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Mike Holt’s Illustrated Guide to

iLECTRICAL XAM PREPARATION Theory •Calculations •Code Includes Mike’s NEC® Exam Practice Questions Workbook Based on the 2014 NEC Suitable for all electrical exams based on the NEC®, such as: AMP, ICC, LocaS/State Examining Boards, Pearson VUE, Prometric, Prow, PSi

7" Mike Holt Enterprises, Inc. ^ m

888.NEC.CODE (632.2633) • www.MikeHolt.com

NOTICE TO THE READER The publisher does not warrant or guarantee any of the products described herein or perform any independent analysis in connection with any of th product information contained herein. The publisher does not assume, and expressly disclaims, any obligation to obtain and include information otto than that provided to it by the manufacturer. The reader is expressly warned to consider and adopt all safety precautions that might be indicated by the activities herein and to avoid all potenti: hazards. By following the instructions contained herein, the reader willingly assumes all risks in connection with such instructions. The publisher makes no representation or warranties of any kind, including but not limited to, the warranties of fitness for particular purpose ( merchantability, nor are any such representations implied with respect to the material set forth herein, and the publisher takes no responsibility wit respect to such material. The publisher shall not be liable for any special, consequential, or exemplary damages resulting, in whole or part, from th reader’s use of, or reliance upon, this material.

Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC® First Printing: April 2014 Author: Mike Holt Technical Illustrator: Mike Culbreath Cover Design: Madalina lordache-Levay Layout Design and Typesetting: Cathleen Kwas COPYRIGHT© 2014 Charles Michael Holt ISBN 978-1-932685-67-1

Produced and Printed in the USA For more information, call 888.NEC.CODE (632.2633), or e-mail [email protected]. All rights reserved. No part of this work covered by the copyright hereon may be reproduced or used in any form or by any means graphic, electronic, c mechanical, including photocopying, recording, taping, or information storage and retrieval systems without the written permission of the publisher. Yo can request permission to use material from this text by e-mailing [email protected]. NEC®, NFPA 70®, NFPA 70E® and National Electrical Code® are registered trademarks of the National Fire Protection Association.

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TABLE OF CONTENTS About This Textbook................................................. xi Passing Your Exam.................................................. xiv

1.16 1.17 1.18

Power Source............................................................................11 Conductance.............................................................................12 Circuit Resistance.....................................................................12

About the National Electrical Code........................... xix

1.19

Ohm’s Law ............................................................................... 13

1.20 1.21 1.22 1.23 1.24

Ohm’s Law and Alternating Current.........................................13 Ohm’s Law Formula Circle....................................................... 13 PIE Formula Circle.....................................................................15 Formula Wheel..........................................................................16 Using the Formula Wheel......................................................... 17

1.25 1.26

Power Losses of Conductors.................................................... 18 Cost of Power...........................................................................18

1.27

Power Changes with the Square of the Voltage.......................19

About the Author.................................................... xxiii About the Illustrator.............................................. xxiv About the Team.......................................................xxv

CHAPTER 1— ELECTRICAL THEORY (Essential for Journeyman and Master/ Contractor Licensing Exams)................................. 1 Unit 1— Electrician’s Math and Basic Electrical Formulas........................................................................................... 3

C o n c lu s io n ........................................................................................21 U nit 1 R eview Q u e stio n s............................................................. 22 U nit 1 C hallenge Q uestion s........................................................ 28

Part A— Electrician’s M ath................................................................. 3 Introduction............................................................................................ 3

U n it 2 — E le c tric a l C ir c u its ..................................................... 31

1.1

Whole Numbers.........................................................................3

1.2

Decimals...................................................................................3

Part A— Series Circuits......................................................................31 Introduction...........................................................................................31

1.3 1.4

Fractions....................................................................................4 Percentages...............................................................................4

2.1

Practical Uses of the Series Circuit........................................... 31

2.2

1.5

Multiplier...................................................................................4

1.6

Percent Increase........................................................................5

2.3 2.4

Understanding Series Calculations...........................................32 Series Circuit Calculations....................................................... 35 Power Calculations...................................................................36

1.7 1.8

Reciprocals................................................................................6 Squaring a Number................................................................... 6

2.5

Variations.................................................................................. 36

1.9 1.10 1.11 1.12 1.13 1.14

Parentheses...............................................................................7 Square Root...............................................................................8 Volume...................................................................................... 8 Kilo............................................................................................ 9 Rounding Off............................................................................10 Testing Your Answer for Reasonableness...............................10

2.6 2.7

Series Circuit Notes..................................................................36 Series-Connected Power Supplies...........................................36

Part B— Basic Electrical Formulas................................................... 11 Introduction........................................................................................... 11 1.15

Part B— Parallel Circuits................................................................... 37 Introduction.......................................................................................... 37 2.8 Practical Uses of the Parallel Circuit.........................................37 2.9 Understanding Parallel Calculations.........................................38 2.10 Circuit Resistance.....................................................................39 2.11 Parallel Circuit Notes................................................................ 42 2.12 Parallel-Connected Power Supplies..........................................42

Electrical Circuit....................................................................... 11

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Part C— Series-Parallel Circuits..................................................... 42 Introduction.........................................................................................42 2.13 Review of Series and Parallel Circuits....................................43 2.14 Working With Series-Parallel Circuits....................................44

3.16 3.17

2.15

Part D— Multiwire Branch Circuits.................................................45 Introduction.........................................................................................45 2.16 Neutral Conductor..................................................................... 45

3.18 3.19 3.20 3.21

Self-Induction........................................................................... 73 Induced Voltage and AppliedCurrent......................................... 74 Conductor Alternating-Current Resistance.............................. 74 Conductor Shape...................................................................... 75

2.17 2.18

Grounded Conductor.................................................................46 Current Flow on the Neutral Conductor....................................46

3.22 3.23

Magnetic Cores......................................................................... 76 Self-Induced and Applied Voltage.............................................77

2.19 2.20

Balanced Systems.................................................................... 47 Unbalanced Current..................................................................47

2.21 2.22

Multiwire Branch Circuits..........................................................49 Dangers of Multiwire Branch Circuits....................................... 50

3.24 3.25 3.26

Inductive Reactance.................................................................. 77 Phase Relationship................................................................... 78 Uses of Induction...................................................................... 78

2.23

NEC Requirements.....................................................................51

Voltage.................................................................................. 44

C o n c lu s io n ......................................................................................52

Uses of Capacitors.....................................................................71 Phase Relationship................................................................... 72

Part C— Induction................................................................................73 Introduction........................................................................................... 73

Part D— Power Factor........................................................................ 78 Introduction........................................................................................... 78 3.27 Apparent Power (Volt-Amperes)............................................... 79 3.28

True Power (Watts).................................................................... 79

U n it 3— U n d e rs ta n d in g A lte rn a tin g C u r r e n t...............63

3.29 3.30 3.31 3.32 3.33

Power Factor.............................................................................80 Unity Power Factor.................................................................... 80 Power Factor Formulas.............................................................80 Cost of True Power.....................................................................81 Effects of Power Factor.............................................................82

Part A— Understanding Alternating Current..................................63

Part E— Efficiency.............................................................................. 84

Introduction.........................................................................................63

Introduction........................................................................................... 84

3.1 3.2 3.3

Current Flow............................................................................. 63 Why Alternating Current Is Used............................................... 64 How Alternating Current Is Produced.......................................64

3.34

3.4

Alternating-Current Generator.................................................. 64

3.5

Waveform................................................................................. 65

3.6 3.7 3.8 3.9

Sine Wave..................................................................................65 Frequency..................................................................................66 Phase........................................................................................67 Degrees.....................................................................................67

3.10 3.11

Lead or Lag.............................................................................. 67 Values of Alternating Current.................................................... 68

U n it 2 R eview Q u e stio n s..............................................................53 U n it 2 C hallenge Q u e stio n s........................................................ 58

Part B— Capacitance..........................................................................69 Introduction...........................................................................................69 3.12

Charged Capacitor....................................................................70

3.13 3.14

Electrical Field.......................................................................... 70 Discharging a Capacitor............................................................70

3.15

Determining Capacitance.......................................................... 71

Efficiency Formulas................................................................... 84

C o n c lu s io n ........................................................................................86 Unit 3 R eview Q ue stio n s...............................................................87 Unit 3 C hallenge Q u e stio n s......................................................... 95

U n it 4 — M o to rs a n d T r a n s fo r m e r s .....................................99 Part A— Motor Basics........................................................................ 99 Introduction...........................................................................................99 4.1 Motor Principles.................................................................... 100 4.2 Dual-Voltage Alternating-Current Motors..............................100 4.3

Motor Horsepower Ratings...................................................100

4.4 4.5

Motor Current Ratings...........................................................101 Calculating Motor FLA...........................................................102

4.6 4.7

Motor-Starting Current..........................................................103 Motor-Running Current.........................................................103

Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

Table of Contents

4.8 4.9 4.10 4.11

Motor Locked-Rotor Current (LRC).........................................103 Motor Overload Protection..................................................... 103 Direct-Current Motor Principles..............................................104 Direct-Current Motor Types.................................................... 105

4.12 4.13 4.14 4.15

Reversing the Rotation of a Direct-Current Motor..................105 Alternating-Current Induction Motor...................................... 105 Alternating-Current Motor Types............................................106 Reversing the Rotation of an Alternating-Current Motor......106

Part B— Transformers..................................................................... 107 Introduction........................................................................................ 107 4.16 Transformer Basics................................................................ 107

Part B— Outlet Box F ill....................................................................145 Introduction........................................................................................145 5.7 5.8 5.9 5.10

Box Volume Calculations...................................................... 145 Sizing Box— Conductors All the Same Size.........................146 Box Fill Calculations..............................................................146 Outlet Box Sizing...................................................................150

Part C— Pull Boxes, Junction Boxes, and Conduit Bodies........ 152 Introduction........................................................................................152 5.11 5.12

Pull/Junction Box Sizing Requirements............................... 152 Pull/Junction Box Sizing Tips...............................................154

4.17

Secondary Induced Voltage....................................................107

5.13 Pull Box Examples................................................................155 Summ ary..........................................................................................157

4.18

Efficiency................................................................................ 108

C o n c lu s io n .....................................................................................158

4.19 4.20 4.21 4.22

Transformer Turns Ratio......................................................... 108 Power Losses......................................................................... 110 Transformer kVA Rating......................................................... 112 Current Flow...........................................................................112

4.23 4.24

Current Rating........................................................................113 Autotransformers................................................................... 114

C o n c lu s io n ..................................................................................... 114 U nit 4 R eview Q uestion s........................................................... 115 U n it 4 C hallenge Q u e stio n s......................................................120

CHAPTER 2— NEC CALCULATIONS

U nit 5 R eview Q u e stio n s......................................................... 159 U nit 5 C hallenge Q u e stio n s.................................................... 164

U n it 6— C o n d u c to r S iz in g a n d P r o te c tio n ..................167 Part A— Conductor Requirements................................................. 168 Introduction........................................................................................168 6.1 Conductor Insulation............................................................ 168 6.2 Conductor Sizes....................................................................170 6.3

Smallest Conductor Size......................................................170

6.4

Conductor Size— Equipment Terminal Rating......................172

6.5

Overcurrent Protection......................................................... 176

(E s s e n tia l fo r J o u rn e y m a n a n d

Part B— Conductor Am pacity.........................................................179 Introduction........................................................................................179

M a s te r/C o n tra c to r L ic e n s in g E x a m s )............................ 125

6.6

Conductor Heating— l2R....................................................... 179

U n it 5 — R a c e w a y a n d B o x C a lc u la tio n s .......................127

6.7 6.8

Limiting Excessive Temperature.......................................... 179 Conductor Ampacity............................................................. 179

Part A— Raceway Sizing................................................................. 127

6.9

Ambient Temperature Correction........................................ 180

Introduction........................................................................................127 5.1 Insulated Conductors and Fixture Wires Dimensions— Chapter 9, Tables 5 and 8 .....................................................128 5.2 Raceway Properties................................................................ 131 5.3 Raceway Sizing......................................................................137

6.10 6.11 6.12 6.13 6.14 6.15 6.16

Rooftop Temperature Adder.................................................182 Ampacity Adjustment........................................................... 184 Combining Ambient Temperature and Conductor Bundling Adjustments......................................................... 186 Current-Carrying Conductors...............................................188 Wireway— Conductor Ampacity Adjustment........................190 Conductor Ampacity Summary.............................................191 Conductor Sizing...................................................................191

6.17

Feeder Tap Rules..................................................................194

5.4

Sizing Raceways Using Annex C ............................................137

5.5 5.6

Sizing Wireways......................................................................142 Tips for Raceway Calculations.............................................. 144

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C o n c lu s io n ......................................................................................198 U nit 6 R e view Q u e stio n s............................................................199

8.11 8.12

Feeder Short-Circuit and Ground-Fault Protection.................248 Motor VA Calculations.............................................................249

8.13

Adjustable Speed Drives.........................................................251

U nit 6 C hallenge Q u e stio n s.......................................................203

8.14

Fire Pump Motor Circuits........................................................252

U n it 7 — V o lta g e -D ro p C a lc u la tio n s ..................................205

8.15 8.16 8.17

Part B— Air-Conditioning Calculations..........................................252

Part A— Conductor Resistance Calculations.................................205 Introduction.........................................................................................205

Scope of Article 440................................................................252 Multimotor Equipment............................................................253 Motor-Compressor and Other Motors— Short-Circuit and Ground-Fault Protection................................................. 253 Motor-Compressor and Other Motors— Conductor Size..... 254

7.1 7.2 7.3

Conductor Size....................................................................... 206 Conductor Resistance.............................................................206 Alternating-Current Conductor Resistance............................ 209

C o n c lu s io n ......................................................................................256

7.4

Alternating-Current Resistance.............................................. 210

U n it 8 R eview Q u e stio n s............................................................ 257

7.5

AC versus DC Resistance....................................................... 212

Part B— Voltage-Drop Considerations........................................... 213 Introduction.........................................................................................213 7.6 MFCVoltage-Drop Recommendations....................................213 7.7 Determining Circuit Conductors’ Voltage Drop— Ohm’s Law Method................................................................214 7.8 Determining Circuit Conductors’ Voltage Drop— Formula Method.................................................................... 215 7.9 Sizing Conductors to Account tor Voltage Drop...................... 217 7.10 Limiting Conductor Length to Minimize Voltage Drop......... 218 7.11

Limiting Current to Limit Voltage Drop....................................220

C o n c lu s io n ...................................................................................... 221

8.18

U nit 8 C hallenge Q u e stio n s........................................................261 U n it 9 — D w e llin g U n it C a lc u la tio n s .................................263 Part A— Standard Method Load Calculations............................. 263 Introduction.......................................................................................263 9.1 General Lighting and Generai-Use ReceptacleDemand Load.....................................................................................264 9.2 Appliance Demand Load........................................................ 267 9.3 Dryer Demand Load................................................................ 268 9.4 Cooking Equipment Demand Load......................................... 268 9.5

Air-Conditioning versus Heat Demand Load.......................... 270

9.6

Service Conductor Sizing........................................................ 271

U nit 7 R eview Q uestion s............................................................222

9.7

Standard Method Load Calculations Example....................... 274

U nit 7 C hallenge Q uestion s.......................................................227

Part B— Optional Method Load Calculations.................................275 Introduction.........................................................................................275

U n it 8 — M o to r a n d A ir-C o n d itio n in g C a lc u la tio n s . 229 Part A— Motor Calculations.............................................................229 8.1 Scope of Article 430................................................................229 8.2 8.3 8.4 8.5

Motor Full Load Current (FLC)................................................ 230 Motor Nameplate Full Load Amperes (FLA)............................ 231 Branch-Circuit Conductor Sizing............................................ 233 Feeder Conductor Sizing.........................................................236

8.6 8.7 8.8

Overcurrent Protection............................................................237 Overload Protection Sizing......................................................239 Branch-Circuit Short-Circuit and Ground-Fault Protection.. 243

8.9

Branch Circuit Summary.........................................................247

8.10

Combined Branch Circuit Overcurrent Protection..................248

9.8

Dwelling Unit Optional Load Calculation.................................275

9.9 9.10 9.11

Dwelling Unit Optional Load Calculation Example..................275 Existing Dwelling Unit Optional Load Calculation................... 277 Existing Dwelling Unit Optional Load Calculation Example................................................................................. 277

Part C— Neutral Load Calculations................................................ 278 9.12 Service Neutral Calculations— Ranges and Dryers............... 278 9.13 Neutral Service and Feeder Calculation..................................279 C o n c lu s io n ......................................................................................280 U n it 9 R e view Q u e stio n s............................................................. 281 U nit 9 C hallenge Q u e stio n s....................................................... 286

Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

Table of Contents C H A P TE R 3 — A D V A N C E D NEC C A LC U LA TIO N S (Essential for M aster/Contractor Licensing E xam s)............................................................. 289 Unit 10— M ultifam ily Dwelling Calculations............ 291 Part A— Standard Method Load Calculations............................292 Introduction............................................................................... 292 10.1 General Lighting and General-Use Receptacle Demand Load............................................................................. 293 10.2 Appliance Demand Load.................................................. 294 10.3 Dryer Demand Load.........................................................295 10.4 Cooking Equipment Demand Load.................................... 297 10.5 Air-Conditioning versus Heat Demand Load....................... 300 10.6 Service Conductor Sizing..................................................301 10.7 Multifamily Dwelling Calculations— Standard Method Example.........................................................................302 Part B— Optional Method Load Calculations.............................306 Introduction............................................................................... 306 10.8 Multifamily Dwelling Optional Load Calculations.................306 10.9 Multifamily Dwelling Optional Method Examples.................307 10.10 Two-Family Dwelling Units............................................... 309 Part C— Neutral Load Calculations...........................................312 10.11 Dryers and Cooking Equipment......................................... 312 10.12 Service Neutral............................................................... 313

Part B— Examples............................................................................332 11.8 Bank/Office Buildings...........................................................332 11.9

Manufactured Home Parks....................................................334

11.10 Recreational Vehicle Parks....................................................335 11.11

Marinas.................................................................................335

Part C— Optional Method— Feeder/Service Load Calculations..................................................337 Introduction........................................................................................ 337 11.12 New Restaurant— Optional Method..................................... 337 Part D— Welders............................................................................... 339 Introduction........................................................................................ 339 11.13 Arc Welders.......................................................................... 339 11.14 Resistance Welders................................................................341 11.15 Light Industrial Calculation....................................................343 C o n c lu s io n ..................................................................................... 345 U nit 11 R eview Q uestions......................................................... 346 U nit 11 C hallenge Q uestions.................................................... 349 U n it 1 2 — T ra n s fo rm e r C a lc u la tio n s .................................351 Part A-General.................................................................................. 352 Introduction.........................................................................................352 12.1 Transformer Basics...............................................................352 12.2

Delta-Connected Transformers............................................ 353

Conclusion.............................................................................316

12.3 12.4

Wye-Connected Transformers.............................................. 356 Transformer kVA Rating....................................................... 357

Unit 10 Review Questions................................................... 317

12.5

Line Currents.........................................................................357

Unit 10 Challenge Questions............................................... 321

Part B-NEC Requirements.............................................................. 360 12.6 Transformer Overcurrent Protection.....................................360

Unit 11— Commercial Calculations.............................323

12.7 12.8

Primary Conductor Sizing......................................................362 Secondary Conductor Sizing................................................ 364

12.9

Grounding and Bonding........................................................366

Part A— General.......................................................................324 Introduction............................................................................... 324 11.1 Lighting— Demand Factors.............................................. 324 11.2 Lighting Load 100% Demand...........................................326 11.3 Sign Circuit..................................................................... 327 11.4 Show-Window Lighting.................................................... 327 11.5 Multioutlet Receptacle Assembly....................................... 328 11.6 Receptacle Load..............................................................328 11.7 Banks and Offices— ReceptacleDemand Load....................330

C o n c lu s io n ..................................................................................... 368 U nit 12 R eview Q uestions......................................................... 369 U nit 12 C hallenge Q uestions.................................................... 372

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Table of Contents

CHAPTER 4— NEC PRACTICE QUIZZES AND EXAMS (Essential for All Journeyman and Master/Contractor Licensing Exams).................... 375 P ra c tic e Q uizzes Practice Quiz 1— Straight Order [Articles 90-110]...........................376 Practice Quiz 2— Random Order [Articles 90-110]...........................386 Practice Quiz 3— Straight Order [Articles 200-230]......................... 391 Practice Quiz 4— Random Order [Articles 90-230]........................... 401

Practice Exams Practice Exam 1— RandomOrder [Article 90-Chapter 9 ] ............... 557 Practice Exam 2— RandomOrder [Article 90-Chapter 9 ] ............... 567 Practice Exam 3— RandomOrder [Article 90-Chapter 9 ] ............... 577 Practice Exam 4— RandomOrder [Article 90-Chapter 9 ] ............... 588 Practice Exam 5— RandomOrder [Article 90-Chapter 9 ] ............... 598

Practice Quiz 5— Straight Order [Articles 240-285].........................406 Practice Quiz 6— Random Order [Articles 90-285]...........................416 Practice Quiz 7— Straight Order [Articles 300-324].........................422 Practice Quiz 8— Random Order [Articles 90-324]...........................432 Practice Quiz 9— Straight Order [Articles 328-398].........................437 Practice Quiz 10— Random Order [Articles 90-398].........................446 Practice Quiz 11— Straight Order [Articles 400-427]...................... 456 Practice Quiz 12— Random Order [Articles 90-427].........................466 Practice Quiz 13— Straight Order [Articles 430-504].......................476 Practice Quiz 14— Random Order [Articles 90-504].........................486 Practice Quiz 15— Straight Order [Articles 511-625].......................496 Practice Quiz 16— Random Order [Articles 90-625].........................506 Practice Quiz 17— Straight Order [Articles 630-695].......................516 Practice Quiz 18— Random Order [Articles 90-695].........................526 Practice Quiz 19— Straight Order [Article 700-Chapter 9 ]............... 536 Practice Quiz 20— Random Order [Article 90-Chapter 9 ]................ 546

Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

ABOUT THIS TEXTBOOK Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC® Mike Holt’s Illustrated Guide to Electrical Exam Preparation is a textbook designed to give you proficiency in performing electrical calculations. Mastery of electrical calculations is essential for success in the electrical trade, whether you are an Engineer, Contractor or Electrician. If you are studying to take your electrical exam, this textbook will provide you with the practice and confidence you need to pass your exam the first time. The writing style of this textbook, as with all of Mike Holt’s products, is informative, practical, easy to read, and applicable for today’s electrical professional. Also, just like all of his textbooks, it contains hundreds of detailed color-coded illustrations that help you visualize the concept. Passing an important electrical exam is the dream of all those who care about improving themselves and their families. Unfortunately, many don’t pass the exam at all, and few do so the first time. The primary reason people fail their exam is because they’re not prepared on the technical material and/or they don’t know how to take it. Most electrical exams contain questions on electrical theory, basic electrical calculations, the Code, and important and difficult National Electrical Code® calculations. This textbook contains hundreds of illustrations, examples, and almost 3,200 practice questions covering all of these subjects. This textbook is intended to be used with the 2014 NEC. The content of this textbook is divided into four chapters: Chapters 1-3 are the calculations section of the textbook, and Chapter 4 helps you practice answering /VfC-related questions for your electrical exam. Chapter 1. Electrical Theory Chapter 2. NEC Calculations Chapter 3. Advanced NEC Calculations Chapter 4. NEC Practice Quizzes and Exams Chapters 1-3 are divided into 12 units and will help you learn how to use the NEC tables and perform calculations based on the Code. Chapter 3 will help you in answering advanced NEC calculations questions. Each Unit in Chapters 1-3 contains objectives, explanations with graphics, examples, steps for calculations, formulas, and practice calculations.

