and capacitors of a prototype constant k low-pass filter composed of π sections to operate with a terminating load of 600 Ohms and to have a cut off frequency of 3 kHz. 9. Determine the characteristic impedance of the T-network attenuator sections in Figure 26.64. 10. Determine the characteristic impedance of the π-network attenuators section in Figure 26.65. 11. Determine the cut off frequency and the nominal impedance of each of the low-pass filter sections in Figure 26.66. 12. Explain how an m-derived filter is obtained from a constant-k type. What are the advantages and disadvantages of m-derived over constant-k type? 13. Define low, high, band pass and band elimination filters in terms of their pass and attenuation band.

510 Electrical Technology 10 Ω

10 Ω

100 Ω

100 Ω

30 Ω

1 kΩ

1 kΩ

400 Ω

250 Ω

(a)

(b)

(c)

Figure 26.64 For CQ 9

25 Ω

10 Ω

1 kΩ

10 Ω

500 Ω

(a)

400 Ω

500 Ω

300 Ω

(b)

300 Ω

(c)

Figure 26.65 For CQ 10

0.5 H

0.5 H

20 mH

0.5 µ F 27.8 nF

(a)

27.8 nF

(b)

Figure 26.66 For CQ 11

ANSWERS (CQ) 3. R1 = 61.36 Ω, R2 = 15.15 Ω 4. R1 = 41.50 Ω, R2 = 27.55 Ω 5. 65.89 Ω, 4.12 Ω, 596.08 Ω 6. 108 Ω, 411 Ω 8. 63.7 mH, 0.1768 μF 9. (a) 26.46 Ω, (b) 244.9 Ω, (c) 1.342 kΩ 10. (a) 7.45 Ω, (b) 353.6 Ω, 189.7Ω 11. (a)1592 Hz, 5 kΩ, (b) 9545 Hz, 600 Ω.

Transmission Lines

27

OBJECTIVES In this chapter you will learn about:  TThe purpose of a transmission line    Define the transmission line primary constants    Define and calculate transmission line secondary constants    Appreciate current and voltage relationships on a transmission line    Calculate the characteristic impedance and propagation coefficient in terms of primary line constants    Understand distortion and distortion less conditions    Understand reflection and calculate reflection coefficient    Understand standing waves and calculate standing wave ratio

Transmission line parameters

27.1 INTRODUCTION Line transmission is the theory of propagation of electric waves along transmission lines. These transmission lines are assumed to consist of a pair of wires that are uniform throughout their whole length. Provided that this uniformity holds good, it is immaterial for the general theory whether the two wires are air-spaced on telegraph poles or two conductors in an underground cable, or they form a pair in a field quad cable. Telephone lines and power lines are typical examples of transmission lines. An important feature of a transmission line is that it should guide energy from a source at the sending end to a load at the receiving end without loss by radiation. One form of construction that is often used consists of two similar conductors mounted close together at a constant separation. The two conductors form the two sides of a balanced circuit and any radiation from one of them is neutralized by the same from the other. Similar twin-wire lines are used for carrying high r.f. power at transmitters, for example. The coaxial form of construction is commonly employed for low power use, one conductor being in the form of a cylinder which surrounds the other at its centre, and thus acts as a screen. At frequencies above 1000 MHz, transmission lines are usually in the form of a wave guide which may be regarded as a coaxial cable without the centre conductor, the energy being launched into the guide or abstracted from it by probes or loops projecting into the guide.

27.2 THE INFINITE LINE The propagation of electric waves along any uniform and symmetrical transmission line may be deduced in terms of the results for a hypothetical line of infinite length having electrical constants per unit length that are identical to those of the line under consideration. Let an a.c. generator be connected to the input terminals of a pair of parallel conductors of infinite length. A sinusoidal wave will move along the line and a finite current will then flow into the line. The variation of voltage with distance along the line will resemble the variations of applied voltage with time. The Is moving wave, sinusoidal in this case, is called a voltage travelling wave. As the wave moves along the line, the capacitance of the line is charged E to ∞ Zo s up and the moving charges cause magnetic energy to be stored. Thus, the propagation of such an electromagnetic wave constitutes a flow of energy. Refer Figure 27.1. Figure 27.1 Infinite Line

M27_AUTH_ISBN_C27.indd 511

8/13/2012 6:12:46 PM

512 Electrical Technology The ratio of the voltage applied to the current flowing will give the input impedance. This input impedance is known as the characteristic impedance of the line and denoted by Zo. The characteristic impedance of any line is defined as the impedance looking into an infinite length of the line.

27.3 SHORT LINE TERMINATED IN ZO Consider an infinite line having input terminals 1 and 2, as has been illustrated in Figure 27.2 (a). The impedance looking in at terminals 1 and 2 will, by definition, be Zo. Now if we suppose that a short section AB at the near end of the line is removed, as represented in Figure 27.2 (b), so that the line now starts at terminals 3 and 4. The impedance looking in at terminals 3 and 4 will still be Zo, since the 1 removal of the short section does not affect the infinite nature of the line. (a) Zo to ∞ This means that the short section AB, from an electrical point of view, 2 was originally terminated in an impedance Zo at B. If the short section AB B A 1 3 is now terminated in an actual impedance Zo, the current and voltage at to ∞ Zo (b) all points along its length will still remain exactly the same as if it were 4 2 terminated in an infinite length of the line. A B 1 It, therefore, follows that a short line terminated in Zo behaves (c) Zo electrically, at all points along its length, as if it were an infinite line. In 2 particular, this means that the input impedance will be Zo and that there Figure 27.2 Short Line Terminated in Zo will be no reflection.

27.3.1 Determination of ZO (for a Short Line)

A short line may be considered as a complex electrical network, and like any other network it may, at the frequency under consideration, be represented by a T section, as has been illustrated in Figure 27.3 (a). If the short line is terminated in Zo, it will behave as an infinite line, and have an input impedance Zo. Since the equivalent T section represents the line, it must also have an input Zo when it is terminated in Zo. Let the equivalent T section have series arms Z1/2, Z1/2 and shunt arm Z2, as seen in Figure 27.3 (a). Then, Z in = but, Z in = Z o ∴

Z o = Z12 4 + Z1Z 2

Z1 Z 2 ( Z1 2 + Z o ) + 2 Z1 2 + Z 2 + Z o or Z o =

(27.1)

(27.2)

Z12 4 + Z1Z 2

Thus, Zo for the T section and, hence, for the line, may be determined, if Z1 and Z2 can be found. This will require two equations, which may be obtained by measuring the input impedance using two different terminating impedances. For convenience, these terminations will be taken as infinity and zero. Let the input impedance with an infinite-impedance terminations, i.e., open-circuit, be Zoc. Considering the equivalent T section on open-circuit, as illustrated in Figure 27.3 (b) Z Z1 2

Z oc = Z1 2 + Z 2 Let the input impedance with a zero-impedance termination, i.e., short-circuit, be Zsc. Considering the equivalent T section on short-circuit, as seen in Figure 27.3 (c).

Z Z Z 2 Z SC = Z11 + Z11 Z 22 2 (27.3) Z SC = 2 + Z1 2 + Z 2 2 Z1 2 + Z 2 Z12 4 + Z1Z 2 Z SC = Z 2 4 + Z Z (27.4) Z SC = 1Z1 2 + Z1 2 2 Z1 2 + Z 2 Equations 27.3 and 27.4 give two simultaneous equations from which Z1 and Z2 may be determined. Since only Zo is required, it may be obtained directly Z SC =

M27_AUTH_ISBN_C27.indd 512

Z o2

Z oc

(a) Z0

Line

Z0

ZIN (ZIN =Z0) Z1 2

(b) Zoc

Zsc

Z0

Z2 Z1 2 Z2

Zoc Z1 2

(c) Zsc

1

2

Z1 2 Z2

Figure 27.3 Short Line Terminated in Zo (a), and Short Line on Open-circuit (b), Short Line on Short-circuit (c)

8/13/2012 6:12:51 PM

Transmission Lines 513

i.e.,

(27.5)

Z o = Ö ( Z oc .Z sc )

The characteristic impedance of a line is, therefore, the geometric mean of the open-circuit and short circuit impedances. Example 27.1 The following measurements have been made on a line at 1500 Hz.

Z oc = 900 Ω −30°,

Z sc = 400 Ω −10°

What is the characteristic impedance of the line? Solution:

Z o = Ö ( Z oc .Z sc ) =

[ 900

−30° ] [ 400 −10° ]

=

360000 −40°

= 600 −20

27.4 TRANSMISSION LINE PARAMETERS There are four parameters of transmission lines—resistance, inductance, capacitance and conductance. Resistance R, is given by the relation R = pl/A where, p is the specific resistance of the conductor material, A is the cross sectional area of each conductor and l is the length of the conductor. For a two-wire line, l represents twice the length of the line. Resistance is stated in ohms per metre length of the line and represents the imperfections of the conductor. A resistance stated in ohms per loop metre is a little more specific since it takes into consideration the fact that there are two conductors in a particular length of the line. Inductance L is due to the magnetic field surrounding the conductors of a transmission line when a current flows through them. The inductance of an isolated twin line is given by D  µ µ  1 L =  o r   + log e  henry/metre a  π  4

(27.6)

where, D is the distance between the centres of the conductor, and a is the radius of each conductor. In most practical line

µr = 1 An inductance stated in henrys per loop metre takes into consideration the fact that there are two conductors in a particular length of line. Capacitance C exists as a result of the electric field between conductors of a transmission line. The capacitance of an isolated twin line is given by. C =

πε oε r farads/metre log e D a

(27.7)

In most practical lines ε r = 1. Conductance G is due to the insulation of the line allowing some current to leak from one conductor to the other. Conductance is measured in siemens per metre length of the line and represents the imperfection of the insulation. Yet another name for conductance is leakance. Each of the four transmission line constants R, L, C, and G, known as the primary constants, are uniformly distributed throughout the length of the transmission line. A transmission line can be considered to consist of a network of a very large number of cascaded T-sections, each of a very short length (δ ) of the transmission line, as shown in Figure 27.4. This is an approximation of the uniformly distributed line; the larger the number of lumped parameter sections, the nearer it approaches the true distributed nature of the line.

M27_AUTH_ISBN_C27.indd 513

8/13/2012 6:12:55 PM

514 Electrical Technology Is VS

R 2 dl

1 2 dl

1 2 dl

R 2 dl

R 2 dl

Gdl

Cdl

dl

1 2 dl

1 2 dl

R 2 dl

Gdl

Cdl

IR VR

Load

dl

Figure 27.4 The Transmission Line, a Network of a Very Large Number of Cascaded T-sections

27.5 PHASE DELAY, WAVELENGTH AND VELOCITY OF PROPAGATION When a generator VS is connected to the transmission line, as represented in Figure 27.4 a current IS flows which divides between that flowing through the leakage conductance G, which is lost, and that which progressively charges each capacitor C and which also sets up the voltage travelling wave moving along the transmission line. The loss or attenuation in the line is caused by both the conductance G and the series resistance R. Each section of the transmission line shown in Figure 27.4 is a low-pass filter possessing losses R and G. In case losses are neglected and R and G are removed, the circuit becomes simplified. The transmission line with the losses neglected and R and G removed reduces to a repetitive T-section L2 L2 L3 L3 L1 Is L1 low-pass filter network, as shown in Figure 27.5. Let a generator become connected to the line and let the voltage be rising to a maximum value just at the instant when the Vs C1 C2 C3 line is connected to it. A current IS flows through inductance L1 into capacitor C1. The capacitor charges and a voltage develops across Figure 27.5 The Transmission Line with the Losses it. This voltage sends a current through inductance L1 and Neglected and R and G Removed L2 into capacit,or,,, C2 charges and the voltage developed across it sends a current through L3 and C3 and so on. Thus, all of the capacitors will, in turn, charge up to the maximum input voltage. When the input voltage falls, each capacitor is charged, in turn, in the opposite polarity and, as before, the input charge is progressively passed along to the next capacitor. Thus, voltage and current waves travel along the line together and depend on each other. The process takes time. There will, therefore, be a time, and thus a phase difference between the generator input voltage and the voltage at any point on the line. The phase delay β is given by

β = ω Ö ( LC ) radians /metre

(27.8)

for a ladder network of low-pass T-section filters. In one wavelength a phase change of 2π radians occurs. The phase change per metre is 2π λ . Hence, phase change per metre is

β = 2π λ or wavelength

λ = 2π β metres

(27.9)

The velocity of propagation, μ is given by μ = f λ, where f is the frequency and λ the wavelength. Hence,

µ = f λ = f ( 2π β ) =

2π f ω = β β

(27.10)

The velocity of propagation in free space is the same as that of light (300 × 10 6 m/s). The velocity of electrical energy along a line is always less than the velocity in free space. The wavelength λ of radiation in free space is given by λ = c/f, where c is the velocity of light. Since the velocity along a line is always less than c, the wavelength corresponding to any particular frequency is always shorter on the line than it would be in free space.

M27_AUTH_ISBN_C27.indd 514

8/13/2012 6:12:59 PM

Transmission Lines 515

Example 27.2 A parallel-wire air-spaced transmission line operating at 1910 Hz has a phase shift of 0.05 rad / km. Determine (1) the wavelength on the line, and (2) the speed of transmission of a signal. Solution: 1.

λ = 2π β = 2π 0.05 = 125.7 km

2. Speed of transmission

µ = f λ = (1910)(125.7) = 240 × 103 km/s = 240 × 106 m/s

27.6 CURRENT AND VOLTAGE ALONG AN INFINITE LINE Let us consider a current IS applied at the sending end A of an infinite line (or a line terminated in Zo), as shown in Figure 27.6. At a point B, at a distance of 1 km down the line, let the current be I1. Due to the loss introduced by the line the current I1 will obviously be less than the current IS; and since, in addition a phase-shift will be introduced, the ratio IS / I, will be a vector quantity. A convenient way of representing a vector quantity is in the form ey, where, γ is a complex quantity. Hence, let IS / I1 = eγ. Here, γ is known as the propagation constant per kilometre of the line. γ may be determined in terms of the arms Z1/2 and Z2. From Figure 27.7, it will be seen that Is

Z1 2

Z1 2

I1

Z2

Z0

Figure 27.7 Current Along an Infinite Line

Z 22 Z Z2 II11 = ...III SS = I1 = Z + Z + Z o S 22 + Z11 2 2 Z + Z o Z 2 + Z1 2 + Z o Z1 Z oo II S = 1 + Z Z Z  = + Z o = I SS = = 1 + 11 + Hence, eee  = + Z II11 1 + 22 Z Z 22 Z Z 222 2Z I1 2  Z Z11 + Z Z oo = log log ee 11 + + Z 1 + Zo  = =   log 1 + + Z e  and,  2 Z 222 Z Z 22 Z Z 222 

(27.11)

A

I1

B

I2

C

I3

D

I4

E

Is Vs

1 km 1 km 1 km 1 km

Figure 27.6 Current and Voltage Along an Infinite Line

(27.12)    

(27.13)

At a distance two kilometres down the line, at point C, let the current be I2. Since the section of the line between B and C is identical with that between A and B, it follows that it may be represented by the same equivalent T section. Thus, I1 Z Z = 1 + 1 + o = eγ 2Z 2 Z 2 I2

(27.14)

I Similarly, it will be seen that for the section between C and D:- 2 = eγ where, I3 is the current at D, three kilometres down I3 the line. I n −1 = e In

For the nth section

(27.15)

where, In-1 and In are the currents at distances (n - 1) and n kilometres down the line, respectively. Now and, in the general case:-

IS = eγ ; I1

IS I I = S . 1 = e 2γ ; I2 I1 I 2

IS I I I = S . 1 . 2 = e3γ I3 I1 I 2 I 3

(27.16)

IS = e nγ In

From this, it follows that I n = I s .e − nγ

M27_AUTH_ISBN_C27.indd 515

(27.17)

8/13/2012 6:13:05 PM

516 Electrical Technology This is the general equation for the current at a point distant n kilometres down an infinite line, in terms of the sending end current IS, and the propagation constant per kilometre. It applies for any value of n (see Figure 27.8). A similar equation can be derived for voltage, since at all Is max –αx e I points along an infinite line the ratio of voltage to current is equal s max Current to the characteristic impedance Zo. I –αl Is maxe

Distance I

x –Is maxe–αl –Is max

Figure 27.8 Decrease in the Amplitude of Current Along an Infinite Line

ES E E E E2 =EEE3n = Z o EEn (27.18) = 1 = EESS2 = = EE13 = E...... ......== n == Z o I1 I1 II 2 = II13 == I2 == I3I n==...... Zo 1 1 2 3 I1 I1 I2 I3 IInn ES IS = = EeESnSγ = IISS = e nγ Hence, (27.19) nγ En In En = I n = e E I n γn En = ES .e − nE − nγ = E n Therefore, (27.20) E = E S.e.e− nγ

Thus,

n

S

Example 27.3 When operating at a frequency of 2 kHz, a cable has an attenuation of 0.25 Np/km and a phase shift of 0.20 rad/km. If a 5V r.m.s. signal is applied at the sending end, determine the voltage at a point 10 km down the line, assuming that the termination is equal to the characteristic impedance of the line. Solution: VR = VS e − n γ = VS e − n γ − n β Since,

∝ = 0.25 N p / km,

β = 0.20 rad km ,

VS = 5 V and n = 10,, then

VR = (5) e − (10)(0.25) −(10)(0.20) = 5e −2.5 − 2.0 V = 0.41 −2.0 V or 0.41 −114.6° V

27.7 PROPAGATION CONSTANT The propagation constant γ is a complex quantity. Let g be equal to α + jβ. Thus, for one kilometre of line

IS = eγ = eα + j β = eα β I1 Hence,

I IS = eµ and the angle of S is β. It fo llows that I1 I1 µ= log e

IS I1

(27.21)

e∝ gives the ratio of the absolute value of the current sent to that of the current received, while β gives the phase angle of the two currents. Here, ∝ is known as the attenuation constant per kilometre of the line and is measured in nepers per kilometre. β is known as the phase constant or wavelength constant per kilometre of the line and is measured in radians per kilometre. If the length of the line is n kilometres, then IS = e n γ = e nµ + jn β = e n µ n β. In

(27.22)

The attenuation of such a line is thus n∝ nepers, and the phase angle is nβ radians. Note: These results can be converted into more convenient units for practical work, the decibel and the degree, by multiplying by 8.686 and 57.3, respectively, or by using the Conversion Tables. In the general case of an infinite line or a short line terminated in its characteristic impedance and having a propagation constant γ, the current I at any point distance x from the sending ends, will be given by I = I S .e − γx = I S .e − µ x − βx

M27_AUTH_ISBN_C27.indd 516

(27.23)

8/13/2012 6:13:09 PM

Transmission Lines 517

Where, IS is the sending-end current The voltage E at any point distant x from the sending end will be given by E = ES .e − γx = ES .e − µ x − βx

(27.24)

where, ES is the sending-end voltage. Example 27.4 A cable has at 1600 Hz, an attenuation of 3.0 db per km and a phase constant of 0.319 radians per km in dry weather. If 2 volts at 1600 Hz are applied to the sending end, what will be the voltage at a point 10 km down the line when the line is terminated in its characteristic impedance? Solution: Ex = ES .e − γx = ES .e − µ x − βx Attenuation at 3db/km = 3.0 × 0.115 nepers/km = 0.345 nepers/km E x = 2.0.e −0.345 ×10 −0.319 × 10 = 2.0.e −3.45 −3.19 = 0.635 volts −3.19 Thus, the voltage at a point 10 km from the sending end is 0.635 volts, lagging 3.19 radians, i.e. 182°.47′ behind the sending end voltage.

27.8 LINE CONSTANTS A practical line has a characteristic impedance Zo, a propagation constant γ, an attenuation constant ∝, and a phase constant β. These are known as the secondary line constants. Although they are referred to as constants, all will vary if the frequency is changed. The primary line constants which, for the purpose of transmission line theory, are assumed to be independent of frequency, are R, G, L, and C where, R is the resistance per kilometre of the line, G is the leakance per kilometre of the line, L is the inductance per kilometre of the line, C is the capacitance per kilometre of the line. They are measured considering both conductors, i.e., per kilometre loop. These primary constants may be obtained by measurements on a sample of the line.

27.8.1 Relationship Between Primary and Secondary Line Constants Consider a short length of line l kilometre long. This short section will have a resistance Rl, a leakance Gl, an inductance Ll and a capacitance Cl. Its characteristic impedance will be Zo (the same as that of the complete line). Its propagation constant will be γl, (where, γ is the propagation constant of the complete line).This short section of the line may be represented by a |per kilometre| T network, as shown in Figure 27.9. If the length of the section is very small, then Z1 will be approximately equal to the series impedance of the section, i.e., to Rl + jωLl; and Z2 will be approximately equal to the shunt impedance of the section, i.e., to 1/(Gl + jωCl). The accuracy Z1 Z1 of this statement increases as l decreases and in order to 1 1 2 (R + jω L)l 2 (R + j ωL)l 2 2 l obtain an accurate answer, it will be assumed that the section is so small that l tends to be zero. Z2

1 (G + j ωC )l

Example 27.5 At a frequency of 1.5 kHz, the open-circuit impedance of a length of transmission line is 800 −50° Ω and the short-circuit impedance is 413 −20° Ω. Determine the characteristic impedance of the line at this frequency.