As you read through these units, review the author’s comments and graphics, and compare examples with the 2014 Code book. After you’ve read each unit, complete the unit review questions, and then answer the more difficult challenge questions. These questions can be answered using the material covered in that unit. Chapter 4 includes the complete Mike Holt’s NEC Exam Practice Questions textbook to help you become proficient in answering questions about the Code. It includes two types of NEC practice quizzes followed by practice exams. You may want to intersperse the practice quizzes with your study of the first three chapters of the textbook, or you may want to complete Chapters 1-3 first, and then work on the practice quizzes and practice exams found in Chapter 4. Be sure to use your 2014 Code book to answer all of the questions: Practice Quizzes (Straight Order). There are ten practice quizzes with questions that are in the same order as found in the NEC book. If you’re short on time before your exam, complete these straight-order quizzes first, then return to the random-order quizzes as time allows. Practice Quizzes (Random Order). There are ten practice quizzes with questions in random order to give you more practice in finding answers in the Code book. Practice Exams. To complete your review, there are five Practice Exams that will help you evaluate your exam-taking skills. Answer these questions using the 2014 NEC. Practice using your Code book for the straight-order questions, and then as you progress, you may want to see how many of the questions you can answer from memory. Most licensing exams are open book, but check with the testing authority in your area to be sure. If there are NEC questions that you don’t understand after reviewing the answer key, don’t hesitate to ask your co-workers or fellow students about them. Additional help in learning the Code on a chapter-by-chapter basis is available in Mike Holt’s Understanding the National Electrical Code, Volumes 1 and 2 textbooks.

The Scope of this Textbook This textbook explains how to perform the electrical calculations nec­ essary for properly selecting the material required to complete an

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About This Textbook

electrical installation that’s safe and conforms to the requirements of the NEC. This textbook contains the following stipulations:

Cross-References, Notes, and Exceptions

•

Power Systems and Voltage. All power-supply systems are assumed to be solidly grounded and any of the following voltages: 120V single-phase, 120/240V single-phase, 120/208V threephase, 120/240V three-phase, or 277/480V three-phase, unless identified otherwise.

Cross-References. This textbook contains NEC cross-references to other related Code requirements to help you develop a better under­ standing of how the NEC rules relate to one another. These crossreferences are indicated by Code section numbers in brackets, an example of which is “ [90.4].”

•

Electrical Calculations. Unless the question or example speci­ fies three-phase, the questions and examples are based on a single-phase power supply.

Informational Notes. Informational Notes from the NEC contained in this textbook will be identified simply as “ Note.”

•

Rounding. All calculations are rounded to the nearest ampere in accordance with 220.5(B).

•

Conductor Material. Conductors are considered copper, unless aluminum is identified or specified.

•

Conductor Sizing. Conductors are sized based on a THHN copper conductor terminating on a 75°C terminal in accordance with 110.14(C), unless the question or example identifies otherwise.

•

Overcurrent Device. The term “overcurrent device” in this textbook refers to a molded case circuit breaker, unless identified otherwise. Where a fuse is identified, it’s to be of the single-element type, also known as a “one-time fuse,” unless identified otherwise. Author’s Comment: ■ Because the neutral conductor of a solidly grounded system is always grounded to the earth, it’s both a “grounded conduc­ tor” and a “neutral” conductor. To make it easier for the reader of this textbook, we’ll refer to the “grounded” conductor of a solidly grounded system as the “neutral” conductor.

Journeyman or Master/Contractor Study Journeyman. If you’re using this textbook to pass a Journeyman Exam, complete Units 1-9 found in Chapters 1 and 2 of this textbook, including the unit review questions and challenge questions. You should also complete the quizzes and exams in Chapter 4. Master/Contractor. If you’re using this textbook to pass a Master/ Contractor’s Exam, complete Units 1-12 in Chapters 1, 2, and 3 in their entirety, including the unit review questions and challenge ques­ tions. You should also complete quizzes and exams in Chapter 4. For additional help in preparing for an exam, you can expand your study program with DVDs that accompany this textbook. To order these DVDs at a discounted price, call 888.632.2633.

Exceptions. Exceptions contained in this textbook will be identified as “Ex” and not spelled out.

Technical Questions As you progress through this textbook, you might find that you don’t understand every explanation, example, calculation, or comment. Don’t become frustrated, and don’t get down on yourself. Remember, this is technical material and sometimes the best attempt to explain a concept isn’t enough to make it perfectly clear. If you’re still confused, visit www.MikeHolt.com/forum, and post your question on our free Code Forum for help. The forum is a moderated community of electrical professionals where you can exchange ideas and post techni­ cal questions that will be answered by your peers.

Textbook Corrections We’re committed to providing you the finest product with the fewest errors. We take great care in researching the NEC requirements to ensure this textbook is correct, but we’re realistic and know that there may be errors found and reported after this textbook is printed. The last thing we want is for you to have problems finding, communicat­ ing, or accessing this information. Any errors found after printing are listed on our website, so if you find an error, first check to see if it’s already been corrected by going to www.MikeHolt.com, click on “Books,” and then click on “Corrections” (www.MikeHolt.com/bookcorrections.htm). If you believe that there’s an error of any kind (typo­ graphical, grammatical, technical, or anything else) in this textbook or in the Answer Key and it isn’t already listed on the website, e-mail Corrections@MikeHolt. com. Be sure to include the textbook title, page number, and any other pertinent information.

Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

About This Textbook

If you are an instructor and have adopted Mike Holt textbooks for use in your classroom you can register for up-to-date Answer Keys that can be downloaded from our website. To register and receive a log-in password, go to our website www.MikeHolt.com, click on “Instructors” in the side­ bar of links, and then click on “Answer Keys.” On this same page you’ll also find instructions for accessing and downloading these Answer Keys. Please note that this feature will only work after you’ve received a log-in password.

QR Codes What’s this symbol? It’s a QR code and gives you the ability to use your smartphone to scan the image and be directed to a website.

We’ve included these in various places in our textbook to make it easier for you to go directly to the website page referenced. Follow the QR Code! When you see a QR code next to a section in the text, scan it with your smartphone— you’ll be able to watch a video clip that shows Mike and his panel of experts discussing this topic. These video clips are samples from the DVDs that were created for this book. Whether you’re a visual or an auditory learner, watching the DVDs will enhance your knowledge and understanding as you read through the textbook. Try it here! Scan this QR code and watch the Bloopers clip from the recording of the DVDs to accompany this book.

How to Use This Textbook The layout of this textbook incorporates special features designed not only to help you navigate easily through the material, but to enhance your understanding as well. Framed yellow notes contain examples and practi­ cal application questions and answers. Green text within a gray bar or gray “ highlight” signifies an important electrical formula. Colored highlighting in the chapter color identifies the solu­ tion to a calculation (or subtotals in the more complicated cal­ culations) in Chapters 2 and 3. Graphics with an icon and green border contain a change in the Code from 2011 NEC to 2014 NEC. Color coding and a modular format make it easy to navigate through each section of the textbook. Danger, Caution, and Warning icons highlight areas of concern. A QR code under the article number can be scanned with a smartphone app to take you to a sample video clip so you can watch Mike and the DVD panel discuss this topic. ■ Bulleted Author’s Comments are intended to help you understand the NEC material and background information.

Mike Holt Enterprises, Inc. • www.MikeHolt.com • 888.NEC.C0DE (632.2633)

xiii

PASSING YOUR EXAM If you are using this textbook to help you prepare for your electrical exam, these tips can help you succeed in the process.

How to Achieve Your Best Grade In order to pass an exam, a few things need to happen. You need to: • Prepare yourself mentally, emotionally, and physically. • Create a proper study program so that you have the time to learn the technical material to answer all the questions. • Prepare for the exam by knowing what to expect with the exam taking process. • Understand how to take an exam, how to approach ques­ tions, and how to manage your time during the test. ® Have the confidence that you can pass.

Prepare Mentally, Emotionally and Physically Studies have concluded that for students to achieve their best grades, they must learn to obtain the most from their natural abilities. It’s not how long you study or how high your IQ is, it’s what you do and how you study that counts the most. To get your best grade, you must make a decision to do your best and follow as many of the following techniques and suggestions as possible. Reality. These instructions are a basic guide to help you achieve your maximum grade. It’s unreasonable to think that all of the instructions can be followed to the letter all of the time. Day-to-day events and unexpected situations must be taken into consideration.

Support. You need encouragement in your studies and you need sup­ port from your loved ones and colleagues. To properly prepare for your exam, you need to study a few hours per week for several months, depending on your existing knowledge. Communication With Your Family. Good communication with your family is very important because studying every night and on week­ ends can cause much tension and stress. Try to win their support, cooperation, and encouragement during this difficult time. Let them

know the long-term benefits to the family and what passing the exam means. Be sure to plan some special time with them during this preparation period; don’t go overboard and neglect them. Stress. Stress can really take the wind out of you. It takes practice, but develop the habit of relaxing before you your studies. Stretch, do a few sit-ups and pus ups, take a 20-minute walk, or a few slow, deep breaths. Close your eyes for a couple of minutes, and deliberately relax the muscle groups that are associated with tension, such as the shoulders, back, neck, and jaw. Attitude. Maintaining a positive attitude is important. It helps keep you going and helps keep you from becoming discouraged. Eye Care. It’s very important to have your eyes checked! Human eyes weren’t designed to constantly focus on something less than an arm’s length away. Our eyes were designed for survival, spotting food and enemies at a distance. Your eyes will be under tremendous stress because of prolonged reading, which can result in headaches, fatigue, nausea, squinting, or eyes that burn, ache, water, or tire easily. Be sure to tell your eye doctor that you are studying to pass an exam (bring this textbook and the Code book with you) and that you expect to do a tremendous amount of reading and writing. Reading glasses can reduce eye discomfort. Reducing Eye Strain. Be sure to look up occasionally, away from near tasks to distant objects. Your work area should be three times brighter than the rest of the room. Don’t read under a single lamp in a dark room. Try to eliminate glare. Mixing of fluorescent and incandes­ cent lighting can be helpful. Posture. Sit up straight, with your chest up and your shoulders back, so both eyes are an equal distance from what you’re viewing. Training. Preparing for the exam is the same as training for any other event. Get plenty of rest and avoid intoxicating drugs, including alco­ hol. Stretching or exercising each day for at least 10 minutes helps you get in a better mood. Eat light meals such as pasta, chicken, fish, vegetables, fruit, and so forth. Try to avoid red meat, butter, sugar, salt, and high-fat content foods. They slow you down and make you tired and sleepy.

Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

Passing Your Exam

Set Your Study Program Make Sure You Start to Prepare Early Enough. Many people fail an exam because they haven’t given themselves enough time to properly prepare and study. Determine the date of your exam, and then work backwards to be sure that you give yourself the time that you need to thoroughly cover the material you know you will be tested on. Getting Organized. Our lives are so busy that simply making time for homework and exam preparation is almost impossible. You can’t waste time looking for a pencil or missing paper. Keep everything you need together. Maintain folders, one for notes, one for exams and answer keys, and one for miscellaneous items. Study Location. It’s very important that you have a private study area available at all times. Keep your materials there. The dining room table isn’t a good spot, if you have other options. Time Management. Time management and planning are very important. There simply aren’t enough hours in the day to complete everything. Make a schedule that allows time for work, rest, study, meals, family, and recreation. Establish one that’s consistent from day to day. Have a calendar and immediately plan your exam preparation schedule. Try to follow the same routine each week and try not to become overtired. Learn to pace yourself to accomplish as much as you can without the need for cramming. Speak Up in Class. If you’re in a classroom setting, the most import­ ant part of the learning process is participation. If you don’t under­ stand the instructor’s point, ask for clarification. Don’t seek to get attention by asking questions to which you already know the answers. Study With a Friend. Studying with a friend can make learning more enjoyable. You can push and encourage each other. You’re more likely to study if someone else is depending on you. Students who study together perform above average because they try differ­ ent approaches and explain their solutions to each other. This kind of interaction is a significant aid to learning retention. Those who study alone may spend much of their time reading and rereading the text and trying the same approach time after time with less success. Study Anywhere/Anytime. To make the most of your limited time, always keep a copy of the book(s) with you. Any time you find a free minute, study! Continue to study any chance you have. You can study at the

supply house when waiting for your material; you can study during your coffee break, or even while you’re at the doctor’s office. Become creative! You Need to Find Your Best Study Time. For some, it’s late at night when the house is quiet. For others, it’s the first thing in the morning before things get going. Set Priorities. Once you begin your study, stop all phone calls, TV shows, radio, snacks, and other interruptions. You can always take care of it later.

Prepare for the Exam Get Advance Information. Check with your Local/State Examining Boards well in advance so that you understand how your exam will work, and what’s expected of you. Most states require you to first register with the State Electrical Licensing Board, and once you have been approved, you would contact the agency that administers their exams, if it’s not the State Board itself. The Candidate Bulletin will provide you with information about the examination and the applica­ tion process. You can visit the website www.MikeHolt.com/license for state contact information, licensing requirements, and where avail­ able, the Candidate Bulletin. Always verify with your State office to be sure this information is the most up-to-date. Some of the things you’ll want to clarify: • Which edition of the NEC you will be tested on, and if the exam is open book or not. • What reference books you are allowed to bring in and if and how you are allowed to mark them up. • If you will be doing a paper-and-pencil test, or a computer­ ized one; take any sample online test(s) on the testing com­ pany’s website so you’re familiar with the testing process. • What personal items are permitted in the testing room: If they allow you to bring in a watch, or a calculator [and if so, what kind is allowed]. Consider bringing a jacket in case the room is cold. • Meals. It’s a good idea to pack a lunch rather than going out if a lunch break is allowed— it can give you a little extra time to review the material for the afternoon portion of the exam, and it reduces the chance of coming back late. However, many testing companies will not allow you to bring food into the exam room but might provide a locker in which you can store it.

Mike Holt Enterprises, Inc. • www.MikeHolt.com • 888.NEC.C0DE (632.2633)

XV

Passing Your Exam

•

Know where the exam is going to take place and how long it takes to get there. Arrive at least 30 minutes early.

Learn How to Use Your Books. One of the best ways to be prepared is to know how to use your Code book. Check whether or not you can take your own Code book into the exam, if you are allowed to mark it up, and which kind of notes are allowed, if you can highlight informa­ tion in different colors, and if you are allowed to tab it. Become really efficient at using your Code book so you can be ready in a timed test environment. Review the table of contents and graphics in all your study books. This will help you to develop a sense of the material. As you read the text, continually ask yourself questions. This should help you to develop a better understanding of the text’s message.

Taking the Exam Being prepared for an exam means more than just knowing electri­ cal concepts, the Code, and the calculations. Have you felt prepared for an exam, and then choked when actually taking it? Many good and knowledgeable people didn’t pass their exam because they didn’t know “how to take an exam.” Taking exams is a learned process that takes practice and involves strategies. The following suggestions are designed to help you develop your strategies: Relax. This is easier said than done, but it’s one of the most important factors in passing your exam. Stress and tension cause us to choke or forget. Everyone has had experiences where they became tense and couldn’t think straight. The first step is becoming aware of the ten­ sion, and the second is to make a deliberate effort to relax. Make sure you’re comfortable; wear layers so you can remove clothing if you’re hot, or put on a jacket if you’re cold. There are many ways to relax and you have to find a method that works for you. Two of the easiest methods that work very well for many people follow: Breathing Technique: Take a few slow deep breaths every few min­ utes. Don’t confuse this with hyperventilation, which is abnormally fast breathing. Single-Muscle Relaxation: When we’re tense or stressful, many of us do things like clench our jaw, squint our eyes, or tense our shoulders without even being aware of it. If you find a muscle group that does this, deliberately relax that one group. The rest of the muscles will

automatically relax also. Try to repeat this every few minutes, and it will help you stay more relaxed during the exam. Understand the Question. To answer a question correctly, you must first understand it. One word in a question can totally change its meaning. Carefully read every word of every question. Underlining key words in the question will help you focus. Skip the Difficult Questions. Contrary to popular belief, you don’t have to answer one question before going on to the next one. The irony is that the question you become stuck on is one that you’ll probably answer wrong anyway. This will result in not having enough time to answer the easy questions. You’ll become all stressed-out worrying that you won’t complete the exam on time, and a chain reaction starts. More people fail their exams this way than for any other reason. Most states are using computerized testing, and have a way for you to “ mark” a question for later review and return to that question after you’ve answered the easier ones. Make sure you mark question number one for further review, because on some tests they limit you to returning no further than to the first question you marked. The following strategy should be used to avoid getting into this situation: First Pass: Answer the questions you know. Give yourself about 30 seconds for each question. If you can’t find the answer in your refer­ ence book within the 30 seconds, go on to the next question. Chances are that you’ll come across the answers while looking up another question. The total time for the first pass should be 25 percent of the exam time. Second Pass: This pass is done the same as the first one except that you allow a little more time for each question, about 60 seconds. If you still can’t find the answer, go on to the next one. Don’t get stuck. The total time for the second pass should be about 30 percent of the exam time. Third Pass: See how much time is left and subtract 30 minutes. Spend the remaining time equally on each question. If you still haven’t answered the question, it’s time to make an educated guess. Never leave a question unanswered. Fourth Pass: Use the last 30 minutes of the exam for review. Read each question and verify that you selected the correct answer. If it’s a paper and pencil test verify that you you transferred the answers carefully to the answer key. With the remaining time, see if you can find the answer to those questions for which you made an estimated guess.

Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

Passing Your Exam Guessing. When time is running out and you still have unanswered questions, GUESS! Never leave a question unanswered. You can improve your chances of getting a question correct by the process of elimination. Many times, there are a couple of choices that can be easily eliminated as illogical or outside the range of expected answers. How do you pick one of the remaining answers? Some people toss a coin; others count up how many of the answers were As, Bs, Cs, and Ds and use the one with the most as the basis for their guess. Checking Your Work. The first thing to check (and you should be watching out for this during the whole exam) is to make sure you mark the answer in the correct spot, whether taking a paper- or computer-based exam. People have failed the exam by one-half of a point. When they reviewed their exam, they found they correctly answered several questions on the test booklet, but marked the wrong spot on the exam answer sheet. They knew the answer was “(b) False” but marked in “(d)” in error. Another thing to be very careful of, is marking the answer for, let’s say Question 7, in the spot reserved for Question 8.

Things To Be Careful Of • Don’t get stuck on any one question. • Read each question carefully. ® Be sure you mark the answer in the correct spot on the screen, or answer sheet if the test is paper-based. • Don’t become flustered or extremely tense.

Summary • • • •

Changing Answers. When re-reading the question and checking the answers during the fourth pass, resist the urge to change an answer. In most cases, your first choice is best and, if you aren’t sure, stick with that choice. Only change answers if you’re positive that you made a mistake. Paper and pencil multiple choice exams are graded electronically so be sure to thoroughly erase any answer that you changed. Also erase any stray pencil marks from the answer sheet.

• °

Rounding Off. You should always round your answers to the same number of places as the exam’s answers.

•

Example: If an exam has multiple choice of: (a) 2.10

(b) 2.20

(c) 2.30

(d) none of these

And your calculation comes out to 2.16; don’t choose the answer (d) none of these. The correct answer is (b) 2.20, because the answers in this case are rounded off to the nearest tenth. Example: It can be rounded to tens, such as: (a) 50

(b) 60

(c) 70

• • •

Make sure everything is ready and packed the night before the exam. Don’t try to cram the night before the exam— if you don’t know it by then, it’s too late! Have a good breakfast. Get the thermos and energy snacks ready. Take all your reference books. Let the proctors tell you what you can’t use. Know where the exam is to be held and arrive early. Bring identification and your confirmation papers from the license board if this is required. Review your NEC while you wait for your exam to begin. Try to stay relaxed. Determine the time per question for each pass and don’t forget to save time to go back to the ones you skipped over. Remember— in the first pass answer only the easy ques­ tions. In the second, spend a little more time with each question, but don’t become stuck. In the third pass, use the remainder of the time minus 30 minutes to answer the remaining questions. In the fourth pass, check your work.

Watch Mike’s video clip on How to Prepare for an Electrical Exam. Use this QR code to go directly to the video, or visit www.MikeHolt.tv to see this and other video clips.

(d) none of these

For this group, an answer such as 67 will be (c) 70, while an answer of 63 will be (b) 60. The general rule is to check the question’s choice of answers and then round off your answer to match it.

Mike Holt Enterprises, Inc. • www.MikeHolt.com • 888.NEC.C0DE (632.2633)

Passing Your Exam

Additional Products To Help You Learn Electrical Theory DVD Library Ifc T R IC A L Only when you truly know electrical TKEOB? theory can you have confidence in the practical aspects of your electrical work. Build your electrical foundation with Mike’s Electrical Theory Library. When you finish this program you will understand: basics of electricity, how electricity is made, circuit types, electrical formulas, magnetism and electromagnetism, practical applications of electricity (wiring systems and equipment), grounding and bonding, and National Electrical Code rules.

This program includes the Basic Electrical Theory textbook and the following DVDs: ® Electrical Fundamentals and Basic Electricity • Electrical Circuits, Systems and Protection • Alternating Current, Motors, Generators, and Transformers Call our office at 888.NEC.C0DE (632.2633), scan this QR Code, or visit www.MikeHolt.com/theory to review our products.

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Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

ABOUT THE

■

NATIONAL ELECTRICAL CODE m

The National Electrical Code is written for persons who understand electrical terms, theory, safety procedures, and electrical trade prac­ tices. These individuals include electricians, electrical contractors, electrical inspectors, electrical engineers, designers, and other qual­ ified persons. The Code isn’t written to serve as an instructional or teaching manual for untrained individuals [90.1(A)], Learning to use the NEC can be likened to learning the strategy needed to play the game of chess well; it’s a great game if you enjoy mental warfare. When learning to play chess, you must first learn the names of the game pieces, how they’re placed on the board, and how each one is moved. Once you understand the fundamentals, you’re ready to start playing the game. Unfortunately, at this point all you can do is make crude moves, because you really don’t understand how all the information works together. To play chess well, you’ll need to learn how to use your knowledge by working on subtle strategies before you can work your way up to the more intriguing and complicated moves. The Code is updated every three years to accommodate new electrical products and materials, changing technologies, improved installation techniques, and to make editorial refinements to improve readability and application. While the uniform adoption of each new edition of the NEC is the best approach for all involved in the electrical industry, many inspection jurisdictions modify the Code when it’s adopted. To further complicate this situation, the NEC allows the authority having jurisdic­ tion, typically the “ Electrical Inspector,” the flexibility to waive specific Code requirements, and to permit alternative methods. This is only allowed when he or she is assured the completed electrical installation is equivalent in establishing and maintaining effective safety [90.4]. Keeping up with requirements of the Code should be the goal of everyone involved in the safety of electrical installations. This includes electrical installers, contractors, owners, inspectors, engineers, instructors, and others concerned with electrical installations.

About the 2 0 U NEC The actual process of changing the Code takes about two years, and it involves hundreds of individuals making an effort to have

the NEC as current and accurate as possible. Let’s review how this process worked for the 2014 NEC: Step 1. Proposals— November, 2011. Anybody can submit a pro­ posal to change the Code before the proposal closing date. Thousands of proposals were submitted to modify the 2011 NEC and create the 2014 Code. Of these proposals, several hundred rules were revised that significantly affect the electrical industry. Some changes were editorial revisions, while others were more significant, such as new articles, sections, exceptions, and Informational Notes. Step 2. Code-Making Panel(s) Review Proposals— January, 2012. All Code change proposals were reviewed by Code-Making Panels. There were 19 panels in the 2014 revision process who voted to accept, reject, or modify proposals. Step 3. Report on Proposals (ROP)— July, 2012. The voting of the Code-Making Panels on the proposals was published for public review in a document called the “Report on Proposals,” frequently referred to as the “ ROP.” Step 4. Public Comments— October, 2012. Once the ROP was avail­ able, public comments were submitted asking the Code-Making Panel members to revise their earlier actions on change proposals, based on new information. The closing date for “Comments” was October, 2012. Step 5. Comments Reviewed by Code Panels— December, 2012. The Code-Making Panels met again to review, discuss, and vote on public comments. Step 6. Report on Comments (ROC)— March, 2013. The voting on the “Comments” was published for public review in a document called the “ Report on Comments,” frequently referred to as the “ ROC.” Step 7. Electrical Section— June, 2013. The NFPA Electrical Section discussed and reviewed the work of the Code-Making Panels. The Electrical Section developed recommendations on last-minute motions to revise the proposed NEC draft that would be presented at the NFPA’s annual meeting. Step 8. NFPA Annual Meeting— June, 2013. The 2014 NEC m s voted by the NFPA members to approve the action of the Code-Making Panels at the annual meeting, after a number of motions (often called “floor actions” or “ NITMAMs”) were voted on.