M27_AUTH_ISBN_C27.indd 517

(a)

(b)

(c)

Figure 27.9 T Sections Equivalent to a Short Length of Line

8/13/2012 6:13:12 PM

518 Electrical Technology Solution: Z o = Ö ( Z oc )( Z sc ) = Ö [ (800 −50°)(413 −20°) ] = Ö (330400 −70°) = 575 −35°Ω

27.8.2. Characteristic Impedance in Terms of Primary Line Constants I1

ZA/2

P l2

ZA/2

(I1–I2)

Zo

ZB X

Q

Y

Figure 27.10 Zo in Terms of R, L, G and C

The characteristic impedance of a line may also be expressed in terms of primary line constants R, L, G, and C. The measurements of the primary constants may be obtained, for a particular line and manufacturers, by de Moivre’s theorem usually state them for a standard length. Let a very short length of line δℓ metres be, as has been shown in Figure 27.10, comprising a single T-section. Each series arm impedance is 1 1 1 Z1 = ( R + j ω l ) δ l ohms and the shunt arm impedance is Z 2 = . = 2 U2 (G + jω C )δ  The total admittance Y2 is the sum of the admittances of the two parallel arms, i.e.,

 1  G δ  and  δ  1 j ω C  Z o = √ ( Z12 + 2 Z1Z 2 ) 2   1   1   1 Z o = √   ( R + j ω L) δ   + 2  ( R + j ω L) δ       (G + j ω C ) δ     2   2

(27.25)

The term involves δℓ2, and since δℓ is a very short length of the line, δl2 is negligible. Hence, Zo =

R + j L o hms G + j C

(27.26)

If the losses R and G are neglected, then Z o = Ö ( L / C ) o hms

(27.27)

Example 27.6 A transmission line has the following primary constants R = 15 Ω/loop km; L = 3.4 mH/loop km G = 3 µS / km and C = 10 nF/km. Determine the characteristic impedance of the line when the frequency is 2 kHz. Solution: R + j ω L = 15 + j (2 π2000)(3.4 × 10−3 ) = 45.29 70.66° Ω G + j ω C = 3 × 10−6 + j (2 π2000)(10 × 10−9 ) = 125.7 × 10−6 88.63° Zo = Ö =Ö

R + j ωL G + j ωC 45.29 70.66° = Ö  0.360 × 106 −17.9 °  Ω 125.7 × 10−6 88.63°

Z o = 600 −8.99° Ω

M27_AUTH_ISBN_C27.indd 518

8/13/2012 6:13:17 PM

Transmission Lines 519

27.8.3 Propagation Coefficient in Terms of Primary Line Constants Figure 27.11 illustrates a T-section with the series arm impedances each expressed as ZA/2 ohms per unit length and the shunt arm impedance as ZB ohms per unit length. The p.d. between points P and Q is given by Z  VPQ = ( I1 − I 2 ) Z B = I 2  A + Z o   2 

I1

ZA/2

(I1– I2)

X

Z + ( Z A / 2) + Z o I1 I 2 = B ZB

from which

P l2

Zo

ZB

I Z I1Z B − I 2 Z B = 2 A + I 2 Z o 2

i.e.,

ZA/2

Q

Y

Figure 27.11 γ in Terms of R, L, G and C

 Z 2  Z Z0 is also equal to Ö   A  + 2  A  Z B   2    2   see Figure 27.11

(

Z o = Ö Z A2 4 + Z A Z B

)

 Z2  Z  ZB +  A  + Ö  A + Z AZB   2   4  I1 I 2 = ZB

Hence,

1Z  =1+  A  + 2  ZB 

12

2 Z 1Z    A +  A   Z B 4  Z B  

I1 I 2 = eγ From the binomial theorem

where, γ is the propogation coefficient. n(n − 1) n − 2 2 a b + 2!

(a + b) n = a n + na n −1.b + 12

Thus

Hence,

Rearranging gives

2 Z 1Z    A +  A   Z B 4  Z B  

Z  =  A  ZB 

I1 1Z  = eγ = 1 +  A  + I2 2  ZB 

12

+

1  ZA  2  Z B 

−1 2

2

1  ZA  + 4  Z B 

  Z 1 2 1  Z 3 2    A  +  A  +  8  ZB    Z B  

Z  e = 1+  A   ZB  

12

1Z  1Z  +  A +  A 2  ZB  8  ZB 

32

+

Let the length XY in Figure 27.11 be a very short length of line δl and let impedance ZA = Zδℓ, where, Z = R + jωL and ZB = 1 / (Yδl), where, Y = G + jωC 12

Then

 Zδ l  eγ δl = 1 +    1 Yδ l 

+

= 1 + ( ZY δ l 2 )1 2 + = 1 + ( ZY )1 2 δ l +

1  Zδ l  1  Zδ l   +   2  1 Yδ l  8  1 Yδ l 

32

1 1 ZY δ l 2 + ZY δ l 2 2 8

32

(

)

(

+L

)

+L

3

1 1 ( ZY )(δ l)) 2 + ( ZY ) 2 (δ l)3 + L 8 2

= 1 + ( ZY )1 2 δ l ex = 1 + x +

x 2 x3 + +L 2! 3!

eγδ l = 1 + ( ZY )1 2 δ l M27_AUTH_ISBN_C27.indd 519

γ = √ ( ZY )

8/13/2012 6:13:25 PM

12

32

12

32

32  Z δ l  Z δ l1  ZZδδ ll1 1 2Z1δ l1 Z δ Zl1δl Z  δ l1  Z δ l  eγ δl = 1 e+γ δl = 1 e+γ δl  =+1 +  +  + +  +   ++ L + L + L  1 Y δ l  1 Y δ2l  11 YYδδ l2l  1 Y8δ2l1 Y1δ Yl8δl1Y δ 8l  1 Y δ l 

12 1 2 12 1 2 1 3 2 12 3 2 2 2 32 = 1 + ( ZY=δ1l+2 )(1ZY δδll 2 )1ZY +2 +δ l1ZY δ+ lδ2 l 2ZY + + δ lL ZY δ+ lL =+δ1l+ )(ZY ZY+ ZY +L 2 8 2 2 8 8

(

()

( ()

() )

3

520 Electrical Technology

()

)

3

1 1 2 11 2 2 1 3 2 1 3 3 2)( ZY ZY lZY = 1 + ( ZY=)11 2+δ(lZY += )11 2+(δZY )( )+ L (δ(lZY ) )+2 L + δ)1l) (lZY δ l++)(8δ(l) ()ZY)+)((δδ (l) ))2 + (δ l)3 + L 2 8 2 8 2

= 1 + ( ZY=)as δ(lZY 1 +negligible x )1 2 δ l = )1 +δ(leZY If (δl)2, (δl)3 and higher powers of δl are considered may be expressed as a series 12

Comparison with

12

x 2 x3 x 2 x3 x 2 x3 e x = 1 + exx += 1 ++ +x + + L exx += 1+ +L + +L 2! 3! 2! 3!2! 3! 12 eγδ l = 1 e+γδ(lZY=)11 e2+γδδ(lZY = )1 +δ(lZY )1 2 δ l shows that

γ = √ ( ZY γ =) √ ( ZY γ =) √ ( ZY ) γ = √ [ (γR =+ √jω[ L +√+jω[ j(Lω G) ]j+ω jLω)(CG) ]+ jω C ) ] (γR)(=G R)(C+ Thus (27.28) γ The unit of Y is √(Ω)(S), i.e., √(Ω)(1/Ω), thus, γ is dimension less. Since I1/I2 = e , from which γ = loge (I1/I2), a ratio of two currents. For a lossless line, R = G = 0 and γ = jω√(LC). Example 27.7 A line has the following primary line constants R = 100 Ω/km; G = 1.5 × 10-6 S/km; L = 0.00 H/km; Find the characteristic impedance in modulus and angle form at 1000 Hz. Solution:

C = 0.062 μF/km.

R + j  L = 100 + j 6.283 = 100.2 3° 36 ′ G + j  C = (1.5 + j 389.5) × 10−6 = 389.5 89° 48 ′ R + j L G + j C

Zo =

100.2 3° 36 ′ 389.5 × 10−6 89° 48′

=

Z o = 507 −43° 6 ′

27.9 CONDITIONS FOR MINIMUM ATTENUATION The value of ∝ depends upon the four primary line constants in addition to the frequency under consideration. If L is variable, the attenuation will be a minimum when

L = CR

G

henrys/km

(27.29a)

In practice, L is normally less than this desired value, and hence the attenuation of a line can be reduced by artificially increasing L. If C is considered as the only variable, its value to give minimum attenuation is

C = LG

R

farads/km

(27.29b)

In practice, C is normally already greater than the value given by LG/R, and to reduce the attenuation it would be necessary to decrease the capacity. If either R or G is the only variable no minimum is found. However, when R = 0 and G = 0, the attenuation is zero.

27.10 DISTORTION If the received signal is not an exact replica of the transmitted signal, the signal is said to be distorted. There are three main causes of distortion along a transmission line. Distortion occurs when— 1. The characteristic impedance of the line varies with the frequency and the line is terminated in an impedance that does not vary with frequency in an identical manner.

M27_AUTH_ISBN_C27.indd 520

8/13/2012 6:13:28 PM

Transmission Lines 521

2. The attenuation of the line varies with frequency, so that waves of different frequencies are attenuated by different amounts. 3 The velocity of propagation varies with frequency so that waves of different frequencies arrive at different times. Zo =

R + j L o hms G + j C

(27.30)

If the frequency is very low, ω is low and Zo ≈ √ R / G. If the frequency is very high, then ωL >> R, ωC >> G and Zo ≈ √ (L / C). A graph showing the variation of Zo with frequency f is shown in Figure 27.12. If the characteristic impedance is to be constant throughout the entire operating frequency range, then the following condition is required. √ ( L C ) = √ ( R G) LG = RC

i.e.,

(27.31) R G

Thus, in a transmission line, if LG = RC it is possible to provide a termination equal to the characteristic impedance Zo at all frequencies. L The attenuation of a line varies with the operating frequency, since γ = C √ (R + jωL) (G + jωC). Thus, waves of different frequencies and component Frequency frequencies of complex waves are attenuated by different amounts. Figure 27.12 Variation of Zo If LG = CR, ∝ = √ RG, then the attenuation coefficient is independent with Frequency of frequency and all frequencies are equally attenuated. The delay time, or the time of propagation and, thus, the velocity of propagation, varies with frequency and, therefore, waves of different frequencies arrive at the termination with different delays. , β = ω √ (LC) when LG = CR. Thus, in a transmission line when LG = CR, the velocity of propagation and, hence, the time delay is independent of the frequency. The condition LG = CR is appropriate for the design of transmission lines, since under this condition no distortion is introduced. This means that the signal at the receiving end is a true replica of the signal at the sending end, except for the fact that it is reduced in amplitude (attenuated) and delayed by a fixed time. With no distortion, the attenuation on the line is minimal. It is undesirable to increase G since the attenuation and power losses increase. The capacitance is large and not easily reduced. Thus, inductance L is the quantity that needs to be increased and such an artificial increase in the line inductance is called loading. This is achieved either by inserting inductance coils at intervals along the transmission line—called lumped loading—or by wrapping the conductors with a high permeability metal tape—this being called continuous loading.

27.11 LOADING The inductance of a line may be increased by the introduction of loading coils (Figure 27.13) at uniform intervals along the line. Provided that the spacing is uniform, the line behaves—at all frequencies—up to a frequency called the cut-off frequency of the line, as if this added inductance were distributed uniformly along it. Above the cut-off frequency, the attenuation increases rapidly. The line, in fact, behaves as if it were a low-pass filter. Provided that a limited frequency range is permissible, this method of loading is more convenient than continuous loading. There is, however, a practical limit to the amount by which the inductance of the line may be increased to reduce attenuation: the loading coils have a certain resistance, and thus increasing L also increases R. Hysteresis and eddy current losses will also occur in loading coils. These will cause a further apparent increase in R, and unless Figure 27.13 A Single Loading Coil the coil is carefully designed, it may introduce distortion. With continuous loading, as has been represented in Figure 27.14, a tape of iron or some other magnetic material such as mumetal is wound round the conductor to be loaded, thus, increasing the permeability of the surrounding medium and thereby increasing the inductance. Iron wire

Copper conductor

Figure 27.14 Continuous Loading

M27_AUTH_ISBN_C27.indd 521

In this method of loading, the cost is excessive due to the difficulties in construction. Further disadvantages are the large apparent increase in the primary constant R, due to eddy current losses and hysteresis losses in the magnetic material, and the fact that small differences in mechanical treatment or the pressure between the tape and conductor cause large variations in the primary constants.

8/13/2012 6:13:30 PM

522 Electrical Technology Continuous loading at the present time is used only in submarine cables, where the problem of making water-tight joints at loading points renders lumped loading difficult. The continuously loaded cable has the advantage over the lump-loaded cable that its attenuation increases smoothly with increase in frequency; there is no cut-off frequency.

27.11.1 Cut-off Frequency A lumped-loaded coil acts as a low-pass filter, since the inductance is lumped instead of being continuously distributed, as can be seen in Figure 27.15. If the inductance of line plus that of the loading coil is LS Henrys per loading coil section, and the capacity of the line per loading coil section is CS, then the cut-off frequency fc is given by fc = L

C

L

C

L

Figure 27.15 A Loaded Line is Equivalent to a Low-pass Filter Hence, d must be less than

1 f ( LC )

1 p √ ( Ls Cs )

(27.32)

Alternatively, if L is the apparent inductance of the line per kilometre after loading, C the capacity of the line per kilometre, and d the loading coil spacing in kilometres then Ls = Ld and Cs = Cd and fc =

1 π d √ ( LC )

(27.33)

where, f is the highest working frequency.

It will be noted in Eq. 27.33 that 1. If the value of the loading coil inductance remains unchanged, the cut-off frequency is inversely proportional to the loading coil spacing. 2. If the loading coil spacing remains unchanged, the cut-off frequency is inversely proportional to the square root of the inductance of the loading coils. 3. If the loading coil inductance is multiplied by any amount and the spacing is divided by the same amount, there is no change in the cut-off frequency. 4. If the inductance of each coil and the spacing between coils are both divided by a factor n, the cut-off frequency is increased by the same factor n. Example 27.8 A sample of field quad cable has the following primary constants: R = 78 ohms per kilometre loop; G = 62 micro ohms per kilometre; L = 1.75 milli henry per kilometre loop; C = 0.0945 μF per kilometre. Determine at 1600 Hz (ω is approximately 10 000 rad(s) (1) Characteristic impedance Z; (2) Attenuation constant ∝ in nepers and decibels per kilometre; and (3) Phase constant β in radians and degrees per kilometre. Solution R + j  L = 78 + j10000 × 1.75 × 10−3 1. = 78 + j17.5 = 79.94 12° 39 ′ G + j  C = 10−6 × 62 + j10000 × 0.0947 × 10−6 = 10−6 (62 + j 947) = 10−6 × 947 86° 15 ′ Zo =

R + j L = G + j C

79.94 −36° 48 ′ 947 × 10−6

Z o = 290 −36° 48 ′ 2.

γ

3.

β = 0.209 rad/km = 11°58′ per km

M27_AUTH_ISBN_C27.indd 522

′

8/13/2012 6:13:34 PM

Transmission Lines 523

Example 27.9 A typical open wire line has the following line constants. R = 14 Ω/km, L = 4.6 mH/km, C = 0.01 μF/km, G = 0.3 × 10-6 S/km. Calculate Zo, α and β. Solution:

Example 27.10 An underground cable has the following primary constants: R = 10 Ω/loop km; L = 1.5 mH/loop km; G =1.2 μS/km and C = 0.06 μF/km. Determine to how much degree should the inductance be increased to satisfy the condition for minimum distortion? Solution: The condition for minimum distortion is given by L = CR / G (0.06 × 10−6 )(10) = 0.5 H or 500 mH 4 .2 × 10−6 Thus, the inductance should be increased by (500 - 1.5) mH or 498.5 mH per loop per kilometre for minimum distortion. =

27.12 REFLECTION

cos(ω t – β x)

If a line, at any point along its length is joined to some impedance having a value other than Zo part of the wave travelling down the line will be reflected back from the point of discontinuity. This reflection will be a maximum when the line is on open circuit or short circuit and will be zero with proper termination. In general terms, reflection occurs whenever there is an impedance mismatch between the two networks. 1 All lines have a definite length and often the terminating impedance does not have the same impedance as the 0.5 characteristic impedance of the line. When this is the case, ω t = 0 ωt =π /4 ω t =π /2 the transmission line is said to have a mismatched load. 0 The forward travelling wave—represented in Figure 27.16 — moving from the source to the load is called the incident –0.5 wave or the sending end wave. With a mismatched load, the termination will absorb only a part of the energy of the incident wave, the remainder being forced to return back –1 along the line toward the source. The latter wave is called the reflected wave. Reflections commonly occur in nature 0 π π /4 π /2 3π /4 5π /4 3π /2 7π /4 when a change of transmission medium occurs. If the line β X (radians) is terminated in an open-circuit or short-circuit, the result is total reflection. Between the two extremes (match and Figure 27.16 The Forward Travelling Wave on mismatch), all degrees of reflection are possible. a Lossless Transmission Line

M27_AUTH_ISBN_C27.indd 523

8/13/2012 6:13:37 PM

524 Electrical Technology

27.12.1 Reflection Coefficient Consider a generator network A, as shown in Figure 27.17, an impedance Zo. working into a load network B, impedance ZR. According to the concept of reflection, an initial current I1 flows from the generator expecting to find a load equal to Zo. This current in a load ZR produces a reflected current I2 flowing back from the load to the generator. The resultant current I in the steady state is, therefore, I Network

A

ZO ZR

Network

B

(a)

I

ZO

Z ′R

E

ZO

ZR

I1

ZO

–IZ ′R Z O

E

ZO

(b)

I2 –IZ ′R ZO + ZO

E

(c)

ZO

(d)

(e)

Figure 27.17 The Concept of Reflection I = I1 + I2 Applying Thevenin’s theorem replace network A by a generator of e.m.f. E and impedance Zo, we can replace network B by the two impedances Zo and Z R′ , in such a way that Z o + Z R′ = ZR . Applying the compensation theorem, replace impedance E − 1Z R′ Z R′ by a generator having zero internal impedance and an e.m.f. equal at all times to − IZ R′ . Then, =I 2Z o ∴ E = I (2 Z o + Z R′ ) Applying the superposition theorem, as in Figure 27.17 (c), the current may be considered as the sum of two currents: the current I1 produced by the e.m.f. E, and the current I2 produced by the e.m.f. -I Z R′ as illustrated in Figures [27.17 (d) and (e)].

′ ′

and Hence,

′

′

′ ′

′ (27.34)

Zo − Z R is called the reflection coefficient and gives the ratio of the reflected current to the incident current. Zo + Z R I 2 I = − Z R′ / (2 Z o ) =

Zo − Z R 2Z o

(27.35)

gives the ratio of the reflected current to the total current flowing.

Consider a line of length l and characteristic impedance Zo, that is terminated in ZR at the distant end. We can consider a generator of impedance Zo and e.m.f. E connected to the sending end (Figure 27.18). Let the sent current be Is, and the received current be IR. It is required to find the current and voltage at any point P distant x from the sending end. The generator of e.m.f. E may be compared to such a current wave along the line in the direction A to B, while a return current wave can

M27_AUTH_ISBN_C27.indd 524

ISe – γ X

IS

27.12.2 General Line Equations From Reflection Considerations

IR

–2 γ le γ x ZO – ZR le ZO + ZR s

ZO

ZR

E A

x

l

P

B

Figure 27.18 Reflected Current in the Line Terminated in ZR

8/13/2012 6:13:42 PM

Transmission Lines 525

be considered to be reflected back along the line from B to A. The current at P at any instant is the vector sum of these two currents. Let the current at P be I. Also let the incident current be I1, and let the reflected current be I2. Let Y be the propagation constant of the line. Recieved current at B is given by:

Reflected current at B is

i.e., The total current I at point P is therefore:

(27.36)

27.13 OPEN-CIRCUITED TERMINATION

λ

Egen

ZO l

Open

3λ 4

λ 2

λ 4

Capacitive

Inductive

If a length of a transmission line is open-circuited at the termination, no current can flow in it and, thus, no power can be absorbed at the termination. This condition is achieved if a current is imagined to be reflected from the termination, the reflected current having the same magnitude as the incident wave but with a phase difference of 180°. Also, since no power is absorbed at the termination (it is all returned back along the line), the reflected voltage wave at the termination must be equal to the incident wave. Thus the voltage at the termination must be doubled by the open circuit. This has been shown in Figure 27.19. For the open-circuited line, if the line length is less than λ/4, then the input impedance appears capacitive. If the line is λ/4 long, the input appears as a short. From a length of λ/4 to λ/2 it appears to be inductive. The variation of reactance along a lossless open-circuited line is illustrated in Figure 27.20.

(a)

E

2Egen (b) λ 4

Figure 27.19 Voltage and Current Standing Waves on an Open-circuited Line

M27_AUTH_ISBN_C27.indd 525

ZR = ∞

Figure 27.20 Variation of Reactance Along a Lossless Open-circuited Line (a) Reactance (b) Nature of the Line Impedance

8/13/2012 6:13:45 PM

526 Electrical Technology

27.14 SHORT-CIRCUITED TERMINATION

Inductive

If the termination of a transmission line is short-circuited the impedance is zero and, hence, the voltage developed across it must be zero. As is the case with the open-circuit condition, no power is absorbed by the termination. To obtain zero voltage at the termination, the reflected voltage wave must be equal in amplitude but opposite in phase to the incident wave. Since no power is absorbed, the reflected current wave at the termination must be equal to the incident current wave and, thus, the current at the end of the line must be doubled at the short circuit as shown in Figure 27.21. As with the open circuited case the resultant voltage (and current) at any point on the line and at any instant of time is given by the sum of the voltages Short Egen ZO (and currents) due to the incident and reflected waves. The voltage and current vary sinusoidally in magnitude as one moves E l along the transmission line. The voltage is zero at the short and becomes a 2Egen maximum at a distance of λ/4 from the load. The current, however, is maximum at the short and is minimum at a distance of λ/4, from the load. The λ resulting standing wave pattern is illustrated in Figure 27.22. 2 A shorted lossless transmission line is either capacitive or induction, Figure 27.21 Voltage and Current depending on the length of the line. If the line length is less than λ/4, the Standing Waves on a line appears to be inductive. If the line is λ/4 long, the input appears as an Short-circuited Line open. From a length of λ/4 to λ/2 it is capacitive. The impedance repeats every half wavelength along the line. The variation of reactance along a lossless short-circuited line is illustrated in Figure 27.22.

3λ 4

λ 2

λ 4

O (a) Capacitive

l

(b)

ZR = 0

Figure 27.22 Variation of Reactance Along a Lossless Short-circuited Line (a) Reactance (b) Nature of the Line Impedance

27.15 STANDING WAVES Whenever two waves of the same frequency and amplitude travelling in opposite directions are superimposed on each other, interference takes place between the two waves and a standing or stationary wave is produced. The points at which the current is always zero are called nodes. The standing wave does not progress to the left or right and the nodes do not oscillate. Those points on the wave that undergo maximum disturbance are called antinodes. The distance between adjacent nodes or adjacent antinodes is λ/2, where λ is the wave length. A standing wave, therefore, seems to be a periodic variation in the vertical plane taking place on the transmission line without travelling in either direction. Figure 27.23 shows standing waves on a lossy transmission line.

27.15.1 Development of the Standing Wave The transmission line is termed resonant when the load impedance does not match the line impedance or is not purely resistive. The load impedance may be either larger or smaller than the surge impedance of the line and may also be both resistive and reactive. The energy sent down the line from the sending end is not all absorbed by the load. The quantity of energy not absorbed is reflected back up the line. There are now two energy waves moving through the line at the same

M27_AUTH_ISBN_C27.indd 526

8/13/2012 6:13:47 PM

Transmission Lines 527 E

I RMS voltage and current

2λ

3λ/2

λ

λ/2

Figure 27.23 Standing Waves on a Lossy Transmission Line time but in different directions. The incident wave is the wave moving from the generator to the load. The reflected wave moves from the generator to the load in the opposite direction, along the line from the load back towards the generator as shown in Figure 27.24. Both waves are continuous in nature, i.e., they are two sine waves moving through the line in opposite directions at the same time. There will be moments in time when the peaks of the waves align and the voltages add together. There will be other moments when the two waves are completely out of phase and totally cancel each other. Figure 27.25 represents these two waves moving along the line. Figure 27.25 (a) shows the same two voltages at the moment before they are exactly in phase. The sum of the two waves would crest at twice the peak voltage of either wave separately. In Figure 27.25 (b), each wave has travelled another quarter wavelength (90°) in its respective direction. The two waves are now almost exactly out of phase and will completely cancel. Incident wave

(a)

Sum of both waves a

b

c

Reflected wave d

e

f

g

(b)

2Egen

(c)

Figure 27.25 The Development of the Standing Wave on the Transmission Line (a) Load End (b) Load End (c) Voltage Standing Wave and Current Standing Wave

M27_AUTH_ISBN_C27.indd 527

RL

Egen Z0 Incident wave + 0 – Reflected wave

Figure 27.24 Incident and Reflected Waves Moving Along the Line with a Mismatched Load

From Figures [27.25 (a) and (b)] it is evident that the sum of the voltages at each of the time intervals a, b, c, d, e, f and g will be equal to zero. In Figure 27.25 (a), this is because both waves are passing through zero voltage at the same time. In Figure 27.25 (b), the voltage sums are zero because the two waves completely cancel each other. Although the incident wave and the reflected wave are constantly in motion along the line, the positions along the line where the waves add together and cancel each other do not move at all. This creates a standing wave along the line that is stationary. The minimum value of the standing wave voltage may go to ov, and the crest of the standing wave voltage will reach a maximum of twice the generator peak voltage. The magnitude of the standing wave can be damaging to the insulation and cause a short circuit in the cable.

27.15.2 Voltage Standing Wave Ratio (VSWR) The relative magnitude of the reflected wave on a low-loss line is generally expressed by means of the voltage standing-wave ratio, which is defined as VSWR =

Emax Emin

8/13/2012 6:13:50 PM

528 Electrical Technology

RMS voltage

Emin

Where Emax is the r.m.s. (or peak) voltage at the highest point on the standing wave and Emin is the r.m.s. (or peak) value shown in Figure 27.26. VSWR is often expressed in terms of decibels and is denoted by VSWR(db) = 20 log10 VSWR

Emax

(27.37)

The VSWR is widely used because it is one of the more easily measured quantities on a line. A flat line, which has Figure 27.26 Voltage Standing Wave Ratio (VSWR) no reflected wave, has Emax = Emin and a VSWR of unity. The ratio becomes larger without limit as complete reflection is approached. The VSWR cannot be directly measured when the standing wave pattern changes its form markedly from one loop to another, as in the case of high losses. The maximum voltage occurs on the line when Emax = E + + E − The minimum voltage occurs on the line when Emin = E + − E − Thus, VSWR = S = S =

Ir S −1 = Ii S +1

1+ ρ 1− ρ

(27.38)

where, ρ is the reflection coefficient. Example 27.11 A typical cable has the following characteristics. R = 86 Ω/loop km; L = 1 mH/loop km; G = 1.4 μS/loop km C = 0.062 μF/loop km Calculate the amount of inductance that must be added for distortionless transmission. Solution: For ideal loading

Example 27.12 110 V of peak signal is applied to a mismatched line. The line absorbs only 80 V and the remainder 30 V is reflected back. Find the VSWR. Solution: VSWR =

Emax 110 + 30 140V = = 110 − 30 80V Emin

= 1.75 : 1

M27_AUTH_ISBN_C27.indd 528

8/13/2012 6:13:54 PM

Transmission Lines 529

Example 27.13 A transmission line has a characteristic impedance of (500 - j40) Ω and is terminated in an impedance of (1) (500 + j40) Ω and (ii) (600 + j20) Ω. Determine the magnitude of the reflection coefficient in each case. Solution: Z −Z Zo − ZR  = = Zo + Z R 1. o Z o + Z RR ((500 500 − 40)) − 500 + 40)) − 80 − jj 40 − ((500 + jj 40 − jj80  = = (500 − j 40) + (500 + j 40) = = 10 0 00 000 (500 − j 40) + (500 + j 40) 10  = 0..08 08 =0  500 − 40)) − 600 + ((500 − jj 40 − ((600 + =  = (500 − j 40) + (600 + (500 − j 40) + (600 + 116.62 −149.04° = 116.62 −149.04° = 1100..18 18 − 04°° −11..04 1100

2.

20)) − 100 − jj 20 = −100 − jj 20 20)) = 1100 1100 − −

jj 6600 20 jj 20

106 − 148°° = 00..106 −148 = 106  = = 00..106 

Example 27.14 A transmission line has a characteristic impedance of 600 × 0° Ω and negligible loss. If the terminating impedance of the line is (400 + j250) Ω, determine (1) reflection coefficient and (2) the standing wave ratio. Solution: = =

600∠0° − (400 + j 250) Zo − Z R = 600∠0° + (400 + j 250) Zo + Z R 200 − j 250 320.16 −51.34 ° = 1000 + j 250 1030.78 −14.04 °

= 0.3106 −65.38°  = 0.3106 Standing wave ratio

=

1+  1 − 0.3106 = 1−  1 + 0.3106

= 1.901

S UM M A RY 1. The primary constants of a transmission line are R, L, G and C. 2. The characteristic impedance, Zo, of a transmission line is the geometric mean of its open-circuit and shortcircuit impedances Zo √( Zoc. Zsc). 3. The loss or attenuation in a transmission line is caused by both the conductance G and the series resistance R. 4. The phase delay is given by β = ω√(LC) radians/metre. 5. The propagation constant is a complex quantity. 6. Z o = 7.  =

M27_AUTH_ISBN_C27.indd 529

R + jω L G + jω C

( R + jwL)(G + jwC )

8. 9. 10. 11.

Artificial increase in the line inductance is called loading. Loading can be continuous or lumped. A loaded line is equivalent to a low-pass filter. Reflection occurs whenever there is an impedance mismatch. 12. Reflections commonly occur in nature when a change of transmission medium occurs. 13. Whenever two waves of the same frequency and amplitude are superimposed, then standing waves are produced. Z − ZR 14. Reflection coefficient ρ = o Zo + Z R 15. Standing wave ratio, s =

1+ ρ 1− ρ

8/13/2012 6:13:58 PM

530 Electrical Technology

M U LT IP LE C H O I C E Q UE S TI O NS (M C Q ) 1. The primary line constants are

4. Reflections occur on a

2. Zo is equal to

(b) Mismatched line 5. The phase delay β is given by

(b) Zo, γ, α, β

(a) R, L, G, C

(a) √(Zoc)/(Zsc)

(b) √(Zoc).(Zsc)

3. Zo is equal to (a)

R + jω L G + jω C

(a) ω√(L/C) (c) ω√(LC)

(b) ω√(C/L) (d) ω√LC

6. The voltage at any point distant x from the sending end is given by (a) E = ES . e-γn

(b) √(R + jωL)(G + jωC) ANSWERS (MCQ) 1. (a) 2. (b) 3. (a)

(a) Matched line

4. (b)

5. (c)

(b) E = ES / e-γn

6. (a).

CON V E N TI O NA L Q UE S TI O NS (C Q ) 1. A parallel-wire air-spaced line has a phase-shift of 0.03 rad/km. Determine the wave length of the line and the speed of transmission of a signal of frequency 1.2 kHz. 2. When the working frequency of a cable is 1.35 kHz, its attenuation is 0.40 Np/km and its phase shift is 0.25 rad/km. The sending end voltage and current are 8.0V r.m.s. and 10.0 mA r.m.s., respectively. Determine the voltage and current 25 km down the line, assuming that the termination is equal to the characteristic impedance of the line.

3. At a frequency of 5 kHz the primary constants of a transmission line are R = 12 Ω/loop km, L = 0.50 mH/ loop km, C = 0.01 μF/km and G = 60 μS/km. Determine for the line (a) the characteristic impedance, (b) the propagation constant, (c) the attenuation coefficient and (d) the phase-shift coefficient. 4. A loss-free transmission line has a characteristic impedance of 600∠0° Ω and is connected to a load impedance of (400 + j300) Ω. Determine (a) the magnitude of the reflection coefficient and (b) the magnitude of the sending-end voltage if the reflected voltage is 14.60 V.

ANSWERS (CQ) 1. 209.4 km, 251.3 × 106 m/s 2. VR = 0.363 -6.25° mV or 0.363 1.90 mV; IR = 0.454 -6.25° μA or 0.454 1.90° μV 3. (a) 248.6 -13.29° Ω, (b) 0.0795 65. 9° (c) 0.0324 Np/km (d) 0.0726 rad/km 4. (a) 0.345 (b) 42.32 V

M27_AUTH_ISBN_C27.indd 530

8/13/2012 6:13:59 PM

First and Second Order Systems

28

OBJECTIVES In this chapter you will learn about:  Natural and forced response  The significance of t = 0 –, t = 0 – , and t = 0 +  Signal waveforms most commonly encountered  The characteristic equation  Initial and final conditions  Steady state d.c. behaviour  The impedance and admittance concepts  The S plane  Natural, forced, and complete response  Simple problems on the above  First-order and second-order systems

0

t

0

(a) Continuous

t (b) Step

Buildup t

0 (d) Sawtooth

0

t

0 (c) Pulse

Decay t

0

t

(e) Exponentials (f) Sinusoidal

Signal waveforms most commonly encountered

28.1 INTRODUCTION In predicting the behaviour of electrical or mechanical systems, we must take into account two different sources of energy. The behaviour determined by an external energy source, we call a forced response; the behaviour due to internal energy, we call a natural response. In an electrical circuit, the external energy source or forcing function may be continuous or sinusoidal, or it may have any of the other waveforms shown above. Energy may be stored internally in the electric field of a capacitor or in the magnetic field of an inductor. In a chemical system, energy may be stored internally in the temperature of a liquid, in the potential energy of a compressed gas, or in the kinetic energy of a moving fluid. In the general problem, both internal and external sources of energy are present and both forced and natural responses must be considered. Forced responses can be maintained indefinitely by the continual input of energy. In contrast, because of unavoidable dissipation in electrical resistance or mechanical friction, natural responses tend to die out. After the natural behaviour has become negligibly small, conditions are said to have reached a steady state. The period that begins with the initiation of a natural response and ending when the natural response becomes negligible is called the transient period.

28.2 FIRST ORDER SYSTEMS The behaviour of an electrical circuit or any other physical system can be described by an integrodifferential equation. In electrical circuits, the governing equation is obtained by applying the experimental laws for circuit elements and combinations of elements, taking into account both the external and internal energy sources. In the circuit of Figure 28.1, for example, a coil of inductance L and resistance R carrying a current Io is suddenly shorted by throwing switch S at time t = 0. The equation governing current i

IO

i S

+ υL + υR

L R

Figure 28.1 LR Example

532

Electrical Technology

(after the switch is thrown) as a function of time t can be obtained by applying KVL to the shorted series circuit. For all time t > 0, the sum of the voltages around the closed path (clockwise) must be zero or Συ = 0 = −υ L − υ R = − L di − Ri S dt (28.1) di L + Ri = 0 i + + dt + V v O

C

−υ c

R

R

Figure 28.2 CR Example

As another example, illustrated in Figure 28.2, capacitor C with initial voltage Vo (or initial charge Qo = CVo) is suddenly shorted across a resistance R at time t = to. For all time t > 0, current i (t) must satisfy KVL or Συ = 0 = υC − υ R = Vo − 1

t

C

∫0 dt − Ri

Differentiating to eliminate the integral and the constant and rearranging the terms, R di

+ 1 i=0 C

dt

(28.2)

These examples result in analogous, first-order linear homogeneous ordinary differential equations; analogous because they are of the same mathematical form, first-order because that is the order of the highest derivative present, linear because the variable and its derivatives appear only to the first power, homogeneous because there are no constant terms or forcing functions, and ordinary because there is a single independent variable t. The homogeneous equation contains only terms related to the circuit itself. The solution of this equation is independent of any forcing function and, therefore, it is the natural response of the circuit. –

Note: The quantity t = 0 refers to the instant just before the switching action occurs. + The quantity t = 0 refers to the instant just after the switching action has occurred.

28.3 SOLVING THE EQUATION The solution of a differential equation consists in finding a function that satisfies the equation. Consider, for example, the circuit of Figure 28.2. R di

dt

+ 1 i=0 C

(28.3)

This equation says that the combination of a function i(t) and its derivative di/dt must equal zero. This implies that the form of the function and that of its derivative must be the same and suggests an exponential function. If we let i = Ae st then di

dt

= sAe st

(28.4)

where A is the amplitude and s is the frequency. Substituting these expressions into the homogeneous Eq. 28.2 RsAe st + 1

C

Ae st = 0

Therefore, for all values of time t,

(R

S

)

+ 1 C Ae st = 0

(28.5)

In looking for solutions to this equation, A = 0 is one possibility, but this is the trivial case. The other possibility is to let

( Rs + 1C ) = 0. This yields s = − 1 RC and

i = Ae( −

1

RC

)t

(28.6)

To evaluate the amplitude of A, we return to the original equation V0 =

1 C

t

∫0 idt − Ri = 0

and consider a time t = 0+, the first instant of time after t = 0.

(28.7)

First and Second Order Systems

533

The integral of current over time represents the flow of charge and it takes a finite time for the charge to be transferred t At t = 0+, no time has elapsed and ∫ idt = 0. Therefore, the initial conditions are that 0

From Eq 28.6 at t = 0

+

Vo − 0 − Ri (0+ ) = 0

or

0 − Ri (0+ ) = 0 i (0+ ) =+ Vo RVo or i (0+ ) = VVoo R− Vo − 0 − Ri (0+ ) = 0 or i (0 ) = R

(28.8)

i (0+ ) =+ I o = Ae S (0) S = Aeo = A i (0 ) = I o = Ae (0) = Aeo = A A after = I o the = Vo R is closed Therefore A = I o = Vo R and for all values of time A = I o switch = Vo R Vo − t RC Vo − t RC i= eV e i= i R= o e − t RC R R i (0+ ) = I o = Ae S (0) = Aeo = A

(28.9)

Note: The units of s are per second. The physical explanation of this mathematical expression is based on the fact that resistor current is proportional to voltage, whereas the capacitor current is proportional to rate of change of voltage. Upon closing the switch, the full voltage of the capacitor is applied across the resistor and initial current is high (Io = Vo/R). But this current flow removes the charge and reduces the voltage (dvc /dt = i/C) across the capacitor and resistor and, therefore, the current is reduced. As the voltage decreases, the current is reduced, the voltage changes more slowly and the current becomes smaller. As another approach to evaluating io, we recall that the energy stored in a capacitor is 1 2 Cv 2 , and this energy cannot be changed instantaneously since that would involve an infinite power. Therefore, the voltage on a capacitor cannot be changed instantaneously and we conclude that the voltage just +after switching (at t = 0+) must equal the voltage just before Vo = υappears ) R the resistance and C =voltage R = i (0 across switching, Vo. When the switch is closed, υthis + υC = Vo = υR = i (0 ) R (28.10) i (0+ ) = I o = Vo R Hence, i (0+ ) = I o = Vo R

Example 28.1

S

Switch S in Figure 28.3 is arranged to disconnect the source and simultaneously at time t = 0, short circuit the coil of 2 H inductance and 10 Ω resistance. If the initial current in the coil is 20 A, predict the current i after 0.2 s has elapsed. How long will it take for the natural behaviour current to become zero? Solution: By KVL Συ = 0 = −υ L − υ R = − L di L di

The homogenous equation is,

dt

dt

IO +

i

L R

Figure 28.3 For Example 28.1

− Ri = 0

+ Ri = 0

Assuming an exponential solution, we write i = Aest where s and A are to be determined. Substituting into the homogeneous equation, LsAe st + RAe st = ( sL + R) Ae st = 0 sL + R = 0, s = − R L

and

R i = Ae − ( L ) t

1 2 Li , cannot change instantaneously. Therefore, 2 the current in the coil just after the switch is thrown, must equal the current just before. The energy stored in an inductance,

At t = (0+ ), i = I o = Aeo = A or Hence, the solution is

A = I o = 20.

20 16 i 12 (A) 8 4

i = I o e − ( L ) t = 20e − ( R

10

2

) t = 20e −5t

After 0.2t, the current is (Figure 28.4), i = 20e −5× 0.2 = 20 × 0.368 = 7.36 A The current decreases continually but never becomes zero.

0

0.2 0.4 0.6 0.8 1.0 1.2 t (s)

Figure 28.4 Exponential Current Decay in the RL Circuit of Figure 28.3

534

Electrical Technology

28.4 GENERAL PROCEDURE A suggested general procedure for determining the natural behaviour of an electrical circuit is 1. 2. 3. 4. 5.

Write the governing equation using Kirchhoff’s laws. Reduce this to a homogeneous differential equation. Assume an exponential solution with undetermined constants. Determine the exponents from the homogeneous equation. Evaluate the coefficients from the given conditions.