Mike Holt Enterprises, Inc. • www.MikeHolt.com • 888.NEC.C0DE (632.2633)

xix

About the National Electrical Code

Step 9. Standards Council Review Appeals and Approves the 2014 NEC— July, 2013. The NFPA Standards Council reviewed the record of the Code-making process and approved publication of the 2014 NEC. Step 10. 2014 NEC Published— September, 2013. The 2014 National Electrical Code was published, following the NFPA Board of Directors review of appeals.

Small Words, Grammar, and Punctuation It’s not only the technical words that require close attention, because even the simplest of words can make a big difference to the appli­ cation of a rule. The word “or” can imply alternate choices for wiring methods, while “and” can mean an additional requirement. Let’s not forget about grammar and punctuation. The location of a comma can dramatically change the requirement of a rule.

Author’s Comment: a

Proposals and comments can be submitted online at the NFPA website (www.nfpa.org). From the homepage, click on “Codes and Standards”, then find NFPA 70 (NationalElectrical Code). From there, follow the on screen instructions to down­ load the proposal form. The deadline for proposals to create the 2017 National Electrical Code will be around November of 2014. If you would like to see something changed in the Code, you’re encouraged to participate in the process.

Not a Game Electrical work isn’t a game, and it must be taken very seriously. Learning the basics of electricity, important terms and concepts, as well as the basic layout of the NEC gives you just enough knowledge to be dangerous. There are thousands of specific and unique applica­ tions of electrical installations, and the Code doesn’t cover every one of them. To safely apply the NEC, you must understand the purpose of a rule and how it affects the safety aspects of the installation.

Slang Terms or Technical Jargon Electricians, engineers, and other trade-related professionals use slang terms or technical jargon that isn’t shared by all. This makes it very difficult to communicate because not everybody understands the intent or application of those slang terms. So where possible, be sure you use the proper word, and don’t use a word if you don’t under­ stand its definition and application. For example, lots of electricians use the term “pigtail” when describing the short conductor for the connection of a receptacle, switch, luminaire, or equipment. Although they may understand it, not everyone does.

NEC Style and Layout Before we get into the details of the NEC, we need to take a few moments to understand its style and layout. Understanding the struc­ ture and writing style of the Code is very important before it can be used and applied effectively. The National Electrical Code is organized into ten major components.

The A/FC contains many technical terms, so it’s crucial for Code users to understand their meanings and their applications. If you don’t understand a term used in a Code rule, it will be impossible to prop­ erly apply the NEC requirement. Be sure you understand that Article 100 defines the terms that apply to two or more Code articles. For example, the term “ Dwelling Unit” is found in many articles; if you don’t know what a dwelling unit is, how can you apply the require­ ments for it?

1. Table of Contents 2. Article 90 (Introduction to the Code) 3. Chapters 1-9 (major categories) 4. Articles 90-840 (individual subjects) 5. Parts (divisions of an article) 6. Sections and Tables (NEC requirements) 7. Exceptions (Code permissions) 8. Informational Notes (explanatory material) 9. Annexes (information) 10. Index

In addition, many articles have terms unique for that specific article and definitions of those terms are only applicable for that given arti­ cle. For example, Section 250.2 contains the definitions of terms that only apply to Article 250— Grounding and Bonding.

1. Table of Contents. The Table of Contents displays the layout of the chapters, articles, and parts as well as the page numbers. It’s an excellent resource and should be referred to periodically to observe the interrelationship of the various NEC components. When attempting to

NEC Terms and Concepts

Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

About the National Electrical Code

locate the rules for a particular situation, knowledgeable Code users often go first to the Table of Contents to quickly find the specific NEC Part that applies. 2. Introduction. The NEC begins with Article 90, the introduction to the Code. It contains the purpose of the NEC, what’s covered and what isn’t covered along with how the Code is arranged. It also gives information on enforcement and how mandatory and permissive rules are written as well as how explanatory material is included. Article 90 also includes information on formal interpretations, examination of equipment for safety, wiring planning, and information about format­ ting units of measurement. 3. Chapters. There are nine chapters, each of which is divided into articles. The articles fall into one of four groupings: General Requirements (Chapters 1-4), Specific Requirements (Chapters 5-7), Communications Systems (Chapter 8), and Tables (Chapter 9). Chapter 1— General Chapter 2— Wiring and Protection Chapter 3— Wiring Methods and Materials Chapter 4— Equipment for General Use Chapter 5— Special Occupancies Chapter 6— Special Equipment Chapter 7— Special Conditions Chapter 8— Communications Systems (Telephone, Data, Satellite, Cable TV and Broadband) Chapter 9— Tables-Conductor and Raceway Specifications 4. Articles. The NEC contains approximately 140 articles, each of which covers a specific subject. For example: Article 110— General Requirements Article 250— Grounding and Bonding Article 300— General Requirements for Wiring Methods and Materials Article 430— Motors and Motor Controllers Article 500— Hazardous (Classified) Locations Article 680— Swimming Pools, Fountains, and Similar Installations Article 725— Remote-Control, Signaling, and Power-Limited Circuits Article 800— Communications Circuits 5. Parts. Larger articles are subdivided into parts. Because the parts of a Code article aren’t included in the section numbers, we have a tendency to forget what “ part” the NEC rule is relating to. For exam­ ple, Table 110.34(A) contains working space clearances for electrical equipment. If we aren’t careful, we might think this table applies to all electrical installations, but Table 110.34(A) is located in Part III, which

only contains requirements for “Over 600 Volts, Nominal” installations. The rules for working clearances for electrical equipment for systems 600V, nominal, or less are contained in Table 110.26(A)(1), which is located in Part II— 600 Volts, Nominal, or Less. 6. Sections and Tables. Sections. Each NEC rule is called a “ Code Section.” A Code section may be broken down into subsections by letters in parentheses (A), (B), and so on. Numbers in parentheses (1), (2), and so forth, may further break down a subsection, and lowercase letters (a), (b), and so on, fur­ ther break the rule down to the third level. For example, the rule requir­ ing all receptacles in a dwelling unit bathroom to be GFCI protected is contained in Section 210.8(A)(1). Section 210.8(A)(1) is located in Chapter 2, Article 210, Section 8, Subsection (A), Sub-subsection (1). Many in the industry incorrectly use the term “Article” when refer­ ring to a Code section. For example, they say “Article 210.8,” when they should say “Section 210.8.” Section numbers in this textbook are shown without the word “ Section,” unless they begin a sentence. For example, Section 210.8(A) is shown as simply 210.8(A). Tables. Many NEC requirements are contained within tables, which are lists of Code rules placed in a systematic arrangement. The titles of the tables are extremely important; you must read them carefully in order to understand the contents, applications, limitations, and so forth, of each table in the NEC. Many times notes are provided in or below a table; be sure to read them as well since they’re also part of the requirement. For example, Note 1 for Table 300.5 explains how to measure the cover when burying cables and raceways, and Note 5 explains what to do if solid rock is encountered. 7. Exceptions. Exceptions are Code requirements or permissions that provide an alternative method to a specific rule. There are two types of exceptions— mandatory and permissive. When a rule has several exceptions, those exceptions with mandatory requirements are listed before the permissive exceptions. Mandatory Exceptions. A mandatory exception uses the words “shall” or “shall not.” The word “shall” in an exception means that if you’re using the exception, you’re required to do it in a particular way. The phrase “shall not” means it isn’t permitted. Permissive Exceptions. A permissive exception uses words such as “shall be permitted,” which means it’s acceptable (but not mandatory) to do it in this way. 8. Informational Notes. An Informational Note contains explanatory material intended to clarify a rule or give assistance, but it isn’t a Code requirement.

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xxi

About the National Electrical Code

9. Annexes. Annexes aren’t a part ot the NEC requirements, and are included in the Code for informational purposes only. Annex A. Product Safety Standards Annex B. Application Information for Ampacity Calculation Annex C. Raceway Fill Tables for Conductors and Fixture Wires of the Same Size Annex D. Examples Annex E. Types of Construction Annex F. Critical Operations Power Systems (COPS) Annex G. Supervisory Control and Data Acquisition (SCADA) Annex H. Administration and Enforcement Annex I. Recommended Tightening Torques Annex J. ADA Standards for Accessible Design 10. Index. The Index at the back of the Code book is helpful in locat­ ing a specific rule. Changes to the NEC since the previous edition(s), are identified by shading, but rules that have been relocated aren’t identified as a change. A bullet symbol is located on the margin to indicate the location of a rule that was deleted from a previous edition. New arti­ cles contain a vertical line in the margin of the page.

Different Interpretations Some electricians, contractors, instructors, inspectors, engineers, and others enjoy the challenge of discussing the NEC requirements, hopefully in a positive and productive manner. This give-and-take is important to the process of better understanding the Code require­ ments and application(s). However, if you’re going to participate in an NEC discussion, please don’t spout out what you think without having the actual Code book in your hand. The professional way of discuss­ ing an NEC requirement is by referring to a specific section, rather than talking in vague generalities.

How to Locate a Specific Requirement How to go about finding what you’re looking for in the Code book depends, to some degree, on your experience with the NEC. Code experts typically know the requirements so well they just go to the correct rule without any outside assistance. The Table of Contents might be the only thing very experienced NEC users need to locate the requirement they’re looking for. On the other hand, average Code users should use all of the tools at their disposal, including the Table of Contents and the Index.

Table of Contents. Let’s work out a simple example: What NEC rule specifies the maximum number of disconnects permitted for a ser­ vice? If you’re an experienced Code user, you’ll know Article 230 applies to “Services,” and because this article is so large, it’s divided up into multiple parts (actually eight parts). With this knowledge, you can quickly go to the Table of Contents and see it lists the Service Equipment Disconnecting Means requirements in Part VI. Author’s Comment: ■ The number 70 precedes all page numbers because the NEC is NFPA Standard Number 70. Index. If you use the Index, which lists subjects in alphabetical order, to look up the term “service disconnect,” you’ll see there’s no listing. If you try “ disconnecting means,” then “services,” you’ll find that the Index indicates that the rule is located in Article 230, Part VI. Because the NEC doesn’t give a page number in the Index, you’ ll need to use the Table of Contents to find it, or flip through the Code book to Article 230, then continue to flip through pages until you find Part VI. Many people complain that the A/fConly confuses them by taking them in circles. As you gain experience in using the Code and deepen your understanding of words, terms, principles, and practices, you’ll find the NEC much easier to understand and use than you originally thought.

Customizing Your Code Book One way to increase your comfort level with the Code book is to cus­ tomize it to meet your needs. (Be aware that if you’re using it to take an exam, you should check to see which kind of markings are permitted.) Highlighting. Highlight those requirements in the Code that are the most important or relevant to you. Use one color for general inter­ est and a different one for important requirements you want to find quickly, including the Index and the Table of Contents. Underlining. Underline or circle key words and phrases with a red pen (not a lead pencil) and use a short ruler or other straightedge to keep lines straight and neat. This is a very handy way to make important requirements stand out. A short ruler or other straightedge also comes in handy for locating specific information in the many Code tables. Tabbing the NEC. By placing tabs on Code articles, sections and tables, it will make it easier for you to use the NEC. Too many tabs defeat the purpose. Order a set designed by Mike Holt, online at www. MikeHolt.com/14tabs, or by calling 888.632.2633

Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

ABOUT THE AUTHOR Special Acknowledgments— First, I want to thank God for my godly wife who’s always by my side and my children, Belynda, Melissa, Autumn, Steven, Michael, Meghan, and Brittney.

Mike Holt— Author Founder and President Mike Holt Enterprises Groveland, FL www.MikeHolt.com Mike Holt worked his way up through the electrical trade. He began as an apprentice electrician and became one of the most rec­ ognized experts in the world as it relates to electrical power installations. He’s worked as a journeyman electri­ cian, master electrician, and electrical contractor. Mike’s experience in the real world gives him a unique understanding of how the NEC relates to electrical installations from a practical standpoint. You’ll find his writing style to be direct, nontechnical, and powerful.

A special thank you must be sent to the staff at the National Fire Protection Association (NFPA), publishers of the NEC— in particular Jeff Sargent for his assistance in answering my many Code ques­ tions over the years. Jeff, you’re a “first class” guy, and I admire your dedication and commitment to helping others understand the NEC. Other former NFPA staff members I would like to thank include John Caloggero, Joe Ross, and Dick Murray for their help in the past. A personal thank you goes to Sarina, my long-time friend and office manager. It’s been wonderful working side-by-side with you for over 25 years nurturing this company’s growth from its small beginnings.

Did you know Mike didn’t finish high school? So if you struggled in high school or didn’t finish at all, don’t let it get you down. However, realizing that success depends on one’s continuing pursuit of edu­ cation, Mike immediately attained his GED, and ultimately attended the University of Miami’s Graduate School for a Master’s degree in Business Administration. Mike resides in Central Florida, is the father of seven children, has five grandchildren, and enjoys many outside interests and activi­ ties. He’s a nine-time National Barefoot Water-Ski Champion (1988, 1999, 2005-2009, 2012-2013). He’s set many national records and continues to train year-round at a World competition level (www. barefootwaterskier.com). What sets him apart from some is his commitment to living a bal­ anced lifestyle; placing God first, family, career, then self.

Mike Holt Enterprises, Inc. • www.MikeHolt.com • 888.NEC.C0DE (632.2633)

xxiii

ABOUT THE ILLUSTRATOR Mike Culbreath— Illustrator Graphic Illustrator Alden, Ml www.MikeHolt.com Mike Culbreath devoted his career to the electrical industry and worked his way up from an apprentice electrician to master electrician. He started in the electrical field doing residential and light commercial construction. He later did service work and custom electrical instal­ lations. While working as a journeyman electrician, he suffered a serious on-the-job knee injury. As part of his rehabilitation, Mike com­ pleted courses at Mike Holt Enterprises and then passed the exam to receive hisMasterElectrician’s license. In 1986, with a keen interest in continuing education for electricians, he joined the staff to update material and began illustrating Mike Holt’s textbooks and magazine articles. He started with simple hand-drawn diagrams and cut-and-paste graphics. When frustrated by the limitations of that style of illus­ trating, he took a company computer home to learn how to operate some basic computer graphic software. Becoming aware that com­ puter graphics offered a lot of flexibility for creating illustrations, Mike took every computer graphics class and seminar he could to help develop his computer graphic skills. He’s now worked as an illustrator and editor with the company for over 25 years and, as Mike Holt has proudly acknowledged, has helped to transform his words and visions into lifelike graphics.

Special Acknowledgments— I would like to thank Ryan Jackson, an outstanding and very knowledgeable Code guy, and Eric Stromberg, an electrical engineer and super geek (and I mean that in the most complimentary manner, this guy is brilliant), for helping me keep our graphics as technically correct as possible. I also want to give a special thank you to Cathleen Kwas for making me look good with her outstanding layout design and typesetting skills and Toni Culbreath who proofreads all of my material, i would also like to acknowledge Belynda Holt Pinto, our Chief Operations Officer and the rest of the outstanding staff at Mike Holt Enterprises, for all the hard work they do to help produce and distribute these out­ standing products. And last but not least, I need to give a special thank you to Mike Holt for not firing me over 25 years ago when I “borrowed” one of his computers and took it home to begin the process of learning how to do computer illustrations. He gave me the opportunity and time needed to develop my computer graphic skills. He’s been an amaz­ ing friend and mentor since I met him as a student many years ago. Thanks for believing in me and allowing me to be part of the Mike Holt Enterprises family.

Originally from South Florida, Mike now lives in northern lower Michigan where he enjoys kayaking, photography, and cooking, but his real passion is his horses. Mike loves spending time with his children Dawn and Mac and his grandchildren Jonah and Kieley.

Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

ABOUT THE TEAM Editorial and Production Team special thanks goes to Toni Cuibreath for her outstand­ ing contribution to this project. She worked tirelessly to proofread and edit this publication. Her attention to I detail and dedication is irreplaceable. Many thanks to Cathleen Kwas who did the design, layout, and pro­ duction of this book. Her desire to create the best possible product for our customers is greatly appreciated. Also, thanks to Paula Birchfield who was the Production Coordinator of the textbook. She helped keep everything flowing and tied up all the loose ends. Thanks to Bruce Marcho for doing such an excellent job recording, editing, and producing our DVDs. Bruce has played a vital role in the production of our products for over 25 years.

video Team Victor Ammons Senior Electrical Engineer Hillsborough, NJ Victor Amnions was born in North Carolina, but has lived primarily in New Jersey since the age of five. He spent his high school and college vacation times working in a small, family owned electrical contracting firm. Vic married his high school sweetheart after sophomore year at North Carolina State University, from which both graduated with honors two years later. After four years in the Army, during which time their first child was born in Germany, Vic and Anne alternated getting MBA degrees. They now have three married chil­ dren and five grandchildren. After a number of years of running the family contracting business while his father’s health declined, Vic trained and turned the business over to his brother-in-law. He then returned to the engineering field in

which he had trained. Vic is a registered Professional Engineer in New Jersey and has served in an engineering management role for several engineering and architectural/engineering firms. He is now a senior electrical engineer for Jacobs in Morristown, New Jersey. Vic and Anne now reside in west central New Jersey in an area with a few dozen families among over a thousand acres of woodlands. Yes, there are still woodlands in New Jersey; they haven’t gotten around to paving them all yet. Vic and Anne, who will celebrate their 45th anni­ versary in 2014, are avid square and round dancers, and are active in several clubs. Vic is also a woodworker and an amateur beekeeper. Both are active in several capacities in their church.

Steve Arne Training Director AETech Electrical Training Center Rapid City, SD www.ElectricianClass.com Steve Arne has worked as an electrician, electrical contractor, and electrical instruc­ tor for 40 years. Nineteen years of his career were spent as a teacher and department head in post-secondary education. Since Steve retired from Western Dakota Tech in 2003, he has provided industry-direct training special­ izing in continuing education Electrical Code classes and Electrician Exam Preparation classes for Journeyman and Master’s level exams. In 2012, Steve and Deb’s son Ryan returned to Rapid City South Dakota to help Steve expand his training business at AETech Electrical Training Center to include a 640 hour full-time pre-employment Apprentice Electrician Training Program. Steve is currently a Board Member of the South Dakota Electrical Commission and serves on the State and Local Chapter Boards of the South Dakota Electrical Council. Steve worked with Mike Holt as a technical editor and video team participant beginning in 2002, and uses Mike’s books in his classes. He is very thankful to have been associated with an industry leader like Mike who has a real heart to help others and provides excellent training products to help students in the electrical industry.

Mike Holt Enterprises, Inc. 3 www.M ikeHolt.com • 888.NEC.C0DE (632.2633)

Jj

About the Team

Steve and his wife Deb live in Rapid City, South Dakota where they’re both active in their church and community, and love to spend time with their children and grandchildren. The highest priority for both of them is putting God first in both their home and business.

Dennis Carlson Master Electrician/Electrical Trainer Rocky Mountain Electrical Training Steamboat Springs, CO www.RMETtech.com Dennis Carlson has been in the electrical trade for over 30 years, and has held the positions of Project Superintendent, Field Engineer, Start-up-Coordinator, General Foreman, Foreman, Journeyman, and Craft Training Manager. He’s certified with NCCER as a Master Trainer, and is a professor for Victoria University, Australia as a Master Trainer. In April 2013, he graduated from Salem International University with a Bachelor’s degree in Business Administration. Dennis is a Master Electrician/Contractor licensed in several states, and a member of the NFPA and IAEI. He’s trained thousands of elec­ tricians in the last 20 years using Mike Holt’s material and has been working with Mike for the last 13 years on his Exam Prep courses. Dennis and his wife, Cindy, live in Steamboat Springs, Colorado, where they’re both active in enjoying God’s great outdoors, their church, and the community. They have two children, Brandy, a graphics arts designer who lives in Greeley Colorado, and Josh who is an electrical estimator in Denver Colorado.“ Life is too short to do only one thing so you have to prepare yourself for the next challenge. Never stop learn­ ing and striving to achieve those goals that you have.”

Michael P. Pillarella Owner, Pillarella School of Electrical Code & Theory Assonet, MA www.pillarellaschool.com Mike Pillarella holds a Master’s Degree in Education, a BS in Industrial Arts, and an AE in Electric and Electronics Engineering Technology. He became a Certified Electrical Related & Shop Instructor in the Massachusetts School system. In 1988, Mike established EMI Electrical Contractors, Inc. providing

electrical maintenance and installation services in the New England South Shore area. In 2007, Mike established the proprietary Pillarella School of Electrical Code & Theory. While serving as an electrical inspector, he became increasingly aware of the areas in which more education was needed and structured his curriculum accordingly. His proudest moment as an instructor was when his son, Tim, earned his Master’s license— the first attendee of the school to do so. Mike holds master licenses in Massachusetts, Rhode Island, New Hampshire, Vermont and Maine, and is an approved CEU provider and OSHA Outreach Construction instructor in these states. He’s President of the Massachusetts Electrical Contractors Association, Bristol Chapter, and President and Education Chairman of the International Association of Electrical Inspectors, Roger Williams Chapter. Mike served for sev­ eral years as an instructor for the Massachusetts Firefighters’ Academy teaching Electrical Safety, Fire, and Arson Investigation. He served his community for 17 years as a firefighter, EMT, and finished as Fire Chief. Mike is extremely proud of his wife, Valerie, and their three children, Amy, Michael, and Tim.

Eric Stromberg Electrical Engineer/Instructor Stromberg Engineering, Inc. Los Alamos, NM www.strombergengineering.com. Eric Stromberg worked as a journeyman electrician, before and during the time he attended college. When he graduated with a degree in Electrical Engineering in 1982, he took a job as an electronics technician. Eric became a licensed fire alarm installation superintendent and spent the next seven years installing and maintaining life safety systems in high-rise buildings. In 1989, he went to work for Dow Chemical, where he designed power distribution systems for world-class industrial facilities. Eric began teaching National Electrical Code classes to engineers in 1997. He received his professional engineering license, for the State of Texas, in 2003 and, in 2005, started Stromberg Engineering. In 2013, Eric retired from Dow Chemical and now lives in the moun­ tains of northern New Mexico. Eric’s oldest daughter, Ainsley, lives in Boston, Massachusetts with her husband Nathan. His son, Austin, is in the Air Force and is stationed at Minot, North Dakota. His youngest daughter, Brieanna, is a singer/songwriter who lives in Austin, Texas.

Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

CHAPTER

1

ELECTRICAL THEORY

ESSENTIAL FOR JOURNEYMAN AND MASTER/CONTRACTOR LICENSING EXAMS Unit 1 Electrician’s Math and Basic Electrical Formulas Unit 2 Electrical Circuits Unit 3 Understanding Alternating Current Unit 4 Motors and Transformers Note: This chapter is a practice review of theory as it relates to basic electrical formulas and calculations. Most State electri­ cal exams include questions on general electrical theory and you need a solid foundation to pass your exam. If after working through this Chapter you feel like you need additional in-depth training on Theory, then Mike Holt’s Basic Electrical Theory Library (textbook and DVDs) will give you a comprehensive overview of this topic. ELECTRICAL

THEORY

For additional help, visit www.MikeHolt.com/theory or call 888.632.2633 to order your copy.

Mike Holt Enterprises, Inc.

Figure 2-5

Series Circuits - Resistance is Additive Conductor 1

R-i R3 Rt Rt

Appliance (Load) Copyright 2014 wAw.Mike.Holt.com

= 0.05 ohms, R2 = 0.15 ohm s = 7 .1 5 ohms, R4 = 0.15 ohms = 0.05 ohms + 0.15 ohm s + 7.15 ohm s + 0.15 ohm s = 7.50 ohms

Figure 2-7

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33

I

Unit 2 I Electrical Circuits

Voltage The voltage, also called “electromotive force” (EMF), provides the pres­ sure necessary to move electrons through the circuit. However, the power supply, the conductors, and the appliance all have resistance that opposes the current flow (although the power supply’s resis­ tance is usually ignored). The opposition (resistance) to current flow (amperes) results in a drop of the circuit voltage (voltage drop), and is calculated by the formula:

In addition, the voltage of the power supply is distributed or divided among the circuit resistors according to the Resistance Law of Propor­ tion. The Resistance Law of Proportion means that the supply voltage is distributed among ail the resistors, according to the proportion of resistance each resistor has relative to the total resistance. Figure 2-9 Law o f Proportion Voltage - R esistance P o w er S ource 0.67% C onductors 4% total (2% per conductor)

EVD = 1 x R

Power Source: Conductor 1: Conductor 2: Load:

Kirchoff’s Voltage Law Kirchoff’s Voltage Law states that in a series circuit, the sum of the voltages across all of the resistors (or “voltage drops”) in a series circuit is equal to the applied voltage. The voltage drop across each resistor is determined by the formula: Figure 2-8 Evo = I x R Copyright 2014 www.MikeHolt.com

I = Current of circuit R = Resistance of resistor 0.80V 2.40V 114.40V + 2.40V 120.00V

S eries C ircuits - V oltag e D rop is A d d itive ~ ' C onductor 1 - 0.15 ohm s 0.80V Power Supply 2.40V Conductor 1 114.40V Load 2.40V Conductor 2 .^Power S upply l i f i i a ------------x 120.00V , 0.05 ohm s v x C onductor 2 - 0.15 o h m s " ----------------------------- -------------------------------------------

A ppliance (Load) 7.15 ohm s

Copyright 2014

Determine the voltage drop of each resistor. VAVW.MikeHolt.com Formula: Evd = I x R Knowns: I = 16A (given) - Use given resistance of each component

Or...