Any such lumped linear system can be described by an ordinary differential equation that can be reduced to the homogeneous equation by eliminating constant terms and forcing functions. The evaluation of the coefficients, which may be the most difficult step, is usually based on consideration of the energies stored in the system. If current is the rate of transfer of charge and if the current in Example 28.1 flows forever, will an infinite charge be transferred? To answer this question express the total charge as the integral of current obeys an infinite time, or q=

∞

∫0

idt =

∞

∫0

20e −5t dt

4 3

∞

q 20 −5t  =− e  (C) 2 5 0 1 20 =− (0 − 1) = 4 C 0.2 0.4 0.6 0.8 1.0 1.2 0 5 t (s) Although the current flows ‘forever’, the rate of flow is decreasing in such a way Figure 28.5 Total Charge Transfer that the total charge transferred approaches a finite limit (see Figure 28.5). The term first order system refers to a system involving only a single energy storage element for which the governing equation is a first-order differential equation. The natural behaviour obtained by solving the homogeneous equation will always be similar in form to the decay of current or voltage in an RL or RC circuit where,

i = I o e − T or ν = Vo e − T –1 The current and voltage reach e or 0.368 times the initial values at T = RC. t

t

(28.11)

28.5 SIGNAL WAVEFORMS Certain patterns of time variation or waveform are of special significance because they are encountered frequently. A direct or continuous voltage, as illustrated in Figure 28.6 is supplied by a storage battery or a direct current (d.c.) generator. A step of current flows when a switch is thrown, suddenly applying a direct voltage to a resistance (Figure 28.7). A pulse of current flows if a switch is turned ON and then OFF. The flow of information in a computer, for example, consists of very short pulses (Figure 28.8).

t

0

Figure 28.6 Continuous Waveform

0

t

t

0

Figure 28.7 A Step Waveform

Figure 28.8 Pulse Waveform

A sawtooth wave increases linearly with time and then resets; this type of voltage variation causes the electron beam to move repeatedly across the screen of a television picture tube (Figure 28.9). A decaying exponential current flows if energy is stored in the electric field of a capacitor and allowed to leak off through a resistor (Figure 28.10). In an unstable system, a voltage may build up exponentially (Figure 28.10). A sinusoidal voltage is generated when a coil is rotated at a constant speed in a uniform magnetic field; also, oscillating circuits are frequently characterized by sinusoidal voltages and currents (Figure 28.11). Buildup

0

t

Figure 28.9 Sawtooth Waveform

Decay 0

0

t

t

Figure 28.10 Exponential Waveform

Figure 28.11 Sinusoidal Waveform

First and Second Order Systems

535

Note: Exponential and sinusoidal waveforms are easy to generate and equally easy to analyze because of their simple derivates and integrals. S

28.6 SECOND-ORDER CIRCUITS

i If there is more than one energy-storage element, the behaviour of the circuit is L R more complicated but the analysis follows the same procedure as for the simple C systems. In an RLC circuit, energy may be stored either in the inductance or in + the capacitance. In the series RLC circuit of Figure 28.12, we can assume that an Vo initial voltage Vo exists on the capacitance C. With the switch S open, there is no current in the inductance L and, therefore, no energy storage in its magnetic field. Figure 28.12 A Series RLC Circuit The resistance R, of course, is incapable of energy storage. Following the general procedure for determining natural behaviour, we first write the governing equation using KVL. Traversing the loop (after the switch is closed) in a clockwise direction, we have di 1 t (28.12) Συ = 0 = − L − Ri + V0 − ∫ idt dt C 0

By differentiating and rearranging certain terms the homogeneous equation is L

di 2 di 1 +R + i=0 dt C dt 2

(28.13)

The presence of the second energy storage element has increased the order of the highest derivative appearing in the homogeneous equation. This equation is typical of a second-order system. Assuming an exponential solution, let i = Aest. Substituting in the homogeneous equation, 1 s 2 LAe st + sRAe st + e st = 0, which is satisfied when Ls 2 + Rs + 1 C = 0 (28.14) C It is the sum of three terms, each related to a circuit element. It contains no voltages or currents, but only terms characteristic of the circuit. Also, from this equation, we expect to obtain information about the character of the natural behaviour, referred to as characteristic equation, it is very useful in circuit analysis.

28.7 THE CHARACTERISTIC EQUATION For this RLC circuit (Figure 28.13) Ls 2 + Rs + 1

C

=0

The roots of the characteristic equation are 2

s1 = −

2

R 1  R  +   −  2L 2L LC

and s 2 = −

1 R  R  −   −  2L 2L LC

(28.15)

If either, i1 = A1e s1t or i 2 = A2 e s 2 t satisfies a linear homogeneous equation, i.e., it makes it zero, then the sum of the two terms also satisfies the equation. The most general solution then is (28.16) i = A es1t + A es2t 1

2

Where A1 and A2 are determined by the initial conditions and s1 and s2 are determined by the circuit constants. The values of R, L, and C are all real and positive, but the values of s1 and s2 may be real, complex, or purely imaginary. The clue to the character of the natural response of a second-order system is found in the quantity under the radical sign—the discriminant. The discriminant determines the nature of the roots. If the discriminant is positive, the roots of the characteristic equation will be real, negative, and distinct. If the discriminant is zero, then the two roots will be real, negative and identical. If the discriminant is negative, in that case the roots will be complex numbers or, in the special case where R = 0, purely imaginary.

28.7.1 Roots Real and Distinct If the discriminator in Eq. 28.15 is positive, then s1 and s2 are real, negative and distinct, and the natural behaviour is the sum of the two decaying exponential responses.

536

Electrical Technology

Example 28.2

S

Given the circuit values in Figure 28.13, determine and plot the current response as a function of time after the switch is closed. Solution: Roots of the characteristic equation are 2

s1 , s 2 = −

4  4  ±  −  2 × 1  2 ×1

1 1×

1 3

L

υL +

i

υC + + C Vo

+ υR

R

L = 1 H, C = 13 F, R = 4 Ω

= −2 ± 1

Figure 28.13 For Example 28.2

The general solution for current (Eq. 28.16) is i = A1e −t + A2 e −3t

(28.17)

To evaluate the constants, we can consider the initial conditions. At the instant the switch was closed, the voltage across the capacitance and the current in the inductance i L = i L (0− ) = 0 . Because energy cannot be changed instantaneously, these values must hold just after closing the switch at, say, t = 0+. At t = 0+, the current in each element in the series circuit must be zero. i = i 0 = 0 = A1eo + A2 eo = A1 + A2 Therefore, A2 = -A1 Also, the voltage iR across the resistance is zero. Hence, υ L = −υC = Vo di di Vo and = L V= o or dt dt L From Eq. (28.17), at t = 0+ di Vo = = − A1eo − 3 A2 eo = − A1 − 3 A2 dt L = − A1 + 3 A1 = + 2 A1 Solving

A1 = +

V Vo Vo and A2 = − A1 = − o + 2L 2 2

The specific equation for the current response, plotted in Figure 28.14 becomes V V i = o e −t − o e −3t 2 2

Figure 28.14 Natural Response of an Over Damped Second-order System

28.7.2 Roots Complex If the discriminant in Eq. 28.15 is negative, s1 and s2 are complex conjugates and the natural behaviour is the exponentially damped sinusoid. Let us consider the case of complex roots in a general way. Three new terms need to be defined (28.18) 4 Now, (28.19) Note: Complex roots always appear in conjugate pairs. By Euler’s theorem e jθ = cos θ + j sin θ , therefore e jω t = cos ω t + j sin ω t and e − jω t = cos ω t − j sin ω t e jω t = cos ω t + j sin ω t and e − jω t = cos ω t − j sin ω t Factoring out the e –∝t and substituting for−eαjωt yields i = e t [( A1 + A2 ) cos ω t + j ( A1 − A2 ) sin ω t ] i = e −α t [( A1 + A2 ) cos ω t + j ( A1 − A2 ) sin ω t ]

First and Second Order Systems

537

where A1 and A2 are constants that may be complex numbers. Since these are physical currents, the coefficients of the cos ωt and sin ωt terms, (A1 + A2) and j (A1 – A2) must be real numbers that could be called B1 and B2, so that i = e −α t ( B1 cos ω t + B 2 sin ω t )

(28.20)

But the sum of a cosine function and a sine function must be another sine function (properly displaced in phase); so Eq. 28.20 can be rewritten to give a natural response of i = Ae −α t sin(ω t + θ )

(28.21)

Note: To obtain Eq. 28.21 from Eq. 28.20, let B1 = A sinq and B2 = A cosq. This is evidently a sinusoidal function of time with an exponentially decaying amplitude, a so-called damped sinusoid. ω is seen to be the natural frequency of the oscillation in radians per second. Factor a is the damping coefficient (per second) and Ae −α t defines the envelope of the response. If a is large, then the response dies out rapidly. Amplitude A and phase angle q are constants to be determined from initial conditions. Example 28.3

S

In Figure 28.15, L = 1 H, C = 1 / 17 F, and R = 2 Ω. Derive an expression for the natural response. Rewrite the equation as a damped sinusoid and then determine the natural response after closing the switch. Solution: 2

s=

R 1  R  ±   −  LC 2L 2L 2

=−

2  2  ±  −  2 × 1  2 ×1

Therefore,

C

R

+ Vo

Figure 28.15 For Example 28.3 1 1×

1 17

= −1 ± −16

s1 = −1 + j 4

where j is equivalent to

i L

and

s 2 = −1 − j 4

−1 and the natural response is i = A1e( −1+ j 4)t + A2 e( −1− j 4)t

α =

For, the natural response (Eq 28.21) is,

R =1 2L

ω =

and

R2 1 − 2 =4 LC 4 L

i = Ae −t sin(4t + θ )

di = Ae −t 4 cos(4t + θ ) − Ae −t sin(4t + θ ) dt

and

Just after the switch is closed, the current must be zero (since iL = 0) and the rate of change of current must be. Vo = V since − υ = υ = υ = L di + o o L C L dt , At t = 0 , i = 0 = Aeo sin(0 + θ) = A sin θ

(

)

A is finite, therefore,

θ = 0 and di

Therefore, A = Vo i=

4

dt

= Vo = Aeo 4 cos(0) − Aeo sin(0) = 4 A

and as shown i n Figure 28.16

Vo −t e sin 4t 4

Figure 28.16 Response of an Oscillatory Second-order System

538

Electrical Technology

The period is

T = 1

f

= 2π

ω

= 2π

4

= 1.57 s

The damping coefficient is a = 1 s –1; in one second the envelope is down to1/e times its initial value.

28.7.3 Roots Real and Equal A system described by an equation with complex roots is said to be oscillatory or underdamped, in contrast, when the roots are real, the system is said to be overdamped. The limiting condition for oscillation, called the critically damped case, occurs when the roots are real and equal. For this case, the discrimiant is equal to zero. Physically, this is not an important case since it merely represents the borderline between the two regimes. Mathematically, it is an interesting case requiring a special form of solution. If the roots are equal S2 = S1 = S. Following the general procedure i = A1e st + A2 e st = ( A1 + A2 )e st = Ae st

(28.22)

For this case, a second term must be included and the general solution is i = A1e st + A2 te st

(28.23)

28.8 THE COMPLEX PLANE The character of the natural response of a second order system is determined by the roots of the characteristic equation. Where the roots are real, negative and distinct, the response is the sum of two decaying exponentials and the system is said to be overdamped. Where the roots are complex conjugates, the natural response is an exponentially decaying sinusoid and the system is said to be underdamped. In general, roots are located in the complex plane, the location being defined by coordinates measured along the real or σ (sigma) axis and the imaginary or jω axis. This is referred to as the s plane, since s has the units of frequency, as the complex frequency plane. Let us consider the locus of the roots for a series RLC circuit as R varies from zero to infinity. For this circuit, the char1 acteristic equation is Ls 2 + Rs + = 0 and the roots are C (28.24) 1. For

defining the values of the undamped natural frequency, ωn. In other words, for

R = 0, a = 0 the response is oscillatory with no damping. Hence, the response will not die out as time increases. The roots corresponding to the undamped case are located on the imaginary axis of Figure 28.17 (o). 2. For R =Ra, small but finite, a real part of s appears and the values of s are complex conjugates, as has been illustrated in Figure 28.17 (a). 3. As R increases to Rb, the roots move along a curved path to points sb. The radial distance from the origin to either root is given by 2  1 R2   R  α 2+ ω2 =  − − 2 =  +  2L   LC 4 L 

1 = ωn LC

(28.25)

a constant. Therefore, the locus is a circular arc of radius ωn, as has been shown in Figure 28.17 (b). 4. The critical value of resistance is defined by ω = 0 or, in other words, where the discriminant is zero. For this condition, R σ = −α C = − = −1 LC . At this value of R = Rc, the roots coincide on the real axis, as seen in Figure 28.17 (c). 2L 5. As R increases to Rd, sd1 moves towards the origin and sd2 moves out along the true negative real axis (Figure 28.17 (d)).

First and Second Order Systems jω so1

jω

jω + ωn

sa1

+ ωa sb1

σ so2

– ωn

R=0 Roots imaginary (o)

σ sa2

ωn

jω

σ – ωb

sb2

jω

sc1 = sc 2

+ ωb

σ

539

sd1 sd2

σ

– ωa

R = Ra Roots complex (a)

R = Rb >Ra Roots complex (b)

R = R critical Roots identical (c)

R = Rd >Rc Roots unequal (d)

Figure 28.17 The Root Locus for an RLC Circuit with R as the Parameter As R increases without limit, s1 approaches the origin and s2 increases without limit. With a different circuit parameter, the locus takes on a different shape. In any such plot, called a root locus, the location of the roots determines the character of the natural response.

28.9 IMPEDANCE CONCEPTS The range of exponential functions for positive and negative, large and small, and real values of s are illustrated in Figure 28.18. The key property of an exponential function is that its time derivative is also an exponential, for example, if

s Large, positive e st s Small, positive

(28.26) or if

s Zero s Small, negative s Large, negative

(28.27) –t

0

+t

Figure 28.18 The Range of Exponential This property greatly facilitates calculating the response Functions for Real Values of s of circuits containing resistance, inductance and capacitance because of the simple voltage-current relations that result. The ratio of voltage to current for exponential waveforms is defined as the impedance Z. For a resistance, υ = Ri

and

ZR =

υ Ri = = R in o hms i i

and

ZL =

(28.28)

For an inductance,

υ sLi = = sL in o hms i i

(28.29)

1 υ υ  dυ  i =C = in o hms  = sCυ and Z C = = i sCυ sC  dt 

(28.30)

 di  υ = L   = sLi  dt  For a capacitance,

The impedances ZR, ZL and ZC are constants of proportionality between voltages and currents that are exponential functions of time t and frequency s. In each case, the dimensions of impedance are the same as those of resistance. The general relation, υ = Zi corresponds to Ohm’s law for purely resistive circuits. Impedance is defined only for exponentials and waveforms that can be represented by exponentials. With this restriction, impedances can be combined in series and parallel just as resistances are, and network theorems can be extended to circuits containing L and C.

540

Electrical Technology i

Example 28.4

R1

In the circuit of Figure 28.19, R1 = 2 Ω, C = 0.25 F and R = 4 Ω. Using the impedance concept, find the currents i and ic for a voltage υ = I o e −2t V, Solution: Following the rules for resistive networks

ic

υ+

R

C

Figure 28.19 For Example 28.4  1  R 1  Z RZC  sC Z = Z1 + Z Z = R1 + R   sC  R C Z = Z 1 + Z R + Z C = R1 + R +  1  Z R + ZC  1  R +  sC   sC  4 −8 (−2 ×4 0.25) = 2+ = −2 Ω = 2+ −28 4 +(−12/(×−20.×250).25) = 2+ = −2 Ω = 2+ 2 4 + 1 /(−2−t 2 × 0.25) 6e i = υ Z = −2t = −3e −2t A Then 6e−2 i= υZ = = −3e −2t A − 2 . Z i Using the current divider and 4impedance concepts, .i R iC = = = −6e −2t A 44−.i 2 Z RZ R+.iZ C iC = = = −6e −2t A Z R + ZC 4−2

Z (Ω) 8 6

Note: Because the impedance function Z (s) contains the same information as the characteristic equation, it is a useful concept in predicting the natural behaviour of a system and it can be extended to include the prediction of steady state response.

2 –4

–3

–2 Z (s)

–1

1

0 –2

2

3

s

–4

Example 28.5

–6

Given the circuit in Figure 28.20 with υ = Vo e , determine and plot Z(s) for real values of s. Solution: A great advantage of the impedance concept is the ease with which impedances may be combined. st

Z ( s ) = Z1 +

For,

Z (s)

4

Z RZC Z R + ZC

–8

Figure 28.20 The Impedance Function for the Circuit of Figure 28.19

 1  R sR1 RC + R1 + R  sC  R = R1 + = R1 + = 1 sRC + 1 RS C + 1 R+ sC

R1 = 2 Ω, R = 4 Ω, and C = 0.25 F, Z ( s ) = 2

s+3 s +1

From the graph in Figure 28.20, it can be seen that Z(s) = 0 at s = –3 and Z(s) increases without limit as s approaches –1. t = 0+

28.10 INITIAL AND FINAL CONDITIONS Any physical variable resulting from an integration process cannot change instantaneously when the process being integrated has a finite level. Let us consider the case of initially relaxed energy storage components, i.e., a capacitor C with no initial voltage and an inductor L with no initial current, as can be seen in Figure 28.21. 0

0

(28.31)

C

t = 0+

L

(a)

(b)

Figure 28.21 Initial Behaviour of Uncharged Capacitor and Unfluxed Inductor

First and Second Order Systems

541

This condition results in the following conditions for initially relaxed components. υυcc ((00++ )) == 00 and and iiLL ((00++ )) == 00,, i.e., an uncharged capacitor initially acts like a short circuit and an unfluxed inductor initially acts like an open circuit. Where initial energy is present, a charged capacitor can be represented as an uncharged capacitor in series with a d.c. voltage source and an unfluxed inductor in parallel with a d.c. current source, as shown in Figure 28.22. In a sense, the models in Figure 28.21 can be thought of special cases of models in Figure 28.22; a voltage source of zero value reduces to a short circuit and a current source of zero value reduces to an open circuit. As a result of the algebraic relationship (Ohm’s law) between resistive voltage and current, the form is always the same. Thus, the model for a resistor at t = (0+) is the same for all time. t = 0+ + C

+ –

V0 –

t = 0+

V0

L

I0

I0

(a)

(b)

Figure 28.22 Initial Behaviour of a Charged Capacitor and a Fluxed Inductor Example 28.6 The initially relaxed circuit of Figure 28.23 (a) is excited at t = 0 by the source shown. Determine all the element voltages and currents at t = 0+. Solution: The equivalent circuit at t = 0+ is shown in Figure 28.23 (b). The initially uncharged capacitor is replaced by a short circuit, and the initially unfluxed inductor is replaced by an open circuit. The circuit at t = 0+ reduces to a simple single-loop configuration for which a single current i may be used to characterize the solution. We have 20 V i= = 2A ( 4 + 6) Ω i R = iC = i R 2 = i = 2 A

Figure 28.23 For Example 28.6. (a) Circuit and (b) Equivalent Circuit at t = 0+

542

Electrical Technology

The initial capacitor voltage and the initial inductive current are zero υ C = 0, i L = 0. The two resistive voltages and uL are determined as υ R1 = 2 A × 4 Ω = 8 V and uL = 2 A × 6 Ω = 12 V. These values are correct only for a brief instant after the switch is closed. The circuit variables will change with the passage of time. Example 28.7 The switch in the circuit of Figure 28.24 (a) is initially in position 1. As a result of excitation uS (t), the initial values of the capacitor voltage and inductor current are shown. At t = 0, the switch is changed to position 2. Determine all the voltages and currents at t = 0+. Solution: It is understood that all the quantities are evaluated at t = 0+. The equivalent circuit is shown in Figure 28.24 (b). The charged capacitor C2 acts initially like a 6 V d.c. voltage source and the fluxed inductor acts initially like a 2 A current source. However, C1 is initially uncharged, and so it acts like a short circuit.

Figure 28.24 For Example 28.7 (a) Circuit and (b) Equivalent Circuit for t = 0+ The simplest approach to this problem is a step-by-step solution. We first note that

υ C1 = 0,

υ C 2 = υ L = 6 V,

iL = 2 A

The resistive voltage is υ R = 16 V − 6 V = 10 V The current in C1 is the same as that in R and is i C1 = i R =

10 V = 5A 2Ω

The current in C2 is determined by the application of KCL to the right-hand node i C 2 = iR − iL = 5 A − 2 A = 3 A The condition when all voltages and currents have ceased to vary with time and have assumed constant values is referred to as a steady state condition. It is customary to use t = ∞, as a reference in many problems in this situation. Thus, a voltage υ (∞) and a current i(∞) refer to circuit variables under these limiting conditions.

543

First and Second Order Systems

Note: t = ∞ is only a mathematical way of indicating that a sufficiently long time has passed to ensure the assumed conditions. Circuits in which steady state conditions are valid include those having d.c. sources, initial conditions (i.e., charged capacitors and fluxed inductors) and certain types of bounded inputs (e.g., single-pulse input and a decaying exponential input). The primary criterion is that all voltages and currents must eventually assume constant values. The concept would be invalid for circuits with voltages and currents that continue to vary with time (e.g., sinusoidal sources). Let iC (t) represent the current flow in a capacitor C, and let nL(t) represent the voltage across an inductor L. We may then assume a voltage v(t) across the capacitor and a current i(t) in the inductor. In general, the relationships are i C (t ) = C

dυ (t ) dt

and

υ L (t ) =

di (t ) dt

(28.32)

In the steady state d.c. case, all the voltages and currents have settled down to constant values and there are no changes; thus, all derivatives C have zero values and iC (∞) = 0 and υ (∞) = 0. L As such, in the d.c. steady state, the current flow in a capacitor is zero, and the voltage across an inductance is zero. These are shown in Figure 28.25. Figure 28.25 Steady State d.c. Behaviour In many problems, the time t = ∞ is used to indicate the reference of Capacitor and Inductor for steady state conditions. In some cases, however, a steady state condition for one part of a circuit will serve as the basis for the initial condition for a different part of a circuit. If t = 0 is used as a switching time, t = 0 – is an appropriate symbol for the steady state condition just prior to switching. Example 28.8 The circuit of Figure 28.26 is initially in a steady state condition in position 1. At t = 0, the switch is moved to position 2 instantaneously. (1) Determine all the element voltages and currents immediately following the switching operation (2) Determine all the element voltages and currents after steady state conditions are established again. Solution: Initial conditions are not explicitly given. Such conditions (in this problem) are determined by analyzing the initial circuit form. The variables to be determined prior to the switching operation are the two capacitor voltages and the inductor current. The steady state d.c. circuit prior to switching is shown in Figure 28.27. However, the time is indicated as t = 0–, since this is just before the switching operation. Since steady state conditions exist, the two capacitors are represented as open circuits, and the inductance is represented as a short circuit. R1 = 2 Ω

1

R2 = 3 Ω

2 C1

10 V

2Ω

+ –

L

C2

10 V

+ VC1(0–) –

+ –

+ 15 V –

Figure 28.26 For Example 28.8

3Ω iL (0–)

+ VC 2 (0–) –

Figure 28.27 Prior to the Switching Operation at Time t = 0–

iR1(0+) +

However, since a short appears across C2, its initial voltage is. 15 V

1. The equivalent circuit just after the switching operation is shown in Figure 28.28. The inductor acts as a d.c. current source of 2 A, the capacitor C1 acts like a 6 V source, and the capacitor C2 acts like a short circuit. The input source voltage is now the 15 V source.

+ –

2Ω

vR1(0+) – iC1(0+)

iR2(0+)

3Ω

+ vR2(0+) – iL(0+) + 6 V + vC1(0+) – – 2A

iC2(0+) + vL(0+) –

+ vC2(0+) –

Figure 28.28 Just After the Switching Operation at Time t = t(0+)

544

Electrical Technology

υR1 (0+ ) = 15 V − 6 V = 9 V 9V 2Ω 9V + + υ Because of the short on the right iRL1 ((00+ )) == υC 2 (0= )4= .5 0A, 2Ω 6V iυR2 ((00++ )) == υ (=0+2) A= 0, L C2 3Ω

υiRR11 (0++ )) == 15 V=−46.5VA= 9 V υR2 (0+ ) = υC1 (0+ ) = 6 V

and

υR2 (0+ ) = υC1 (0+ ) = 6 V

and

By KCL

+ + 6 V+ R2 (0 ) = 4 . 5 A − 2 A = 2 . 5 A iiRC12 ((00+ )) == iR1 (0 =) −2 iA 3Ω iL (0+ ) = 2 A iC (0+ ) = iR (0+ ) − iR (0+ ) = 4.5 A − 2 A = 2.5 A iC21 (0+ ) = iR21 (0+ ) − iL (20+ ) = 2 − 2 = 0 iL (0+ ) = 2 A

By KCL

iC2 (0+ ) = iR2 (0+ ) − iL (0+ ) = 2 − 2 = 0

2. The steady state circuit based on the new source voltage is shown in Figure 28.29. The two capacitors have been replaced by open circuits once again, and the inductor has been replaced with a short circuit. iR1 (∞) = iR 2 (∞) = i L (∞) =

iR1 (∞)

15 V

15 V = 3A 2Ω + 3Ω

i C1 (∞) = ic 2 (∞) = 0

+ –

2Ω

iR2 (∞)

+vR1 (∞) – iC1 (∞)

3Ω

+vR2 (∞) – iL (∞)

+ vC1(∞) –

+ vL(∞) –

iC2(∞)

+ vC2(∞) –

Figure 28.29 The Steady State Circuit Based on the New Source Voltage

υ R1 (∞) = 3 A × 2 Ω = 6 V υ C 1 (∞ ) = υ R 2 (∞ ) = 3 A × 3 Ω = 9 V υ L (∞ ) = υ C 2 (∞ ) = 0

28.11 THE ADMITTANCE CONCEPT Impedance is defined for exponentials as the ratio of voltage to current. The reciprocal of impedance is admittance; a useful property defined as the ratio of exponential current in amperes to voltage in volts so that

Y=

i = v

1 Z

(28.33)

measured in siemens. For an ideal resistance, the admittance is just the conductance or Y =

1 iR i = = =G υR Ri R

(28.34)

For exponential voltages and currents, the admittances of ideal inductive and capacitive circuit elements are YL = YC =

iL

υL iC

υC

=

i i 1 = = L ( di dt ) Lsi sL

=

C ( dυ dt ) sC υ = = sC υ υ

(28.35) (28.36)

Admittance is particularly useful in analyzing those circuits that contain elements connected in parallel. Since current is directly proportional to admittance (i = Yu), admittances in parallel can be added directly just as conductances in parallel are added. For example, the total admittance of a parallel GCL circuit is Y L ( s ) = G + sC

1 sL

First and Second Order Systems

The admittance functions Y ( s ) =

1 1 ( s − sa )( s − sb )( s − sc ).....( s − sm ) = . Z ( s ) K ( s − s1 )( s − s2 )( s − s3 )......( s − sn )

The impedance functions Z ( s ) = K

545

(28.37)

( s − s1 )( s − s2 )( s − s3 )......( s − sn ) ( s − sa )( s − sb )( s − sc ).....( s − sm )

(28.38)

where K is a scale factor. Note: 1. Poles and zeros must occur in complex conjugate pairs. 2. The admittance function has poles when the impedance function has zeros. 3. Both the functions can be reduced to the ratio of two polynomials in s. 4. Both give the same information but the diagrams are labelled differently. 5. Every point in the complex frequency plane defines an exponential function. 6. The complete plane represents all such functions. 7. Pole-zero diagrams contain the essential information of the impedance function. 8. The poles are marked (X) and the zeros (0). 9. Poles and zeros tell us a great deal about the natural response. Example 28.9

a

Given the circuit in Figure 28.30, determine the poles and zeros of the impedance. If energy is stored in the circuit in the form of an initial voltage Vo on the capacitor, predict the current i that will flow when the switch S is closed. Solution:

R1 = 2Ω i

S

V0

R = 4Ω

If the impedance is zero, a current can exist with no external forcing voltage. I1 is evaluated from the initial data. At the instant the switch is closed. Vo appears across R1 (tending to cause a current opposite to that assumed) and i o = I1eo = I1 = − Vo R1

28.12 FORCED RESPONSE

F

}} jw

−3t

Vo −3t e is the natural response. R1

1 4

Figure 28.30 For Example 28.9

{{

Hence i = −

C=

b

ZR Z C R (1 sC ) Z ( s ) = Z1 + Z R Z C = R1 + R (1 sC ) Z ( s ) = Z1 + Z R + Z C = R1 + R1 + 1 sC ZR + Z C R + 1 sC R sR RC 1+ R + R = R1 + = sR11 RC + R11 + R for R1 = 2 Ω, R = 4 Ω and C = 0.25 F R for R1 = 2 Ω, R = 4 Ω and C = 0.25 F = R1 + R S C + 1 = sRC + 1 R C +1 sRC + 1 2s + 6 S s+3 Z ( s) = 2s + 6 = 2 s + 3 Z (s) = s + 1 = 2 s + 1 s + 1 s = −s1,+t1he denominator zero and Z ( s ) = ∞; therefore s = −1 is a pole. where where s = −1, t he denominator zero and Z ( s ) = ∞; therefore s = −1 is a pole. where s = −3, the numerator is zero and Z ( s ) = 0; therefore s = −3 is a zero. where s = −3, the numerator is zero and Z ( s ) = 0; therefore s = −3 is a zero. The natural current behaviour is defined by s = s1 = −3 or i = I1e

+

2 1 –2

s –1 –2

Figure 28.31 Solution For Example 28.9

Internal energy storage gives rise to natural behaviour; regardless of the manner in which the energy is stored or where it is stored. The natural response of a circuit is determined by the characteristics of the circuit itself. In contrast, the form of a forced response due to an external energy source is dependent of the form of the forcing function.

28.12.1 RESPONSE TO EXPONENTIALS In general, υ = Z(s)i, a special form of Ohm’s law when impedance Z(s) is in ohms. For resistance,

υ R = Ri and Z R ( s ) = υ i = R

(28.39)

546

Electrical Technology

For inductance,

υ L = L di

For a capacitance,

υC = C d υ

Since

Z R ( s )i = RI o e − αt

dt

= sLi and Z L ( s ) = υ i = sL

dt

(28.40)

= sCυ and Z C ( s ) = υ i = 1 sC

(28.41)

where α > 0

υ R = Z L ( s ) = sL = − αL

(28.42)

υ L = Z L ( s )i = − αLI o e − αt

(28.43)

What is the forced response of an RL circuit to an exponential current? In Figure 28.32 υ = υR + υL = ( R − α L) I o e −α t As expected, the response to an exponential function is itself exponential of the same frequency. Since current is possible with no forcing voltage, this is a natural response current. Example 28.10 In the circuit of Figure 28.32 (a), R = 1 kW, and L = 0.1 H. Determine the time constant T of the circuit. 1. The voltage υ for i = 2e–5000t A. 2. The voltage υ for i = 2e–20000t A. i υ

i υR

0

t υ = υR + υL

i = SOe–αt

υ+

υR+

R

υL+

L

(a)

υL

(b)

Figure 28.32 Forced Response of an RL Circuit to an Exponential Current Solution: 1.

0.1 T = LR = s 1000 = 10−4 s = 0.1ms

2.

υ R = Ri = 1000e −5000t V

3.

υ L = L di dt = −2000e −20000t V

28.12.2 Response to Direct Currents ot For the special case of s = 0, the general exponential becomes = i I= I dc . For resistance, inductance and capacitance, oe those impedances are:

Z R (0) = R, υ = RI , V I = R

(28.44)

Z L (0) = 0, υ = L di dt, υ i = 0

(28.45)

First and Second Order Systems

The inductance looks like a short circuit to a d.c. current: When a direct voltage is applied to an inductance carrying no initial current 1 t V i = ∫ υ dx = t L o L 1 t V i = ∫ υ dx = t dυ and the current increases linearly withZtime 0) =o ∞, υ =L V , = 0, υ = ∞ C (L i dt dυ υ Z C ( 0) = ∞ , υ = V , = 0, =∞ i dt

547

(28.46)

(28.47)

A capacitance looks like an open circuit to a d.c. voltage. When a direct current i = 1 flows in an initially uncharged capacitor 1 t I υ = ∫ idx = t C o C

(28.48)

and the voltage increases linearly with time. Example 28.11

R2

Switch S in Figure 28.33 has been closed for a long time. Determine the current i and the voltages across R1, R2 and R3.

R1

S +

i

V R3 Solution: C L We assume that after the specified long time all natural response has died away and only a forced response exists. For a direct forcing voltage, the response is a direct current. To a direct current, inductance L looks like a short circuit, Figure 28.33 For Example 20.11 effectively removing R2 from the circuit, therefore the voltage across R2 is zero. For a direct voltage, capacitance C looks like an open circuit and it can be removed R1 from the circuit; therefore, all the current i flows through R3. For the d.c. case, the circuit looks like Figure 28.34. + I V

R3

Figure 28.34 The Circuit in Figure 28.33 Simplified

RT = R1 + R3 By the voltage-divider rule V1 = R1 I =

R1 R1 + R3

V

and V3 = R3 I =

R3 R1 + R3

Example 28.12 For the circuit in Figure 28.35, V = 5 V, R1 = R2 = 1 kΩ; R3 = R4 = 4 kΩ, L = 1 mH, C1 = C2 = 10 μF. Assume that the switch S has been closed for a long time. R2

Determine 1. Current i; 2. The voltage across each resistor R1, R2, R3 and R4. Solution: L looks like a short circuit whereas C1 and C2 look like open circuits. 1.

i = 5 V 5 kΩ =1 A

2.

υ R1 = 1 mA •1 kΩ =1 V υ R2 = 0 V υ R 3 = 1 mA • 4 kΩ = 4 V υ R4 = 0 V

S i V+

R1

L C1

C2 R4

Figure 28.35 For Example 28.12

R3

548

Electrical Technology

28.13 COMPLETE RESPONSE

S

The natural response of a circuit is due to energy stored i i i R R R in inductances or capacitances. The forced response is + υ + υ I0 produced by external energy sources, such as batteries or L L L generators. In determining natural response, as shown in Figure 28.36 (a), we assumed that energy had been stored (a) Natural (c) Complete (b) Forced by one external source and then that source was removed. In determining the forced response, as seen in Figure Figure 28.36 Components of Circuit Response 28.36 (b), we assumed that a sufficient time had elapsed so that all natural response components had died away or at least had become negligibly small. In general, however, there is a transient period during which the behaviour is not so simple. To determine the current that flows after switch S, as illustrated in Figure 28.36 (c), is closed, we must determine the complete response.

28.14 COMPONENTS OF THE COMPLETE RESPONSE A natural response tends to decay exponentially as the stored energy is dissipated. The forced response continues indefinitely because energy is supplied to make up any losses. Both effects are present in the transient period so that the complete response is a combination of natural and forced responses. Let us assume a linear circuit and a sinusoidal forcing function at all times after closing the switch S in Figure 28.37. di L + Ri = Vm cos ω t dt di =current, Vmt +cos iLf response = I+mRi cos( ω φ i)fω t a solution of this caption is the forced dt di L + Ri = Vm cos ω t di i f dt = I m cos(ω t + φ ) L di + Ri = 0 =ω Vmt +cos = I+mRi cos( φ )ω t iLf dt di for the related homegeneous equations di + Ri L dt t − ( R=L 0 ) = Vm cos ω t == II+x eRi iiLx dt di cos( Lf dt +mRi =ω 0t + φ) −( R L )t −( R L )t = I e I cos( + = iixff dt i m ω t +ω = Ixxm cos( φt) + φ ) + I x e di − ( R=L 0 )t iL = Ii+x eRi a solution is the natural response current, + ixx = I m cos(ω t + φ ) + I x e − ( R L ) t ixf dt di + Ri = 0) t L ω t + φ ) + I x e− ( R L ) t ixf dt =+ iIxx e=− (IRmL cos( iixf =+ The governing equation is satisfied by

−( R L )t iIxx e= I m cos(ω t

S i + V cos wt m

R L

Figure 28.37 Component Responses (28.49)

(28.50)

(28.51)

+ φ ) + I x e− ( R L ) t

i f + ix = I m cos(ω t + φ ) + I x e − ( R L ) t

(28.52)

forced + natural = complete We conclude that the complete response is the sum of the forced and natural responses. Note: These two components can add only if the circuit is linear.

28.15 CHARACTERISTICS OF THE COMPONENTS Each component has a distinctive form and an amplitude to be determined. 1. For any excitation that can be described as an exponential, the form of the forced response is the same as the form of the forcing function; a direct voltage causes a direct current and a sinusoidal current produces a sinusoidal voltage. The amplitude of the forced response is determined by the magnitude of the forcing function and the impedance of the circuit. For direct currents it is Z(0) and for sinusoids it is Z(jω). 2. The form of the natural response is governed by the circuit itself. The form is obtained from the homogeneous differential equation. The amplitude of the natural response is obtained from energy considerations. The amplitude is required to primarily provide for the difference between the actual initial energy storage (in L and C) and that indicated by the forcing function.

First and Second Order Systems

549

28.15.1 Procedure The complete response of any linear two-terminal circuit to exponentials is obtained in the following manner. 1. We can write the appropriate impedance or admittance function. The function Z(s) or Y(s) carries all the information for the integrodifferential equations. The next step is to choose the terminals across which the desired voltage appears (as an open-circuit voltage) or into which the desired current flows (as a short circuit current). 2. We need to determine the forced response from the forcing function and the proper immitance. The form of the forcing function indicates the value of s in Z(s) or Y(s). 3. The next stage is to identify the natural components from poles or zeros of Z(s) or Y(s). Using the poles to obtain possible components natural response voltage and the zeros to obtain current components, we may write these components with undetermined amplitudes. 4. Finally, we need to add the forced and natural responses, evaluate the undetermined constants and obtain the necessary information from initial conditions. The energy distribution just after the switch is closed must be the same as that just before the switch was closed. Note: The word immitance is a combination of impedance and admittance. Example 28.13 The network in Figure 28.38 is an RLC series n-port network with transform impedances marked for each element. Write the driving-point impedance of the network. Solution: Z ( s ) = R + Ls + 1 Cs Ls R

LCs 2 + RCs + 1 Z (s) = Cs

( )

s 2 + R L s + 1 LC Z (s) = L s

Figure 28.38 Network for Example 28.13

28.16 NETWORK FUNCTIONS FOR THE ONE-PORT AND TWO-PORT

2

2 1

1 Cs

Z(s)

3

In Figure 28.39 (a) is shown as a representation of a one-port network. The pair of terminals is customarily connected to an 2′ 1′ 4 energy source which is the driving force of the network, so (b) (c) (a) that the pair of terminals is known as the driving point of the Figure 28.39 (a) One-port Network, (b) Two-port network. Figure 28.39 (b) shows a two-port network. The port designated 1-1' is assumed to be connected to the driving force Network, and (c) A Representation (or the input), and the port 2-2' is connected to a load (as an for the n-port Network output). Figure 28.39 (c) shows a representation of an n-port network for the general case. Note: Two associated terminals are given the name terminal pair or port. The transform impedance at a port is defined as the ratio of the voltage transform to the current transform for a network in the zero state (no initial conditions) with no internal voltage or sources except controlled sources. Thus, Z (s) = V (s) I (s) (28.53) 1

Similarly, the transform admittance is defined as the ratio = Y ( s ) I= (s) V (s) 1 Z (s)

(28.54)

The voltage transform and current transform that define transform impedance and transform admittance must relate to the same port, 1-1' or 2-2' as shown inn Figure 28.40. The impedance or admittance found at a given port is called the driving-point impedance (or admittance).

1 + V1 – 1′

I1

I2

2 + V2 – 2′

Figure 28.40 The Two-port Network with Reference Directions for the Port Voltages and Currents

550

Electrical Technology

Due to the similarity of impedance and admittance, the two quantities are assigned one name: immittance. An immittance is, thus, an impedance or an admittance (a combination of both). The transfer function is used to describe networks which have atleast two ports, and these functions relate the transform of a quantity at one port to the transform of another quantity at another port. Thus, transfer functions which relate voltages to currents have the following possible forms. 1. The ratio of one voltage to another voltage or the voltage transfer ratio. 2. The ratio of one current to another current or the current transfer ratio. 3. The ratio of one current to another voltage or one voltage to another current. It is conventional, although not universal, to define transfer functions as the ratio of an output quantity to an input quantity. In terms of the two-port network of Figure 28.40, the input quantities are V1(s) and I1(s) and the output quantities are V2(s) and I2(s) (see Table 28.1). Table 28.1 Transfer Functions for the two-port Denominator

Numerator V2(s)

I2(s)

V1(s)

G12(s)

g12(s)

I1(s)

Z12(s)

α12(s)

Example 28.14 Figure 28.41 shows a series RL network shunted by a capacitor. Find the driving point impedance. Solution: 1 Z ( s) = Z(s) ( s + 1 ( R + Ls )

=

1 s+R L C s 2 + ( R L) s + 1 LC

1 Cs

R Ls

Figure 28.41 For Example 28.14

In the driving-point impedance function, the numerator is of the first degree and the denominator is of the second degree.

S UM M A RY 1. Forced response can be maintained indefinitely by the continual input of energy. 2. When the natural response becomes negligible, it is called the transient period. 3. The homogeneous equation contains only terms related to the circuit itself. 4. The quantity t = 0 – refers to the instant just before the switching action occurs. 5. The quantity t = 0 + refers to the instant just after the switching action has occurred. 6. Resistor current is proportional to the voltage. 7. Capacitor current is proportional to the rate of change of voltage. 8. The voltage on a capacitor cannot be changed instantaneously. 9. The term first order system refers to a system involving only a single energy storage element. 10. A step of current flows when a switch is thrown, suddenly applying a direct voltage to a resistance.

11. A pulse of current flows when a switch is turned ON and then OFF. 12. A sawtooth current increases in a linear manner with time and then resets. 13. A decaying exponential current flows if energy is stored in the electric field of a capacitor and allowed to leak off through a resistor. 14. A sinusoidal voltage is generated when a coil is rotated at a constant speed in a uniform magnetic field. 15. Exponential and sinusoidal waveforms are easy to generate and also easy to analyze because of their simple derivatives and integrals. 16. The presence of a second energy-storage element increases the order of the highest derivative appearing in the homogeneous equation. 17. The values of s1 and s2 may be real, complex, or purely imaginary. 18. The discriminant determines the nature of the roots. 19. A sinusoidal function of time with an exponentially decaying amplitude is called a damped sinusoid.

First and Second Order Systems

20. S has the units of frequency; the s plane is referred to as the complex frequency plane. 21. The ratio of voltage to current is defined as the impedance Z. 22. For R it is R ohms, for L it is sL ohms, and for C it is 1/sC ohms. 23. An uncharged capacitor initially acts like a short circuit. 24. An unfluxed inductor initially acts like an open circuit. 25. In the steady-state d.c. cases, all the voltages and currents have settled down to constant values and there are no changes. 26. A steady state condition for one part of a circuit will serve as the basis for the initial condition for a different part of a circuit. 27. Admittance is particularly useful in analyzing circuits that contain elements connected in parallel. 28. Admittances in parallel can be added directly just as conductances in parallel.

551

29. Internal energy storage gives rise to natural behaviour. 30. Forced response due to an external energy source depends on the form of the forcing function. 31. Internal energy storage gives rise to natural behaviour regardless of how it is stored and where it is stored. 32. Poles and zeros tell us a great deal about the natural response. 33. The complete response is obtained by adding the forced and natural responses. ic = i f + in 34. Two associated terminals are given the name terminal pairs or port and are the points of entry into the network. 35. An immittance is an admittance or an impedance; a combination of both. 36. The transfer function relates the transform of a quantity at one port to the transform of another quantity at another port.