Figure 2-8

E Vd

= I x Rt ,

= = = =

V oltage 120 x 0.0067 1 2 0 x 0 .0 2 1 2 0 x 0 .0 2 1 2 0 x 0 .9 5 3

= = = =

= = = =

0.67% 2.00% 2.00% 95.33%

0.8V 2.4V 2.4V 114.4V

16A x 16A x 1 6A x 16A x

E Vd

0.05 0.15 7.15 0.15

ohms ohms ohms ohms

= = = = =

Resistance Power Source Conductor No. 1 Appliance Conductor No. 2 Total

0.05 ohms 0.15 ohms 7.15 ohms + 0.15 ohms 7.50 ohms

Percentage

Voltage

0.67% 2.00% 95.33% 2.00% 100%

0.80V 2.40V 114.40V 2.40V 120.00V

Kirchoff’s Current Law

03

Power Supply EVd Conductor 1 EVd Appliance Load EVd Conductor 2 EVd Total Voltage Drop

Power Source: Conductor 1: Conductor 2: Load:

0.0067 0.02 0.02 0.953

Figure 2-9

16Ax 0.05 ohms 16Ax 0.15 ohms 16Ax 7.15 ohms 16Ax 0.15 ohms 16A x 7.50 ohms

Power Supply Conductor 1 Appliance Conductor 2 Total

R esistance 0.05£}/7.50n = 0.15Q/7.50f2 = 0.15a/7.50Q = 7 .1 5 fi/7 .5 0 n =

0.80V 2.40V 114.40V 2.40V

Kirchoff’s Current Law states that the sum of currents flowing into a junction equals the sum of currents flowing away from the junction. Another way to say it is that, current flowing through each resistor of a series circuit is the same. Figure 2-10 To calculate the current in a series circuit, the power-supply voltage (Es) and the total circuit resistance (RT) are required. The current of the circuit is determined by the formula: '=

= 16A x 7.50 ohms = 120V

es/Rt

In the following example, the current of the circuit is equal to 120W7.50 ohms = 16A. Since this is a series circuit, 16A is flowing through every component of the circuit.

Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

Electrical Circuits I Unit 2

Series Circuit - Current Remains the Same C o n d u c to r 1 = 0 .1 5 ohm s

------------------------ vA-

The same am ount of current is in every part of a series circuit. 120 V P ow er S up ply 0 .05 oh m s

A pp lia nce (lo ad) 7.15 ohm s

C o n d u c to r 2 = 0.15 ohm s

Formula: = Evd/R Proof: Determine the j Power Supply: 0.80V/0.05 ohms = 16A ' current in each resistor. Conductor 1: 2.40V/0.15 ohms = 16A Copyright 2014 Appliance Load: 114.40V/7.15 ohms = 16A www.MikeHalt.eom Conductor 2: 2.40V/0.15 ohms = 16A Or... use total resistance, I = Es/RT E = 120V, RT = 0.05 + 0.15 + 7.15 + 0.15 = I = 120V/7.50 ohms = 16A

Power can also be calculated according to the Resistance Law of Proportion. Resistance Percentage Power Power Source 0.05 ohms 0.67% 12.80W Conductor No. 1 2.00% 38.40W 0.15 ohms Appliance 7.15 ohms 95.33% 1.830.00W Conductor No. 2 + 0.15 ohms 2.00% 38.40W Total Resistance 7.50 ohms 100% 1.920.00W

2.3 Series Circuit Calculations When performing series circuit calculations, follow these steps: Figure 2-11

Figure 2-10 Author’s Comment: ■ The current flowing through each resistor can be calculated by I = E/R, where E (voltage) represents the voltage drop across the individual resistors, not the voltage source. Power Source Conductor No. 1 Appliance Conductor No. 2 Total Resistance

I= I= I= I= I=

0.80V/0.05 ohms 2.40V/0.15 ohms 114.40V/7.15 ohms 2.40V/0.15 ohms 120V/7.50 ohms

16A 16A 16A 16A 16A

Series Circuit Calculations Step 1 Determ ine the resistance o f each resistor in the circuit.

Conductor 1 ---------R2 = 0 .1 5 ft

Ri r 0 .0 5 ft , — la , A .rv 120 V Power Supply

R 3 = 7 .1 5 ft R4 = 0 .1 5 ft

Copyright 2014 www.MikeHort.com

Appliance -vF (Load) Conductor 2 Step 2: Calculate the total resistance of the circuit. R t = 0.05 + 0.15 + 7.15 + 0.15 ohm s = 7.50 ohm s

Power

Step 3: Determine the current of the circuit.

The power consumed in a series circuit equals the sum of the power consumed by all of the resistors in the series circuit. The Law of Con­ servation of Energy states that the power supply (a battery, and so forth) will only produce as much power as that consumed by the circuit elements. Power is a result of current flowing through a resistance and is calculated by the formula: P = I2 x R

I = E s /R t = 120V/7.50 ohm s = 16A Figure 2-11

Step 1: Determine the resistance of each resistive element in the circuit. Often, the resistance of each element is given in the problem. If you know the nameplate voltage and power (wattage) rating of the appliance or equipment, determine its resistance by the formula:

► Example Power Source Conductor No. 1 Appliance Conductor No. 2

R = EVP P= P= P= P=

16A2 x 0.05 ohms 16A2x 0.15 ohms 16A2x 7.15 ohms 16A2 x 0.15 ohms

12.80W 38.40W 1.830.40W 38.40W

E = Nameplate voltage rating P = Nameplate power rating Step 2:

Calculate the total resistance (RT) of the circuit: RT= R, + R2 + R3 + R4

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35

I

Unit 2 I Electrical Circuits

Step 3: The current of the circuit is determined by the formula.

2.5 Variations

I = Es/Rt Es = Voltage Source Rt = Total circuit resistance (Step 2)

There are often many different ways to solve an electrical circuit prob­ lem involving voltage, current, resistance, and power. It’s also often possible to verify or check one’s calculations by solving the problem different ways.

2.4 Power Calculations If you know the current of the circuit and the resistance of each resistor, the power of each resistor is determined by the formula: Figure 2-12

2.6 Series Circuit Notes

P = I2x R

Note 1: The total resistance of a series circuit is equal to the sum of all of the resistances of the circuit.

I2 = Current of circuit (squared) (Step 3) R = Resistance of the resistor (Step 1)

Note 2: Current is the same value through all of the resistances.

The power of the circuit is determined by adding up the power of all of the resistors, or by the formula:

Note 3: The sum of the voltage drops across all resistances equals the

P = I2 x RT

Note 4: The sum of the power consumed by all resistors equals the total power consumed by the circuit.

I2 = Current of the circuit (squared) Rt = Resistance total of the circuit

2.7 Series-Connected Power Supplies

S eries C ircu it - P ow er C alculations 120V Power ---J--j-—■—---------------- ------------------------

Supply 0.05 ohms

Copyright 2014 www.MikeHolt.com

f l6 p

Conductor 1 = 0.15 ohms

Conductor 2 = 0.15Aohms -------- / ----------Formula: P = I2 x R or P = I2 x RT

Power Supply: 16A 2 x 0.05 ohm s C onductor 1: 16A2 x 0.15 ohm s Appliance Load: 16A2 x 7.15 ohm s C onductor 2: 16A2 x 0.15 ohm s Total Power of C ircuit

= = = =

voltage of the source.

Appliance (load) 7.15 ohms

When power supplies are connected in series with each other, the total voltage of all of the power supplies connected in series is equal to the sum of all of the power supplies (provided the polarities are connected properly). Figure 2-13 and Figure 2-14 S e rie s-C onnected P o w er S u pply ► I I 1 1.50vj 1 > i ; ii. 5 o v l

12.80W 38.40W 1.830.40W 38.40W 1,920.OOW

Or... use total resistance, P = I2 x R T l = 16A, R T = 0 .0 5 + 0 .1 5 + 7 .1 5 + 0.15 = 7.50 ohm s P = 16A2 x 7.50 ohm s = 1.920W

I

f

6

1 V olts

Voltage is Additive.

K l

J I 150V I y jd is o v j // 1 —«------- :

All fo u r 1.50V one cell batteries are installed in series. I

Figure 2-12

Voltage is Additive. O ne battery is installed backw ards.

Figure 2-13

Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

Electrical Circuits I Unit 2

Series-Connected Power Supply D e lta /D e lta T ra n s fo rm e r S e ries-conn ected | seco n da ry w indings

S e ries-conn ected prim ary w indings

A parallel circuit has different characteristics than a series circuit. For one, the total resistance of a parallel circuit isn’t equal to the sum of the resistors. The total resistance in a parallel circuit is always less than any of the branch resistances. Adding more parallel resistances to the paths causes the total resistance in the circuit to decrease.

2.8 Practical Uses of the Parallel Circuit Parallel circuits are used for most building wiring. Secondary

Three transform ers connected in series form the basis of the delta-connected transformer. Figure 2-14

Receptacles When wiring receptacles on a circuit, they’re connected in parallel to each other. Figure 2-16

Part B— Parallel Circuits

Practical Use - Receptacles in Parallel Note: Pigtails on hots and neutral aren’t required.

Introduction “ Parallel” is a term used to describe a method of connecting electri­ cal components so there are two or more paths through which current may flow. A parallel circuit is one with several different paths over which the electricity can travel. It’s like a river that’s been divided up into smaller streams and all the streams come back to form the river once again. Figure 2-15

Ao (V )

mm ao 101

Branch 1

1 Ac

v

All three parts of this diagram represent three receptacles wired in parallel. 5A Load

Figure 2-16

Branch 2 ;; ^

Figures “A ” (top) and “B” (bottom ) are the sam e parallel circuit draw n in tw o different styles.

Q

T

Copyright 2014 wvw.MikeHoU.ccra

Parallel Circuit

o

T

< 0 A1 A2l p P Branch 1 | Branch 2

Lighting Another example is lights connected in parallel to each other. Figure 2-17A The major advantage of a parallel circuit is that if any branch of the circuit is opened or turned off, the power supply continues to provide voltage to the remaining parts of the circuit. Figure 2 -1 7B

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Figure 2-15

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Unit 2 I Electrical Circuits

Practical Uses - Parallel Connections in Equipment

Practical Use - Lights in Parallel

E lectric Furnace Heat Bank

(V )

O D ual-V oltage

L1 P r L2-

Copyright 2014 WAv.MikeHolt.com

An advantage of parallel circuits is that if any part of the circuit is opened, the remaining portions of the circuit still operate.

Copyright 2014 www.MikeHolt.com

115V

115 V ~ * 'N 4 X A A A ^ x

115/230V — • - \ L / Y Y Y \ J ) 1-P hase M otor

/ J . A A A_ -L. p I

7

L1_/■

L1 = T1,7 L2 = T2.8 L3 = T3,9 Tie Together T4,5,6 P a rallel-C onnected, D ual-Voltage, 3-P hase M otor

Low Voltage Connections

Figure 2-19

Figure 2-17

Other Uses Parallel circuits, also called “open-loop systems,” are used for fire alarm pull stations and smoke detectors. If any initiating device (pull station or smoke detector) closes, the signal circuit is complete and the alarm sounds. Figure 2-18

1

Practical Use of Parallel Connection

Voltage In a pure parallel circuit (one with no resistors in series with the par­ allel resistors), the voltage drop across each resistance is equal to the voltage supplied by the power source (ignoring any voltage drop in the source and conductors). Figure 2-20

! N.o. Pull Smoke I Heat | Relay ; £ OpenStation I Detector {Detector i Contacts 1 Loop . V Relay i ^ C ir c u it A J 1 Coil P arallel-C onnected. N.O. Sensing D evices Alarm-Sounding Circuit

t ! N.O. | y y ! Relay ! Contacts f | Close I Relay energized by pull station.

Alarm Activated © Smoke I Heat Detector Detector

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Figure 2-18

Often appliances, such as electric furnace heat banks and motors have their internal electrical components connected in parallel. Figure 2-19

I

It’s important to understand the relationship between voltage, current, power, and resistance in parallel circuits.

Alarm-Sounding Circuit

t________

1

2.9 Understanding Parallel Calculations

For the moment, we’ll ignore the voltage drop and power loss effects of the conductor and power supply as it’s usually much, much smaller than the voltage drop across the resistive elements.

Kirchoff’s Current Law In a parallel circuit, current from the power source will branch in dif­ ferent directions and magnitudes. The current in each branch is dependent on the resistance of each branch. Kirchoff’s Current Law states that the total current provided by the source to a parallel circuit is equal to the sum of the currents of all of the branches. The current in each branch is calculated by the formula: Figure 2-21

Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

Electrical Circuits I Unit 2

Author’s Comment:

Voltage - Parallel Circuits

■ The resistances and currents have been rounded off.

Power When current flows through a resistor, power is consumed. The power consumed by each branch of the parallel circuit is determined by the formulas: Vo indicates the ! I The voltage drop across voltage at the s o u r c e . each resistor is equal ------------------------------------ 1 to the source voltage. *The voltage drop of the conductors is ignored.

P = I2xR or P = E x l or P = E2/R The total power consumed in a parallel circuit equals the sum of the power in each branch. Figure 2-22

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Figure 2-20

P arallel C ircuits - P o w er is A d ditive Parallel C ircuits - C urrent is A dditive

_ J \_

// R i: Coffee Pot |i R 2 : Skillet R 3 : Blender r notarm i l i I Rated 900W ! Rated 1,100W Rated 400W at 120 V. | current of | ! at 120V ;| at 120V. I each branch. I E = 120V in each branch, R of each resistor is given. lo f R i = 120V/16 ohms = 7.50A Formula: I = E/R I of R2 = 120V/13 ohms = 9.20A I of R3 = 120V/36 ohms = 3.30A Copyright 2014 www.MikeHott.com Total amperes = 20.00A

Determine total power of the branch circuit. Formula: P = I2 x R

R 3 : Blender Rated 400W at 120 V. I = Current of each branch R = resistance of each branch. P of R 1 = 7.50A2 x 16 ohms = 900W P of R2 = 9.17A2 x 13 ohms = 1,100W P of R3 = 3.33A2 x 36 ohms = 400W Total wattage = 2.400W

Figure 2-22

Figure 2-21 Using the formula P = I2 x R, determine the power of each resistor: Coffee Pot Skillet Blender Total Circuit Power

I = E/R E = Voltage of each branch. R = Resistance of each branch (appliance).

P = 7.50A2 x 16900W ohms P = 9.20A2 x 1,1 13 OOW ohms P = 3.30A2 x+36400W ohms 2,400W

The current of each branch is as follows: I = E/R Coffee Pot (R,) Skillet (R2) Blender (R3) Total Current (RT)

I = 120V/16 ohms I = 120V/13 ohms I = 120V/36 ohms

7.50A 9.20A + 3.30A 20.00A

2.10 Circuit Resistance Calculating total circuit resistance is different in series and parallel cir­ cuits. In a series circuit, the total circuit resistance is equal to the sum of the resistances. Figure 2-23A

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Unit 2 I Electrical Circuits

Series

Resistance Calculations JR1 16Q

Calculating Parallel Resistance Equal Resistors in Parallel

Resistance Total = R, + R 2 + R3 RT = 1 6 + 36 + 13 = 65 ohms

--------- 1

R2 13ft

i

Ri

R2

r3

ohms

ohms

ohms

10

10

[V ;

; R3 36 ft

I

10

€>

V jy

R1

The total resistance is less than the sm allest resistor of 13 ohms.

6 ohms

R2

R3

Copyright 2014, www.MikeHolt.com

R _ 10 ohms _ 3^33 Qhms 3 resistors

Copyright 2014, www.MikeHolt.com

Figure 2-24

Figure 2-23 In a parallel circuit, the total circuit resistance is always less than the smallest individual resistance. Figure 2-23B There are three basic methods of calculating the total resistance of a parallel circuit: the Equal Resistance method, the Product-Over-Sum method, and the Reciprocal method.

► Example 2 Question: The total resistance of ten 10-ohm resistors in paral­ lel is (a)1ohm

(b) 10 ohms

(c) 50 ohms

(d) 100 ohms

Answer: (a) 1 ohm

Equal Resistance Method

Rr = Resistance o f One Resistor/Number o f Resistors

When all of the resistors of the parallel circuit have the same resis­ tance, the total circuit resistance is found by dividing the resistance of one resistive element by the total number of resistors in parallel.

► Example 1

Product-Over-Sum Method

Question: The total resistance of three 10-ohm resistors in par­ allel is____ . Figure 2-24 (a) 3.33 ohms

RT= 10 ohms/10 resistors Rr = 1 ohm

(b) 10 ohms

(c) 20 ohms

RT= (R ,xR 2)/(R1 + R2)

(d) 30 ohms

Answer: (a) 3.33 ohms Rt = Resistance o f One Resistor/Number o f Resistors Rr = 10 ohms/3 resistors RT= 3.33 ohms

This method is used to calculate the resistance of two resistors at a time:

Author’s Comment: ■ The term “product” means the answer obtained when num­ bers are multiplied. The term “sum” means the answer obtained by adding a group of numbers.

Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

Electrical Circuits I Unit 2

► Example 1

► Example 2

Question: The resistance of a 900W coffee pot is 16 ohms and the resistance of a 1,100W skillet is approximately 13 ohms. The appliances are connected in parallel. What’s the total resistance of the two appliances? Figure 2-25A and Figure 2-25B

Question: What’s the total resistance of a 16-ohm resistor, a 13-ohm resistor, and a 36-ohm resistor connected in paral­ lel? Figure 2-26

(a) 7.20 ohms

(b) 13 ohms

(c) 16 ohms

(d) 29 ohms

Answer: (a) 7.20 ohms

Rt = (16 ohms x 13 ohms)/(16 ohms + 13 ohms) Rt = 208ohms/29 ohms Rt = 7.20 ohms

R2 13ft

(c) 43 ohms

(d) 65 ohms

Answer: (a) 6 ohms

R r = (R1,2 x ^ 3 ^ 1 ,2 + ^3)

Rt = (7.20 ohms x 36 ohms)/(7.20 ohms + 36 ohms) Rr = 259.2 ohms/43.20 ohms Rr = 6 ohms

Calculating Parallel Resistance “Product-Over-Sum ” Method Ri 16ft

(b) 26 ohms

The 16-ohm and 13-ohm resistors are treated as an “equivalent” single resistor of 7.20 ohms (previous example). The resistance of the circuit will be calculated as follows: Figure 2-26B and Figure 2-26C

Rr = (R ,xR J/(R l + R2)

A,

(a) 6 ohms

R3 ! 3 6 ft:

Calculating Parallel Resistance “Product-Over-Sum ” Method

Ri,

( \ I

1 6 ft x

1 3 f t . 7 . 2

1 6 ft

+

ft

r3

(V

R 2

7.2CI

3 6 ft

;

1 3 ft

I

el %

R __ 7 .2 ft x 3 6 ft _ 6Q T 7 .2 ft + 3 6 ft

& L

R t , 1 6 0 X 13 0 , j 2si 16 ft + 13ft

Rs

7 2a

3 6 ft

_L

Copyright 2014, www.MikeHott.com

el

Figure 2-25

%

Ri,

R 2

p /a 6 £ 1

1

R 3d

_

7 .2 f t x 3 6 f t

K

m>m>O L2

t

7 .2 f t

+

3 6 ft

Copyright 2014, www.MikeHoll.com

The total resistance of a parallel circuit is always less than the small­ est resistance. The Product-Over-Sum method can be used to determine the resis­ tance total for more than two resistors in a parallel circuit, but only two resistors can be dealt with at a time. If more than two resistors are in parallel, the Product-Over-Sum method must be applied several times, each time considering the equivalent resistance of the last two resis­ tors looked at as a “new” resistance for the equation.

Figure 2-26

Author’s Comment: ® The answer must be less than the smallest resistor of the cir­ cuit (13 ohms).

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Unit 2 I Electrical Circuits Reciprocal Method

2.12 Parallel-Connected Power Supplies

The advantage of the Reciprocal method in determining the total resis­ tance of a parallel circuit is that this formula can be used for as many resistors as the parallel circuit contains.

RT= 1/[(1/R,) + (1/R2)+ (I/R3) •■■]

When power supplies are connected in parallel, the voltage remains the same, but the current (or in the case of batteries the amp-hour capacity) increases. To place batteries in parallel to each other, connect them with the proper polarity, which is positive (+) to positive (+) and negative (-) to negative (-). Figure 2-27A

► Example Question: What's the resistance total of 16-ohm, 13-ohm, and 36-ohm resistors connected in parallel? (a) 6 ohms

(b) 13 ohms

(c) 16 ohms

(d) 36 ohms

Answer: (a) 6 ohms Rr = Rr = Rt = Rt =

l/[(Yie ohms) + (V13 ohms) + (% ohms)] 1/[0.0625 ohms + 0.0769 ohms + 0.0278 ohms] 1/0.1672 ohms 6 ohms

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2.11 Parallel Circuit Notes

Figure 2-27

A parallel circuit has the following characteristics: Author’s Comment: Note 1: A parallel circuit has two or more paths through which current flows. Note 2: Total resistance is less than the smallest individual resistor. Total resistance in a parallel circuit is calculated with the following formula:

a When jumping a car battery, place the red cables on the pos­ itive (+) terminals and the black cables on the negative (-) terminals. Figure 2-27B Batteries are often connected in parallel in radios, toys, and other appliances that operate on direct-current power. Figure 2-27C

RT= 1/[(1/R1) + (1/R2) + (1/R3)...] Note 3: The sum of the currents through each path is equal to the total current that flows from the source.

Part C— Series-Parallel Circuits Introduction

Note 4: Total power is equal to the sum of the branches’ powers. Note 5: Voltage is the same across each component of the parallel circuit.

A series-parallel circuit is a circuit that contains some resistors in series and some in parallel to each other. That portion of the series-parallel cir­ cuit that contains resistors in series must comply with the rules for series circuits. That portion of the series-parallel circuit that contains resistors in parallel complies with the rules for parallel circuits. However, it’s good to remember that Ohm’s Law always prevails. Figure 2-28

Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

Electrical Circuits I Unit 2

Note 2: Current is constant.

Series-Parallel Circuits

Note 3: The sum of the voltage drop of all resistors must equal the volt­ age of the source. Note 4: The sum of the power consumed by all resistors equals the total power of the circuit.

Parallel Circuit Review, Figure 2-30

Parallel Circuit Review

Figure 2-28

2.13 Review of Series and Parallel Circuits To understand seriesparallel circuits, we must review the rules for series and parallel circuits. Series Circuit Review, Figure 2-29

Parallel Circuit: • Resistance is less than the sm allest resistor. • Current is additive. • Power is additive. Copyright 2014 www.MikeHolt.com • Voltage is constant. • Multiple paths for current to flow. Figure 2-30

Note 1: A parallel circuit has two or more paths through which current flows.

R2

Note 2: Total resistance is less than the smallest individual resistor. Total resistance in a parallel circuit is calculated with the following formula: Rt = 1/[(1/R,) + (1/R2) + (1/R3)...]

• Resistance is additive, Ri + R 2 + R3... • Current remains the same. • Voltage is additive. Voltage Source (Vo) = Vi + V2 + V 3... • Power is additive, Pi + P2 + P3...