M U LT IP LE C H O I C E Q UE S TI O NS ( M C Q ) 9. The key property of an exponential function is

1. Capacitor current is proportional to (a) Voltage (b) Rate of change of voltage

(a) Its time derivative is also an exponential (b) Its time derivative is not an exponential

2. The energy stored in a capacitor

10. Admittance is particularly useful in analyzing circuits that contain elements

(a) Can be changed instantly (b) Cannot be changed instantly

3. A direct or continuous current is supplied by (a) A battery

(b) An alternator

(c) A d.c. generator

4. Exponential and sinusoidal waveforms are (a) (b) (c) (d)

Easy to generate but difficult to analyze Easy to generate and analyze Difficult to generate and difficult to analyze Difficult to generate and analyze

(a) (b) (c) (d)

Connected in series Connected in series-parallel Connected in parallel-series Connected in parallel

11. Internal energy storage gives rise to (a) Natural response (b) Forced response (c) Complete response

5. First-order systems have

12. Forced response is produced by

6. In a decaying exponential the current decreases continually but

13. Complete response is the

7. The presence of a second energy storage element

14. The form of the forced response is

8. The character of the natural response of a second-order system is

15. Natural response is

(a) A single energy storage element (b) More than one energy-storage elements

(a) Never becomes zero

(b) Ultimately becomes zero

(a) Does not affect the order of the highest derivative (b) Increases the order of the highest derivative by one

(a) Determined by the roots of the characteristic equation (b) Not determined by the roots of the characteristic equation

ANSWERS (MCQ) 1. (b) 2. (b) 3. (a) and (c) 5. (a) 6. (a) 7. (b) 8. (a)

4. (b) 9. (a)

(a) Internal energy (b) External energy sources

(a) Sum of forced and natural response (b) Difference of forced and natural response (a) Governed by the form of the forcing function (b) Not governed by the form of the forcing function (a) Governed by the circuit itself (b) Not governed by the circuit itself

10. (b) 11. (a) 12. (a) 14. (a) 15. (a).

13. (a)

552

Electrical Technology

CON V E NTI O NA L Q UE S TI O NS ( C Q ) 1. The presence of a source function in a network causes a response to occur in the various parts of the network. Distinguish between the source function and the forcing function in producing the network response. 2. What is meant by the forced solution of a linear network? How is it characterized generally? 3. In a linear network identify the general form of the network response function for an exponential source. Repeat for a sinusoidal source. 4. Explain why the transient solution of a linear network is always characterized by the exponential function. R1 = 4 Ω

5. The initially relaxed circuit of Figure 28.42 is excited at t = 0 by the source shown. Determine all element voltages and currents at t = 0+. 6. The switch in the circuit of Figure 28.43 is in position 1. As a result of some excitation vs(t), the initial values of the voltage and current are as shown. At t = 0, the switch is thrown to position 2. Determine all element voltages and currents at t = 0+. 7. Determine the steady state voltage and current for each element of the circuit in Figure 28.44.

C1

C

R=2Ω

2 1

t=0 20 V

+ –

L

+ –

R2 = 6 Ω

16 V

vs(t )

Figure 28.42 For CQ 5

24 V

C2

+ 6V –

Figure 28.43 For CQ 6

R1= 5 Ω

+ –

+ –

C1

L2

L1

C2

R2 = 3 Ω

Figure 28.44 For CQ 7 ANSWERS (CQ) 5. iR1 = i C = iR 2 = 2 A; υ C = 0, i L = 0, υ R1 = 8 V υ L = υ R 2 = 12 V 6. υ C1 = 0, υ C 2 = 6 V, i L = 2 A, VR = 10 V, i C1 = 5 A, i c 2 = 3 A

7. υ L1 = υ L 2 = i C1 = i C 2 = 0, i R1 = i L1 = i L 2 = i R 2 = 3 A, υ R1 = 15 V, υ R 2 = 9 V, υC1 = υC 2 = υ R 2 = 9 V. υ R1 = 15 V, υ R 2 = 9 V, υC1 = υC 2 = υ R 2 = 9 V.

L

2A

29

Laplace Transform OBJECTIVES In this chapter you will learn about:    Define the Laplace transform of a function    Use the table of Laplace tra nsforms    Use  partial  fractions  to  deduce  inverse  Laplace transforms     Deduce  expressions  for  compo nent  and  circuit impedances    Deduce  Kirchhoff’s  law  equ ations  in  the  s  plane  for  deter mining  the  response  of  networks given initial conditions    Predict the circuit response of a network  Classify poles and zeros    Analyse a circuit with Laplace tran sform

Time domain Resistor ν (t) = R i(t)

Frequency domain ν (s) = R I(s) I(s)

i(t) + ν (t) −

+ V(s) −

R

Inductor di ν (t) = I dt i(t)

ν (s) = sL I(s) − Li(0−)

I(s) +

+ ν (t) L − Capacitor dν ν(t) = C dt i(t) + ν (t) −

C

Z(s) = R

V(s) −

Z(s) = sL + −Li(0−) −

I(s) = 1 V(s) R I(s) + V(s) −

Y(s) = 1 R

I(s) = V(s) + i(0 ) s sL I(s) −

+ V(s) Y(s) = 1 sL −

i(0−) s

ν − I(s) = sC V(s) − Cν (0−) V(s) = I(s) + (0 ) s sC I(s) I(s) + + + Cν (0−) Z(s) = 1 sC V(s) Y(s) = sC V(s) − − − + ν (0 ) − s − s-domain impedance

29.1 INTRODUCTION The Laplace transform is a so-called operational method for solving differential equations by algebraic methods. Developed  by the French mathematician Pierre-Simon de Laplace, this transform is an important part of the study of applied mathematics. It is frequently necessary to determine the solutions of linear differential equations. In this way, information is obtained  about the dynamic and forced responses of a circuit, device, or system to a driving function. The Laplace transformation method for solving differential equations offers a number of advantages over the classical methods.  For example,   1.  The solution of differential equations is routine and progresses systematically.   2.   The method gives the total solution—the particular integral and the complementary function—in one single operation.   3.  Initial conditions are automatically specified in the transformed equations.  Furthermore, the initial conditions are incorporated into the problem as one of the steps rather than as the last step.

29.2 FLOWCHART FOR MATHEMATICAL PROCEDURE It  is  the  nature  of  a  transform  to  simplify  the  analytical  procedure  of  a  problem.  Specifically,  the  Laplace  transform  is a mathematical tool for transforming functions. The logarithm is an example of a transformation that we use so often.  Logarithms greatly simplify operations such as multiplication, division, extracting roots, and raising quantities to powers.  Rather than multiply two numbers together, we transform these numbers by taking their logarithms. These logarithms are  added (or subtracted in the case of division). The resulting sum itself has little meaning. However, if we perform an inverse  transformation (if we find the antilogarithm), then we have the desired numerical result. The direct process looks more straight  forward, but our experience dictates that the use of the logarithm often saves time.

554 Electrical Technology A flowchart of the operation of using logarithm to find a product or quotient is illustrated in Figure 29.1(a). The individual  steps are: (1) find the logarithm of each separate number; (2) add or subtract the numbers to obtain the sum of logarithms; and  (3) take the antilogarithm to contain the product or quotient. This can be to an extent compared with direct multiplication or  division, yet we use logarithm to advantage. The  flowchart  idea  may  also  be  used  to  illustrate  what  we  will  do  in  using  the  Laplace  transformation  to  solve  a  differential equation. The flowchart for Laplace transformation is shown in Figure 29.1(b) with a block corresponding to  every block of the logarithm flowchart considered above. The steps will be as follows: (1) The first step is to start with  an integro differential  equation  and  find the corresponding Laplace transform. This is a mathematical process, but there are tables of transforms just as there are tables of logarithms (see Table 29.1). (2) In the second stage of the transform  is manipulated algebraically after the initial conditions are inserted. The result is a revised transform. (3) The final step  involves performing an inverse Laplace transformation for a solution. In this step, we also can use a table of transforms  just as we use the table of antilogarithms in the corresponding step for logarithms. 

Numbers

Logarithm

Direct multiplication or division

Product or quotient

Logarithms of numbers

Addition of numbers

Antilogarithm

Sum of logarithms

(a)

Integrodifferential equation

Laplace transformation

Classical solution

Solution

Initial conditions Transform

Algebraic manipulation Inverse Laplace transformation

Revised transform Frequency domain

Time domain (b)

(c)

Figure 29.1 Comparison of the Logarithms and the Laplace Transformation

Laplace Transform 555

The flowchart reminds us that there is another way, namely, the classical solution. It looks more direct (and sometimes  it is for simple problems). For complicated problems, an advantage will be found for the Laplace transformation, just as an advantage is found for the use of logarithms.

Table 29.1 Table of Transforms f (t)

F(s)

1 s 1 s2 1 sn

u(t) t t n −1 n = integer (n − 1) !′ eat

1 (s − a)

teat

1 ( s − a)2 1

1 t n −1e at (n − 1)!

( s − a )n 1

1 (e at − ebt ) a−b

(s − a) (s − b)

e − at (b − a )(c − a ) +

− bt

1

− ct

e e + (a − b)(c − b) (a − c)(b − c)

(s + a) (s + b) (s + c) −a s (s − a)

1 − e + at

1 (s + ω 2) s (s 2 + w 2 )

1 sin ω t ω

2

cos ω t

ω2 s ( s2 + ω 2 )

1 − cos ω t sin(ω t + θ ) cos(ω t + θ ) e −∝t sin ω t e −∝ t cos ω t sinh α t cosh α t

(s sin θ + ω cos θ ) (s 2 + ω 2 ) (s cos θ − ω sin θ ) (s2 + ω 2 ) ω (s + α )2 + ω 2 (s + α) (s + α )2 + ω 2 α 2 (s − α 2) s 2 (s − α 2 )

All f (t) should be thought of as being multiplied by u(t), i.e., f (t) = 0 for t  0 are shown in Figure 29.5. Both sine and cosine are referred to as sinusoidal  functions, or sinusoids. The following definitions relating to the sinusoidal functions are important. sin ω t 1

t

T

cos ω t 1

t

T

Figure29.5 Forms of Sine and Cosine Functions ω = angular frequency in radians/second (rad/s) = 2π f = 2π/T f = cyclic frequency in Hertz = 1/T T = Period in seconds (s) The basic sinusoidal functions have an amplitude (peak value) of unity. When a different amplitude is required, the  function is multiplied by that factor. For example, the voltage υ (t) = 10 sin 1000 t describes a sinusoid with an amplitude  or peak value of 10 V, an angular frequency ω = 1000 rad/s, a cyclic frequency f = 1000/2 π = 159.15 Hz and a period  T = 1/159.15 = 6.283 ms. The Laplace transform of cos ωt is

  Similarly, it can be proved that for f (t) sin wt

(sin t ) =

s   s + ω2 2

Example 29.1 Use Table 29.1 to determine the Laplace transforms of the following waveforms (see Figure 29.6). 1. A step voltage of 10 V which starts at time t = 0 2. A step voltage of 10 V which starts at time t = 5 s 3. A ramp voltage which starts at zero and increase at 4 V/s 4. A ramp voltage which starts at time t = 1 s and increases at 4 V/s

(29.8)

(29.9)

Laplace Transform 559 V 10

V 4

0

0

t

1

t (c)

(a) V 10

V 4 5

0

0

t

2

1

(b)

t

(d)

Figure 29.6 For Example 29.1 Solution: (10 ) = 10 (1)

1.   

 

 

 

 

 

 1  10 = 10   = s s

2. A step function of 10 V which is delayed by T = 5s is given by  e − st   e −5 s  10   = 10    s   s  =

10 −5 s e s

    This is, in fact, the function starting at t = 0 multiplied by e–sT, where, T is the delay in seconds. 3. The Laplace transform of the unit ramp   (t) = 1/s2. Hence, the Laplace of a ramp voltage increasing at 4 V/s given by 4 (t) = 4/s2 4. For a delayed function, the Laplace transform is the undelayed function multiplied by e–sT, where T in this case is 1s. Hence, the Laplace transform is given by (4/s2)e–s Example 29.2 Using the results provided in Table 29.1, determine the Laplace transform of f (t) = 10 + 3e–4t + 12 sin 3t + 4e–2t cos 5t. Solution: 10 3 12(3) 4( s + 2 ) F ( s) = + + 2 + 2 s s + 4 s + (3) ( s + 2 )2 + 52 = Example 29.3 Find the Laplace transform of  Solution: 1.

4( s + 2 ) 10 3 36 + 2 + + 2 s s + 4 s + 9 s + 4s + 29

560 Electrical Technology 2. 

 4e3t co s 5t  = 4  e3t cos 5t    4( s − 3) ( s − 3) = 4 = 2 2 2  − 3 + 5 s − 6s + 9 + 25 ( s ) ( )   =

4( s − 3 ) s 2 − 6s + 34

Example 29.4

V 8

Determine the Laplace transform of the following waveforms (Figure 29.7): 1. An impulse voltage of 8 V which starts at time t = 0  2. An impulse voltage of 8 V which starts at time t = 2s 3. A sinusoidal current of 4 A and angular frequency 5 rad/s which starts at time t = 0. Solution: 1. An  impulse  is  an  intense  signal  of  very  short  duration. This  function  is  also  known as the Dirac function.     From Table 29.1, the Laplace transform of an impulse starting at time t = 0 is  given by   (δ) = 1, hence, an impulse of 8 V is given by 8 (δ) = 8  (Figure 29.7(a)) 2. Delaying the impulse by 2s involves multiplying the undelayed impulse by e–ST where, T = 2s     Hence, the Laplace transform of the function is given by          8e–2S  (Figure 29.7(b)) ω 3. Again from Table 29.1,  (sin t ) = 2  and when the amplitude is 4 A  ω2 s + and ω = 5, then 5   (4 sin t ) = 4  2   s + 52  20 = 2 s + 25

0

t (a)

V 8

0

2

t (b)

i 4

0

π 5

2π t 5

–4

 (Figure 29.7(c))

(c)

29.8 LAPLACE TRANSFORM OPERATIONS

Figure 29.7 For Example 29.4

For a given functional operation in the time domain (e.g., differentiation), the operation pair indicates the corresponding operation that is performed in the s-domain. The most common operation pairs of interest are listed in Table 29.2.  The column designated f (t) indicates an operation that can be performed in the time domain. The column designated f (s)  indicates the corresponding operation that must be performed on the Laplace transform of the original function. Table 29.2 Laplace Transform Operations Encountered in Circuit Analysis f (t)

f ¢(t ) t

F(s)

sF ( s ) − f (0)

∫0 f (t )dt

F (s) s

e −α t f (t )

F (s + α )

f (t − T )u (t − T )

e-sT F ( s )

f ( 0) lim s f (t )

t →∞

* Poles of sF(s) must be in left-hand half-plane

lim s F ( s )

s →∞

lim s F ( s )* s®0

Laplace Transform 561

29.8.1 Derivative of a Time Function

[ f ′(t ) ] = sF ( s) −

(29.10) f ( 0)   This operation indicates that to find the Laplace transform of the derivative of a function, the transform of the original function is multiplied by s and the initial value of the time function is subtracted.  If we momentarily consider time functions with an initial value of zero, that is, f (0) = 0, we can say that differentiation  in  the  time  domain  corresponds  to  multiplication  by  s in the s-domain.  The  differentiation  process  is  replaced  by  the  simpler process of multiplication by s.

29.8.2 Integral of Time Function  t f (t )dt  = F ( s )   (29.11)  ∫ 0  s This operation indicates that to find the Laplace transform of the definite integral of a time function (from t = 0 to an  arbitrary  time  t),  the  transform  of  the  original  function  is  divided  by  s.  Integration  in  the  time  domain  corresponds  to  division by s in s-domain. The integration process is replaced by the simpler process of division by s.

29.8.3 Multiplication by e–  at The theorem also states that if the original function is multiplied  by e – ∝ t then the transform of the product function is  obtained by replacing s in the transform of the original function by s + µ.  e∝ t f (t )  = F ( s + ∝)   (29.12) The major benefit of this theorem is to simplify the correlation between transforms of functions with and without exponential factors.

29.8.4 Initial Value Theorem The initial value theorem states that  limit t →0

for example, If  f (t) = υ = Ve−t/CR and if, say. V = 10 V and CR = 0.5s, then

[ f (t ) ] =

limit  s s →∞

{ (t )} 

f (t) = υ = 10e−2t 1  [ f (t ) ] = 10  s + 2   s

[ f (t ) ] = 10 

s   s + 2

∞ From the initial value theorem, the initial value of f (t) is given by  10  ∞  +

 = 10(1) = 10 2 

29.8.5 Final Value Theorem ∞  The final value theorem states that  10   = 10(1) = 10 ∞ + 2 limit [ f (t ) ] = limit  s { f (t ) }  t→∞ s→0 In the above example, f (t) = 10−2t, the final value is given by  0  10  =0  0+2  Note: 1. The initial and final value theorem can often considerably reduce the work of solving electrical circuits. 2. These theorems are used in pulse circuit applications where the response of the circuit for small periods of time, or the  behaviour immediately after the switch is closed, are of interest. 3. The final value theorem is particularly useful in investigating the stability of the system and is concerned with the  steady state response for large value of time, i.e., after all the transient effects have died away.

562 Electrical Technology Example 29.5 Starting with the Laplace transform of the sine function, derive the Laplace transform of the cosine function by employing  the differentiation transformation operation. Solution: ω f (t) = sin wt whose Laplace transform is  F ( s ) = 2 s + ω2 Differentiating both sides first f '(t) = w cos wt The Laplace transform of f '(t) can be expressed as sω [ f ’(t ) ] = sF ( s) − F (0) = 2 2 − 0 s +ω

[ f ’(t ) ] = α [ ω cos ω t ] = ωα [ cos ω t ] [ cos ω t ] =

s s2 + ω 2

Example 29.6 Starting with the Laplace transform of the constant 1, derive the Laplace transform of the linear function t making use of  the integration transform operation. Solution: f (t) = 1 whose Laplace transform is F(s) = 1/s Integrating both sides of the equation yields t

t

∫0 f (t )dt = ∫0 (1)dt = t The Laplace transform of the integral function can be expressed as  t f (t )dt  = F ( s ) = 1  1  = 1    ∫ 0  s s  s  s2 However, 

 

 

 

 

 t f (t )dt  =  ∫ 0 

(t ) and (t) =

1 s2

29.9 INVERSE LAPLACE TRANSFORMATION To determine the Laplace transform of common time functions, tabulated results such as those given in Table 29.1 may be  readily employed. In a few simple cases, these tables can also be used to determine inverse Laplace transforms directly.  In the majority of cases, however, transforms usually appear in forms in which these tables may not be directly applied  without considerable reduction. The process of inverse Laplace transformation consists of finding the time function corresponding to a given s domain  function. Most transforms of interest in circuit analysis turn out to be expressible as ratios of polynomials in the Laplace variable s, such as F ( s) =

N (s) D( s)

(29.13)

Where, N(s) is a numerator polynomial of the form N ( s ) = an s n + an −1s n −1 + an − 2 s n − 2 + ..... + an and D(s) is a denominator polynomial of the form  m

D( s ) = bm s + bm −1s

m −1

+ bm − 2 s

(29.14) (29.15)

m−2

+ ..... + bm

(29.16)

29.9.1 Poles and Zeros The  integer  n  in  the  numerator  of  the  polynomial  is  the  degree  or  order  of  the  numerator  polynomial  and  m is the corresponding degree or order of the denominator polynomial. A fundamental theorem of algebra states that the number of  roots of a polynomial is equal to the degree of the polynomial. This means that N(d) is assumed to have n roots, and D(s)  is assumed to have m roots.

Laplace Transform 563

The roots of the numerator polynomial N(s) are called the zeros of F(s), and the roots of the denominator polynomial  D(s) are called the poles of F(s). The value of the polynomial is zero when the variable s assumes the value of any one  of its roots. Since N(s) is in the numerator, F(s) = 0 when N(s) = 0. However, D(s) is in the denominator and the function  F(s) increases without limit (i.e., approaches infinity) when D(s) = 0. jω The form of the s-plane is illustrated in Figure 29.8. The s-plane  is  a  rectangular  coordinate  system  from  which  certain  properties  s-plane of the transfer function may be visualized. The concept is based on  representing s as a complex variable of the form  s = σ + jω. σ where, σ is the real part of s and it can be interpreted as a damping  factor. ω is the imaginary part of s and it can be interpreted as angular  frequency. The quantity s is often denoted as the complex frequency. The horizontal axis is the σ-axis, and the vertical axis is the jω-axis.  The first axis is often called the real-axis and the latter axis is called the  imaginary axis. Finite poles are denoted by xs and finite zeros by 0s. Figure 29.8 Form of the s-plane Poles  and  zeros  are  both  critical  frequency. At  poles,  as  has  been  illustrated in Figure 29.9, the network function becomes infinite. At zeros, the network function becomes zero. At other finite  frequencies, the network has a finite non-zero value. A three-dimensional representation of the magnitude of the transfer function  as a function of complex frequency is given in Figure 29.9. Pole Pole

Magnitude of network function +jσ

Zero –jω

Figure 29.9 The Magnitude of a Network Function Shown as a Function of Complex Frequency with Two Poles and One Zero The impedance function in three dimensions has the appearance of a tent pitched on the s-plane. The height of the tent  magnitude of Z becomes very great (approaches infinity) for particular values of s appropriately called poles. The tent  touches the ground at particular values of s called zeros. The locations of the poles (X) and zeros (0) tell us a great deal  about the natural response. The order of a circuit is the total number of non redundant inductors and capacitors after the circuit is reduced to its  simplest form. Thus, simple series or parallel circuits of more than one inductance or capacitance are each equivalent to  only one inductance or capacitance.

29.9.2 Classification of Poles The poles of a particular transform function may be classified in two particular ways. The first way is to classify a given pole  as either real, imaginary or complex. A real pole is an ordinary real number, either positive or negative. An imaginary pole is  one preceded by √−1, the basis for imaginary numbers. A complex pole is one having both a real part and an imaginary part. The  second  way  of  classifying  is  their  order,  which  is  the  number  of  times  a  root  is  repeated  in  the  denominator  polynomial. The most common case is that of the first-order root, in which the root appears only once. Higher-order roots  are referred to as multiple-order roots.

564 Electrical Technology Example 29.7 10 15 20 = 2 + Determine the inverse transform of  F ( s ) = s s+3 s Solution: f (t) is readily written as           ( f ) = 10 + 15t + 20e−3t Example 29.8 8s + 30 Determine the inverse transform of  F ( s ) = 2 s + 25 Solution: F(s) can also be written as     s 5 +6 2 F (s) = 8  2 2  2   s + (5)   s + (5)  = 8 cos 5t + 6 sin 5t Example 29.9 Find the following inverse Laplace transforms. 1. Solution: 1.   

 

−1

 1   s2 + 9  =   =

2.   

−1

−1

 1  = 2  s + 9

2.

  1 1 =  2 2   s + (3)  3

−1

−1

 5  =   3s − 1 

 3   s 2 + 32   

1 sin 3t 3

 

Example 29.10 Determine the inverse transform of  F (s) =

2 s + 26 s + 6 s + 34 2

Solution: The first step is to write the denominator polynomial in the form  ( s + ∝) 2 + ω 2 ,  which is accomplished by the process of  completing the square. s 2 + 6 s + 34 = s 2 + 6 s + 9 + 34 − 9 = ( s + 3) 2 + 25 = ( s + 3) 2 + (5) 2 ; a = 3 and ω = 5 2 s + 26 2( s + 3) 20 F ( s) = = + 2 2 2 2 2 ( s + 3) + (5) ( s + 3) + (5) ( s + 3 ) + (5) 2   2( s + 3) 5  +4 = 2 2 2 2   ( s + 3) + (5)  ( s + 3) + (5)  f (t ) = 2e−3t cos 5t + 4e−3t sin 5t

Laplace Transform 565

29.10 USE OF PARTIAL FRACTIONS FOR INVERSE LAPLACE TRANSFORMS Sometimes the function whose inverse is required  is not recognizable as a standard type as those listed in Table 29.1.  In such cases, it may be possible by using partial fractions to resolve the function into simpler functions which may be  inverted on sight. 2( s − 3) For example, the function  F ( s ) =  cannot be inverted on sight, using partial fractions s ( s − 3) 2( s − 3) A B A( s − 3) + Bs ≡ + = s ( s − 3) s s−3 s ( s − 3)  

 

 

    2(s−3) = A(s−3) + Bs

Letting   

   

 

 

s = 0 gives: A = 1

Letting   

   

 

 

s = 3 gives: B = 1

2( s − 3) s ( s − 3)

Hence,    

1 s

≡ +

1 and s− 3

−1

  

 

 2s − 3  s ( s − 3)

≡

−1

  

From 

1 1 = 1 + e3t  + s s − 3 

Example 29.11 −1

  

Determine 

 4s − 5 2 s − s − 2

Solution: 4s − 5 4s − 5 A B = = + s − s − 2 ( s − 2)( s + 1) s − 2 s + 1 2

Hence,   

   

 

 

4s − 5 = A(s + 1) + B(s − 2) 

when,   

   

 

 

    s = 2, A = 1; 

when,   

   

 

 

    s = −1, B = 3

−1

 4s − 5   s2 − s − 2  =  

−1

3   1  s − 2 + s + 1  

f (t ) = e 2t + 3e −t

29.11 INVERSE LAPLACE TRANSFORMS AND THE SOLUTION OF DIFFERENTIAL EQUATIONS Using integration by parts, it may be shown that 1. For the first derivative [ f ' (t ) ] = s or

 dy   dx  = s  

[ f (t ) ] −

f (0)

{ y } − y(0)

 

(29.17)

where, y (0) is the value of y at x = 0

2. For the second derivative or  dy  where, y' (0) is the value of    at x = 0.  dx 

' '

566 Electrical Technology Procedure to solve differential equations by using Laplace Transforms consists of the following stages. 1. Take the Laplace Transform of both sides of the differential equation by applying the formulae for the Laplace transform of derivates. 2. Put in the given initial conditions, i.e., y (0) and y′ (0). 3. Rearrange the equation to make  (y) the subject. 4. Determine y by using, where necessary, partial fractions, and taking the inverse of each term. Example 29.12 Use Laplace transform to solve the differential equation 2

d2y dy +5 − 3 y = 0, dx dx 2

given that where x = 0, y = 4 and dy/dx = 9. Solution:  d2y   dy  1.      2  2 +5  −3  dx   dx 

[ y ] = [0]

2  s 2 ( y ) − sy (0) − y (0)  + 5 [ s ( y ) − y (0) ] − 3 ( y ) = 0 2.   

 

  Thus,  

y(0) = 4

2  s 2 ( y ) − 4 s − 9  + 5 [ s ( y ) − 4 ] − 3 ( y ) = 0

i.e., 

( y ) − 8s − 18 + 5s y − 20 − 3 ( y ) = 0

2s 2

3. Rearranging gives  

(2 s 2 + 5s − 3)α ( y ) = 8s + 38 ( y) =

i.e., 4.   

 

Let   

 

y′(0) = 9

and

 

y=

−1

8s + 38 2 s 2 + 5s − 3

 8s + 38   2 s 2 + 5s − 3   

A B A( s + 3) + B (2 s − 1) 8s + 38 8s + 38 = = + = 2 (2 s − 1)( s + 3) 2s + 5s − 3 (2s − 1)( s + 3) 2s − 1 s + 3

Hence,   

   

 

8s + 38 = A(s + 3) + B(2s − 1)

Where,   

 

 

s = 1/2, A = 12 and when s = −3, B = −2

Hence,    

 

= Hence,   

   

 

−1

y = 6e

1

 12  2 s − 1  2

2x

(

)

 −  

−1

 2   s + 3  

− 2e − 3 x

29.12 CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS The use of Laplace Transform operations allows the derivative and integral relationships to become converted into algebraic  relationships. This is achieved by transforming voltages and currents to the s-domain. The block shown in Figure 29.10 (a) is assumed to contain only passive circuit components. The instantaneous voltage  across the circuit is ν(t) and the current flowing into the circuit is i(t). With the exception of a resistive circuit, there is in  general, no simple algebraic relationship between the voltage and current because of the derivate/integral relationships involved. 

Laplace Transform 567 i(t)

I(s)

+ v(t)

+ Passive RLC circuit

–

V(s)

Z(s)

–

(a)

(b)

Figure 29.10 Passive RLC Circuit and its Transform or s-domain Impedance

29.12.1 Transform Impedance and Admittance Let V(s) = [ν(t)] represent the Laplace transform of the voltage, and let [(s) =  [i(t)] represent the Laplace transform of  the current, A function Z(s) called the s-domain or transform impedance is defined as V ( s) Z (s) = (29.18a) I ( s) The reciprocal of the transform impedance is called the transform admittance. Y(s) and is defined as I (s) 1 Y ( s) = = Z ( s) V ( s)

(29.18b)

The use of transform impedances and admittances permits circuit analysis techniques to extend to complex circuits. 

29.12.2 Resistance The time-domain voltage-current relationship for a resistance is u(t ) = Ri (t ) Applying Laplace transform to both sides of the equation results in  V(s) = RI(s) Thus, the transform impedance for the resistance is V (s) Z (s) = =R (29.19a) I (s)

i(t)

I(s)

+ v(t) –

+ R

V(s)

R

–

The s-domain  impedance  of  a  resistance  is  simply  the  value  of  the  resistance and no change is required in converting the resistance to the  Figure 29.11 Conversion from Timedomain to s-domain s-domain. This is shown in Figure 29.11. Resistance

29.12.3 Capacitance Consider an uncharged capacitor C, in which the voltage current relation is 1 t u(t ) = ∫ i (t )dt C 0 Applying Laplace transform to both sides of the equation gives t 1  t i (t )dt  [ (t ) ] =  1C ∫0 i(t )dt  =  i(t ) I(s)   C  ∫ 0 1 + + V (s) = I (s) and  sC The  integral  relationship  in the  domain, thus,  becomes  an  algebraic  v(t ) V(s) C equation in the s-domain. This has been shown in Figure 29.12. – – The impedance Z(s) for the capacitor is 1 V ( s) Z (s) = = (29.19b) Figure 29.12 Conversion from TimeI ( s) sC domain to s-domain Capacitance The s-domain impedance of a capacitance is 1/sC.

1 sC

568 Electrical Technology

29.12.4 Inductance Consider an unfluxed inductance L, in which the voltage-current relationship is u(t ) = L

di (t ) dt

Applying Laplace transform to both sides of the equation gives

[

 di (t )   di (t )   L dt  = L  dt      V ( s ) = sLI ( s ) (t ) ] =

The impedance Z(s) for the inductance is  Z (s) =

V ( s) = sL I ( s)

(29.19c)

This has been shown in Figure 29.13. The s-domain impedance of an inductance is sL. i(t )

I(s)

+

+ L

v(t ) –

sL

V(s) –

Figure 29.13 Conversion from Time-domain to s-domain Inductance Example 29.13 The following components are shown on the left in Figure 29.14: (1) a 1 kΩ resistor, (2) a 0.5 mF uncharged capacitor, and  (3) a 29 mH unfluxed inductor. Show the corresponding s-domain impedance models. Solution: Given component

s - domain impedance

1k Ω

0.5 µ F

30 mH

1000

2 X 106 s

0.03s

Figure 29.14 For Example 29.13

(a)

(b)

(c)

Laplace Transform 569

Example 29.14 Show two s-domain models for the charged capacitor shown in Figure 29.15(a). Solution: Thevenin’s form

Norton’s form •

5 X 106 s

+ 0.2 µ F

5 X 10–6 s

60V –

12 X 10–6

+ –

60 s

• (a)

(b)

(c)

Figure 29.15 For Example 29.14 Example 29.15 Show two s-domain models for the fluxed inductor, as shown in Figure 29.16. Solution: Norton’s form •

Thevenin’s form

0.05s 50 mH

0.4 A

0.4 s

0.05s

– +

0.02

• (a)

(b)

(c)

Figure 29.16 For Example 29.15

S UM M A RY 1.  I  t is the nature of a transform to simplify the analytical  procedure of a problem. 2.   Most time functions of interest in circuit analysis are voltages and currents. 3.   It is convenient in  studying    electric  circuits to  sub-divide  time into three parts: t = 0−, t = 0, and t = 0+. 4.   Once  a  transform  is  found,  it  can  be  tabulated  for  future use. 5.   The Laplace transform of constant times a function is the constant times the transform of the function. 6.   The Laplace transform of the sum of two functions is the sum of the respective Laplace transforms. 7.   One  of  the  most  important  mathematical  functions  is  the exponential function. 8.   For a delayed function, the Laplace transform is the undelayed function multiplied by e−st. 9.   To find the transform of the derivative of a function,  the  transform  of  the  original  function  is  multiplied 

by s  and  the  initial  value  of  the  time  function  is  subtracted. 10.   Integration in the time domain corresponds to division  by s in the s-domain. 11.   The  number  of  roots  of  a  polynomial  is  equal  to  the  degree of the polynomials. 12.   The roots of the numerator polynomial N(s) are called  the zeros of F(s). 13.   The  roots  of  the  denominator  polynomial  D(s)  are  called the poles of F(s). 14.   The  order  of  a  circuit  is  the  total  number  of  nonredundant inductors and capacitors after the circuit is  reduced to its simplest form. 15.   The s-domain impedance of a resistance is simply the  value of the resistance. 16.  The s-domain value of a capacitance is 1/sC. 17.  The s-domain value of an inductance is sL.

570 Electrical Technology

M U LT IP LE C H O I C E Q UE S TI O NS (M C Q ) 1.   Most time functions of interest in circuit analysis are     

  (a) Voltages   

(b) Currents 

  (c) Power 

(d) Impedances

 

2.  I  t  is  convenient  in  studying  electric  circuits  to  subdivide time into    

  (a) Two parts      (c) Four parts   

(b) Three parts (d) Five parts

3.   The Laplace transform of the sum of two functions is         

       

(a) The sum of their respective transforms (b) The product of their respective transforms (c) The difference of their respective transforms (d) The quotient of their respective transforms

4. The s-domain impedance of a capacitance is   

  (a) C/s (c) Cs

(b) 1/sC (d) s/C

ANSWERS (MCQ) 1. (a) and (b)  2. (b)  3. (a)  4. (b).

CON V E NTI O NA L Q UE S TI O NS (C Q ) Determine the Laplace transforms of 1. 2t −3 2. 5t2 + 4t − 3 3.

t3 − 3t + 2 24

t5 t2 − 2t 4 + 15 2 5. 5e3t 6. 2e−2t 7. 4 sin 3t 8. 3 cos 2t 9. 5e–2t cos 3t 10.  4e–5t sin t Determine the inverse Laplace transforms of 11.  7/s 12.  2 / (s – 5) 4.

13.  3 / (2s + 1) 14.  2s / (s2 + 4) 15.  6 / s2 Use partial fractions to find the inverse Laplace transforms  of the functions 11 − 3s 16.  2 s + 2s − 3 17. 

2s + 3 ( s − 2) 2

18. 

3s 2 + 16 s + 15 ( s + 3)3

19. 

5s 2 − 2 s − 19 ( s + 3)( s − 1) 2

20. 

26 − s 2 s ( s + 43 + 13) 2

ANSWERS (CQ) 2 3 1. -   2. 10 + 4 − 3   3. 1 − 3 + 2   4.  8 − 48 + 1   5.  5   6.  2   7. 12  8. 3s 2 s s s s s-3 s3 s2 s2 s6 s5 s3 4s 4 s2 + 4 s+2 s2 + 9 1

(- )t 5( s + 2)   10.  4   11. 7  12. 2e5t 13.  3 e 2   14. 2 cos 2t  15. 6t 16. 2et – 5e–3t  17. 2e2t + 7te2t 2 2 2 s + 10 + 26 s + 4 s + 13 2  2 -3t 2 18. [2e (3−2t−3t )] 19.  2 - 3e-2t cos 3t - e-2t sin 3t 20. 2 − 3e−2t cos 3t − e−2t sin 3t . 3  3

  

9. 

30

Coupled Circuits OBJECTIVES In this chapter you will learn about:       

The need for coupling Classification of coupled circuits Performing calculations based on M = k√L1L2 Response of different types of coupled circuits The effect of variations of coupling elements Mutual coupling

Loose coupling Tight (over) coupling Critical coupling

Is

Critical coupling Tight coupling

Ip

Loose coupling

Frequency

Frequency

Transfer of energy through magnetic couplings: (loose, critical and tight coupling)

Flux

30.1 INTRODUCTION When the interaction between two loops of a circuit takes place through a magnetic field instead of through common elements, the loops are said to be inductively or magnetically coupled. The windings of a transformer, for example, are magnetically coupled. This is shown in Figure 30.1. Two a.c. circuits are said to be coupled when they are linked in such a way that energy can be transferred from one circuit to the other. The two circuits coupled are usually tuned to resonance at the same frequency, i.e., the product LC is the same for each. With the coupled circuits it is found that resonance occurs at two frequencies. The existence of two resonant frequencies arises from the fact that in a coupled circuit there are two paths in which the current can oscillate.

Primary coil Alternating current

Secondary coil – +

Primary source of power

+ – Alternating voltage

Figure 30.1 The Primary Coil is Magnetically Coupled to the Secondary Coil or a Magnetic Flux

572 Electrical Technology One of these frequencies may be that at which each of the component of the component circuits would resonate on its own.

ωr   1 =  fr =  2π 2π LC  

(30.1)

The other frequency is dependent on the degree of coupling employed.

30.2 DEGREE OF COUPLING The degree of coupling is expressed by the coupling coefficient denoted by the symbol K: it is equal to the ratio of the amount of coupling actually present to the maximum possible coupling. As a consequence, the value of K cannot exceed 1. The coupling coefficient is defined as the ratio of the mutual or common impedance component of the two circuits, to the square root of the product of the totals, in the two circuits, of the impedance components of the same kind (impedance components may be inductive, capacitive or resistive).

30.3 CLASSIFICATION OF COUPLED CIRCUITS Coupled circuits may be conveniently classified, according to the nature of the path connecting one to the other, into three categories: 1. Direct coupling through an impedance not included in either circuit. RM 2. Direct coupling through an impedance common to both circuits. In the first two categories the coupling impedance may be a resistance, an L1 C1 C2 L2 impedance, or even a capacitance. 3. Indirect coupling through a field of force common to both circuits. The E (a) field may be either magnetic or electric but the latter is infrequently used.

30.4 CATEGORY (I)

LM L1

C1 E

C2

L2

(b) CM

L1

C1 E

C2

L2

(c)

Figure 30.2 Circuits Coupled Through an Imped­ ance Not Included in Either Circuit

The three types of circuits in the first category are illustrated in Figure 30.2. The coupling impedance is the resistance RM, the inductance LM, or the capacitance CM. The resistance of the inductors is assumed to be negligible in comparison with their reactance. The total impedance is that of the impedance of L2 and C2 in parallel, with the coupling element added in series together with C1 in parallel with L1 in series. If this impedance is evaluated and equated to zero (for resonance), the result is a quadratic equation with two roots f1 and f2. With resistive coupling, as represented in Figure 30.2(a), the resistance value is usually very small compared with the reactance and the resonant frequency of the circuit is but little affected by the coupling element. For the inductive and capacitive couplings, the values of the two resonant frequencies depend upon the relationships assumed for the LC values of the separate circuits. For the inductive coupling, as shown in Figure 30.2(b), provided that

 1  L1C1 = L 2 C2 = LC  = 2  − ω1 =  ωr    ω2 =

1 LC

(30.2)

1   LC 

 L1 + L 2 1 − LM L1 + L 2 

    

(30.3)

Of the two resonant frequencies, f1 = ω1/2p is the same as that of either circuit taken separately. The other, f2 = ω1/2p is equal to f1 modified by the term within the curved brackets, and is a variable frequency, depending upon the value of the coupling inductance LM: the frequency f2 may vary between f1 and ∞ as the value of LM decreases from ∞ to zero.

Coupled Circuits

573

For capacitive coupling, as seen in Figure 30.2(c), provided that L1C1 = L 2 C2 = LC − ω1 =

ω2 =

1 LC

1   LC 

(30.4)

 CM CM   + 1 +  C1 C2   

Here again, f1 = ω1/2p is the resonant frequency of the separate circuits and is independent of the coupling capacitance CM. The other frequency f2 = ω2/2p depends on the values of CM: it is variable between f1 (when CM is zero) and zero (when CM is infinite). These coupled circuits will each have a double-peaked resonance curve with peaks occurring at f1 and f2. For inductive coupling (LM), a set of resonance curves is given in Figure 30.3, corresponding to three increasing values of LM. The value of f1 is the same in each case, being independent of LM. The value of f2 increases with decreasing value of LM. Corresponding resonance curves are given in Figure 30.4 for the capacitance coupled circuits with three different values of CM. The value of f1 is the same for each resonance curve; values of f2 decrease with increasing values of CM. f2 (CM = 0.00225 µ F) f2 (LM = 5780 µ H)

f2 (CM = 0.00144 µ F)

f2 (LM = 1350 µ H)

10 0

40 Current (mA)

Current (mA)

20

f1

30 20 10 0

200 300 Frequency (kHz)

L1 = L2 = 840 µ H

C1 = C2 = 0.000862 µ F

Figure 30.3 Resonance Curves for Ind­ uctively Coupled Circuits

f2 (CM = 0.00067 µ F)

50

f2 (LM = 530 µ H)

30

75

100 150 Frequency (kHz)

L1 = L2 = 840 µ H

f1

C1 = C2 = 0.000612 µ F

Figure 30.4 Resonance Curves for Cap­ acitatively Coupled Circuits

For inductive coupling as illustrated in Figure 30.2(b), provided that LM L1 LM + L1

C1 =

LM L 2 LM + L 2

C2 = LC = ω r2

(30.5)

then

ω = when K =

1± K LC

( LM

(30.6)

L1 L 2

+ L1 ) ( LM + L 2 )

(30.7)

For capacitive coupling, as can be seen in Figure 30.2(c), provided that L1 ( CM + C1 ) = L 2 ( CM + C2 ) = LC = ω 2r

ω =

1 LC ( 1 ± K )

(30.8) (30.9)

574 Electrical Technology where, + C1 ) ( CM + C2 )

With these total LC relationships, there are two resonant frequencies—one higher and one lower than the individual equal resonant frequencies. The resonant curves for them are shown in Figure 30.5.