Copyright 2014 www.MikoHolt.com

Note 3: The sum of the currents through each path is equal to the total current that flows from the source. Note 4: Total power is equal to the sum of the branches’ powers.

Figure 2-29

Note 5: Voltage is the same across each component of the parallel circuit.

Note 1: The total resistance of a series circuit is equal to the sum of all of the resistances of the circuit.

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43

I

Unit 2 I Electrical Circuits

2.14 Working With Series-Parallel Circuits When working with series-parallel circuits, keep breaking the circuit down from series to parallel to series to parallel, and so on, until you have only one resistance. It’s best to redraw the circuit so you can see the series components and the parallel branches. Each circuit should be examined to determine the best plan of attack. Some turn out to be easier to analyze if you tackle the parallel elements first and then com­ bine them with the series elements. Figure 2-31

calculating Series-Parallel Circuit Resistance -vA Series R-, R3 Resistance Total Rn re ­ R3 = 0.05Q r2 R4 = 13.09Q R5 = JX05Q Re 13.19Q V ------------------* -------------------1 Ri R2 ^ R °


16 a

Parallel Resistance Total

^ 3 ,4 ,5 1 3 . 1 |

^ 2 ,3 ,4 ,5 1

Re

720 .

------------------/ ■ ----------------- 1------------- 1

R of R2 and Rsas = (16Q x, = 211Q = 7.23ft (16Q + 1 3 .1 9 ft) 29.19ft

W orking Series-Parallel Circuits R t = 0 .0 5 ft -^Ro = 0 .0 5 ft

Calculation continued in next Figure,

R3 = 0 .0 5 ft 3

13.09ft

1 6 ft R6 = 0 .0 5 ft R5 = 0 .0 5 ft ------------------/ ------------- 1--------------/ ------------R 1t R3, R5, and R6 are each 25 ft of 12 AW G, 0.05 ohm s N E C Chapter 9, Table 9 per ft resistance: (2 ohm s/1,000 ft) x 25 ft = 0.05 ohm s per 25 ft. R2 is a coffee pot rated 900W at 120V. R2 = E2/P = 120V2/900W = 16 ohm s R4 is a skillet rated 1,100W at 120V. R4 = E2/P = 120V2/1,100W = 13.09 ohm s

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Author’s Comment: ■ The total resistance of the two branches will be less than that of the smallest branch (13.19 ohms). Since we’re only trying to determine the total resistance of two parallel branches, the Product-Over-Sum method can be used to determine the resistance. Rt = (R2 x R345)/(R2 + R345) Rt = (16 ohms x 13.19 ohms)/(16 ohms + 13.19 ohms) RT= 211.04 ohms/29.19 ohms RT= 7.23 ohms

Figure 2-31

Other circuits are best worked by combining series elements first and then combining the result with the parallel resistances. In Figure 2-31, the series combination of R3, R4, and R5 is the first step. Step 1: Series: Determine the total resistance of each series branch using the formula: Figure 2-32A

Author’s Comment: ■ When working with series-parallel circuits, keep breaking the circuit down from series to parallel to series to parallel, and so forth, until you have only one resistor. Figure 2-33

R? = R3 + R4 + r5 R3 Conductor (25 ft of 12 AWG) R4 Skillet (1.100W) R5 Conductor (25 ft of 12 AWG)

0.05 ohms 13.09 ohms + 0.05 ohms 13.19 ohms

The circuit can now be redrawn showing the relationship between the two conductors and the two parallel branches. Figure 2-32B Step 2:

2.15 Voltage Even though the current is different in the different resistors, remem­ ber that Ohm’s Law always works. Every complicated problem is really just a series of easy problems waiting to be worked out. To calculate the voltage of each resistor, consider each one on a case-by-case basis and multiply its value by the current flowing through it.

Parallel: Determine the total resistance of the two parallel branches. Figure 2-32B

Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

Electrical Circuits I Unit 2

Calculating Series-Parallel Circuit Resistance R0, Power Supply R 1t 12A W G R 2345 Rs, 12AW G Resistance Total

t

= 0.05Q = 0.05Q = 7 .2 3 ft = 0 .0 5 ft = 7 .3 8 ft

2.16 Neutral Conductor The NEC defines a “neutral conductor” as the conductor connected to the neutral point of a system that’s intended to carry current under normal conditions [Article 100]. Figure 2-35 Neutral Conductor Article 100 Definition

Rt = 7 .3 8 ft i___ !_________ i

Wye 3-phase, 4-wire System

Copyright 2014 www.MikeHolt.com

1-phase, 3-wire System Neutral Point

Calculation continued from previous Figure. Prim ary N ot S how n

Prim ary Not Show n

Figure 2-33

Delta 3-phase, 4-wire System

P rim ary N ot Shown,

X1

I

Part D— Multiwire Branch Circuits The conductor connected to the neutral point of a system that’s intended to carry current under normal conditions.

Introduction Understanding series, parallel, and series-parallel circuits is the foun­ dation for understanding multiwire branch circuits. A multiwire branch circuit is a circuit consisting of two or more ungrounded conductors (hot wires) that have a voltage difference between them, and an equal voltage between each ungrounded conductor and the neutral conduc­ tor [Article 100]. A typical 3-wire, 120/240V, single-phase circuit is an example. Figure 2-34

Multiwire Branch Circuit Conductor Voltage Relationships

Figure 2-35

Author’s Comment: ■ The neutral conductor of a solidly grounded system is required to be grounded to the earth; therefore, this conduc­ tor is also called a “grounded conductor.” The term “Neutral Point” in the Code is defined as the common point on a 4-wire, three-phase, 120/208V or 277/480V wye-connected system, the midpoint of a 3-wire, 120/240V single-phase system, or the midpoint of the single-phase portion of a 120/240V, three-phase delta-connected system [Article 100]. Figure 2-36 The definition of “ Neutral Point” describes the point to which the neu­ tral conductor is connected.

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Figure 2-34

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45

Unit 2 I Electrical Circuits

Neutral Point A rticle 100 Definition Wye 3-phase, 4 -w ire S y s te m

1-phase, 3-wire System

2-W ire Circuit Current on Neutral Conductor Delta 3-phase, 4-wire System X 1 fl

o P rim ary N ot S how n

A - Comm on point of a wye 3-ph, 4-wire system. B - M idpoint of a 1-ph, 3-wire system. C - M idpoint of one phase of a delta 3-ph, 4-wire system.

The current flowing on the neutral conductor will be the same as the current on the ungrounded (hot) conductor. Copyright 2014, www.MikeHolt.com

Copyright 2014, www.MikeHolt.com

Figure 2-37

Figure 2-36

3-Wire, 120/240V, Single-Phase Circuit

2.17 Grounded Conductor The “grounded conductor,” according to the NEC, is a circuit or system conductor that’s intentionally grounded [Article 100], In the case of home wiring (3-wire, 120/240V, single-phase), the grounded conduc­ tor is often called the “neutral conductor,” and it will typically be white in accordance with the 200.6 of the NEC.

The current flowing in the neutral conductor of a 3-wire, 120/240V, single-phase circuit equals the difference in current flowing in the ungrounded conductors (lN= LT - L2). Figure 2-38

3-Wire, 1-Phase Circuit Current on Neutral Conductor

Author’s Comment: ■ When a grounded conductor is also used as a neutral con­ ductor, this textbook will refer to it as a “neutral conductor” except in the Code questions at the end of each unit, which will follow the NEC wording.

2.18 Current Flow on the Neutral Conductor

The current flowing on the neutral conductor is the difference between the current flowing on Line 1 and Line 2. Copyright 2014, www.MikeHolt.com

To understand the current flow on the neutral conductor, review the following circuits.

2-Wire Circuit The current flowing in the neutral conductor of a 2-wire circuit is the same as the current flowing in the ungrounded conductor. Figure 2-37

Figure 2-38

The reason is that at any instant the currents on the two ungrounded conductors oppose each other. Figure 2-39

Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

Electrical Circuits I Unit 2

Neutral Current on Multiwire Circuits

Author’s Comment:

Correct Connection 20A - 1 5A = 5A

■ This is one reason white neutral conductors sometimes turn brown or black.

2.19 Balanced Systems Currents on the neutral conductor cancel because the current flowing through the neutral conductor at any instant from the two phase conductors oppose each other.

If the current in each ungrounded conductor of a multiwire branch circuit is the same, the neutral conductor carries OA. This applies to 3-wire, 120/240V, single-phase and all three-phase circuits, regard­ less of configuration or voltage. Figure 2-41

Copyright 2014. www.MikeHolt.cxKn

Figure 2-39

Balanced Systems Q

120/240V L1

/j\

CAUTION: If the ungrounded conductors of a m ulti' > wire branch circuit are incorrectly terminated so they’re not on different lines, the currents on the ungrounded conductors won’t cancel, but w ill add on the neutral conductor. This can cause the neutral current to be in excess o f the neu­ tral conductor ampacity, causing overheating of the neutral insulation. Figure 2-40 Caution

Connection of Ungrounded Conductors of Multiwire Circuit

120/208V or 277/480V

If the current flow on each line of a multiwire circuit is the same, the neutral conductor will carry zero amperes. Copyright 2014, www.MikeHoU.com

DANGER Improper Connection 2 0 A + 15A = 35A

Figure 2-41

2.20 Unbalanced Current

Copyright 2014, www.MikeHoH.com

If the ungrounded conductors of a multiwire circuit aren’t term inated to different phases, the current on the neutral conductor add instead of cancel, which can overload the neutral conductor. Figure 2-40

The current flowing in the neutral conductor of a multiwire branch cir­ cuit is called “ unbalanced current.”

3-Wire, 120/240V, Single-Phase Circuit The neutral conductor of a 3-wire, 120/240V, single-phase circuit only carries current when the current in the ungrounded conductors isn’t identical. The unbalanced current is determined by the following for­ mula: Figure 2-42 lN= L1

L2

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47

Unit 2 I Electrical Circuits

Unbalanced 120/240V, 3-W ire Circuit

Unbalanced Current - 3-Wire Circuit on 4-Wire System

20A

The neutral conductor of a 3-wire, 120/240V circuit will carry current only when the ungrounded conductors don’t have identical current flow. The unbalanced current is the difference between line 1 and line 2. Copyright 2014, www.MikeHott.com

The neutral of a 3-wire circuit from a 4-wire wye system carries about the same current as the phase conductors. IN = 7 (L12 + L22 + L32) - [(L1 x L2) + (L2 x L3) + (L1 x L3)] lN = v (202"+ 202 + O2) - [(20 x 20) + (20 x 0) + (20 x 0)] lN = v' (400 + 400 + 0) - (400 + 0 + 0) 800 - 400

In

In

=

'/4 0 0 "

In

= 20A

Figure 2-42

Figure 2-43

3-Wire Circuit from a 4-Wire, Three-Phase System

4-Wire, Three-Phase Circuit

The neutral conductor of a 3-wire, 120/208V or 277/480V, three-phase circuit derived from a 4-wire, three-phase system always carries neu­ tral current. The current in the neutral conductor of a 3-wire circuit supplied from a 4-wire, three-phase system is determined by the fol­ lowing formula:

The neutral conductor of a 4-wire, 120/208V or 277/480V, three-phase system carries neutral current when the ungrounded conductors aren’t identically loaded. The current in the neutral conductor of a 4-wire circuit supplied from a 4-wire system is determined by the following formula:

lN= V(L1fe+ L2S+ L35) - [(L1 X L2) + (L2 X L3) + (L1 x L3)]

lN= V(L1J + L25 + L34) - [(L1 X L2) + (L2 X L3) + (L1 x L3)]

► Example

► Example

Question: What's the neutral current for a 3-wire, 120/208V, single-phase circuit, if each ungrounded conductor carries 20A, and the circuit is supplied from a 4-wire, 120/208V, three-phase system? Figure 2-43

Question: What’s the neutral current for a 4-wire, 120/208V, three-phase circuit if Line 1 = 100A, Line 2 = 100A, and Line 3 = 50A? Figure 2-44

(a)0A

(b) 20A

(c)80A

(d)100A

Answer: (b) 20A

(a)0A

(b) 50A

(c) 100A

(d)125A

Answer: (b) 50A lN= V ( W + l W + 5 t f f :ftl00~x T00)T(T00xW fhJW 0x5d)]

lN= { W 2 + 2CF + W) - W X 20) + J20 w + W> x 0)]

lN= - ) 22,500 - 20,000

iN= V(40d + m + o)~ J400 T o T o )

lN= 42500A

iN= {M T W o

lN= 50A

I , = {400 lN= 20A

Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

Electrical Circuits I Unit 2

Neutral Current — 4-W ire W ye Circuit Unbalanced W ye 4-W ire Circuit (Linear Loads)

► Example Question: What's the voltage drop of two 12 AWG conductors, each 75 ft long, supplying a 2-wire, 20A load? The resistance for 1,000 ft of 12 AWG copper is 2 ohms. Figure 2-45 (a)2V

(b)3V

(c)4V

(d)6V

Answer: (d) 6V Copyright 2014 www.MikeHolt.com

L3

Neutral Conductor

100A 100A L3 = 50A

In = s/(L12 + L22 + L32) - [(L1 x L2) + (L2 x L3) + (L1 x L3)] In = J 22,500 - 20,000

In = J 2,500

lN = 50A

EVD= l x R I = 20A R = (2 ohms per 1,000 ft/1,000) x 75 ft x 2 wires R = 0.30 ohms Ew = 20A x 0.30 ohms Evd- 6 V

Figure 2-44

2.21 Multiwire Branch Circuits W ye System - 2-W ire Circuit Voltage Drop Multiwire branch circuits are more cost-effective than 2-wire circuits in that they have fewer conductors for a given number of circuits, which enables the use of a smaller raceway. In addition, multiwire branch circuits result in lower circuit voltage drop. Reduced Number of Conductors. Normally it requires four wires to install two 120V single-phase circuits. By using a multiwire branch circuit, which allows sharing the neutral between the two circuits, an installation can be made using three wires. Likewise, the six con­ ductors required for three 120V circuits on a three-phase system can be replaced by a multiwire branch circuit of four conductors; three ungrounded conductors sharing a neutral conductor between them. Reduced Raceway Size. If the number of circuit conductors is reduced, the size of the raceway can often be reduced. Reducing the number of conductors and installing a smaller raceway is very cost-effective in terms of savings in both material and labor. Reduced Circuit Voltage Drop. The voltage drop of the circuit con­ ductors is dependent upon the magnitude of current and conductor resistance: Ev0 = I x R

2-Wire Circuit Voltage Drop

R = (2 ohm s per 1,000 ft/1,000) x 75 ft x 2 w ires R = 0.30 ohm s Evd = 20A x 0.30 ohm s E vd = 6 V

Figure 2-45

Multiwire Branch-Circuit Voltage Drop A balanced 3-wire, single-phase or 4-wire, three-phase multiwire branch circuit has current flowing only in the ungrounded circuit con­ ductors. Therefore, the circuit voltage drop for the line-to-neutral only includes the voltage drop of one conductor. The line-to-line voltage drop is found by multiplying the line-to-neutral voltage drop by 1.732.

A typical 2-wire circuit has current flow in both the ungrounded and neutral conductors. Therefore, the circuit voltage drop includes the voltage drop of both conductors.

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I

Unit 2 I Electrical Circuits

2.22 Dangers of Multiwire Branch Circuits

► Example Question: What’s the line-to-neutral voltage drop over each line conductor of a balanced 4-wire multiwire branch circuit? Each conductor is 12 AWG, 75 ft long, supplying a 20A load. Figure 2-46 (a)2V

(b)3V

(c) 4V

(d) 6V

Multiwire branch circuits offer fewer conductors, reduced race­ way size, and reduced voltage drop. However, the improper wiring or mishandling of multiwire branch circuits can cause a fire hazard because of conductor overloading and/or the destruction of connected equipment.

Answer: (b) 3V The neutral conductor in a balanced 4-wire system effectively has OA of current flow (the three return currents cancel each other out because of their phase relationship). Thus, by Ohm’s Law, the voltage drop over this conductor is OV. The remaining phase conductor’s voltage drop can be calculated as follows: Em = lx R I = 20A R = (2 ohms per 1,000 ft/1,000) x 75 ft x1 wire R = 0.15 ohms

Fire Hazard Failure to terminate the ungrounded conductors to separate phases or lines may cause the neutral conductor to become overloaded from excessive neutral current, and the insulation may be damaged or destroyed. Conductor overheating is known to decrease insulating material service life, potentially resulting in a fire from arcing faults in hidden locations. It’s difficult to predict just how long conductor insulation will last under normal operating conditions, but heat does decrease its life span. Figure 2-47 Fire H azard - O verload on N eutral

Em = 20A x 0.15 ohms Evo = 3 V

;

, , 1— 0 .

Correct Connection 20A - 1 5A = 5A

;L2

lN = 5A L1 = 20A

L1 i2 0 A | ___ r\ 3 L___i

f1 i 1

Balanced M ultiwire Circuit - Voltage Drop L1

it h

I

L2 = 15A

Balanced 4-W ire, 120/208V M ultiwire Circuit

DANGER Im proper Connection 20A + 15A = 35A

U J S

»

► 15A

Copyright 2014 www.MikeHolt.com

The neutral conductor in a balanced 4-wire system effectively has no current flow. Thus, the voltage drop on the neutral conductor is OV. The remaining phase conductor voltage drop can be calculated as follows: E v d = I x R, I = 20A, R = 2 ohms per 1,000 ft = 0.15 ohm s for 75 ft Evd = 2 0 A x 0.15 ohm s = 3 volts dropped Copyright 2014, •yvww.MikeHoH.com

Figure 2-46

Failure to term inate th e ungrounded (hot) conductors to different phases can cause the neutral conductor to be overloaded, w hich can cause a fire.

Figure 2-47

Destruction of Equipment as Well as Fire Hazard The opening of the ungrounded or neutral conductor of a 2-wire circuit during the replacement of a device doesn’t cause a safety hazard, so the pigtailing of these conductors isn’t required. If the continuity of the neutral conductor of a multiwire branch circuit is interrupted (opened), there can be a fire and/or destruction of electrical equipment resulting from overvoltage or undervoltage.

Mike Holt’s illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

Electrical Circuits I Unit 2 Danger of an Open Neutral on a Multiwire Circuit ► Example H

A 3-wire, 120/240V circuit supplies a 1,200W, 120V hair dryer and a 600W, 120V television. If the neutral conductor is inter­ rupted, it will cause the 120V television to operate at 160V and consume 1,067W of power (instead of 600W) for only a few sec­ onds before it burns up. Figure 2-48

Proper Connection

O ka y

L11 L2 1 12Q:

■

i ------------600W l TV L2L _ i H-VAr-J

24Q: H.D.

—

120V Hair Dryer X

TV

I 120

N P.

P arallel C ircu it

R esistan ce o f R esistan ce o f ha ir d ry e r = 12 o h m s te le visio n = 24 ohm s

Step 1: Determine the resistance of each appliance.

D anger

Open Neutral

'T ia if ft r y e r

80V 240V Series

R = E2/P

S eries C ircuit

TV 160 V

—wv---

Hair Dryer R = 120V2/1,200W R = 12 ohms

O pe n N eutral S eries C irc u it coovrioht2014, www.MikeHoit.ccm

Television R = 120V2/600W R = 24 ohms

V oltage dro p o f h a ir drye r = 80V V oltage d ro p o f te le visio n = 160V

P o w e r c onsum ed by ha ir d ry e r = 53 3W P ow e r c° nsum ed by te levision = 1.067W

Figure 2-48

Splicing Neutral Conductors of a Multiwire Circuit

Step 2: Determine the current of the circuit. I = E/R I = 240V/(12 ohms + 24 ohms) 1= 6.70A Step 3: Determine the operating voltage for each appliance. E= Ix R

M ultiwire Circuit Pigtail required for neutral.

Copyright 2014 www.MikeHolt.com

This portion of the circuit is not a multiwire circuit. Pigtails are not required for any of the conductors

Hair dryer operates at = 6.70Ax 12 ohms Hair dryer operates at = 80V Television operates at = 6.70A x 24 ohms Television operates at = 160V

Because of the dangers associated with an open neutral conductor, the NEC requires that the neutral conductors of a multiwire circuit be spliced together. Figure 2-49

2.23 NEC Requirements Author’s Comment: Because of the dangers associated with an open neutral conductor, the NEC specifies that the continuity of the neutral conductor of a multiwire branch circuit can’t be dependent upon any wiring device [300.13(B)], In other words, the neutral conductors of a multiwire branch circuit must be spliced together, and a wire (pigtail) brought out to the device. This way, if the receptacle is removed, it doesn’t result in an open neu­ tral conductor. Figure 2-49 Each multiwire branch circuit must have a means to simultaneously disconnect all ungrounded conductors at the point where the branch circuit originates [210.4(B)],

® Individual single-pole breakers with handle ties identified for the purpose, or a breaker with a common internal trip, can be used for this application [240.15(B)(1)],

AI \

CAUTION: This rule is intended to prevent people from working on energized circuits they thought were disconnected.

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Unit 2 I Electrical Circuits

Conclusion to Unit 2— Electrical Circuits In completing this unit, you’ve laid another block in the foundation of your understanding of electricity. You’ve learned about series circuits, parallel circuits, and combinations of series and parallel circuits. It’s important to be able to look at a circuit and visualize the relationship between the components and be able to understand why there are different levels of voltage or cur­ rent in various locations of the circuit. This ability will greatly enhance your troubleshooting success on real branch circuits. In this unit, you learned about a special circuit called a “multiwire branch circuit.” There are special Code considerations for multiwire branch circuits. You now know how important it is to correctly install multiwire branch circuits with neutral connec­ tions that are very securely made. A neutral must never be opened on an energized multiwire branch circuit, as this can result in placing both undervoltages and overvoltages on circuit components, which can result in damage. You learned about Kirchoff’s Current Law in this unit. If you understand the application of Kirchoff’s law, you’re less likely to be misled by some of the incorrect “myths” that are commonly accepted in the electrical industry concerning fault current flow, particularly as it relates to the practices of grounding and bonding. For instance, does current always take only the path of least resistance? Is current always seeking a path to the earth? Ponder these questions as you continue your study of the intriguing field of electricity.

Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

UNIT 2 REVIEW QUESTIONS Part A— Series Circuits

6.

Introduction 1.

A series circuit is a circuit in which a specific amount of current leaves the voltage source and flows through every electrical device in a single path before it returnsto the voltage source.

Kirchoff’s Voltage Law states that in a series circuit, the sum of the voltage drops across all of the resistors equals the applied voltage. (a) True (b) False

7.

2.1 Practical Uses of the Series Circuit

No matter how many resistances there are in a series circuit, the sum of the voltages across all of the resistances equals the voltage of the source according to the Resistance Law of Pro­ portion.

2.

(a) True (b) False

(a) True (b) False

Series circuits are often used fo r__ applications. (a) signal (b) control (c) a and b (d) none of these

3.

8.

(a) proportional (b) distributed (c) additive (d) the same

A 115/230V rated motor connected to a 230V circuit must have the windings connected in series so that each winding will receive at least 230V. (a) True (b) False

9.

2.2 Understanding Series Calculations 4.

Resistance opposes the flow of electrons. In a series circuit, the total circuit resistance is equal to the sum of all of the resis­ tances in series.

5.

The opposition to current flow results in a voltage drop of the circuit voltage.

The power consumed in a series circuit is equal to the power consumed by the largest resistance in the series circuit. (a) True (b) False

2.3 Series Circuit Calculations 10.

(a) True (b) False

Kirchoff’s Current Law states that in a series circuit, the current is ____ through the transformer, the conductors, and the appli­ ance.

To determine the resistance of each resistive element in the cir­ cuit, use the formula R = E2/VA. E is the rated voltage of the resistance, and VA is the rated power of the resistor. (a) True (b) False

(a) True (b) False

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53

Unit 2 I Review Questions

11.

To calculate the total resistance of the circuit, use the formula:

RT= R, + R2 + R3+ ...

12.

2.9 Understanding Parallel Calculations

(a) True (b) False

In a parallel circuit, the voltage drop across each resistance is equal to the sum of the voltage drops of each of the resistors in parallel.

The current of the circuit can be determined by the formula

(a) True (b) False

i

17.

= es/ rt .

(a) True (b) False

18.

2.4 Power Calculations 13.

If you know the current of the circuit and the resistance of each resistor, the power of each resistor can be determined by the formula P = l2xR .

According to Kirchoff’s Current Law, the total current provided by the source to a parallel circuit equals the sum of the currents of all of the branches. (a) True (b) False

19.

(a) True (b) False

The total power consumed in a parallel circuit equals the sum of the power of all branches. (a) True (b) False

2.5 Variations 2.10 Circuit Resistance 14.

There can never be variations in the formulas used or the order in which they’re used for series circuits.

20.

(a) True (b) False

In a parallel circuit, the total circuit resistance is always greater than the smallest resistance. (a) True (b) False

2.7 Series-Connected Power Supplies 21. 15.

When power supplies are connected in series, the circuit voltage remains the same as when only one power supply is connected to it, provided that all the polarities are connected properly.

(a) equal resistance (b) product-over-sum (c) reciprocal (d) any of these

(a) True (b) False

Part B— Parallel Circuits Introduction 16.

A parallel circuit has two or more paths in which current can flow.

The total resistance of a parallel circuit can be calculated by the method.

22.

According to the equal resistance method, when all the resis­ tances of the parallel circuit have the same resistance, divide the resistance of one element by the largest resistor in parallel. (a) True (b) False

(a) True (b) False

Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

Review Questions i Unit 2

23.

The product-over-sum method is used to calculate the resis­ tance o f____ resistance(s) at a time.

29.

(a) one (b) two (c) three (d) four 24.

2.14 Working With Series-Parallel Circuits

(a) True (b) False

The advantage of the reciprocal method is that the formula can be used for as many resistances as the parallel circuit contains.

Part D— Multiwire Branch Circuits Introduction

(a) True (b) False

30.

2.12 Parallel-Connected Power Supplies 25.

When power supplies are connected in parallel, the voltage remains the same, but the current or ampere-hour capacity is increased. (a) True (b) False

(a) True (b) False

31.

Introduction

32.

28.

33.

That portion of the series-parallel circuit that contains resis­ tances in parallel must comply with the rules for parallel cir­ cuits. (a) True (b) False

An ungrounded delta system contains a neutral point. (a) True (b) False

That portion of the series-parallel circuit that contains resis­ tances in series must comply with the rules for series circuits. (a) True (b) False

A neutral conductor is the conductor connected to th e _____ of a system that’s intended to carry neutral current under normal conditions. (a) grounding electrode (b) neutral point (c) intersystem bonding termination (d) earth

A ____ circuit is a circuit that contains some resistances in series and some resistances in parallel with each other. (a) parallel (b) series (c) series-parallel (d) of these

27.

A multiwire branch circuit has two or more ungrounded conduc­ tors having a potential difference between them, and having an equal difference of potential between each ungrounded conduc­ tor and the grounded conductor.

2.16 Neutral Conductor

Part C— Series-Parallel Circuits

26.

When working with series-parallel circuits, it’s best to redraw the circuit so you can see the series components and the paral­ lel branches.

The common point on a 4-wire, three-phase, 120/208V wye-connected system is defined by the Code as the “___ point.” (a) wye (b) neutral (c) bridge (d) ungrounded

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Unit 2 I Review Questions 2.17 Grounded Conductor 34.

40.

The grounded conductor is a conductor that’s intentionally grounded to the earth.

(a) OA (b) 10A (c) 20A (d) 60A

(a) True (b) False

2.18 Current Flow on the Neutral Conductor 35.

The current in the neutral conductor of a 2-wire circuit is ____ of the current in the ungrounded conductor. (a) 0 percent (b) 70 percent (c) 80 percent (d) 100 percent

36.

37.

2.20 Unbalanced Current 41.

The neutral conductor of a 3-wire, 120/240V, single-phase circuit only carries the unbalanced current when the circuit isn’t balanced.

42.

If the ungrounded conductors of a multiwire branch circuit aren’t terminated to different phases, this can cause the neutral current to exceed the neutral conductor’s ampacity. (a) True (b) False

43.

If the current in each ungrounded conductor of a multiwire branch circuit is the same, the neutral conductor carries OA. (a) True (b) False

The neutral conductor of a 3-wire, 120/208V or 277/480V circuit supplied from a 4-wire, three-phase system never carries neu­ tral current. (a) True (b) False

44.

The neutral conductor of a 4-wire, 120/208V or 277/480V, threephase system has neutral current flow when the ungrounded conductors are equally loaded. (a) True (b) False

2.21 Multiwire Branch Circuits 45.

Multiwire branch circuits have more conductors for a given number of circuits, which requires the use of a larger raceway. (a) True (b) False

2.19 Balanced Systems 39.

The neutral conductor of a 3-wire, 120/240V, single-phase circuit only carries current when the current in the ungrounded conductors isn’t identical. (a) True (b) False

(a) True (b) False 38.

The current flowing in the neutral conductor of a multiwire branch circuit is called “ unbalanced current.” (a) True (b) False

A balanced 3-wire, 120/240V, single-phase circuit is connected so that the ungrounded conductors are from different trans­ former lines (Line 1 and Line 2). The current in the neutral con­ ductor is ____ of the ungrounded conductor current. (a) 0 percent (b) 70 percent (c) 80 percent (d) 100 percent

What’s the neutral current for a 4-wire, 120/208V circuit, where L1 = 20A, L2 = 20A, and L3 = 20A?

46.

A balanced multiwire branch circuit has current flow only on the ungrounded conductors. (a) True (b) False

Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

Review Questions I Unit 2

47.

What’s the voltage drop of two 12 AWG conductors, each 100 ft in length, supplying a 2-wire, 16A load? The resistance of 12 AWG conductors is 2 ohms per 1,000 ft.

50.

(a) 2-wire (b) 3-wire (c) 4-wire (d) all of these

(a) 3.20V (b) 6.40V (c) 7.20V (d) 9.60V 48.

The opening of the ungrounded or neutral conductor of a ____ circuit during the replacement of a device doesn’t cause a safety hazard, so the pigtailing of these conductors isn’t required.

What’s the voltage drop of each ungrounded conductor of a 4-wire multiwire branch circuit? Each conductor is 12 AWG, 100 ft in length, supplying a 16A load. The resistance of 12 AWG conductors is 2 ohms per 1,000 ft. (a) 3.20V (b) 6.40V (c) 7.20V (d) 9.60V

2.23 NEC Requirements 51.

Because of the dangers associated with an open neutral con­ ductor, the continuity of t h e ____ conductor in a multiwire branch circuit can’t depend upon the receptacle. (a) ungrounded (b) grounded (c) a and b (d) none of these

2.22 Dangers of Multiwire Branch Circuits 49.

The improper wiring or mishandling of multiwire branch circuits can cause____ connected to the circuit. (a) overloading of the ungrounded conductors (b) overloading of the grounded conductors (c) destruction of equipment (d) b and c

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UNIT 2 CHALLENGE QUESTIONS Part A— Series Circuits 2.2 Understanding Series Calculations 1.

A series circuit contains two resistors, one rated 4 ohms and the other rated 8 ohms. If the total voltage drop across both resis­ tors equals 12V, then the current that passes through either resistor will be____ . (a) 1A (b)2A (C)4A (d)8A

2.

A series circuit has four 40-ohm resistors and the power supply is 120V. The voltage drop of each resistor will be____ . (a) one-quarter of the source voltage (b) 30V (c) the same across each resistor (d) all of these

3.

R2 = • 7.5 ohms ■

Copyright 2014, www.MikeHolt.com

Figure 2-50

5.

The voltmeter connected across the switch reads. 2-51 (a) 3V (b)6V (c) 12V (d) 18V

The power consumed in a series circuit is ____ . (a) the sum of the power consumed by each load (b) determined by the formula PT= I2 x RT (c) determined by the formula PT= E x I (d) all of these The reading on Voltmeter 2 (V2) is ____ . Figure 2-50

(a) 5V (b) 6V (c) 7V (d) 10V

Figure 2-51

Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

Figure

Challenge Questions I Unit 2

Part B— Parallel Circuits

9.

2.9 Understanding Parallel Calculations 6.

In general, when multiple light bulbs are wired in a single luminaire, they’re connected in ____ to each other. (a) series (b) series-parallel (c) parallel (d) order of wattage

(a) series (b) series-parallel (c) parallel (d) any of these

2.10 Circuit Resistance

7.____A single-phase, dual-rated, 120/240V motor will have its wind­ ing connected in ____ when supplied by 120V.

10.

(a) series (b) parallel (c) series-parallel (d) parallel-series The voltmeters shown in Figure 2-52 are connected. of the loads. (a) in series to (b) across (c) in parallel to (d) b and c

If the supply voltage is 120V, the total energy consumed for four 10-ohm resistors will be more if all of the resistors are con­ nected in ____ .

A parallel circuit has three resistors. One resistor is rated 2 ohms, one is rated 3 ohms, and the other is rated 5 ohms. The total resistance of the parallel circuit is ____ Remember, the total resistance of any parallel circuit is always less than the smallest resistor. (a) 0.50 ohms (b) 1 ohm (c) 2 ohms (d) 3 ohms

each

Figure 2-53 applies to the next three questions:

A1

A2

Figure 2-53

Figure 2-52

11.

In Figure 2-53, the total current of the circuit can be measured by ammeter____ . (a) A, (b) A2 (c)A 3 (d) none of these

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59

H

Unit 2 I Challenge Questions

12.

If Bell 2 consumes 12W of power when supplied by two 12V batteries (connected in series), the resistance of this bell is ____ . See Figure 2-53.

14.

(a) Ammeter 1 (b) Ammeter 2 (c) Ammeter 3 (d) all of these

(a) 12 ohms (b) 24 ohms (c) 36 ohms (d) 48 ohms 15. 13.

Determine the total circuit resistance of the parallel circuit shown in Figure 2-53 based on the following facts:

16.

The reading of V4 is __ _ . See Figure 2-54. (a) 1.50V (b)3V (c)5 V (d)8V

(a) 22 ohms (b) 32 ohms (c) 42 ohms (d) 60 ohms

Part C— Series-Parallel Circuits

The reading of Voltmeter 2 (V2) is ____ . See Figure 2-54. (a) 1.50V (b)4V (c) 5V (d)8V

1. The current on ammeter 1 reads 0.75A. 2. The voltage of the circuit is 30V. 3. Bell 2 has a resistance of 48 ohms. Tip: The total resistance of a parallel circuit is always less than the smallest resistor.

The total current of the circuit shown in Figure 2-54 can be read on____ .

Figure 2-55 applies to the next two questions:

2.14 Working With Series-Parallel Circuits Figure 2-54 applies to the next three questions:

Figure 2-55

17.

Figure 2-54

Resistor R, has a resistance of 5 ohms and resistors R2, R3, and R4 have a resistance of 15 ohms each. The total resistance of this series-parallel circuit is ____ . See Figure 2-55. (a) 10 ohms (b) 25 ohms (c) 35 ohms (d) 50 ohms

Mike Holt’s illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

Challenge Questions I Unit 2

18. What’s the voltage drop across R,, if R, is equal to 5 ohms and the total resistance of R2, R3, and R4 is 5 ohms? See Figure 2-55. (a) 33V (b) 40V (c) 60V (d) 120V

Part D— Multiwire Branch Circuits 2.22 Dangers of Multiwire Branch Circuits 19. If the neutral of the circuit in the diagram is opened, the cir­ cuit becomes one series circuit of 240V. Linder this condition, the current of the circuit is ____ . Tip: Determine the total resis­ tance. Figure 2-56 (a) 0.25A (b) 0.58A (c) 0.67A (d) 2.25A

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Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

UNIT

3

UNDERSTANDING ALTERNATING CURRENT

Introduction to Unit 3— Understanding Alternating Current Direct-current (dc) circuits and alternating-current (ac) circuits have similarities as well as differences. In resistive circuits, direct-current and alternating-current circuits have many similarities. The calculations of resistors used in series, parallel, or series-parallel combination circuits discussed in the preceding unit apply to both direct-current and alternating-current circuits. In contrast, this unit focuses on the special characteristics of alternating current and how it’s generated, distributed, and utilized. The practical use of electric power depends on the efficient generation and distribution of power to the point of utilization. The principles of magnetism and induction are important in the generation and transformation of alternating-current power. The flow of direct current in a circuit is opposed by resistance, but because of the changing current flow of an alternating-current circuit, alternating current is opposed by a property called “ impedance.” Impedance is a combination of resistance, inductive reactance, and capacitive reactance. Be sure to study this unit carefully so you’ll understand the differences between alternating-current and direct-current circuits. There are some types of equipment that rely on the characteristics of the changing magnetic field developed by alternating-current power to func­ tion, such as the alternating-current motors and transformers that will be covered in Unit 4.

Part A— Understanding Alternating Current Introduction Because alternating current is inexpensive to transmit compared to direct current, alternating current has become the dominant form of electricity in our modern infrastructure. In the early days of commer­ cially available electric power, direct current was more common. But, economics won out. Applying alternating current safely or effectively however, requires an understanding of certain concepts that border on the complex. All of those concepts build on what you’ve already learned.

3.1 Current Flow In order for current to flow in a circuit, the power supply must apply sufficient electromotive force (voltage) to cause the electrons to move. The movement of the electrons themselves doesn’t produce any useful work. It’s the effects that the moving electrons have on the loads they flow through that are important. The effects of electron movement are the same regardless of the direction of the current flow. Figure 3-1

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Unit 3 ! Understanding Alternating Current R e c tify in g a c to d c fo r E le c tro n ic s

C urrent Flow A lternatin g curren t alte rn ate ly flow s in both directions.

J

F u ll-W a v e R e c tifie r j

" T i' 1 C losed Sw itch

C losed Sw itch

A ltern atin g cu rre nt (ac) rapidly ch an g es polarity and m agnitude. T he co n sta ntly changing po larity causes the cu rre nt to alte rn ately flo w in both directions. D irect c u rre n t flow s in one direction. D ire ct curre nt (dc) flow s from the negative term inal o f the pow er source to the positive term inal o f the pow er source. The polarity o f the voltage alw ays rem ains the sam e.

Figure 3-1

3.2 Why Alternating Current Is Used Alternating current is primarily used because it can be transmitted inexpensively by transforming it to high-transmission level voltages and then converting this voltage back to lower distribution voltage levels. High-voltage transmission lines allow for the transmission of alternating-current at high voltage, and relatively low current. High-voltage transmission, with the outcome of lower current, results in reduced voltage drop in the transmission lines. Conductors and electrical equipment are sized based on the current in the lines, so the lower current levels also allow the use of smaller wire and smaller electrical equipment.

Direct-Current Use. There are other applications however, particu­ larly in electronic equipment, where only direct current can perform the desired function. This is accomplished by rectifying alternating cur­ rent to direct current to power these electronic loads. Figure 3-2

3.3 How Alternating Current Is Produced in 1831, Michael Faraday discovered that electricity could be produced from a source other than a battery. He knew that electricity produced magnetism, and wondered why magnetism couldn’t produce electric­ ity. Faraday discovered that when he moved a magnet inside a coil of wire, he got a pulse of electricity. When he pulled the magnet out, he

"

ac S u pply'

F ilte re d d c

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D ire c t c u rre n t fro m a fu ll-w a v e re c tifie r (s u p p lie d by a c ) is v e ry c o m m o n in s id e e le c tro n ic e q u ip m e n t w h e re o n ly d c can p e rfo rm th e d e s ire d fu n c tio n .

Figure 3-2 got another pulse. He also got the same reaction when he moved the coil toward and away from the magnet. Faraday’s experiments revealed that when a magnetic field moves through a coil of wire, the lines of force of the magnetic field cause the electrons in the wire to flow in a specific direction. When the magnetic field moves in the opposite direction, electrons in the wire flow in the opposite direction. Electrons flow only when there’s motion of the con­ ductors relative to the magnetic field. Figure 3-3

3.4 Alternating-Current Generator A simple alternating-current generator consists of a loop of wire rotat­ ing between the lines of force between the opposite poles of a magnet. The halves of each conductor loop travel through the magnetic lines of force in opposite directions, causing the electrons within the conductor to move in a given direction. The magnitude of the voltage produced depends upon the number of turns of wire, the strength of the mag­ netic field, and the speed at which the coil rotates. Figure 3-4 The rotating conductor loop is called a “ rotor” or “armature.” “ Slip” or “collector” rings and carbon brushes are used to connect the output voltage from the generator to an external circuit.

Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

Understanding Alternating Current I Unit 3

Magnet

Producing Alternating Current

Single-Phase Alternating-Current Generator | Stationary Armature | Windings (conductor)

J k i W hen the m agnet is inserted into the coil, an electric pulse m easures positive on a galvanometer. Magnet \

W hen the magnet is removed from the coil, an electric pulse measures negative on a galvanometer. Copyright 2014, www.MikeHolt.com

Figure 3-3

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Most ac generators have a rotating magnetic field inside stationary armature windings. The voltage induced in the armature windings add to produce the ac output voltage. Note: All ac generators are separately excited. Batteries or a small dc generator are typically used with the ac generator. Figure 3-5

Alternating-Current Generator

3.5 Waveform

Motive Force (e.g., engine, turbine) A conductor loop rotating in a m agnetic field is a single-phase ac generator Excitation voltage input / from a separate dc source ^ produces a m agnetic field, j

s

A waveform image is used to display the level and direction of current and voltage.

f

Slip Rings

^-ac Output — to Load Small single-phase ac generators have a conductor loop rotating inside a magnetic field. Note: Larger ac generators have a rotating m agnetic field inside stationary conductors. Copyright 2014, www.MikeHolt.com

Figure 3-4 In generators that produce large quantities of electricity, the conduc­ tor coils are stationary and the magnetic field revolves within the coils. The magnetic field is produced by an electromagnet, instead of a per­ manent magnet. Using electromagnets permits the strength of the magnetic field (and thus the lines of force) to be modified, thereby con­ trolling the output voltage. Figure 3-5

Direct-Current Waveform A direct-current waveform displays the direction (polarity) and magni­ tude of the current or voltage. Figure 3-6

Alternating-Current Waveform An alternating-current waveform displays the level and direction of the current and voltage for every instant of time for one full revolution of the rotor. Figure 3-7

3.6 Sine Wave Sinusoidal Waveform The waveform for alternating-current circuits is symmetrical, with positive above and negative below the zero reference level. For most alternating-current circuits, the waveform is called a “sine wave” or a “sinusoidal waveform.”

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Unit 3 I Understanding Alternating Current

S in e W a v e s - S in u s o id a l W a v e fo rm

D ire c t-C u rre n t W a v e fo rm s

o

II

Basic Direct-Current W aveform

f + ■.DO' *TO+ O

--------------------- i--------------------

N

S

N _

S

N

S

;1 R e fe re n c e L in e s in.

22

40

Above 31/2 in. to 12 in.

17

30

Above 12 in. to 36 in.

14

25

Note to Table 310.15(B)(3)(c): The temperature adders in Table 310.15(B)(3)(c) are based on the measured temperature rise above local climate temperatures ambient temperatures due to sunlight heating.

Note: See ASHRAE Handbook— Fundamentals (www.ashrae.org) as a source for the average ambient temperatures in various locations. Author’s Comment: ■ This handbook may be available from the Copper Development Association (www.copper.org/applications/electrical/building/ derating.html).

8 THWN-2, rated 55A 90°C [Table 310.15(B)(16)] Corrected Ampacity = 55A x 0.76 Corrected Ampacity = 41.80A

Raceways and Cables Exposed to Sunlight on Roofs Am bient Temperature Adjustments 310.15(B )(3)(c) _______________________ . . . Am bient Temperature The raceway is 3/4 in. above j is 90°F the roof, so add 40°F to the j ambient temperature. 8 THW N-2 Am pacity?

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Table 310.15(B)(16) ampacity, 8 THWN-2 = 55A at 90°C Adjusted Temperature: | 90 of + 40°F [Table 310.15(B)(3)(c)] = 130°F Temp Correction Factor = 0.76 [Table 310.15(B)(2)(a)] New Ampacity = 55A x 0.76 = 41 80A Figure 6-30

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183

Unit 6 I Conductor Sizing and Protection Calculations

► Rooftop Temperature Adder Example 2 Question: What’s the ampacity of a 6 THWN-2 conductor installed 3A in. above the roof, where the ambient temperature is 90°F? Figure 6-31 (a) 20A

(b) 43.50A

(c) 56.20A

(d) 65.40A

Answer: (b) 43.50A Corrected Ampacity = Table 310.15(B)(16) Amperes x Ambient Temperature Correction Factor from Table 310.15(B)(2)(a) Ambient Temperature Includes Roof Top Temperature Adder [Table 310.15(B)(3)(c)] 105°FAmbient + 40°F Rooftop Adder = 145°F Ambient Temperature Correction Factor for 145°F= 0.58 [Table 310.15(B)(2)(a)] 6 THWN-2, rated 75A 90°C [ Table 310.15(B)(16)] Corrected Ampacity = 75A x 0.58 Corrected Ampacity = 43.50A

Raceways and Cables Exposed to Sunlight on Roofs Am bient Temperature Adjustm ents 310.15(B )(3)(c) Am bient Temperature is 105°F The raceway is 3/4 in. above the roof, so add 40°F to the ambient temperature.

7Z77~

7~7Ti:T~~~~

6 THW N-2 Ampacity? Copyright 2014, www.MikeHolt com

Table 310.15(B)(16) ampacity, 6 THWN-2 = 75A at 90°C Adjusted Temperature: 105°F + 40°F [Table 310.15(B)(3)(c)] = 145°F Temp Correction Factor = 0.58 [Table 310.15(B)(2)(a)] New Ampacity = 75A x 0.58 = 43.50A Figure 6-31

6.11 Ampacity Adjustment [310.15(B)(3)(a)] Conductor Bundle. Where the number of current-carrying conductors in a raceway or cable exceeds three, or where single conductors or multiconductor cables are installed without maintaining spacing for a continuous length longer than 24 in., the allowable ampacity of each conductor, as listed in Table 310.15(B)(16), must be adjusted in accor­ dance with the adjustment factors contained in Table 310.15(B)(3)(a).

Table 310.15(B)(3)(a) Conductor Ampacity Adjustment for More Than Three Current-Carrying Conductors Number of Conductors1

Adjustment

4-6

0.80 or 80%

7-9

0.70 or 70%

10-20

0.50 or 50%

21-30

0.45 or 45%

31-40

0.40 or 40%

41 and above

0.35 or 35%

1Number of conductors is the total number of conductors, including spare conductors, including spare conductors, adjusted in accordance with 310.15(B)(5) and (B)(6). It doesn’t include conductors that can’t be energized at the same time. Author’s Comment: ■ Conductor ampacity reduction is required when four or more current-carrying conductors are bundled because heat gen­ erated by current flow isn't able to dissipate as quickly as when there are three or fewer current-carry conductors. Figure 6-32

Conductor Bundling Ampacity Adjustment Formula: Adjusted Ampacity = Table 310.15(B)(16) Ampacity x Bundled Ampacity Adjustment Factor from Table 310.15(B)(3)(a)

m

Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

Conductor Sizing and Protection Calculations I Unit 6

Conductor Am pacity Adjustm ent Factor 310.15(B )(3)(a) No Am pacity Adjustm ent Am pacity Adjustm ent Three or Fewer Conductors Factor = 70%

12 THWN-2 is rated 30A at 90°C [Table 310.15(B)(16)] Adjustment factor for four current-carrying conductors is 0.80 [Table 310.15(B)(3)(a)] Corrected Ampacity = 30A x 0.80 Corrected Ampacity = 24A

Conductor Am pacity Adjustm ent Factor 310.15(B )(3)(a) and Table 310.15(B )(16) Conductors have m ore surface area for heat dissipation. Copyright 2014, www.MikeHolt.com

Bundled conductors have heat held in by other conductors.

ii m .wimp,

»

1

1

Raceway contains 4 current-carrying conductors. The Table 310.15(B)(3)(a) adjustment factor is 0.80.

: igure 6-32 r*^

Author’s Comment: ■ The neutral conductor might be a current-carrying conduc­ tor, but only under the conditions specified in 310.15(B)(5). Equipment grounding conductors are never considered cur­ rent carrying [310.15(B)(6)]. ■ When correcting or adjusting conductor ampacity, the ampacity Is based on the temperature insulation rating of the conductor as listed in Table 310.15(B)(16), not the tempera­ ture rating of the terminal [110.14(C)]. ■ Where more than three current-carrying conductors are pres­ ent and the ambient temperature isn’t between 78°F and 86°F, the ampacity listed in Table 310.15(B)(16) must be cor­ rected and adjusted for both conditions.