C Resonance

( CM

CM

Current (I)

K =

30.4.1 Category (ii)

A

B

Frequency

The three types of impedance coupling in the second category of coupled circuits are shown in Figure 30.6. Figure 30.5 Resonant Curves Showing The resistive coupling, as can be seen in Figure 30.6(a) produces neglithe Effect of Coupling. A: gible frequency changes unless RM is abnormally higher and the frequency Loosely Coupled; B: Tightly is very low. Coupled; and C: Compro­ With the inductive (Figure 30.6(b)) or capacitive (Figure 30.6(c)) mise between A and B coupling element, the total impedance is made up from L1 and C1 is series together with LM (or CM) in parallel with L2 and C2 in series. Evaluating this impedance and equating it to zero for resonance provides two solutions for the resonant frequency. Provided that C2

C1 L1

RM

L1C1 = L 2 C2 = LC = L2

1 ω 2r

then, for the inductive coupling (Figure 30.6(b)) E

ω1 = C2

C1 L1

ω2 =

E

C2

C1 CM

L1

(30.10)

LC

and L2

LM

1

L2

  LM L  + M   LC  1 + L L2   1  

(30.11)

This is a similar form of result to that for capacitive coupling in the first case and the resonance curves relating to this case are similar to those in Figure 30.4. For capacitive coupling, as shown in Figure 30.6(c)

ω1 =

E

Figure 30.6 Circuits Coupled Through an Imped­ ance Common to Both Circuits

1

1

(30.12)

LC

and

ω2 =

1   LC 

 C1 + C2 1 − C  M + C1 + C2

  

(30.13)

This case is an analogous to the inductive coupling in the first category and the resonant curves relating to the present case are similar to those in Figure 30.4. The equality is between the total LC products of the separate circuits, i.e., including coupling element with each of the separate circuits. For the inductive coupling, as seen in Figure 30.6(b), provided that

( LM

+ L1 ) C1 = ( LM + L 2 ) C2 = LC =

1 ω r2

Coupled Circuits

ω =

1

575

(30.14)

LC ( 1 ± K )

where, LM

K =

(30.15)

 ( LM + L1 ) ( LM + L 2 )   

For the capacitive circuits in Figure 30.6(c), provided that  C M C2   CM C1  1 L1   = LC = 2  = L2  + + C C C C ω  M  M 1  2  r

ω = Where, K =

(30.16)

1± K LC

(30.17) (30.18)

C1C2 + C1 ) ( CM + C2 )

( CM

In the directly coupled circuits, K may be increased by varying the values of the coupling impedance (LM or CM) and adjusting the circuit elements L1C1 or L2C2 in order to maintain resonance.

30.4.2 Category (iii) The most commonly employed form of coupling is that of category (iii) where two resonant circuits are coupled electromagnetically by mutual inductance M henrys. Typical examples have been illustrated in Figure 30.7. With this arrangement, two resonant frequencies again result whose values may be derived from the following expressions (see Figure 30.8).

ω =

1 LC ( 1 ± K )

,

K =

where,

M

(30.19)

L1 L 2

With electromagnetic coupling, the only applicable relationship is L1C1 = L 2 C2 = LC =

C1 L1 C2

1 ω 2r

C3

C1

L1

L2

C2

C4

C3

L3

L4

C4

L2

Figure 30.7 The Most Commonly Employed Form of Coupling: Electromagnetic Coupling M

C1

L1

L2

C2

Figure 30.8 Circuits Coupled by Mutual Inductance

576 Electrical Technology The double-peaked resonance curves for three different values of the coupling inductance M are given in Figure 30.9. Values of M and, therefore, of K also, will vary according to the degree of coupling circuits that are said to be tightly or loosely coupled together according to whether the proportion of energy transferred from one to the other is large or small, i.e., depending upon whether K is large or small. For electromagnetic coupling, M and K may be increased by moving the coupled coils more closely together so that an increased proportion of the primary flux links with the secondary coil. The expression

M = 40 µ H M = 420 µ H

Current (mA)

40

M = 85 µ H

30 20 10 0

200

300

Frequency (kc/s) L1 = L2 = 840 µ H

C1 = C2 = 0.0007 µ F

Figure 30.9 Resonance Curves for Circuits Coupled Electromagnetically

M2 = =

M =

N12 Aµo µ r N 22 Aµ o µ r ×  

(30.22) (30.23)

i1

(30.24)

L1 L 2

This is the maximum possible mutual inductance between the circuits and M cannot be greater that L1 L 2 . In any practical case, it is impossible to avoid flux leakage and M < so that M

Linkage flux

(30.21)

2

= L1 × L 2 M =

(30.20)

l

assumes that all the primary flux links with secondary coil. On this assumption

N12 N 22 A2 µ 2o µ 2r 

N1 N 2 Aµ o µ r

L1 L 2

L1 L 2 = K is less than 1 (see Figure 30.10).

One advantage arising from the use of coupled circuits is the increased frequency range or band width over which the impedance is low when compared with the resonance curve of a single tuned circuit. With suitable design, the trough between the peaks may be smoothed out to give an almost linear response over the desired frequency range. This point should be noted with reference to Figure 30.11. dI E1 = − M 2 dt

dI E 2 = −M 1 , dt

(30.25)

Leakage flux

Coil L1 N1 Turns Circuit 1

Coil L2 E 2 N 2 Turns Circuit 2

M

Figure 30.10 Mutual Coupling: the Value of K is Always Less Than 1 Because of Leakage Flux

When E2 is the e.m.f. in circuit 2 due to current I1 is circuit 1 and E1 is the e.m.f. in circuit 1 due to current I2 in circuit 2. Also A

I1

R1

R2

I2

M E

L1

Z

E1 B

C1

E2 = −N 2

dφ 2 dt

and

where, M = N 22

L2 E2 C2

Figure 30.11 Impedance of Inductively Coupled Circuits

By similar reasoning

E1 = − N 1 d φ 22 dI11

→ M = N11

d φ11 dI 22

d φ1 dt

(30.26)

(30.27) (30.28)

If the coils are linked with iron as the medium the flux and current are linearly related and

Coupled Circuits

M =

N 2φ 2

M =

and

I1

N1φ1 I2

577

(30.29)

Example 30.1 A and B are two coils in close proximity. A has 1200 turns whereas B has 1000 turns. When a current of 0.8 A flows in coil A, a flux of 100 μWb links with coil A and 75 per cent of this flux links coil B. Determine (1) the self inductance of coil A and (2) the mutual inductance. Solution:

1. Self inductance of coil A, LA =

N Aφ A IA

LA =

(1200)(100 × 10 −6 ) = 0.15 H 0.80

2. Mutual inductance M = =

N Bφ B IA (1000)(0.75 × 100 × 10 −6 ) 0.80

= 0.09375 H

or

= 93.75 µ H

Example 30.2 Two coils have self inductances of 250 mH and 400 mH, respectively. Determine the magnetic coupling coefficient of the pair of coils if their mutual inductance is 80 mH. Solution: K = = =

M √L1 L 2 80 × 10 −3 √  (250 × 10 −3 )(400 × 10 −3 )  80 × 10 −3 = 0.2253 √ (0.1)

30.5 IMPEDANCE OF COUPLED CIRCUITS When two a.c. circuits are inductively coupled, the effective impedance of the primary circuit is modified by the influence of the secondary circuit components. Two circuits, electromagnetically coupled by their mutual inductance M, are shown in Figure 30.11. The primary circuit comprises of the elements R1 ohms L 1 henrys, and C 1 Farads; it is energized from a generator of e.m.f. E volts at a frequency of f Hz. The secondary circuit components have values of R2 ohms, L 2 henrys, and C2 farads. The primary impedance without regard to the secondary circuit is Z1 ohms; the secondary impedance is Z 2 ohms. The circuit currents are I 1in the primary and I 2in the secondary. The e.m.f. induced in the secondary coil is E 2 volts; the back e.m.f. in the primary circuit has a magnitude of E1 volts. The e.m.f. induced in secondary circuit by the current I1 flowing in the primary circuit is ωMI1. Since it is 180° out of phase with the current I1 it may be written as E2 = -jωMI1. The current flowing in the secondary circuit is I 2 = E2 Z 2 = − jω MI1 Z 2 The back e.m.f. induced in L1 by the current I2 is E2 = jω MI 2 = jω M (− jω MI1 Z 2 ) = ω 2 M 2 I1 Z 2

(30.30)

578 Electrical Technology The applied e.m.f. E must be equal to the sum of the two voltage components: (1) the p.d. due to the current I1 flowing in the impedance Z1 and (2) the back e.m.f. E1 induced by the secondary. Accordingly, E = I1Z1 + E1 = I1Z1 + ω 2 M 2 I1 Z 2 (30.31)

= I1 ( Z1 + ω 2 M 2 Z 2 )

The effective impedance Zeff of the whole circuit viewed from the input terminals AB is the ratio of the applied e.m.f. E to the resulting current I1 so that Z eff = E I1 = Z1 + ω 2 M 2 Z 2

(30.32)

In other words, the impedance looking into the primary terminals (AB) is equal to the impedance of the primary circuit itself  Z 1 = R1 + jω L1 + 1 jω C1 = R1 + jX1  together with an impedance equal to ω 2 M 2/ Z 2 reflected from the secondary circuit. Writing Z 1 = R1 + jX1 and Z 2 = R 2 + jX 2 in the above expression (Eq. 30.32) Z eff = Z 1 + ω 2 M 2 Z 2 = R1 + jX 1 +

ω2M 2 R 2 + jX 2

  ω2M 2  ω2M 2 = R1 + R 2   + j  X 1 − X 2 2 Z 22  Z2  

   

(30.33)

2

Effective resistance

 ωM R eff = R1 + R 2   Z2 

  

Effective reactance,

 ωM X eff = X1 − X 2   Z2 

  

2

It can be seen that the effect of the secondary is always to increase the resistive component of Zeff the , reactive component may be increased or decreased, depending upon the frequency. At the resonant frequency for the secondary circuit X2 = 0

and

Z eff = R1 + ω 2 M 2 / R2 + jX1

For direct coupled circuits of the type shown in Figure 30.6, the values of Reff, Xeff and Zeff may be obtained by substituting ωLM (inductive coupling) or 1 ω CM (capacitive coupling) for ωM in the above expressions. Example 30.3 Two mutually coupled coils X and Y, as shown in Figure 30.12, are connected in series to a 240 V d.c. supply. Coil X has a resistance of 5 Ω and an inductance of 1 H. Coil Y has a resistance of 10 Ω and an inductance of 5 H. At a certain instant after the circuit is connected, the current is 8 A and increasing at a rate of 15 A/s. Determine (1) the mutual inductance between the coils and (2) the coefficient of coupling. Solution: 1. From KVL Coil X

V = iR + L di dt 240 = 8(5 + 10) + L(15) L= Also

240 − 120 = 8H 5

V = 240 V

Rx = 5 Ω Lx = 1 H

M Coil Y

Ry = 10 Ω Ly = 5 H

L = L X + LY + 2 M M = 1H

Figure 30.12 For Example 30.3

Coupled Circuits

K =

2.

579

M 1 = = 0.447 √ ( L X LY ) √ [ (1)(5) ]

Example 30.4 The inductances of two inductively coupled coils A and B are 0.002 and 0.032 H, respectively, and the coupling coefficient is 0.75 as shown in Figure 30.13. Regarding the resistance of the coil A as negligible, what will be the open circuit voltage across the terminals of coil B when 0.5 V is applied across the terminals of coil A? Solution: IA = =

IA

E ( R + jω L A ) 0.5 250 = (ω × 0.002) ω A

IB

B

LA 0·002 H

E1 = 0·5 V

M = K √ L A L B = 0.75 √ (0.002) × (0.0322)

LB 0·032 H

EB

K = 0·75

= 0.006 Η E B = ω MI A =

A

Figure 30.13 For Example 30.4 (ω × 0.06 × 250) = 1.5 V ω

Example 30.5 A high-frequency ammeter is coupled to an oscillatory circuit by means of an air-cored transformer. The primary winding carries 100 A at 105 Hz and the secondary circuit has a total inductance of 20 mH and a natural inductance with the primary of 1 mH. If the resistance of the secondary circuit including the meter is 4 Ω, what current will flow through the meter? Solution: f

4 76 A Example 30.6 For the circuit shown in Figure 30.14 determine the p.d. E2 that appears across the open circuited secondary winding, given that E1 = 8 sin 2500t volts. Solution: Impedence of primary, Z1 = R1 + jω L1

= 15 + j (2500)(5 × 10 −3 )

I1

j Primary current

I1 = E1 / Z1 =

E2 = jω MI1 = =

15 Ω

8 0° 19.53 39.81° jω ME1 j = ( R + jω L1 )

15 Ω E2

E1 −

5 µH

5 µH

19.53 39.81°

2 90° = 0.102 50.19° V 19.53 39.81°

M = 0.1 µ H

Figure 30.14 For Example 30.6

580 Electrical Technology Example 30.7 A circuit consisting of a capacitor of 0.001 µF and an inductor of 500 µH is coupled magnetically with a second circuit consisting of a capacitor of 0.0005 µF and an inductor of 1000 µH. The mutual inductance between the coils is 100 μH and the resistances of the primary and secondary coils are 15 Ω and 35 Ω, respectively. What is the effective series impedance of the primary circuit at a frequency of 200 kHz? Solution: f

Primary Circuit X1 = ω L1 −

1 (ω C1 )

= 4π × 105 × 5 × 10−4 −

1 4π × 105 × 10−9

= 628 − 796 = 168 Ω Secondary Circuit X 2 = ω L2 −

1 (ω C2 )

= 4π × 105 × 10−3 −

1 5 π × × 4 10 5 × 10−10

1

1

Z 2 = (352 + 3342 ) 2 = (00.122 × 104 + 11.15 × 104 ) 2 1

= (11.27 × 104 ) 2 = 336 Ω Effective Resistance = R1 + R 2 (

ωM 2 ) Z2

 16π 2 × 1010 × 10−8  = 15 + 35   3362   = 15 + 4.88 = 19.88 Ω Effective Reactance

ωM 2 ) Z2 ωM 2 X1 − X 2 ( ) Z 2  16π 2 −168 − (−334 )  16π 2 −168 − (−334)  −168 + 46.6 

= X1 − X 2 ( = = = =

× 1010 × 10−8 3362 × 1010 × 10−8 3362

=− −168 121.+ 4 46.6 = Effective Impedance

1

1

= −121.4 2 2 =(R 2 + X 2 ) = (19.882 + 121.42 ) 1 2

1

1

1 2

=(R 2 + X 2 ) 2 = (19.882 2+ 2121.42 ) = (3.98 × 10 × 147 × 10 ) = (3.98 × 1022 × 147 × 102 ) = 10(150.9) = 123 Ω = 10(150.9)

1 2

= 123 Ω

1 2

    

581

Coupled Circuits

Example 30.8 A search coil having 50 turns, enclosing a cross-sectional area of 2 cm2 is placed at the middle of a solenoid having 500 turns of wire uniformly wound on a length of 12 cm and carrying an alternate current of 0.2 A at a frequency of 5 kHz. Calculate the mutual inductance between the coils and the e.m.f. induced in the search coil. Solution: Flux density per ampere = B = μ0H =

4π × 10−7 × 500 × 1 = 5.24 × 10−4 Wb/m 2 0.12

Mutual inductance, M 5.24 × 10−4 × 2 × 10−4 × 50 = 52.4 µ H 1 = ω MI = 2π × 5 × 103 × 52.4 × 106 × 0.2 = Voltage induced

= 0.329 V Example 30.9 Two coupled coils of self inductances 0.8 H and 0.2 H have a coefficient of coupling of 0.9. Find the mutual inductance and turns ratio. Solution: M = K L1 L 2 = 0.9 √ (0.8)(0.2) = 0.36 H M = N2

K φ1 I1

= KN 2

L1 N1

 L1 =

N1φ1 I1

KL1 0. 9 × 0. 8 N1 = = =2 M 0. 36 N2 Example 30.10 For the circuit shown in Figure 30.19, determine the value of the secondary current I2 if E1 = 2∠0° volts and frequency is 103/p Hz. Solution: R1(eff ) = R +

ω 2 M 2 R2 R22 + ω 2 L22

I1 16 Ω 4Ω 10 µ H

2

 103  −3 2  2π  (2 × 10 ) (16 + 50) π 1056  = (4 + 16) +  = 20 + = 20.222 Ω 2 4756  103  2 −3 2 66 +  2π  (10 × 10 ) π   X1(eff) is the imaginary part of Z1(eff) i.e, R1(eff) is the real part of Z1(eff)

16 Ω 10 µ H

I2 50 Ω

E1 M=2 µ H

Figure 30.15 For Example 30.10

582 Electrical Technology Hence, primary current,

Hence, secondary current, I 2 =

=

E2 = Z2

0.282 45.41° 0.282 45.41° = 3 (66 + j 20)  10 66 + j  2π (10 × 10 −3 )  π  

0.282 45.41° = 4.089 × 10 −3 28.55° A 68.964 16.86 °

Example 30.11 Two 1000 turn air-cored coils, with a length of 100 cm, having a cross-sectional area of 5 cm2, are placed side by side. The mutual inductance between them is 25 mH. Find the self inductance of the coils and the coefficient of coupling. Solution: As the two coils are same µ o µ r AN 2 4π × 10 −7 × 1 × 5 × 10 −2 × 10002 = L1 = L 2 =  100 × 10 −2

= 0.0628 H = 62.8 µ H K =

M = √ ( L1 L 2 )

25 × 10 −3 (62.8 × 10 −3 )(62.8 × 10 −3 )

= 0.3981 Example 30.12 A solenoid consists of 1500 turns of wire wound in a length of 60 cm. A search coil of 500 turns consisting of a mean area of 20 cm2 is placed centrally on the solenoid. Find (1) the mutual inductance of the arrangement, and (2) the e.m.f. induced in the search coil when the current in the solenoid is changing uniformly at the rate of 250 A per second. Solution: 4π × 10−7 × 1500 I 1. B = µoH = = 10−3 π I 0.6 M = B × area of search coil × No. of turns 10−3 π I × 20 × 10−4 × 500 = I = 3.14 µ H 2.

di = 3.14 × 10−3 × 256 V dt = 0.786 V

Voltage induced = M

Example 30.13 Two coils, whose self inductances are 75 μH and 125 μH, have a mutual inductance of 15 μH. What is the coupling factor? Solution: 15 M K = = √ ( L1 L 2 ) √ (75 × 125) = 0.155

Coupled Circuits

583

Example 30.14 Two circuits, each consisting of an inductance of 160 μH a resistance of 10 Ω and a capacitor of 0.001 μF are coupled by a mutual inductance between the inductors of 2 μH. If an e.m.f. of 100 V r.m.s at 400 000 Hz is applied in series with one circuit, what will be the current in the circuits? Solution:

ω = 2π f = 2.51 × 106 radu/sec,

f = 4 × 105 Hz, E = 100 V,

R1 = R 2 = 10 Ω,

L1 = L 2 = 1.6 × 10 −4 H,

C1 = C2 = 10 −9 F,

M = 2 × 10 −6 H

X1 = X 2 = ω L1 − 1 (ω C1 ) = 2.51 × 106 × 1.6 × 10 −4 −

1 = 401 − 398 = 3 Ω 2.51 × 106 × 10 −9

Z 22 = 102 + 32 = 109 Ω

ω 2 M 2 = (2.51 × 106 ) 2 × (4 × 10 −12 ) = 25.4 Ω Effect ive resitance = R1 + R 2 (ω 2 M 2 Z 22 ) = 10 + 10 (25.4 109) = 12.33 Ω Effective reactance = X1 − X 2 (ω 2 M 2 Z 22 ) = 3 − 3 (25.4 109) = 2.3 Ω Effective impedance = (12.332 + 2.32 ) I1 = E1 Z1 =

1

2

100 = 84 A 1 (12.33 + 2.32 ) 2 2

I 2 = E2 Z 2 = (ω MI1 ) Z 2 = (25.4 109)

1

2

×8

= 3.88 A Example 30.15 Two coils, X of 12000 turns and Y of 15000 turns, lie in parallel planes so that 45 per cent of the flux produced by coil X links with the coil Y. A current of 5 A in X produces 0.05m Wb, when the same current in Y produces 0.075m Wb. Calculate (1) the mutual inductance and (2) the coupling coefficient. Solution: 0.05 × 12000 1. LX = = 120 µ H 5

2.

LY =

0.075 × 15000 = 225 µ H 5

M =

0.45 × 0.05 × 15000 = 67.5 µ H 5

K =

67.5 (120 ×

1 22.5) 2

= 0.41

S UM M A RY 1. Two a.c. circuits are said to be coupled when they are so linked that energy can be transferred from one to the other. 2. K, the coupling coefficient, is the ratio of the amount of coupling actually present to the maximum possible coupling.

3. Coupled circuits may be conveniently classified according to the nature of the path connecting one to the other. 4. Direct coupling is through an impedance not included in either circuit. 5. Values of M and, therefore, of K also, will vary according to the degree of coupling.

584 Electrical Technology 6. M and K may be increased by moving the coupled coils more closely together. 7. In many practical cases, it is not possible to avoid flux leakage and K1

(b)