«F>4>4.y

m

jM Hm

i -

*

Copyright 2014, www.MikeHolt.com

12 THW N-2 rated 30A, Table 310.15(B)(16) at 90°C Ampacity = 30 A x 0.80 [Table 310.15(B)(3)(a)] Ampacity = 24A Figure 6-33

► Ampacity Adjustment Example 2 Question: What’s the adjusted ampacity of four 1 THWN con­ ductors in a raceway? Figure 6-34 (a) 120A

(b) 104A

(c)29A

(d) 32A

Answer: (b) 104A

► Ampacity Adjustment Example 1 Question: What’s the adjusted ampacity of four 12 THWN-2 conductors in a raceway? Figure 6-33 (a)20A

(b) 24A

(c) 29A

(d) 32A

Answer: (b) 24A Adjusted Ampacity = Table 310.15(B)(16) Ampacity x Bundled Ampacity Adjustment Factor from Table 310.15(B)(3)(a)

Adjusted Ampacity = Table 310.15(B)(16) Ampacity x Bundled Ampacity Adjustment Factor from Table 310.15(B)(3)(a) 1 THWN is rated 130A at 90°C [Table 310.15(B)(16)] Adjustment factor for four current-carrying conductors is 0.80 [Table 310.15(B)(3)(a)] Corrected Ampacity = 130Ax0.80 Corrected Ampacity - 104A

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Unit 6 I Conductor Sizing and Protection Calculations

Conductor Am pacity Adjustm ent Factor 310.15(B )(3)(a) and Table 310.15(B )(16)

► Ampacity Adjustment— Short Raceways Example Question: What’s the adjusted ampacity of four 3/0 THWN-2 con­ ductors in a raceway not exceeding 24 ” in length? Figure 6-35 (a) 150A

The raceway contains four current-carrying 1 THWN. 1 THWN rated 130 A at 75°C Table 310.15(B)(16) Adjustm ent Factor = 0.80

(b) 195A

(c) 205A (d) 225A

Answer: (d) 225A The Table 310.15(B)(16) ampacity for 3/0 THWN-2 is 225A at 90°C.

Adjusted Ampacity: 1 30 A x 0.80 = 104A

C onductor Am pacity - Adjustm ent Factor 310.1 5(B) (3) (a) (2)

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Figure 6-34

► Ampacity Adjustment Example 3 Question: What’s the adjusted ampacity of 10 THHN conductors when nine current-carrying conductors are installed in a raceway or cable? (a) 20A

(b) 28A

(c) 29A

(d) 32A

Answer: (b) 28A Adjusted Am pacity = Table 310.15(B)(16) Ampacity x Bundled Ampacity Adjustment Factor from Table 310.15(B)(3)(a) 10 THHN is rated 40A at 90°C [Table 310.15(B)(16)] Adjustment factor for nine current-carrying conductors is 0.70 [Table 310.15(B)(3)(a)] Corrected Ampacity = 40A x 0.70 Corrected Ampacity = 28A

Ampacity Adjustment— Short Raceways Conductor ampacity adjustment doesn’t apply to conductor bundling in raceways having a length not exceeding 24 in.

>6

Four current carrying 3/0 THW N-2 Conductors Table 310.15(B)(3)(a) does not apply Am pacity = 2 2 5 A a t 90°C [Table 310.15(B)(16)] Figure 6-35

6.12 Combining Ambient Temperature and Conductor Bundling Adjustments Ambient and Conductor Bundling Adjustments When conductors are bundled together, the ability of the conductors to dissipate heat is reduced. The NEC requires that the ampacity of a conductor be reduced whenever four or more current-carrying con­ ductors are bundled together. The higher insulation temperature rating of 90°C rated conductors pro­ vides a greater conductor ampacity for use in ampacity adjustment, even though conductors must be sized based on the column that cor­ responds to the temperature listing of the terminals [110.14(C)(1)].

Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

Conductor Sizing and Protection Calculations I Unit 6

► Conductor Ampacity Example 1

► Conductor Ampacity Example 2

Question: What's the ampacity of four current-carrying 10 THWN-2 conductors installed in a raceway less than 1A in. above a rooftop; ambient temperature 90°F? Figure 6-36

Question: What’s the ampacity of four current-carrying 8 THHN conductors installed in an ambient temperature of 90°F?

(a) 15.30A

(b) 18.56A

(c)25.34A

(d)31.25A

(a) 21A

(b) 42A

(c) 51A

(d) 64A

Answer: (b) 42A

Answer: (b) 18.56A Adjusted/Corrected Ampacity = Table 310.15(B)(16) Ampacity x Temperature Factor x Bundled Adjustment Factor

Adjusted/Corrected Ampacity = Table 310.15(B)(16) Ampacity x Temperature Factor x Bundled Adjustment Factor Ambient Temperature Includes Roof Top Temperature Adder [Table 310.15(B)(3)(c)]

Ambient Temperature Correction Factor for 90°F = 0.96 [Table 310.15(B)(2)(a)] 8 THHN is rated 40A at 90°C [Table 310.15(B)(16)]

90°FAmbient + 60°F Rooftop Adder = 150°F Ambient Temperature Correction Factor for 150°F = 0.58 /Table 310.15(B)(2)(a)] 10 THWN-2 is rated 40A at 90°C [Table 310.15(B)(16)] Adjustment factor for four current-carrying conductors is 0.80 [Table 310.15(B)(3)(a)] Corrected Ampacity = 40A x 0.58 x 0.80 Corrected Ampacity = 18.56A

Adjustment factor for four current-carrying conductors is 0.80 [Table 310.15(B)(3)(a)] Corrected Ampacity = 40A x 0.96x0.80 Corrected Ampacity = 42.24A

Lower Ampacity Rule [310.15(A)(2)] Where more than one ampacity applies for a given circuit length, the lowest ampacity value must be used. Figure 6-37

Conductor Am pacity After Correction and Adjustm ent 310.15(B )(2)(a) and (3)(a) Conductor Am pacity - Lower Rating 310.15(A )(2)

Raceway on roof with four 10 THW N-2 Am bient Temperature is 90°F ■■■•m

in Copyright 2014 www.MikeHolt.com

m

m

iliu i

0 1

I

I

m

10 THW N-2 rated 18.56A after am pacity correction (40A x 0.58 x 0.80) can be protected by a 20A protection device [240.4(B)], Figure 6-36

Lower Am pacity on Roof Higher Am pacity 12 THW N-2 = 17.40A (30A x 0.58) on Wall ~ ..... 7 / r 12 THW N-2 = 30AJ

Copyright 2014 www.MikeHolt.coim

11 i i a i w cm .

W here more than one am pacity rating applies to a single conductor length, the lower am pacity must be used for the entire circuit. Entire circuit is rated 17.40A. Figure 6-37

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187

Unit 6 I Conductor Sizing and Protection Calculations Higher Ampacity Exception [310.15.(A)(2) Ex]

6.13 Current-Carrying Conductors

When different ampacities apply, the higher ampacity can be used for the entire circuit if the reduced ampacity length doesn’t exceed 10 ft or 10 percent of the length of the higher ampacity, whichever is less. Figure 6-38 and Figure 6-39

Conductor Am pacity - Higher Rating 310.15(A )(2) Ex

Two-Wire Circuits

60 ft 55 ft

5 ft

Higher Am pacity 8 AWG = Rated 55A at 90°C • [Tbl 310.15(B)(16)]

Lower Ampacity ■8 AWG = 31.9A* (55A x 0.58)

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The Table 310.15(B)(3)(a) adjustment factors only apply when there are more than three current-carrying conductors bundled together. Nat­ urally, all phase conductors are considered current carrying, and the following factors help determine if the neutral conductor is considered a current carrying conductor.

Both the neutral and ungrounded conductors of a 2-wire circuit carry current and both are considered current carrying. Figure 6-40

2-W ire Circuits 310.15(B )(5)

Entire Circuit ’ Am pacity = 5 5 A "

Both the neutral and ungrounded conductors of a 2-wire circuit carry current and both are considered current-carrying.

W hen different am pacities apply, the higher am pacity is perm itted fo r the entire circuit if the reduced am pacity length do esn ’t exceed 10 ft or 10 percent o f the length o f the higher am pacity, w hichever is less.

Figure 6-38

Neutral = 10A Neutral COUNTED Conductor Am pacity - Higher Rating 310.15(A )(2) Ex

Higher Am pacity on Wall 8 THW N-2 = 55A at 90°C

Lower Am pacity on Roof 8 THW N-2 = 31.90A (55A x 0.58)

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Figure 6-40

m m

\ I 1 f m Copyright 2014 www.MikeHolt.com

11 j

m

12m 1— r—

E3 1 T * * lL !

W hen different am pacities apply, the higher am pacity is perm itted fo r the entire circuit if the reduced am pacity length d o esn’t exceed the lessor of 10 ft or 10 percent of the length o f the higher ampacity. j

Figure 6-39

j

Neutral Conductor [310.15(B)(5)(a)] The neutral conductor of a 3-wire, single-phase, 120/240V system, or 4-wire, three-phase, 120/208V or 277/480V wye-connected system, isn’t considered a current-carrying conductor for conductor ampacity adjustment of 310.15(B)(3)(a). Figure 6-41

Neutral Conductor— Unbalanced 3-Wire Wye Circuit [310.15(B)(5)(b)] The neutral conductor of a 3-wire circuit of a 4-wire, three-phase, wye-connected system carries approximately the same current as the line-to-neutral load currents of the other conductors and is considered a current-carrying conductor. Figure 6-42

i8

Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

Conductor Sizing and Protection Calculations I Unit 6

Neutral Conductors From the Same Circuit 310.15(B)(5)(a)

Question: What’s the neutral current for two 16A, 120V circuits with a common neutral? The system is a 120/208V, threephase, 4-wire, wye-connected system that supplies fluorescent lighting. Figure 6-43

120/208V or 277/480V

120/240V

► Neutral Conductor Current Example

Primary Not Shown

(a) 8A

(b) 16A

(c) 32A

(d) 40A

Answer: (b) 16A N NOT Counted

^Neutral ~ ^ ^Linel^ u iie ^

^Llm 1

^

^Lifted

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Neutral conductors that carry only unbalanced current from other conductors from the same circuit aren’t considered current-carrying conductors.

iNeM= { W T W F ( m T 6 ) 1 ^ = ^ 1 2 - 256) V256 11Neutral ~

rigure 6-41

Neutral Conductor of a 3-W ire Circuit From a W ye 4-W ire System 310.15(B)(5)(b) 120/208V or 277/480V

Neutral Current - Wye Systems 310.15(B)(5)(b)

Circuit 2

C irc u it 1

Current-Carrying C o n d u cto r! The neutral conductor of a 3-wire circuit from a 4-wire, wye system is considered a current-carrying conductor.

Two 16A, 120V circuits w ith a com m on neutral. T he system is 120/208V, 4-w ire, 3-phase. I o f N eutral = v (L 1 2 + L22) - (Li x l 2)

Determine the neutral current. I N E U T R A L = : V(L22 + L32) - (L2 x L3) Copyright 2Q14 www.MikeHolt.com

I

NEUTRAL

I NEUTRAL

= v (i6 2 + 162) - (16 x 16) = x 1256+ 256)-(256) 256" = v' 256

= ./ (1 002 + 1002) - (1 00 X 100) =

1 0 0 A

o f Neutral = 16A

:igure 6-42

Figure 6-43

n such a situation, one of the line-to-neutral currents isn’t present and :an be zeroed out of the neutral current formula, resulting in the folowing formula:

Neutral Conductor— Nonlinear Loads [310.15(B)(5)(c)] Author’s Comment:

Inbalanced 3-Wire Wye Secondary Neutral Current Formula: ^Neutral

^ L in e l

+ ^Une2 )

^ L ir> e 1 L in e 2 ^

■ According to Article 100, a nonlinear load is a load where the current waveform doesn’t follow the applied sinusoidal volt­ age waveform.

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Unit 6 I Conductor Sizing and Protection Calculations The neutral conductor of a 4-wire, three-phase, 120/208V or 277/480V wye-connected system is considered a current-carrying conductor for conductor ampacity adjustment [310.15(B)(3)(a)] if more than 50 per­ cent of the neutral load consists of nonlinear loads. Figure 6-44 Neutral Conductor of a W ye 4-W ire Circuit Supplying Nonlinear Loads 310.15(B)(5)(c)

6.14 Wireway— Conductor Ampacity Adjustment Wireways are commonly used where access to the conductors within the raceway is required to make terminations, splices, or taps to sev­ eral devices at a single location.

Conductor Ampacity Adjustment Factors When more than 30 current-carrying conductors are installed in any cross-sectional area of the wireway, the conductor ampacity, as listed in Table 310.15(B)(16), must be adjusted in accordance with Table 310.15(B)(3)(a). Figure 6-46

Odd triplen harm onic currents from nonlinear loads add on the neutral conductor and the actual current can be twice the ungrounded conductor’s current. Figure 6-44

Grounding and Bonding Conductors [310.15(B)(6)] Grounding and bonding conductors aren’t counted when adjusting conductor ampacity for the effects of conductor bundling [310.15(B) (6)]. Figure 6-45

Conductor Am pacity - Grounding and Bonding Conductors 310.15(B)(6)

Copyright 2014

W hen m ore than 30 current-carrying conductors are installed in any cross-sectional area o f the w ireway, ? the conductor ampacity, as listed in Table 3 1 0 .1 5 (B )(1 6 ),1 1 m ust be adjusted in accordance w ith 310.15(B)(3)(a). ■I.i«i ..... i.ji.......i ........1—...—i ni

Figure 6-46

► Wireway— Conductor Ampacity Adjustment Example Question: What’s the ampacity of 8 THHN if there are thirty-one conductors in a cross-sectional area of a wireway? Figure 6-47 (a) 20A Copyright 2014, www.MlkeHolt.com

Equipm ent grounding and bonding conductors aren’t current-carrying and aren’t counted when applying the provisions of Table 310.15(B)(3)(a). Figure 6-45

(b) 22A

(c)29A

(d) 32A

Answer: (b) 22A Adjusted Ampacity = Table 310.15(B)(16) Ampacity x Bundled Ampacity Adjustment Factor from Table 310.15(B)(3)(a) 1THHN is rated 55A at 90°C [Table 310.15(B)(16)]

Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

Conductor Sizing and Protection Calculations I Unit 6 C onductor A m pacity - C orrection and A djustm ent

Adjustment factor for thirty-one current-carrying conductors is 0.40 [Table 310.15(B)(3)(a)] Corrected Ampacity = 55A x 0.40 Corrected Ampacity = 22A

310.15(B) and Table 310.15(B)(16) D

This racew ay contains only 3 current-carrying conductors,

*7 ^

Table 310.15(B )(16) am pacity is based on an am bient tem perature o f 86 °F and no m ore than 3 current-carrying conductors bundled together.

W ire w a y - C o n d u c to r A m p a c ity

376.22(B)

Am pacity C orrection

Ambient

\ Tem perature

..,ID

If the am bient tem perature is above 86 °F or below 78°F, the conductor am pacity changes. [Table 310.15(B)(2)(a)].

T h irty -o n e 8 THHN

Copyright 2014, www.MikeHolt.com

A m pacity A djustm ent C onductor Bundling If the num ber o f currentcarrying conductors exceeds 3, the conductor am pacity decreases [Table 310.15(B)(3)(a)].

Figure 6-48

8 TH H N is rated 5 5 A a t 90°C [Table 310.15(B )(16)] A d justm ent fo r 31 conductors is 40% [Tbl. 310.15(B )(3)(a)]. A djusted A m p a city = 5 5A x 0.40 = 22A

6.16 Conductor Sizing Branch-Circuit Overcurrent Protection [210.20(A)]

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Branch-circuit overcurrent devices must have a rating of not less than 125 percent of the continuous loads, plus 100 percent of the noncon­ tinuous loads.

:igure 6-47

6.15 Conductor Ampacity Summary ► Branch Circuit Protection Sizing Example \ccording to the Article 100 definition, the ampacity of a conductor s based on the “conditions of use.” Table 310.15(B)(16) contains the illowable ampacities for insulated conductors, where no more than hree current-carrying conductors bundled together, based on ambient emperature of 30°C (86°F). 'he allowable ampacity as listed in Table 310.15(B)(16), must be corected if the ambient temperature isn’t 86°F from Table 310.15(B)(2)(a), ind adjusted if there are four or more current-carrying conductors bunlled together according to the factors contained in Table 310.15(B)(3)(a). :igure 6-48

Question: What size overcurrent protection will be required for a branch circuit supplying a 45A continuous nonlinear load? Figure 6-49 (a)50A

(b)60A

(c)70A

(d) none of these

Answer: (b) 60A, 45A x 1.25 = 56A [240.6(A)]

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191

I

Unit 6 I Conductor Sizing and Protection Calculations Branch Circuit Sizing 210.19(A )(1)(b}

Branch-Circuit O vercurrent Protection 210.20(A) 45A x 1.25 = 56A [240.4(B), 240.6] O vercurrent Protection - 60A

I I Overcurrent protection must be sized no less than 125 percent of the continuous load, plus 100 percent of the noncontinuous load. Copyright 2014, www.MlkeHolt.com

Conductors must be sized to the maximum load to be served after applying any adjustment or correction factors.

45 A Continuous Load \wmmmmmmmmmmm

!

31

m

Copyright 2014, www.MikeHolt.u

a

\% ___

Figure 6-51

Figure 6-49

Branch Circuits Conductor Sizing [210.19(A)(1)]

► Branch Circuit Conductor Sizing Example

Branch-circuit conductors must have an ampacity of not less than the maximum load to be served. The conductor must be the larger of (a)

or (b). (a) Conductors must be sized no less than 125 percent of the contin­ uous loads, plus 100 percent of the noncontinuous loads, based on the terminal temperature rating ampacities as listed in Table 310.15(B)(16). Figure 6-50

Question: What size branch circuit conductor (THHN) is required for a 45A continuous nonlinear load that requires three ungrounded conductors and a neutral (four current-carrying conductors)? Figure 6-52 (a) 8 AWG (c) 4 AWG

(b) 6 AWG (d) none of these

Answer: (c) 4 AWG Branch Circuit Sizing 210.19(A )(1)[a}

Conductors must be sized no less than 125 percent of the continuous load, plus 100 percent of the noncontinuous load.

210.19(A)(1)(a)— Since the load is 45A continuous, the conduc­ tor must be sized to have an ampacity of not less than 56A (45A x 1.25). According to Table 310.15(B)(16), 75°C column, a 4 AWG conductor Is suitable, because it has an ampere rating of 70A at 60°C before any conductor ampacity adjustment and/or correc­ tion is applied.

H t~

210.19(A)(1)(b)— Because the neutral is considered a currentcarrying conductor per 310.15(B)(5)(c), there are four currentcarrying conductors. Therefore we must apply Table 310.15(B)(3) adjustment factor of 80%.

Copyright 2014, www.MikoHoit com

J

4 THHN is 95A at 90°C [ Table 310.15(B)(16)] IK

Adjustment factor for four current-carrying conductors is 0.80 [Table 310.15(B)(3)(a)].

Figure 6-50 (b) Conductors must be sized to carry the load after the application of correction or adjustment factors. Figure 6-51

92

Corrected Ampacity = 95A x 0.80 Corrected Ampacity = 76A, which is adequate for the 45A load.

Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

Conductor Sizing and Protection Calculations I Unit 6

I Branch C ircuit Sizing

Feeder O vercurrent Device 215.3

1210.19(A)(1)(a) and(b) 210.19(A)(1)(a): 4 5 A x 1.25 = 56A 4 AWG Rated 70A at 60°C [Table 310.15(B)(16)] 210.19(A)(1)(b):

2 0 0 A x 1.25 =

Data Processing Equipm ent With Nonlinear Load

4 Current Carrying Conductors [310.15(B)(5)(c)] [Table 310.15(B)(3)(a)] 80% Adjustm ent Factor Use 90°C rating for adjustment 4 AWG Rated 95A at 90°C [Table 310.15(B)(16)] 95A x 0.80 = 76A (Ok for 45A Load [210.19(A)(1)(b)]

Equipment Rated 60°C

45 A C ontinuous Load

Copyright 2014 wvvwMikeHolt.com

250A Overcurrent Device [240.6(A)]

a /

i

200A Continuous Load

I

Feeder overcurrent devices must have an am pacity not less than 125 percent of the continuous loads, plus 100 percent of the noncontinuous loads. Copyright 2014, www.MikeHolt.com

Lit

Figure 6-52

Figure 6-53

Feeder Overcurrent Protection [215.3] ■

Feeder Conductor Size 215.2(A )(1)(a}

Feeder overcurrent devices must have a rating of not less than 125 percent of the continuous loads, plus 100 percent of the noncontinu­ ous loads.

► Feeder Protection Sizing Example Question: What size overcurrent protection will be required for a feeder supplying a 200A continuous nonlinear load? Figure 6-53 (a) 250A

(b) 200A

Answer: (a) 250A, 200A x 1.25 = 250A [240.6(A)]

(c)150A(d)noneofthese The feeder ampacity must be not less than 125% of the continuous load, plus 100% of the noncontinuous load. Figure 6-54

Feeder Conductor Sizing [215.2(A)(1)] Feeder conductors must be sized to the larger of 215.2(A)(1)(a) or 215.2(A)(1)(b). (a) The minimum feeder conductor ampacity must be no less than 125 percent of the continuous load, plus 100 percent of the noncontin­ uous load, based on the terminal temperature rating ampacities as listed in Table 310.15(B)(16) [110.14(C)(1)], Figure 6-54

(b) The feeder ampacity must be not less than the maximum load to be served after the application of any adjustment or correction factors. Figure 6-55

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193

Unit 6 I Conductor Sizing and Protection Calculations

Feeder C o nductor Size H

250 kcmil THHN is rated 255A at 90°C [Table 310,15(B)(16)]

215.2(A)(1) (t).

II ■ I

A\T |* t|| I)

fs i

8

.............

kTfiT IVZLZJ ^

.

ifjjl

The fee de r am pacity m ust be not less than the m axim um load to be served after the application o f anv adjustm ent or correction factors.

Corrected Ampacity = 255A x0.91 x 0.80 Corrected Ampacity = 211A The conductor if 250 kcmil THHN is rated 211A conductor after ampacity correction and adjustment and it is capable of supplying the 200A continuous load in accordance with 215.2(A)(1)(b), but it’s not permitted to be protected by the required 250A protection device as required by 240.4. A 300 kcmil conductor is rated 285A before correction and adjustment at 75°C according to Table 310.15(B)(1), it’s rated 233A after correction and adjustment [320A x 0.91 x 0.80], and it’s permitted to be protected by a 250A overcurrent device as per the next size up rule contained in 240.4(B).

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Figure 6-55

► Feeder Conductor Sizing Example

„ Feeder C onductor Size

S

Question: What size feeder conductor (THHN) is required for a 200A continuous nonlinear load at an ambient temperature of 100°F (four current-carrying conductors)? Figure 6-56 (a) 2/0 AWG

(b) 250 kcmil

(c) 4/0 AWG

A m bient Tem perature 1 00°F

200A Continuous Load

Ungrounded Conductor 250 kcmil THHN Rated 255Aat 75°C, 290A at 90°C [Tbl 310,15(B)(16)] 100°F Correction Factor = 0.91 for 90°C insulation, [Tbl 310.15(B)(2)(a)] 4 hot conductors adjustment = 0.80 [Table 310.15(B)(3)(a)] 290A x 0.91 x 0.80 = 211 A, okay for the 200A load but...

(d) 300 kcmil

Answer: (d) 300 kcmil 215.2(A)(1)(a)— Since the load is 200A continuous, the conduc­ tors must be sized to have an ampacity of not less than 250A (200A x 1.25). According to Table 310.15(B)(16), 75°C column, a 250 kcmil conductor is suitable, because it has an ampere rating of 255A at 75°C before any conductor ampacity adjustment and/ or correction is applied.

215.2(A)(1)(b)

VIO LA TIO N : T he 250 kcm il conductor is rated 211A after am pacity adjustm ent isn’t perm itted to be protected by the 250A protection device [240.6(A)], Copyright 2014, www.MikeHolt.com

210.19(A)(1)(b)— Because the ambient temperature is not 86°C and there are more then three current-carrying conductors we need to determine the corrected ampacity to ensure that it has the ability to supply the 200A load as well as being protected by a 250A protection device. Ambient Temperature Correction Factor for 100°F= 0.91 [Table 310.15(B)(2)(a)] Because the neutral is considered a current-carrying conductor per 310.15(B)(5)(c), there are four current- carrying conductors. Therefore we must apply Table 310.15(B)(3)(adjustment factor of 80%.

Figure 6-56

6.17 Feeder Tap Rules Tap. A conductor, other than a service conductor, that has overcurrent protection rated higher than normally allowed in 240.2. Figure 6-57

Feeder Taps Not Over 10' [240.21(B)(1)] Feeder tap conductors up to 10 ft long are allowed without overcurrent protection at the tap location if installed as follows:

Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

Conductor Sizing and Protection Calculations I Unit 6

(3) The tap conductors must be installed in a raceway if they leave the enclosure.

Tap Conductors 240.2 D efinition

S e rv ic e C o n d u c to rs

Legend O ; ^ Not | ft Tonci .

P

i

— S ervice — F eeder ‘— F e ede r Tap _ B ranch C kt

(4) The tap conductors must have an ampacity not less than onetenth of the rating of the overcurrent device that protects the feeder. Figure 6-59 Inside Feeder Taps N ot O ver 10 Ft Tap A m pa city -1 0 % R ule

240.21(B)(1)(4)

Branch Circuit Copyright 2014. www.MikeHolLcom

Conductors, other than service conductors, that have overcurrent protection ahead of the point of supply that exceeds the value permitted for sim ilar conductors.

H

400A Feeder Protection D evice (1/10th = 40A )

11

Figure 6-57 (1) The ampacity of the tap conductor must not be less than: Figure 6-58 a. The calculated load in accordance with Article 220, and

The am pacity o f tap conductors c a n ’t be less than 1/10th the rating o f the device protecting the feeder. Copynght 2014, www.MikeHolt.com

b. The rating of the device or overcurrent device supplied by the tap conductors. Inside F eede r Taps N ot O ver 10 Ft Tap A m pa city

Figure 6-59

► 10-Foot Feeder Tap Example 1

240.21(B)(1)(1) Question: Using the 10-foot tap rule, what’s the minimum size conductor required to supply a 200 overcurrent device, if the tap is from feeder conductors protected by a 400A circuit breaker? Figure 6-60

N N

'I

(a) 3 AWG

(b) 1/0 AWG

(c) 3/0 AWG

(d) 250 kemii

Feeder

Answer: (c) 3/0 AWG Copyright 2014 www.MikeHoltcom

\ x / r - : T he fe e d e r tap c o n d ucto r am pacity m ust not be less than: a. T he calculated load. b. T he rating o f the o v ercu rren t device supplied by the tap conductors.

:igure 6-58

The tap conductors must have an ampacity no less than the rating of the 200A overcurrent device that it terminates into and no less than one-tenth the rating of the overcurrent protection device protecting the feeder conductors. A 3/0 AWG conductor is rated 200A at 75°C, and it has an ampacity of at least 40A (onetenth the rating of the 400A feeder overcurrent device (400/10) [Table 310.15(B)(16)].

2) The tap conductors must not extend beyond the equipment they supply.

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Unit 6 I Conductor Sizing and Protection Calculations

Inside Feeder Taps Not O ver 10 Ft I | 240.21 (B )(1)(4) 400A Protection Device (1/1 Oth = 40A)

The tap conductors must have an ampacity no less than the 40A rating of the overcurrent device that it terminates into and no less than one-tenth the rating of the overcurrent protection device protecting the feeder conductors. An 8 AWG conductor is rated 40A at 60°C, and it has an ampacity of at least 40A (one-tenth the rating of the 400A feeder overcurrent device (400/10) [Table 310.15(B)(16)].

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3/0 AWG Rated 200A at 75°C

1/0 AWG Rated 150A at 75°C

8 AWG Rated 40A at 60°C

The a m pacity o f tap conductors c a n ’t be less than 1/1 Oth the rating o f the device protecting the feeder.

Figure 6-60

Feeder Taps Over 10 Feet but Not Over 25 Feet [240.21 (B)(2)] Feeder tap conductors over 10 ft in length but not over 25 ft are per­ mitted without overcurrent protection at the tap location if installed as follows: Figure 6-61 Inside Feeder Taps Not Over 25 Ft 240.21(B)(2)

► 10-Foot Feeder Tap Example 2 Question: Using the 10-foot tap rule, what’s the minimum size conductor required to supply a 150A overcurrent device, if the tap is from feeder conductors protected by a 400A circuit? See Figure 6-60. (a) 3 AWG

(b) 1/0 AWG

(c) 3/0 AWG

400A D evice:

(d) 250 kcmii

B

j

Answer: (b) 1/0 AWG

. .

-

^

r.r

The tap conductors must have an ampacity no less than the rating of the 150A overcurrent device that it terminates into and no less than one-tenth the rating of the overcurrent protection device protecting the feeder conductors. A 1/0 AWG conductor is rated 150A at 75°C, and it has an ampacity of at least 40A (onetenth the rating of the 400A feeder overcurrent device (400/10) [Table 310.15(B)(16)].

copynght^cm

www.MtkeHolt.com

The tap conductors must: • Have am pacity of not less than 1/3 the rating of the overcurrent device. _ . . . . r. • Terminate in a single circuit breaker or set of fuses rated not more than the am pacity of the conductor.

Figure 6-61

(1) The ampacity of the tap conductors must not be less than onethird the rating of the overcurrent device that protects the feeder.

► 10-Foot Feeder Tap Example 3 Question: Using the 10-foot tap rule, what’s the minimum size conductor required to supply a 30A overcurrent device, if the tap is from feeder conductors protected by a 400A circuit? See Figure 6-60. (a) 12 AWG

(b) 8 AWG

(c) 1/0 AWG

(2) The tap conductors terminate in a single circuit breaker, or set of fuses rated no more than the tap conductor ampacity in accor­ dance with 310.15 [Table 310.15(B)(16)]. (3) The tap conductors must be protected from physical damage by being enclosed in a manner approved by the authority having juris­ diction, such as within a raceway.

(d) 3/0 AWG

Answer: (b) 8 AWG

Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

Conductor Sizing and Protection Calculations I Unit 6

► 25-Foot Feeder Tap Example 1

► 25-Foot Feeder Tap Example 2

Question: Using the 25-foot tap rule, what’s the minimum size conductor required to supply a 200A overcurrent device, if the tap is from feeder conductors protected by a 400A circuit? Figure 6-62

Question: Using the 25-foot tap rule, what’s the minimum size conductor required to supply a 150A overcurrent device, if the tap is from feeder conductors protected by a 400A circuit? See Figure 6-62.

(a) 3 AWG

(b) 1/0 AWG

(c) 3/0 AWG

(d) 250 kcmil

Answer: (c) 3/0 AWG

Inside Feeder Taps Not Over 25 Ft N

II

30A

N

(d) 250 kcmil

The tap conductors must have an ampacity no less than the 150A rating of the overcurrent device that it terminates into and no less than one-third the rating of the overcurrent protection device pro­ tecting the feeder conductors. A 1/0 AWG conductor is rated 150A at 75°C, and it has an ampacity of at least 133A (one-third the rating of the 400A feeder overcurrent device (400/3) [Table 310.15(B)(16)].

Question: Using the 25-foot tap rule, what’s the minimum size conductor required to supply a 30A overcurrent device, if the tap is from feeder conductors protected by a 400A circuit? See Figure 6-62. (a) 3 AWG

Copyright 2014, www.MikeHoH.com

*4E-

w

^0

r -------- --------- ^ r - ------------------^ : 3/0 AWG R ated: 1/0 AWG Rated 1/0 AWG Rated 200A at 75°C 150Aat75°C 150A at 75°C The am pacity of tap conductors can’t be less than 1/3 the rating of the device protecting the feeder. zigure 6-62

(c) 3/0 AWG

► 25-Foot Feeder Tap Example 3

240.21(B)(2)

r -

(b) 1/0 AWG

Answer: (b) 1/0 AWG

The tap conductors must have an ampacity no less than the 200A raprfg of the overcurrent device that it terminates into and no less than one-third the rating of the overcurrent protection device pro­ tecting the feeder conductors. A 3/0 AWG conductor is rated 200A at 75°C, and it has an ampacity of at least 133A (one-third the rating of the 400A feeder overcurrent device (400/3) [Table 310.15(B)(16)].

400A Device (400/3 = 133.3A)

(a) 3 AWG

(b) 1/0 AWG

(c) 3/0 AWG

(d) 250 kcmil

Answer: (b) 1/0 AWG The tap conductors must have an ampacity no less than the 30A rating of the overcurrent device that it terminates into and no less than one-third the rating of the overcurrent protection device protecting the feeder conductors. A 1/0 AWG conductor is required because it has an ampacity of at least 133A (one-third of the rating of the 400A feeder overcurrent device (400/3) /Table 310.15(B)(16)].

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Unit 6 I Conductor Sizing and Protection Calculations Outside Feeder Taps

Outside Feeder Taps [240.21(B)(5)] 240.21(B)(5)

Outside feeder tap conductors can be of unlimited length, without overcurrent protection at the point they receive their supply, if they comply with the following: Figure 6-63 (1) The tap conductors are suitably protected from physical damage in a raceway or manner approved by the authority having jurisdiction.

Tap conductors must: • Be protected from physical damage. • Terminate in a single breaker or set of fuses that limit the load to the am pacity of the conductors. • Remain outdoors, except at the point of entrance. ~1 • Have the disconnect located near the point of entrance.

(2) The tap conductors must terminate at a single circuit breaker or a single set of fuses that limits the load to the ampacity of the conductors. (3) The overcurrent device for the tap conductors is an integral part of the disconnecting means, or it’s located immediately adjacent to it. (4) The disconnecting means is located at a readily accessible loca­ tion, either outside the building, or nearest the point of entry of the conductors.

Outside Feeder Taps Any Length Copyright 2014 www.MikeHolt.com

Point °* ; Entrance

Figure 6-63

Conclusion to Unit 6— Conductor Sizing and Protection Calculations The sizing of conductors and overcurrent protection is much more complicated than it appears at first. This unit has provided an insight into the numerous factors that must be taken into account in properly sizing conductors and providing for their protection. Temperature is a key factor in the resistance of a conductor, and consequently temperature affects a conductor’s currentcarrying capacity. High temperature on conductors can lead to the breakdown of conductor insulation and damage to equipment. Therefore, when sizing conductors, the ampacity values given in the NEC tables must be corrected for ambient temperatures that differ from the ambient of 86° F, on which the Code tables are based. A bundle of current-carrying conduc­ tors will have increased operating temperatures, so the NEC also requires that an adjustment be made when bundling more than three current-carrying conductors together. This unit also taught you that conductors must not be sized to a higher temperature column of the ampacity table than the rating that the equipment terminals allow. This means that even though you install a 90°C rated conductor such as THHN, a lower ampacity column must be used to match the terminal rating. This confusing requirement results in calculation mistakes, so review this part of the unit again if you’re not sure how to apply Section 110.14(C) of the Code. Overcurrent protection must be sized correctly to protect the circuit wiring and equipment from damage that can result from overheating. This unit covered common overcurrent sizing requirements as well as some of the exceptions to the general rules of Article 240. A number of specific examples of overcurrent protection were explained along with examples of how to size the overcurrent devices in order to help you understand that there are specific rules for many types of equipment or situ­ ations that don’t follow the general overcurrent protection requirements.

Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

UNIT 6 REVIEW QUESTIONS Please use the 2014 Code book to answer the following questions.

Part A— Conductor Requirements

6.4 Conductor Size— Equipment Terminal Rating [110.14(C)]

6.1 Conductor Insulation [310.104(A)]

4.

1.

THHN can be described as ___ (a) thermoplastic insulation with a nylon outer cover (b) suitable for dry and wet locations (c) having a maximum operating temperature of 90°C (d) a and c

6.2 Conductor Sizes 2.

Conductor sizes are expressed in American Wire Gage (AWG) from 40 AWG through 4/0 AWG. Conductors larger than ____ are expressed in circular mils.

The smallest size conductor permitted for branch circuits, feed­ ers, and services for residential, commercial, and industrial locations is ____ . (a) 14 AWG copper (b) 12 AWG aluminum (c) 12 AWG copper (d) a and b

What’s the minimum size THHN conductor permitted to termi­ nate on a 70A circuit breaker or fuse if the circuit breaker and equipment terminals are listed for 60°C conductor sizing? (a) 8 AWG (b) 6 AWG (c) 4 AWG (d) 2 AWG

6.

6.3 Smallest Conductor Size [310.106] 3.

(a) True (b) False 5.

(a) 1 AWG (b) 1/0 AWG (c) 3/0 AWG (d) 4/0 AWG

Equipment terminals rated 100A or less (receptacles, switches, circuit breakers, fuses, and so on) and pressure connector ter­ minals for 14 AWG through 1 AWG conductors must have the conductor sized according to the 60°C temperature rating, as listed in Table 310.15(B)(16).

What’s the minimum size THHN conductor that’s permitted to terminate on a 50A circuit breaker or fuse if the circuit breaker and equipment terminals are listed for 75°C conductor sizing? (a) 10 AWG (b) 8 AWG (c) 6 AWG (d) 4 AWG

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Unit 6 I Review Questions

7.

Terminals for equipment rated over 100A and pressure connec­ tor terminals for conductors larger than 1 AWG must have the conductor sized according to the 75°C temperature rating, as listed in Table 310.15(B)(16).

6.5 Overcurrent Protection [Article 240] 12.

(a) True (b) False

(a) True (b) False 8.

What size THHN/THWN conductor is required for an airconditioning unit if the nameplate requires a conductor ampac­ ity of 34A? The terminals of all the equipment and circuit breakers are rated 75°C and the raceway is installed outdoors in a wet location. (a) 14 AWG (b) 12 AWG (c) 10 AWG (d) 8 AWG

9.

What’s the minimum size THHN conductor required for a 150A circuit breaker or fuse for a nonmotor circuit? Be sure to comply with the requirements of 110.14(C)(1)(b).

13.

In general, THHN (90°C) conductor ampacities can’t be used when sizing conductors. When more than three current-carrying conductors are bundled together, or if the ambient temperature is greater than 86°F, the allowable conductor ampacity must be decreased. THHN provides a greater ampacity for conductor adjustment purposes.

(a) 5,000A, 5,000A (b) 5,000A, 10,000a (c) 10,000A, 5,000A (d) 10,000A, 10,000A 14.

What size conductor is required to supply a 190A noncontinuous load in a dry location? The terminals are rated 75°C. (a) 2/0 AWG (b) 3/0 AWG (c) 4/0 AWG (d) 300 kcmil

i0

Which of the following is a standard size for circuit breakers and fuses? (a) 25A (b) 90A (c) 350A (d) any of these

15.

Where a circuit supplies continuous loads or any combination of continuous and noncontinuous loads, the rating of the overcur­ rent device must not be less than the noncontinuous load plus _____percent of the continuous load. (a) 80 (b) 100 (c) 125 (d) 150

(a) True (b) False 11.

Overcurrent devices must be designed and rated to clear fault current and must have a short-circuit interrupting rating suffi­ cient for the available fault levels. If the minimum interruption rating isn’t marked on the overcurrent device the minimum for circuit breakers is ____ and_____ for fuses, according to the

NEC.

(a) 1/0 AWG (b) 2/0 AWG (c) 3/0 AWG (d) 4/0 AWG 10.

One of the purposes of conductor overcurrent protection is to protect the conductors against excessive or dangerous heat.

16.

The maximum overcurrent device size for 14 AWG is 15A, 12 AWG is 20A, and 10 AWG is 30A. This is a general rule, but it doesn’t apply to motors or air conditioners according to 240.4(G). (a) True (b) False

Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

Review Questions i Unit 6

17.

If the ampacity of a conductor doesn’t correspond with the stan­ dard ampere rating of a fuse or circuit breaker, the next size up overcurrent device is permitted. This applies only if the conduc­ tors supply multioutlet receptacles for portable cord-and-plugconnected loads.

6.9 Ambient Temperature Correction [310.15(B)(2)(a)] 21.

(a) True (b) False 18.

What size conductor is required for a 70A breaker that supplies a 70A noncontinuous load where the terminals are rated 75°C? (a) 8 AWG (b) 6 AWG (c) 4 AWG (d) any of these

19.

(a) True (b) False 22.

23.

What size THHN conductor is required to feed a 16A noncontin­ uous load when the conductors are in an ambient temperature of 100°F in a dry location? The circuit is protected with a 20A overcurrent device. (a) 14 AWG (b) 12 AWG (c) 10 AWG (d) 8 AWG

Part B— Conductor Ampacity 6.8 Conductor Ampacity 20.

What’s the ampacity of an 8 THHN conductor when installed in a walk-in cooler (dry location) if the ambient temperature is 50°F? (a) 40A (b) 50A (c) 55A (d) 63A

What size feeder overcurrent device is required for a 100A con­ tinuous load? (a) 100A (b) 125A (c) 150A (d) 200A

The ampacities listed in Table 310.15(B)(16) apply only when the ambient temperature is 40°C and there are no more than two current-carrying conductors bundled together. If the ambi­ ent temperature isn’t 40°C, or there are more than two currentcarrying conductors in a raceway, the allowable ampacities must be adjusted to reflect the ampacity under the conditions of use.

The temperature rating of a conductor is the maximum operat­ ing temperature the conductor insulation can withstand (with­ out serious damage) over a prolonged period of time. The____ provide guidance for adjusting conductor ampacities for differ­ ent conditions. (a) conductor allowable ampacities (b) ambient temperature correction factors (c) over three current-carrying conductors adjustment factors (d) all of these

6.11 Ampacity Adjustment [310.15(B)(3)(a)] 24.

When four or more current-carrying conductors are bundled together for more than . the conductor allowable ampac­ ity must be reduced according to the factors listed in Table 310.15(B)(3)(a).

(a) 12 in. (b) 24 in. (c) 36 in. (d) 48 in.

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201

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Unit 6 I Review Questions

25.

What’s the ampacity of 1/0 THHN conductors when there are four current-carrying conductors installed in a raceway in a dry location? (a) (b) (c) (d)

26.

30.

111A 136A 153A 171A

A raceway contains eight current-carrying conductors. What size conductor is required to feed a 21A noncontinuous lighting load in a dry location? The overcurrent device is rated 30A.

(a) True (b) False 31.

(a) 14 THHN (b) 12 THHN (c) 10 THHN (d) any of these

6.12 Combining Ambient Temperature and Conductor Bun­ dling Adjustments 27.

What’s the ampacity of 10 THHN conductors when there are eight current-carrying conductors installed in the same raceway in a dry location and the ambient temperature is 100°F?

The ampacity of a conductor can be different along the length of the conductor. The higher calculated ampacity can be used if the length of the lower ampacity is no more than 10 ft, or no more than 10 percent of the length of the higher ampacity cir­ cuit conductors, whichever is less. (a) True (b) False

6.13 Current-Carrying Conductors 29.

The neutral conductor of a 3-wire wye circuit from a 4-wire, three-phase, wye system isn’t considered a current-carrying conductor for the purpose of applying bundle adjustment factors. (a) True (b) False

6.17 Feeder Tap Rules 32.

Feeder tap conductors not over 25 ft are permitted, providing the____ . (a) ampacity of the tap conductors isn’t less than one-third the rating of the overcurrent device protecting the feeder con­ ductors being tapped (b) tap conductors terminate in a single circuit breaker or set of fuses that limit the load to the ampacity of the tap conductors (c) tap conductors are suitably protected from physical damage (d) all of these

(a) 21A (b) 25A (c) 32A (d) 40A 28.

The neutral conductor of a balanced 4-wire, three-phase wye circuit, that’s at least 50 percent loaded with nonlinear loads (electric-discharge lighting, electronic ballasts, dimmers, con­ trols, computers, laboratory test equipment, medical test equip­ ment, recording studio equipment, and so forth) isn’t considered a current-carrying conductor for the purpose of applying bundle adjustment factors.

33.

Outside feeder tap conductors can be of unlimited length with­ out overcurrent protection at the point they receive their supply if the conductors____ . (a) are suitably protected from physical damage (b) terminate at a single circuit breaker or a single set of fuses that limits the load to the ampacity of the conductors (c) a and b (d) none of these

The neutral conductor that carries only the unbalanced current from a 3-wire delta circuit, or 4-wire, three-phase wye circuit, is considered a current-carrying conductor for the purpose of applying the adjustment factors of Table 310.15(B)(3)(a). (a) True (b) False

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Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

UNIT 6 CHALLENGE QUESTIONS Please use the 2014 Code book to answer the following questions.

Part B— Conductor Ampacity

6.12 Combining Ambient Temperature and Conductor Bundling Adjustments

6.9 Ambient Temperature Correction [310.15(B)(2)(a)] 4. 1.

A 2 TW conductor is installed in a location where the ambient temperature is expected to be 102°F. The temperature correc­ tion factor for conductor ampacity in this location is ___ (a) 0.71 (b) 0.82 (c) 0.88 (d) 0.96

2.

If the ambient temperature is 71 °C, the minimum insulation temperature rating that a conductor must have and still have the capacity to carry current is ____ .

The ampacity of six current-carrying 4/0 XHHW aluminum con­ ductors installed in a ground floor slab (wet location) is .

A(n)____ THHN conductor is required for a 19.70A noncontinuous load if the ambient temperature is 75°F and there are nine current-carrying conductors in the raceway in a dry location. (a) 14 (b) 12 (c) 10 (d) 8

6.

6.11 Ampacity Adjustment [310.15(B)(3)(a)] 3.

(a) 12A (b) 15A (c) 22A (d) 30A 5.

(a) 60°C (b) 90°C (c) 105°C (d) 120°C

The ampacity of 15 current-carrying 10 RHW aluminum conduc­ tors in a raceway in an ambient temperature of 75°F is ____ .

The ampacity for each of nine current-carrying 10 THW conduc­ tors installed in a 20 in. long raceway is . (a) 25A (b) 30A (c) 35A (d) 40A

(a) 135A (b) 144A (c) 185A (d) 21OA

Mike Holt Enterprises, Inc. 3 www.MikeHoit.com • 888.NEC.C0DE (632.2633)

Unit 6 I Challenge Questions

7.

The ampacity for each of 10 current-carrying 6 THHW conduc­ tors installed in an 18 in. long conduit in a dry location having an ambient temperature of 39°C is ____ .

6.16 Conductor Sizing 10.

(a) 40A (b) 68A (c) 75A (d) 90A

A continuous load of 27A requires the circuit overcurrent device to be sized a t____ . (a) 20A (b) 30A (c) 35A (d) 40A

6.13 Current-Carrying Conductors 11. 8.

A raceway contains the following: One 4-wire, multiwire branch circuit that supplies a balanced incandescent 120V lighting load; one 4-wire, multiwire branch circuit that supplies a balanced 120V fluorescent lighting load; two conductors that supply a receptacle; and one equipment grounding conductor. The system is 120/208V, three-phase. Taking these factors into con­ sideration, how many of these conductors are considered cur­ rent carrying?

(a) 45A (b) 50A (c) 60A (d) 70A 12.

(a) 7 conductors (b) 8 conductors (c) 9 conductors (d) 11 conductors 9.

There’s a total of nine 10 THW conductors in a raceway. The system voltage is 120/208V, three-phase. One conductor is an equipment grounding conductor; four conductors supply a 4-wire, multiwire branch circuit for balanced electric-discharge luminaires; and the remaining conductors supply a 4-wire, mul­ tiwire branch circuit for balanced incandescent luminaires. Taking all of these factors into consideration, how many of these conductors are considered current carrying?

What size overcurrent device is required for a 45A continuous load? The circuit is in a raceway with 14 current-carrying con­ ductors.

A 65A continuous load requires a ____ overcurrent device. (a) 60A (b) 70A (c) 75A (d) 90A

13.

A department store feeder (continuous load) supplies a lighting load of 103A. The minimum size overcurrent device permitted for this feeder is ____ . (a) (b) (c) (d)

110A 125A 150A 175A

(a) 6 conductors (b) 7 conductors (c) 9 conductors (d) 10 conductors

m

Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC