P.E. Electrical & Computer Power Practice Exam: Complete Set (Volumes 1-4) [3 ed.] 9781635870510

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volumes 1-4

3rd edition

RE. ELECTRICAL & COMPUTER:

POWER PRACTICE EXAM COMPLETE SET VOLUMES 1-4 3rd edition

Com W * n a y 932 S Ainsworth Ave | Tacoma | WA 98405 www, compleximaginary.com

RE. ELECTRICAL & COMPUTER: POWER PRACTICE EXAM COMPLETE SET VOLUMES 1-4 3rd edition

Published by Complex Imaginary, LLC 932 S Ainsworth Ave Tacoma, WA 98405 ISBN 978-1-63587-051-0

© 2017 Complex Imaginary, LLC

All Rights Reserved. Copyright under Berne Copyright Convention, Universal Copyright Convention, and Pan-American Copyright Convention. No Part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise without prior written consent of the publisher. This text is composed and intended to aid candidates taking the California Professional Engineer License Examination for the Electrical discipline. Information, references, and content provided herein is intended to be useful for test preparation, yet is not intended to be completely exhaustive or authoritative in any way. The NCEES has not authorized, nor has it approved, this text and its content. Complex Imaginary, LLC and its publications are in no way affiliated with the NCEES or any official test-providing agency. The authors, publishers, and distributors neither assume nor accept any liability for negligence for the content provided herein. This text does not claim to be exhaustive, complete, free from error, nor does it guarantee - in any way - its usefulness or effectiveness in test preparation. This manual is not intended for use in preparing any engineering plans, specification, designs, or other professional engineering tasks.

Purpose Okay, let’s get to it. The purpose of this practice test is just that: practice. Simple right? The NCEES Principles and Practice of Engineering PE License Electrical and Com puter— Power (a.k.a Electrical PE) test is a very technically demanding and stamina-straining exam. The wise test-taker will have studied and practiced with an eye toward three goals: (1) know the content of the subjects covered in the exam, (2) know source reference material to be used when taking the exam, and (3) be able to work efficiently for 8 long hours of testing. Numbers one and three will be handled by working on these practice tests, number two is up to you. If you’re smart, you’ll look at these practice questions as an opportunity to gather your sources and organize them for repeated use. Looking up answers and forgetting where you found them is a waste of your time.

Getting to the Gore Question These tests are filled with misdirecting information that may seem critical to the answer to the tester unfamiliar with the subject matter. This misdirecting (or extraneous) information is just as much a part of what is being tested as the pertinent information to the question at hand. You see, knowing what matters to the question is just as important as knowing what doesn’t matter to the question. Only a knowledgeable tester will be able to make the distinction quickly — and that is the same sort of knowledge which will get the answers right. You need to have both areas covered. Ignorance of what is important vs unimportant for a given subject matter is a sure guarantee for failure. Massive amounts of time can be wasted in idle searches for connections that don’t exist and data that doesn’t relate to the actual question. One of the significant uses for this practice test is to prepare you for this inevitability. The misdirects and useless stuff is not there to be tricky, it’s there to make sure you’re understanding where the question really lies.

Sources This cannot be stressed enough: as you work through these practice problems, take note of your sources. Highlight, tag, dog-ear, index, and cross-reference your sources as you work. You will be coming back to the very same subjects during the actual exam. Intimate familiarity with your reference material will pay huge dividends. And volume of information doesn’t replace quality of information. Too many testers bring whole libraries to the test. What good will that much information do? The more information you have to page through, the more wasted time. Your source and reference material should be very well “ broken in” by the time you take the exam. That way you’ll need less volume, you’ ll have more dense and directlyapplicable material at hand, and your efficiency will be honed. Notate and personalize your references as well: areas that confused you, areas that resolved confusion, references to other sources, anything that sparks the memory. This is a huge deal. Understand this vital issue: the more time spend wandering through texts will not only slow you, it will fatigue you. You need to be concerned about keeping your mind as fresh as possible as well as protecting your use of time. It’s a tw o-part sin to be wandering through sources during the test: wasted time, wasted energy. Both are precious.

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The Nature of Practice The exam is like a marathon; it requires stamina that is most certainly physical. Can you keep a keen eye out and a critical mind functioning intensely for eight total hours? You will need to. This test is full of dry, technical information. Your mind will need to be strong to work through all of it in a condensed and time-sensitive stretch. The weaker you are at working through these questions over a long haul, the weaker your chances for success. This practice test is also meant to hammer on this issue. To run a marathon successfully, you need to practice running, lots of running. You need to run short bursts, run long stretches, train your mind to deal with the fatigue, acclimate your whole body to the experience; then you’ll do well. Test your stamina as well as your subject matter with this test. Don’t just look at the practice test as a key to studying the subject (which it is), look at it as the chance to do mental laps around the track. It is reasonable that a very intelligent and capable electrical engineer will critically fatigue after a couple of hours of intense questioning. Can you last the whole day?! Again, it’s not a matter of knowledge or smarts - it’s a matter of raw mental stamina. These tests should help “get you in shape” mentally to run a very long distance without having a break down.

Repetition One issue that may frustrate the student taking multiple practice tests is the apparent lack of variety of the questions. More often than not, a given question will sound/seem very similar to another. This is not due to a lack of originality on the part of the test writers, rather it is because these questions are of the sort (and deal with the sort of information) that will appear on the actual exam. The point of practice tests is never originality or novelty, it is to drill, drill, drill. It would be foolish for the student to become frustrated or feel discouraged because a bunch of questions are basically the same as another bunch of questions. That’s the point of practice! You should feel encouraged when you see a question and think, “this is exactly the same as that other one I’ve worked on.” Such thoughts indicate that you are remembering, using, and learning the pertinent information for a particular question type. Also, inevitably, there will be areas that you are weak at. So practicing question-types that you are strong in will allow you to answer even more quickly, thus giving yourself more time on the questions that are more difficult to you. Plus, skipping questions because you already know how to answer them defeats the purpose of having to work through the volume of questions that will be on the exam.

Answer Guessing It’s pretty foolish to guess answers to the practice problems. The point of this exercise is to work through the problems, to find an answer, to note it’s source, and to learn the subject in question. Guessing does none of this. Even if you guess right, you’re being foolish. A guessed answer on a practice test w on’t help you during the real exam. Each question gives you the opportunity to study it’s subject in greater depth and breadth than is necessary to simply answer that question. Don’t just know the answer, know the stuff around it and beneath it. You should know why the wrong answers are wrong. You won’t get the same, exact question twice, but you will get ones concerned with the same subject matter. A guessed answer cannot account for this variation.

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M orning S ession A load is connected to the secondary of a transformer, as shown in the diagram. What is the equivalent impedance of the load when reflected to the primary side of the transformer? 1° 2°

0 x" = 0.15

5

A.

8.6 + 5.8/ Q

B.

15 + 1 0 /0

C.

56 + 37/ Q

D.

108 + 7 3 /Q

4 :1 5

210 + 140j £2

In the single-phase circuit shown below, when the switch is in position 1, the current ^ is 437 A. When the switch changes to position 2, a fault occurs at point F. What is

V = 1500 V R =2.4 Q

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B.

358 A

C.

437 A

D.

620 A

The secondaries of tw o 3-phase transformers are connected in parallel to the same feeder. One is wye-wye connected while the other is wye-delta connected. What will be the consequence (assume the secondaries are both the same voltage)? A.

No consequences, as long as only line-line voltages are used.

B.

The current output of both transformers will be 1/3 the rated value.

C.

One of the phases will be lost.

D.

The secondary winding voltages will be out of phase. COPYRIGHT 2017 BY COMPLEX IMAGINARY, LLC

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M orning S ession 7

In the circuit below, the voltage across the capacitor is 465 V. What is the voltage across the ampmeter?

A/VV

8

9

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A.

46.5 V

B.

4.65 V

C.

80.5 V

D.

26.8 V

1 0 :i -

A transformer connected to its full-load measures power losses of 1,572 W. The same transformer, with no load connected, experiences losses of 57 W. If the transformer is only 50% loaded, what will be the power loss? A.

436 W

B.

815 W

C.

1,572 W

D.

1,629 W

The storage capacity rating of a lead-acid battery is described as: A.

Capacitive Hour (CH)

B.

Charge Density (CD)

C.

Amp Hour (AH)

D.

Current Potential (CP)

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M orning S ession 10

Below is a circuit whose waveform across AB looks like: v

a

d2

In one scenario, the capacitor breaks and acts as an open circuit. What does the resulting waveform across AB now look like?

11

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According to NEC 2017, fixed equipment which is supplied by Class 1 circuits must be grounded if they operate at more than what specified voltage? A.

50 Volts

B.

120 Volts

C.

240 Volts

D.

They must

alwaysbe grounded

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M orning S ession 12

13

A 3-phase, 700 MVA, 34.5 kV distribution line is stepped down to 15 kV using three identical single-phase autotransformers. What is the power rating for each autotransformer? A.

134 MVA

B.

233 MVA

C.

404 MVA

D.

700 MVA

The diagram below illustrates a transmission system with 40 kV lines having an impedance (Z) of 0.002 pu per mile (using transformer’s rated base values). When a 3-phase fault occurs at Tower 2, the fault current (A) is most nearly (assume infinite bus conditions):

9 MVA Z = 8.4%

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A.

130 A

B.

940 A

C.

1,120 A

D.

1,940 A

A motor rated for 2,000 rpm at full-load has a no-load speed of 2,077 rpm. The speed regulation for this motor is: A.

1.85%

B.

2.85%

C.

3.85%

D.

4.85%

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M orning S ession 15

16

17

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A 3-phase, ungrounded, 15 kV, wye source is connected to a balanced delta load. The current on leg “a” (L-N) of the source is 45 A. What is the current on the A-B leg of the delta load? A.

26 A

B.

45 A

C.

55 A

D.

78 A

A 2,500 kW inductive load on a 3-phase system has a 0.74 lagging power factor. When a capacitor bank is added to the system, how much reactive power (kvar) will need to be added to bring the lagging power factor up to 0.87? A.

325 kvar

B.

517 kvar

C.

857 kvar

D.

2,344 kvar

According to the NEC®, if a circuit has capacitors installed on it, the conductors for the circuit must be rated at no less than: A.

125% of the total load ampacity

B.

135% of the m otor’s current rating

C.

135% of the capacitor’s current rating

D.

one-third of the capacitor’s current rating

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M orning S ession 18

19

20

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A power distribution panel is serving a single-phase 3.3 kV, 75 kW load, with a 0.90 lagging power factor. The feeder to the load has an impedance of 0.2 + y'0.5 Q. What is the actual voltage for this feeder at the panel? (Hint: set voltage reference at the load) A.

3,242 V

B.

3,297 V

C.

3,307 V

D.

3,310 V

According to the 2017 National Electrical Code®, which one of the following overload protection methods is NOT mentioned as suitable for a continuous duty m otor rated 1 hp or less that automatically starts? A.

Thermal Protector

B.

Impedance-Protected

C.

W ound-Rotor Secondaries

D.

Separate Overload Device

What is the complex power delivered to a single-phase inductive load if the current is 131 L -12° A and the applied voltage is 34.5 A. 7.8° kV. A.

2.46 + y'0.88 MVA

B.

4.25 + y'1.53 MVA

C.

4.51 -y‘0.33 MVA

D.

7.36 + y'2.65 MVA

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M orning S ession 21

22

23

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in the ferromagnetic core shown below, the flux in leg A is 0.004 Wb. If the flux density in leg A is 1.6 Teslas (T), what is the dimension “x” ?

A.

2 cm

B.

3 cm

C.

4 cm

D.

5 cm

Two single-phase transformers are paralleled together, and are supplying a total load current of 11,500 A. Transformer #1 has an impedance of 5.25%, and Transformer #2 has an impedance of 4.75%. What is Transformer #1’s contribution to the total load current? (Assume both transformers have the same turns ratios} A.

608 A

B.

5463 A

C.

6038 A

D.

7888 A

Which of the following would be considered the triplen harmonics? A.

1st, 2nd, and 3rd harmonics

B.

9th, 15th, 21st harmonics

C.

3rd, 6th, 9th harmonics

D.

9th, 12th, 15th harmonics

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M orning S ession 24

25

Page 17

The area of a room with ceiling-hung luminaries measures 35 ft wide, 22 feet long, and 8 feet high. Provided a 0.62 coefficient of utilization for the fixtures which are hung 1 ft from the ceiling, and the work plane being at 36 in, the total lumens needed to produce an illumination level of 75 fc is most nearly: A.

97,500 lumens

B.

93,100 lumens

C.

85,200 lumens

D.

50,100 lumens

For the purpose of sizing a disconnect, the maximum locked-rotor current for a code “ Design C” type 3-phase squirrel-cage induction motor rated 15 hp, 230 V, is most nearly: A.

232 A

B.

257 A

C.

267 A

D.

300 A

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M orning S ession 26

27

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M orning S ession 28

Given the half-wave rectifier shown below, which of the choices represents the expected output waveform across the load?

29

A private utility company has been approved to receive a government grant. The government is offering two options: a lump-sum payment, or $25 million a year for 10 years at an interest rate of 7.5% per-annum. What lump-sum payment amount would be equal to the present value of the second option?

Page 21

A.

$154 million

B.

$172 million

C.

$193 million

D.

$250 million

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M orning S ession 30

31

32

PAGE 23

A short circuit study is being performed on part of a system that includes a wyeconnected 3-phase synchronous generator rated for 15 kV, 10 MVA, 0.78 lagging power factor, and a synchronous reactance of 14 Q per-phase. If the internal generated voltage at rated conditions is 9.14 Z_14° kV, what is the maximum power output of this generator? A.

7.8 MW

B.

13.5 MW

C.

15.4 MW

D.

17.0 MW

Assume that in a 3-phase transmission system the conductors are spaced at irregular distances as shown below: O —— 4 ft-----*-(> « *------5 ft----The equivalent spacing is most nearly: A.

5.6 ft

B.

5.2 ft

C.

4.5 ft

D.

4.1 ft

Resonant effects can occurwhen the ground fault path includes reactance approximately equal to what? A.

the ratio of signal frequency to current magnitude

B.

the sum of the distortion of all harmonic components

C.

5% of total system power consumption

D.

the system capacitive reactance to ground

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M orning S ession 33

34

35

Page 25

The methodology for symmetrical components is most commonly used for: A.

Analysis of DC electrical power systems

B.

Analysis of single-phase electrical power systems

C.

Analysis of three-phase electrical power systems

D.

Analysis of balanced electrical power systems

What is the synchronous speed of a 60 Hz, 3-phase squirrel-cage motor with one polepair? A.

1,800 rpm

B.

72 mph

C.

3,600 rpm

D.

2,400 rpm

A single-phase transformer is rated 12.5 kV/480 V with an impedance of and a lowside full-load current rating of 250 A. If the secondary terminals are shorted together and 625 V is applied to the high-voltage side, what is the current across the secondary terminals? A.

14 A

B.

25 A

C.

150 A

D.

250 A

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M orning S ession 36

37

38

Page 27

According to the 2017 National Electrical Code®, the maximum percentage of full-load current a single-phase m otor with an instantaneous trip breaker is permitted to receive on a branch-circuit is: A.

150%

B.

800%

C.

300%

D.

100%

Transformer core losses are primarily due to: A.

Impurities in the ferromagnetic core

B.

Winding insulation

C.

Eddy currents

D.

Turns ratio mutual induction

A delta connection is shown below. What is the amount of power dissipated by resistor A if the line current is 45 A and the line-line voltage is 208 V?

A.

1.56 kW

B.

3.12 kW

C.

5.41 kW

D.

9.36 kW COPYRIGHT 2017 BY COMPLEX IMAGINARY, LLC

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M orning S ession 39

40

A large 30, AC transmission system has a 34.5/6.6 kV step-down transformer rated 60 MVA. If this transformer has a 6% nameplate impedance, what is the per-unit impedance most nearly on a 10 MVA base (assume base voltage doesn’t change)? A.

2.7%

B.

1%

C.

6%

D.

3.7%

A 3-phase, 208 V system consists of tw o 5 kVA motors paralleled together off of a bus. The first m otor has a power factor of 0.75, and the second motor has a pow er factor of 0.65 (both lagging). What is the system power factor? A.

0.61

B.

0.65

C.

0.70

D.

0.75

STOP

This Completes The Morning Session of the Test

Pa g e 29

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PAGE 30

AFTERlJOON SESSION QUESTIONS 41=80 4 HOUR TIME LIMIT

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Pag e 32

A fternoon S ession 41

42

43

PAGE 33

Which of the relay types listed are particularly useful in setting up zones of protection in transmission applications? A.

Impedance relay

B.

Differential relay

C.

Overcurrent relay

D.

Voltage relay

A 3-phase, 15 kV power transmission line is carrying 270 MVA, with a 0.90 lagging power factor. The real power available (MW) is: A.

156 MW

B.

243 MW

C.

270 MW

D.

468 MW

A pair of two generators produces 1,705 kW on a 3-phase system. Provided that these generators are delta-connected, and both have a rating of 927 kVA, and one has a lagging power factor of 0.94, the power factor for the other generator is: A.

0.94

B.

0.92

C.

0.90

D.

0.88

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A fternoon S ession 44

According to the effective impedance values in the 2017 National Electrical Code®, if the conductors between the generator and the balanced load shown below are three 250-AWG uncoated copper wires in an aluminum conduit, the voltage at the load is

0

400 ft-

480 V 3

II

en

pf = 0.92 V = 12.5 kV S = 200 kVA

L

- 9 .2 A

The answer to this question is in Table 9 of the NEC®. Find the resistance for the indicated cable, and multiply it by the distance, and the answer is (A). See below: NEC per 1000 ' Rl = 0.059 L = 200' R = 0^59 1000

200 = 0.012 Q

This question is only answerable by looking up NEC® code. In table 430.251 (A), you’ll find the answer to be 115 V.

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A fternoon S olutions 50

51

52

Page 229

In transmission lines, which of the following is not true about shield wires? A.

A properly placed shield wire protects the phase conductors from lightning surges.

B.

A shield wire never receives a lightning current if the lightning directly hits a transmission tower.

C.

It is possible for a lightning strike to pass a shielded wire and hit the phase wire.

D.

Shield wires struck by lightning can carry currents as high as 200,000 A.

The impedance of a transformer rated 12 MVA, 240/120 kV, on a 3-phase system is 4.6%. What is the impedance of the transformer on a 50 MVA, 230 kV base used for a short-circuit study? A.

1.2%

B.

21

C.

4.6%

D.

36%

A 3-phase, 4-wire, 460 V, wye source is connected to a balanced wye load. The current on leg “A ” (L-N) of the source is 115 A. What is the current on line A going to the load? A.

38 A

B.

66 A

C.

115 A

D.

199 A

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N otes / W orkspace The right answer is (B). A proper shield wire’s purpose in transmission lines is to protect the phase conductors from lightning strikes by intercepting them. You can see the shield wires on the top of the transmission towers if you look closely. Sometimes, the lightning may directly strike the phase wires, but the idea is to have the shield wire intercept it first. The lightning can cause currents as high as 200,000 A, albeit for only very short durations. Sometimes, the lightning will hit the tow er directly and that current will partly reach the shield wire.

Use the impedance base conversion formula to solve this problem. Note that both the MVA bases and voltage bases change. The is (B), see below: Use conversion formula:

Z1

5 0 /2 4 0 )

Z = 0.21 = 21 %

The answer is (C). In wye connections, the line and phase voltages are different, but the line and phase currents are same. Be careful when applying the ^ 3 in electrical engineering problems. There can be bad consequences to assumptions made here. This problem tests your ability to visualize the circuit described. See below:

A

(leg)

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A

(line)

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A fternoon S olutions 53

54

55

In the circuit shown, what is the peak voltage magnitude, V1?■ Assume the circuit elements have ideal characteristics.

A.

277 V

B.

480 V

C.

679 V

D.

831 V

A 3-phase, 125 kVA, wye-connected synchronous generator has a stator resistance of 0.6 O and a synchronous reactance of 1.3 Q. If the connected load draws 47 A at a power factor of 0.82 lagging, and the internal generated phase voltage is 512 Z 14° V, what is the output phase voltage of the generator? A.

455 Z9.7° V

B.

460 Z 0° V

C.

512 Z 6.5° V

D.

577 Z 16° V

An engineer is using the Polarization Index Ratio to test Insulation Resistance (IR) in order to evaluate the condition of the insulation between the conductors and ground of a large industrial motor. What must be the time marks for his two insulation readings? A.

0 minute (start) mark and 5 second mark

B.

0 minute (start) mark and 5 second mark

1 minute mark and 10 minute mark D.

PAGE 231

10 minute mark and 30 minute mark

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N otes / W orkspace The voltage indicated next to the AC generator is an rms voltage. Use the rms to peak value relationship to calculate the peak voltage. The answer is (C) see below:

V

V = 679 V P

The phase output voltage of the generator is equal to the internal generated voltage minus all the impedances. Use the information given to find the missing variables, and solve. The answer is (A). pf = 0.82, laggingj

0 = c o s ' 1 (pf) =-^34.9° RA = 0.6 Q

relative to 14°, s o :

X S =1.3 Q

IA = 47X - 20.9°

IA =4 7 A

EA4 =512X14° V V* = EA - j X s Ia - R a Ia = 512X14° - j( l,3 ) ( 4 7 X - 20.9°)- 0.6(47X - 20.9°) = 455X9.7° V

The correct answer is (C). The Polarization Index Ratio uses two readings: (i) at the 1 minute mark and (ii) at the 10 minute mark. It has become understood that many insulation systems polarize from 10 minutes to several hours. From this discovery it’s possible use a ratio from the insulation readings at the 1 minute and 10 minute marks. This ratio is then compared against a table, and the condition of the winding is thereby derived.

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A fternoon S olutions 56

Find the 3-phase short-circuit fault current Use the information shown and assume the current. 30>MVA MVA Xd’

at point F in the diagram shown below. fault current is equal to the subtransient A v /Q 0 \ ._____ AA T

r =o.24V_y

57

58

A.

577 A

B.

1,000 A

C.

8,027 A

D.

14,167 A

20 MVA Xd ” = 0.24

X 15/35 kV

An overcurrent protection device is protecting a 3 -phase, 60 Hz, 480 V line. The time delay setting is set for 5 cycles. How long will it take (maximum) for this device to trip during fault conditions? A.

0.017 s

B.

0.083 s

C.

12 s

D.

31 s

What is the foot-candle value for a room with the following lighting design parameters? Ceiling height = 9’ Room length = 50’ Room width = 32’ Total lumens = 100,000 Coefficient of utilization = 0.62

Page 233

A.

38.8 foot-candles

B.

41.2 foot-candles

C.

45.7 foot-candles

D.

51.7 foot-candles

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N otes / W orkspace Use the subtransient Set the base values: impedances given for Vb =15 kV the generators. Convert = 50 MVA everything to a 50 MVA base, because that is the S, = 1925 A total system power (you •J3V can use other bases, but this is probably more convenient). Then proceed with the typical per-unit calculations. The answer is (C), see the calculations:

The given subtransient values are based on the given individual generator MVA ratings. Use this to convert the subtransient values to the new base:

Z. = 0.24 Z 2 = 0.24

(S \

2 ' — 2 n n

new base s \

old base

Z =0.24 f — 1

i = 0 .4

UoJ

Z 2 =0.24 ( — ] = 0 .6

2

UoJ

Z= — + — z; z; 'sc - Y

’ 4 -17

_-1

=0.24

PU

Isc =4.17(l\ b /) = 8027 A The device is set to trip within 5 cycles. One cycle for a 60 Hz system is 0.017 seconds. So multiply that by 5 to get 0.083 seconds. The answer is (B). For one cycle at 60 Hz: At = — = 0.017 s 60 5 (A t) = 0.083 s

The foot-candle calculation is shown below. Use the room’s dimensions to calculate the area and the given lumens and coefficient of utilization. The coefficient of utilization functions as a simplification, removing the complication of working with the details of heights and working planes. lum = 100,000 lumens L = 50' W = 32' A = L • W = 1,600 ft 2 cu = 0.62

fc MM A = 38.8 fc

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Pag e 234

A fternoon S olutions 59

60

According to the NEC®, what is the Locked-Rotor Indicating Code Letter that represents the highest kVA/hp value? A.

V

B.

W

C.

X

D.

Y

In the circuit shown below, what is the per-unit impedance of the transmission line between the two transformers? Assign the base voltage as the generator’s output line voltage, and the base kVA as the rated value of the motor.

©

20 kVA

5 :2

1 :2 ZL = 45 + /137 Q

61

Page 235

A.

0.04 + y'0.49 pu

B.

0.12 + y‘0.35 pu

C.

0.22 + y‘0.51 pu

D.

1.3 + y'3.2 pu

25 kVA 2500 V

What is the only instance that the 2017 National Electrical Code® does not permit conductor overload protection? A.

When the circuit is grounded

B.

When the circuit’s total load exceeds 1,000 Q

C.

When the transformer terminals are welded to the mainplate

D.

When the interruption of the circuit causes a hazard

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N otes / W orkspace The NEC® table for Locked-Rotor Indicating Code Letters can be found at Table 430.7 (B). It’s easy to see that “V” is the highest letter possible, regarding kVA/hp.

The problem instructs you which bases to choose for the per-unit calculation. Use the two turns ratios to figure out the generator’s voltage. Then proceed with the normal per-unit calculations to find the impedance. The answer is (B), see below: S b = 25 kVA V =2500

b

{-)(-)=3125 V

UJUJ

V2 z = — = 391 Q b

S,

ZL

45 + j137

Z “

391

= 0.12 + j0.35 pu

Refer to NEC® 240.4.A for solution, which gives the answer as (D). One of the primary functions of the NEC® is to protect public safety. An instance where a system of protection w ouldn’t be permitted would logically be an instance where that protecting system could hazard public safety.

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PAGE 236

A fternoon S olutions 62

63

64

Page 237

For all motor branch-circuits, the 2017 National Electrical Code® requires short-circuit and ground-fault protective devices to be: A.

rated no greater than 600 A

B.

capable of carrying the starting current of the motor

C.

carried along aluminum conductors

D.

twice the locked m otor current requirement

Which of the following motors would be considered a split-phase motor? A.

shaded pole motor

B.

capacitor start motor

C.

wound rotor motor

D.

repulsion motor

The symmetrical component method is being used to analyze a 3-phase circuit, with the phase rotation reference set to ABC. Which of the following statements about the zero-sequence currents is true? A.

They are derived using only the balanced phase currents.

B.

They represent the magnetic flux line losses.

C.

They are all in-phase with one another.

D.

They are all symmetrical, but opposite in direction relative to the ABC reference.

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N otes / W orkspace Look up NEC® code: in section 430.52 (B), you’ll find the answer to be option B.

A split-phase motor uses a high resistance startup winding to create a high startup torque. A capacitor start m otor is one of these types of motors, which uses a capacitor as part of this mechanism. A shaded pole motor is characterized by a low startup torque, which eliminates it. A wound rotor motor has a totally different winding configuration from a split-phase motor, which eliminates it. A repulsion m otor is a type of wound rotor motor, which eliminates it.

The answer is (C) because all the zero-sequence currents are in the same direction.

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Pag e 238

A fternoon S olutions

65

In the 3-phase rectifier circuit shown below, the silicon-controlled rectifiers have firing angles of 30°. If the incoming voltage waveform is v(t) = 525 sin ( a t ) , where go is 377 rad/s, what is average voltage across the load?

A. B.

66

P a g e 239

168 V

D.

313 V

A certain motor maintains a speed of 1,761 rpm under no-load conditions. Provided this motor has a nameplate rating of 100 hp, 1,750 rpm, and 0.85 lagging power factor, then its speed regulation must be: »

67

C.

A.

0.63%

B.

1 .4%

C.

3.1%

D.

3.9%

Which of the following is considered a load bus in a typical power-flow study? A.

a generation bus

B.

a voltage-controlled bus

C.

a P-Q bus

D.

a slack bus COPYRIGHT 2017 BY COMPLEX IMAGINARY, LLC

N otes / W orkspace Since the SCR’s fire at a non-zero angle, that means a portion of the wave will be chopped off. Integrate over the remaining portion to calculate the average voltage. The answer is (D), see below. Be sure to put your calculators in radian mode!

length of each half-cycle period: T=— = 0.0083 s CD

SCR triggers at 30°: Voltage Across Load:

V

av9

= — f T v (t)d t J J

t/6

V /

v (t) = 525sin(377t) -525 cos Vavg = y-

M

377 T/ 6

For this calculation, don’t forget to put your calculator in radian mode.

= 313 V

This is a speed regulation problem, but there is misdirect information in the question. The full-load speed and no-load speed will alone be sufficient to answer the question. NL = 1761 rpm FL = 1750 rpm SR =

N L -F L

x100

FL 0.63%

All you need to know to answer this question is the basic terminology used for power flow studies. A load bus is synonymous with a “ P-Q” bus, where the “ P” and “Q” represent the traditional letters for real and reactive power components.

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P a g e 240

A fternoon S olutions 68

69

Which of the following lamp types is characterized by its yellowish color?

Page 241

metal halide lamps

B.

tungsten-halogen lamps

C.

cold cathode lamps

D.

high-pressure sodium lamps

An electrical engineer is starting a new consulting business. He has estimated that his cash flow for the first five years will be $35,000 per year. His initial startup cost will be $50,000. Both his borrowing and savings annual interest rates are 7%. Based on this estimation, what will his future worth be (most nearly) at the end of the 5 years? A. $116,000

»

70

A.

B.

$125,000

C.

$131,200

D.

$145,700

The no-load losses of a transformer are due to: A.

The harmonics caused by the alternating current.

B.

Heating of the core due to applied voltage.

C.

The heating of the windings.

D.

Capacitance discharge effects in the windings.

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N otes / W orkspace Of the choices listed, only the high-pressure sodium lamps are strongly characterized by their yellowish color, so the answer is (D). The metal halide lamp is characterized by its bright white color. Tungstenhalogen lamps are also known to be white in color. Cold cathode lamps are those neon, flexible lights used in diners to indicate if they are “Open” or “ Closed” ; and they can be a lot of different colors.

There are two formulas (factors) involved here: the single payment compound amount for the startup cost, and the uniform-series compound amount for the positive cash flow. The single payment compound amount must be subtracted from the uniform-series compound amount to get the answer, which is (C). See the calculations. Initial Startup Cost: Future value of startup cost Total Future Value C = 50000

( f /P , i%,n)

i = 0-07

F =50000(1+ i

F V -^ -F ,

r

(+) Cash Flow:

= 70128

35000 / yr i = 0.07 n=5

Future value of (+) cash flow

= 131148

( f /A , i%, n)

M

F = 35000------

2

i

= 201276

The no-load losses in a transformer are caused by the heating of the core, which is present as soon as voltage is applied. In other words, this heat will be present even if no load is attached to the transformer. The answer is (B).

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PAGE 242

A fternoon S olutions 71

How much clearance is required by the NEC® between conduits entering a switchboard bus enclosure and the uninsulated busbars inside? A.

72

73

Page 243

6”

B.

8”

D.

12”

The relay pickup setting for an overcurrent protection device is set to 420 A. If the sensor being used is a current transformer with a 450:7 ratio, the instrument current that would trigger the pickup setting is most nearly: A.

0.02 A

C.

64 A

D.

27,000 A

A transformer is serving a building load and has a secondary voltage of 480 V. The high-side of the transformer is connected to the utility’s 6.6 kV service, and the source impedance is 0.94%. If the transform er’s rated impedance and current are 4.25% and 1,720 A, respectively, what is the short-circuit current available (most nearly) at the transformer’s secondary? A.

1,720 A

B.

33.100A

C.

40,500 A

D.

183,000 A

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N otes / W orkspace NEC® Table 408.5 states that 10” are required for this situation.

Find the smaller current in the instrument using the current transformer turns ratio formula. Remember that for current transformers, the turns ratio is directly proportional to the two currents. In contrast, potential transformers have an inverse relationship between current and turns ratio. When there are problems with turns ratios, do not automatically always use the same formula. Beware to note whether the problem is describing a current transformer or voltage transformer. This distinction will affect the appropriate formula to be used. 450 420 7

“

x

( j j 1,450 J = 6.5 A

x = 420

The short-circuit current available at the transformer’s secondary needs to take into account not only the transform er’s impedance, but the source impedance as well. The answer is (B), see the calculations below: S C A ------ — — x 100 %Z + % zs FLA = 1720 A % ZT = 4.25%

%ZS = 0.94% SCA = 33,141 A

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PAGE 244

A fternoon S olutions 74

75

A 230/34.5 kV, delta-wye, 3-phase transformer bank is constructed using three single­ phase transformers. What is the turns ratio for each of the individual transformers (primary:secondary)?

PAGE 245

3.85:1

B.

6.67:1

C.

7.7:1

D.

11.6:1

A protection instrument uses a 550:5 current transformer to connect to a 34.5 kV line, and has an internal resistance of 2.4 Q. When a fault of 4,250 A is detected, what is the resulting voltage induced in the instrument?

—

76

A.

►

A.

38.6 V

B.

92.7 V

C.

161 V

D.

314 V

A public utility has decided to build a new generation plant at an initial cost of $75 million. The annual maintenance costs are $2 million and the interest rate is 7% per year, compounded annually. What is the total present worth of the plant m ost nearly (using the capitalized equivalent method)? A.

$75.0 million

B.

$77.0 million

C.

$96.4 million

D.

$104 million

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N otes / W orkspace Since the primary side is delta the phase voltage and line voltage are the same. In contrast, for the secondary, being wye, the phase voltage and line voltage are different. Primary

Secondary

V = V = 230 kV ♦ L

34.5

= 19.9 kV

fs

230 a = -----19.9

11.6

To get the voltage in the instrument, you must first find the current in it. To do this, convert the fault current over to the instrument s de and apply Ohm’s Law. The answer is (B). V

550

H ( R)

= 38.6 A L =■ (550)

il = 4250 A V = 34500 V

l 5 j

R = 2.4 Q

V2 = 38.6

M - 92.7 V

For engineering economics problems, the most important reference to have is a table of the formulas (or “factors”) for doing calculations based on the financial situation described. The present worth of the initial cost is whatever it is, so no calculation needs to be made. However, the annual cost represents additional money needed in the future for the plant to remain in operation. Therefore, we need to know its “capitalized equivalent” (CE). The CE method also assumes that the lifetime of the plant is infinite. So the total present worth will include the initial costs plus the capitalized equivalent of the annual maintenance costs. The answer is (D), see the calculations: Initial Cost:

Capitalized Equivalent of Annual Cost:

C. = 75x10®

CE = —

1

A = 2 x 10 i = 0.07

Present Worth Calculation: PW = C. + CE =103.6x10®

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P a g e 246

A fternoon S olutions 77

The total process time-elapse indicated by the IC C graph below for a 2,000 A shortcircuit current to break the circuit -- beginning from the mechanical unlatching of the breaker to final current interruption (maximum interrupting time) -- is EXACTLY: A.

20 seconds

B.

1,000 seconds 19.97 seconds

D.

PAGE 247

14.75 seconds

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P a g e 248

A fternoon S olutions 78

79

Resistance grounding may be either of tw o classes: high resistance or low resistance, these are distinguished by: A.

the relative proximity to the ground-fault

B.

the use of mechanized ground-fault interrupters

C.

the magnitude of ground-fault current permitted to flow

D.

the resonant effects in the ground-fault path

A 3-phase AC waveform is shown below:

If a 3-phase half-rectification was performed on this input waveform, what would the resulting waveform look like assuming no smoothing or filtering components on the circuit?

Page 249

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N otes / W orkspace If you think about this question in terms of Ohm’s Law, you will see that changing the resistance controls the magnitude of the current. This is a common way for designers to analyze and control ground-fault currents. The correct answer is therefore (C).

Rectified waveforms of sine waves will always involve moving the sine “ peaks” that are below the x-axis and flipping them over to the positive side. So the answer choices come down to (A) and (B). (D) is not possible, and neither is (C) because rectification can’t change the shape of a wave in that way. The answer is (A) because this is a half-wave rectification. See the diagram:

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P a g e 250

A fternoon S olutions 80

PAGE 251

According to the 2017 edition of the NEC®, multiple plate electrodes with individual resistances of 60 ohms must be installed at least how many feet apart from each other? A.

20 feet

B.

10 feet

C.

18 feet

D.

6 feet

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N otes / W orkspace NEC® Section 250.53 (B) gives the answer: 6 feet.

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Pag e 252

PRACTICE EXAM VOLUME 3 80 QUESTIONS 8 HOUR TIME LIMIT

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MORNING SESSION QUESTIONS 1-40 4 HOUR TIME LIMIT

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PAGE 256

M orning S ession 1

2

3

Pa g e 257

Using the data in the diagram provided: what is the steady-state impedance (Q) between terminals A1 to A2 of this lossless AC transformer?

A.

1.7Z15°Q

B.

17.3^15° Q

C.

37.6Z15°Q

D.

138Z300 Q

In systems having a fundamental frequency of 60 Hz, which of the following harmonics may be possible? A.

90 Hz, 135 Hz, 220 Hz

B.

120 Hz, 180 Hz, 240 Hz

C.

120 Hz, 200 Hz, 480 Hz

D.

30 Hz, 15 Hz, 7.5 Hz

A current transformer is used as the sensor for an overcurrent protection relay. The pickup setting is 171 A, and this is transformed to an 8 A current in the instrument. If the turns-ratio for the CT is 427:x, what is x? A.

2.5

B.

20

C.

53

D.

60

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Pag e 258

M orning S ession 4

5

6

Pa g e 259

An important characteristic of an induction motor is: A.

not relying on slip to run the machine

B.

not needing a DC field to run the machine

C.

having an electric com mutator system

D.

the ability to operate with a rotor

A lead-acid battery generates voltage by which of the following means: A.

Water and sulfuric acid chemically reacting with each other

B.

Oxidization of the positive lead plate

C.

Lead-acid batteries don’t generate voltage, they only store charge from another source

D.

Electrolyte density

Two towers for a 3-phase distribution line are spaced 3,900 feet apart. One of the three conductors was recently replaced. This installed conductor has 0.0600 Q per thousand feet, whereas the other two conductors have 0.0500 Q per thousand feet. What is the exact geometric mean of the total line resistances for the three conductors? A.

0.053 O

B.

0.207 O

C.

0.836 Q

D.

208 Q

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PAGE 260

M orning S ession 7

8

Page 261

The following diagram represents what type of grounding system?

A.

High-Impedance Grounding System

B

Low-Impedance Grounding System

C.

Residual Grounding System

D.

Isolated Neutral Grounding System

A potential transformer is connected to a 208 V rated line with a turns-ratio of 222:11. At a certain point in time, the voltage in the instrument is 10.2 V . What is the voltage of the line at this time? A.

201.1 V

B.

205.9 V

C.

208 V

D.

209.5 V

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PAGE 262

M orning S ession In the 75 kV circuit shown below, a 3-phase fault occurs at point F. What is the fault current, assuming infinite bus conditions? __ ----:©o Z = 17.65ZL450 Q

10

11

P a g e 263

A.

2,453 A

B.

1,623 A

C.

1,353 A

D.

937 A

Z=/32 0 >

Pa g e 293

A.

113.3 V

B.

114.5 V

C.

116.1 V

D.

118.0 V

70 ft

Load Phase Conductor and Neutral Specs: Made of Aluminum Size 3 AWG In Aluminum Conduit 75° C (167°F)_____________________

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PAGE 294

A fternoon S ession 50

51

52

When a capacitor is connected to the load side of a 460 V motor circuit’s overload protection device, the setting of this device is based on what according to the National Electrical Code®? A.

The kVA rating of the capacitor

B.

125% the capacitor’s rated ampacity

C.

The improved power factor of the circuit

D.

The m otor’s rated reactance

During which type of fault will the amount of available power in the affected system be the greatest (without losing synchronism)? A.

Three-phase fault

B.

Line-line fault

C.

Line-ground fault

D.

Double line-ground fault

Given the information in the system shown below, what is x : y (turns ratios)? Z = (10 + y'150) Q

—A/W— 30 kVA 1000 V

Page 295

A.

9 : 16

B.

16 : 9

C.

2 :9

D.

9 :2

600 : x

M y :2700

10TVA 8000 V

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PAGE 296

A fternoon S ession 53

54

55

Page 297

To decrease charging current in an energized high-voltage transmission line requires a(n): A.

increase in line length

B.

decrease in line reactance

C.

increase in line voltage

D.

decrease in line capacitance

Using the MVA method, find the 3-phase short-circuit fault current for a transform er’s secondary terminal. The low-side of the transformer is served by a generator. The generator’s nameplate data indicates ratings of: 20 MVA, 23% reactance. The transformer’s ratings are: 35 MVA, 13% impedance, 12.5/34.5 kV. Assume no shortcircuit contributions downstream of the transformer. A.

10,300 A

B.

5,970 A

C.

1,910 A

D.

1,100 A

Which of the following would increase the synchronous speed of a squirrel-cage motor operating at 60-Hz? A.

Remove the motor winding

B.

Decrease the size of poles

C.

Decrease the number of poles

D.

Increase the applied voltage to strengthen the m otor’s magnetic field

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PAGE 298

A fternoon S ession 56

57

58

Pa g e 299

The National Electrical Code® horsepower rating for a 3-phase, 208 V, squirrel-cage induction motor with a 960 A maximum locked rotor current is most nearly: A.

50 hp

B.

75 hp

C.

120 hp

D.

60 hp

A private utility company is considering purchasing portable solar equipment that costs $85,000 for a conceptual project. The life of the project will be 5 years. The project manager estimates that the equipment will generate an annual cash flow of $32,000 to be paid at the end of each year of the project. If the company’s MARR is 22% , what is the net present value of the equipment purchase? A.

-$16,500

B.

$6,636

C.

$75,000

D.

$91,640

What is the power factor for a 3-phase inductive load rated at 22 MVA, 5 kV with a power output of 15 MW? A.

0.59

B.

0.68

C.

0.78

D.

0.89

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PAGE 300

A fternoon S ession 59

60

A 40 in. high working plane is in a 20 x 60 ft. room, with 90,000 lumens provided by fixtures hung 1.25 ft. from the 14 ft. ceiling. If the coefficient of utilization is 0.82, then the illumination level most nearly: 38.8 foot-candles

B.

45.8 foot-candles

C.

52.4 foot-candles

D.

61.5 foot-candles

Which ofthe following circuits represents a bipolar arrangement of a HVDC transmission system?

A.

PAGE 301

A.

B.

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PAGE 302

A fternoon S ession 61

62

What is the line current of a delta-connected, 3-phase, 1,325 kVA, 2.4 kV inductive load? A.

220 A

B.

319 A

C.

452 A

D.

552 A

An electrician is performing a test on a single-phase 480/208 V transformer. following values are measured:

The

Core losses = 57 W Copper losses = 223 W Secondary voltage = 208 V Secondary current = 58 A Load power factor = 0.85 What is the efficiency of the transformer? A. 95.3%

P a g e 303

B.

96.3%

C.

97.3%

D.

97.7%

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PAGE 304

A fternoon S ession 63

64

65

Pa g e 305

In a rectifier circuit, the resulting DC output voltage level is controlled by what? A.

Inducing harmonic effects on the AC input voltage

B.

Changing the firing times of the devices in the rectifier circuit

C.

Applying or limiting external gain to the devices in the rectifier circuit

D.

Reversing the polarity of the DC input voltage

A 3-phase, 3-wire, ABC-delta source is connected to a balanced wye load. The voltage on leg “AB” of the source is 460 V, and the current in the same leg is 37 A. W hat is the current in each leg of the wye load? A.

12 A

B.

37 A

C.

64 A

D.

111 A

In rectification applications what device is commonly used as a voltage regulator due to its ability to produce a constant voltage under large current fluctuations? A.

Capacitor Bank

B.

Zener Diode

C.

LED inverter

D.

Split-Capacitor Motor

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PAGE 306

A fternoon S ession 66

67

Page 307

Symmetrical components analysis is a method which converts three unbalanced current phasors into: A.

three phasors of differing magnitudes

B.

a single current phasor

C.

three sets of balanced currents

D.

three equal currents differing only in phase angles

In the ferromagnetic core shown below, the flux density of the center leg B is 1.5 Teslas (T). What is the flux in each of the two outer legs (A and C)?

A.

0.0053 Wb

B.

0.011 Wb

C.

0.015 Wb

D.

0.15 Wb

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PAGE 308

A fternoon S ession 68

An equivalent circuit for a transformer is shown below. Calculate the efficiency of the

20 kVA 6.6/2.8 kV Core Loss = 75 W Copper Loss = 241 W Is = 6.6Z. - 22° A

69

70

Page 309

B.

98.7%

C.

99.3%

D.

99.8%

An ideal single-phase transformer has a 200:50 turns-ratio, a high-side voltage of 250 V, and a full-load secondary current of 75 A. What is the kVA value of this transformer? A.

6.3 kVA

B.

4.7 kVA

C.

4.3 kVA

D.

1.9 kVA

A single-phase, 277 V load is being served by a 100’ long PVC conduit in an office building. According to the NEC® conductor values, what is the total impedance of the 4 AWG uncoated copper conductor?

JO.0 Q 5 0

A.

0.029 +

B.

0.062 + y'0.005 O

C.

0.062 + y'0.01 Q

D.

0 .0 1 8 + /0 .2 2 Q

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PAGE 310

A fternoon S ession 71

Page 311

The following TCC graph represents an insulated case circuit breaker (ICCB) that also includes a short time delay (STD). For what current magnitude will the instantaneous override mechanism of the ICCB be activated? A.

8,000 Amperes

B.

1,000 Amperes

C.

12,000 Amperes

D.

2,000 Amperes

copyright 201

by complex imaginary, llc

N otes / W orkspace

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PAGE 312

A fternoon S ession 72

73

74

Pa g e 313

A 3-phase motor needs to have its power factor improved by 1,008 kvars. If 48 kvar capacitor units are available, which of the per-phase capacitor bank arrangements shown below will provide the necessary result? (Assume balanced load conditions)

Where should the conductor be positioned when using a donut-type current transformer? A.

At the primary polarity mark

B.

Through the center opening

C.

On the transformer’s bushings

D.

On the transformer’s contact nodes

The presence of hysteresis in a ferromagnetic transformer is indicative of:

A.

A ground fault in the system

B.

Losses due to winding resistance

C.

No-load losses

D.

Secondary feeder line losses COPYRIGHT 2017 BY COMPLEX IMAGINARY, LLC

N otes / W orkspace

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PAGE 314

A fternoon S ession 75

76

A station engineer is considering the purchase of a $125,000 motor for a project. For the two year life of the project, the motor will generate a cash benefit of $75,000 the first year, and $85,000 the second. The company’s minimum acceptable rate of return for investments is 12%. Based on this information, what would be the present worth A.

-$2,294

B.

$5,267

C.

$9,726

D.

$35,000

The diagram below shows an equivalent circuit for part of a transmission system:

How much power is lost in the n line (per-unit)?

77

Pa g e 315

A.

0.0019 pu

B.

0.0029 pu

C.

0.0031 pu

D.

0.0039 pu

3.2 L 23° pu

A 3-phase motor bank consists of 4 motors: 2 of them are induction motors, and the other 2 are synchronous motors. The synchronous motors are running in an overexcited state, while the induction motors operate with their usual lagging power factors. Compared to a situation where all motors running with lagging power factors, what does this indicate for the overall system? A.

The overall system efficiency is increased.

B.

The overall system efficiency is decreased.

C.

The transmission supply line losses are increased.

D.

The transmission supply line currents are increased. copyright 2017 by complex imaginary, llc

N otes / W orkspace

COPYRIGHT 2017 BY COMPLEX IMAGINARY, LLC

Pag e 316

A fternoon S ession 78

Below is an example of a simplified transmission circuit including the line impedance. Calculate the apparent power output of the generator. 12:1

79

80

A.

233 kVA

B.

403 kVA

C.

698 kVA

D.

1,209 kVA

3:20

A capacitor on a 3-phase system corrects the power factor produced by an induction motor to 0.96 by adding 10.8 kvar. The rated voltage for the motor is 230 V, and the power used is 23.9 kW. What was the uncorrected power factor of this inductive load? A.

0.80

B.

0.67

C.

0.87

D.

0.75

A delta-wye transformer is rated for: 34.5 kV/480 V, 125 A primary (high-side) line current. What is the transformer’s power rating (kVA) and the secondary current? A.

4,300 kVA, 8,980 A

B.

4,300 kVA, 15,600 A

C.

7,500 kVA, 8,980 A

D.

7,500 kVA, 15,600 A

STOP

This Completes The Afternoon Session of the Test PAGE 317

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N otes / W orkspace

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PAGE 318

M orning S ession S olutions #

CORRECT SOLUTION

#

CORRECT SOLUTION

1

C

transform er | turns-ratio | transients

21

A

unbalanced loads | complex values

2

B

harmonics

22

D

motor | synchronous speed

3

B

current transform er | turnsratio | overload protection

23

C

power form ula | com plex values

4

B

m otor

24

B

transform er | per-unit | base conversion

5

C

battery

25

A

NEC | grounding

6

B

geom etric mean

26

C

rectification | waveform s

7

D

grounding

27

A

power formula | delta/wye

8

B

transform er | turns-ratio

28

D

m otor | speed regulation

9

D

short-circuit | fault current

29

A

transform er | losses

10

D

NEC | conductor sizing

30

B

autotransform er

11

C

econom ics | salvage value | present worth

31

B

parallel m otors | NEC

12

C

parallel m otors | power factor

32

D

generator | stator resistance | synchronous reactance | armature current

13

A

NEC | separately derived systems

33

D

grounding

14

A

lightning

34

B

generator | per-unit | transients

15

A

m otor

35

D

lighting | zonal cavity method

16

D

parallel transformers

36

A

NEC

17

C

capacitor | power factor

37

A

rectification

| waveform s

18

B

generator | pow er factor | pow er formula

38

B

rectification

| waveform s

19

B

power factor

39

C

power flo w study

20

C

m otor | startup torque | full­ load speed

40

A

lighting

Pa g e 319

TOPIC(S) OF STUDY

TOPIC(S) OF STUDY

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A fternoon S ession S olutions #

CORRECT SOLUTION

41

C

42

#

CORRECT SOLUTION

delta/w ye | pow er formula

61

B

delta/wye

D

relay protection | protection zones

62

C

transformer | losses | efficiency

43

A

circuit analysis | com plex values

63

B

rectification

44

B

relay protection

64

C

delta/w ye

45

A

transform er | per-unit

65

B

rectification | zener diode

46

C

parallel transformers

66

C

symmetrical com ponents

47

A

NESC

67

A

flux

48

B

pulse w idth m odulator | variable frequency drive | waveform s

68

A

transformer | equivalent circuit | losses | efficiency

49

D

NEC | pow er factor | ohm 's law

69

B

transformer | turns-ratio | pow er formula

50

C

m otor | NEC | overload protection

70

C

NEC

51

C

faults

71

C

tim e-characteristic curve | insulated case circuit breaker

52

B

turns-ratio

72

D

capacitor bank | pow er factor

53

D

charging current

73

B

current transform er

54

D

transform er | generator | MVA method

74

C

transform er | losses

55

C

m otor | synchronous speed

75

C

economics | present worth

56

D

NEC | m otor | maximum locked rotor current

76

A

equivalent circuit | per-unit | ohm 's law | pow er formula

57

B

| MARR | net present value

77

A

58

B

power factor

78

C

59

D

lighting

79

A

capacitor | pow er factor

60

C

bipolar transmission

80

C

transformer | pow er formula

TOPIC(S) OF STUDY

econom ics

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TOPIC(S) OF STUDY

parallel motors turns-ratio

| pow er factor

| circuit analysis | pow er formula

PAGE 320

M orning S olutions Using the data in the diagram provided: what is the steady-state impedance (Q) between terminals A1 to A2 of this lossless AC transformer? A,

2

B.

17.3^15° Q

C.

37.6Z.150 Q

D.

138Z300 Q

A.

90 Hz, 135 Hz, 220 Hz

B.

120 Hz, 180 Hz, 240 Hz

C.

120 Hz, 200 Hz, 480 Hz

D.

30 Hz, 15 Hz, 7.5 Hz

A current transformer is used as the sensor for an overcurrent protection relay. The pickup setting is 171 A, and this is transformed to an 8 A current in the instrument. If the turns-ratio for the CT is 427:x, what is x?

'

Page 321

1,7Z.15° Q

In systems having a fundamental frequency of 60 Hz, which of the following harmonics may be possible?

►

3

A.

....>

A.

2.5

B.

20

C.

53

D.

60

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N otes / W orkspace The steady-state impedance refers to the transformer’s impedance during normal operations (as opposed to transient and sub-transient impedance). Use the turns ratio and the secondary impedance shown in the diagram to calculate the reflected impedance on the primary side. n _ __ _1 Z„

n2 13 _

x

6 ~

8/115 2

x=

8/115°

= 3 7 .6 Z 1 5 °

Q

Harmonics, by definition, are multiples of whatever the fundamental frequency is, in this case, 60 Hz. So the answer will have to be multiples of 60, so (B) is the right choice.

Find the smaller term of the ratio using the mathematical ratio transformation: 171

427

8 "

x

x = 427

(_8_\ 117lJ

= 20

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PAGE 322

M orning S olutions 4

5

6

An important characteristic of an induction motor is: not relying on slip to run the machine

B.

not needing a DC field to run the machine

C.

having an electric commutator system

D.

the ability to operate with a rotor

A lead-acid battery generates voltage by which of the following means: A.

Water and sulfuric acid chemically reacting with each other

B.

Oxidization of the positive lead plate

C.

Lead-acid batteries don’t generate voltage, they only store charge from another source

D.

Electrolyte density

Two towers for a 3-phase distribution line are spaced 3,900 feet apart. One of the three conductors was recently replaced. This installed conductor has 0.0600 Q per thousand feet, whereas the other tw o conductors have 0.0500 Q per thousand feet. What is the exact geometric mean of the total line resistances for the three conductors?

►

PAGE 323

A.

A.

0.053 Q

B.

0.207 Q

C.

0.836 Q

D.

208 Q

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N otes / W orkspace The main distinguishing characteristic of an induction motor is the lack of a DC field to run the machine. The power is transferred to the rotor through induction, much like the induction in transformer windings. The other answers are characterisics of other types of motors. Choice (D) is nonsense.

This question tests your understanding of how lead-acid batteries work. A lead-acid battery doesn’t generate voltage on it’s own. Technically speaking, the battery only stores a charge that it received from another power source. The answer is (C).

Use the per-length resistances to calculate the total line resistance for each line, then calculate the geometric mean of the resistances which is (B), see below: d = 3900' R „ ° ^ ® - d = 0.234 Q 1 i1000 nnn

0.207 Q

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PAGE 324

M orning S olutions 7

The following diagram represents what type of grounding system?

D.

8

A potential transformer is connected to a 208 V rated line with a turns-ratio of 222:11. At a certain point in time, the voltage in the instrument is 10.2 V . What is the voltage of the line at this time?

»

PAGE 325

Isolated Neutral Grounding System

A.

201.1V

B.

205.9 V

C.

208 V

D.

209.5 V

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N otes / W orkspace Each line is grounded, but the capacitors allow for isolation since the grounded conductors have no solid connection to the lines. Hence, this is an isolated neutral grounding system.

This is a turns-ratio, voltage conversion problem. Be careful, the 208 V line voltage is a misdirect! The answer is (B), see below: n n2 "

222 11

V2 =10.2 V V,

",

\

"2

/ 222 \ V -1 0 .2 1 l 11 J = 205.9 V

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PAGE 326

M orning S olutions In the 75 kV circuit shown below, a 3-phase fault occurs at point F. What is the fault current, assuming infinite bus conditions? — — — :oo Z = 17.65Z1450 Q

........ »

10

P a g e 327

2,453 A

B.

1,623 A

C.

1,353 A

D.

937 A

Z = /3 2Q

One set of supply conductors serves a m otor with a full-load current rating of 200 A. The same conductors also serve tw o other motors requiring 75 A each. According to the NEC®, what is the total ampacity that the conductors should be capable of delivering?

" ''

11

A.

»

A.

200 A

B.

250 A

C.

350 A

D.

400 A

A building transformer needs to be replaced. The new, more efficient transformer costs $25,000, and is expected to last 30 years. The salvage value of the new transformer will be 10% of the original cost. If the interest rate is 8%, what is the net present worth of the new transformer most nearly? A.

$25,000

B.

-$22,700

C.

-$24,800

D.

-$25,000

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N otes / W orkspace This is a one-step short-circuit current calculations. The answer is (D), see below:

VLL = 75 kV

sc

Z a + Zh b = 937 A

In the NEC® 430.24, it says that for multiple motors served by one set of conductors, the conductors shall be rated for an ampacity of not less than the SUM of: (1) 125% of the highest rated motor, (2) the sum of the full-load current ratings of the other motors in the group, (3) 100% if the noncontinuous non-m otor load, and (4) 125% of the continuous non-motor load. In the question, non-m otor loads were not mentioned, and so are not factored in the solution leaving out consideration of (3) and (4). So the correct answer is the sum of (1) and (2): (200-1.25) + (2-75) = 400 A. The answer is (D).

Since we know the future worth of the transformer from its salvage value, we need to convert that to a present worth. Then subtract the cost of the transformer from that and you have the net present worth of the replacement. Obviously, there isn’t much positive cash flow other than the salvage value, the total present worth is negative. The answer is (C), see: SV = 0.1 ( p ) = 2500

Salvage value present worth

Total present net w o rth :

(P /F , i%, n)

PW = P2 - P

i = 0.08 n = 30

= -24751

P2 = 2500(l + i)~" = 248

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PAGE 328

M orning S olutions 12

13

A 3-phase, 208 V system consists of two 5 kVA motors paralleled together off of a bus. The first motor has a power factor of 0.75, and the second motor has a pow er factor of 0.65 (both lagging). What is the system power factor?

B.

0.65

C.

0.70

D.

0.75

A.

No parallel path for the grounded conductor is established when connecting the bonding conductor.

B.

It must be connected at the source.

C.

It must be connected to the primary side of the source transformer.

D.

It’s not necessary since the voltage is under 600 V.

What is most likely to occur when lightning hits a phase conductor on a transmission line? ►

P a g e 329

0.61

A homeowner builds an extension to his mansion, and requires a new transformer and power distribution panel to accommodate the remodel. He happens to be a retired electrical engineer, and realizes that this extension of his home’s 480 V power system is considered a separately derived system. Based on this information, what does the NEC® require for the equipment bonding conductor? ►

14

A.

A.

The lightning current will divide with half going in either direction.

B.

The velocity of the resulting traveling wave will be significantly less than the speed of sound.

C.

The lightning causes a temporary reduction of phase voltage, but a very large increase in phase current.

D.

The surge of the lightning current will travel to the tower, where the shield wire will transmit the current from the tower to earth.

COPYRIGHT 2017 BY COMPLEX IMAGINARY, LLC

N otes / W orkspace This problem involves finding the system power factor when multiple individual power factors are given. This is simply a matter of adding the apparent power vectors together (i.e. adding complex power terms together): V = 208 V Find complex power for each motor Motor 1

Motor 2

5 kVA

5 kVA

pf = 0.75

pf = 0.65

0 = COS'1 ( p f )

ei = 41.4° 02 = 49.5° S1= 5441.4° S 2 = 5449.5° ST = S1+ S2 = 5441.4° + 5449.5° = 7 .0 + j7.1 = 10.0445°

Now, find pf of ST : pf = cos^s) = cos( 4 5 ) = 0.71

According to the NEC®, there is some flexibility about where to connect the equipment bonding conductor for separately derived systems. NEC® 250.30 (A) (1) states that “This connection shall be made at any single point on the separately derived system from the source to the first system disconnecting means or overcurrent device.” The 2nd Exception to this stipulates, for separately derived systems, there may not be a parallel path for the grounding conductor established. So the answer is (A). The other answers are false.

This question a little tricky. The answer is (A); the current will split with half going in each direction. This is because the lightning power source is coming from the sky and hitting an object and both directions of that object (the conductor) are the same as far as the lightning is concerned. So the current will just go both ways (similar to the “ path of least resistance” in parallel circuits with equal resistances on each branch of a node). As far as the other answers, the lightning wave will approach the speed of light, so the eliminates (B). Lightning is essentially a spark that goes across thousands of feet (between the cloud and the striking point) and very high voltage is necessary to produce that kind of spark. So the phase conductor’s voltage will be very high. (D) is wrong because the phase conductor will send the current to the tower, which sends it to ground. The phase conductor is not going to send the current to the shield, and then the ground.

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PAGE 330

M orning S olutions

15

16

An AC motor that: (i) is excited by an external DC source, (ii) requires slip rings, (iii) is often used to improve power facl :or, is considered to be what type of motor? — ► A. Synchronous B.

Asynchronous

C.

Squirrel Cage

D.

Single Phase

A 4,000 kVA load is on a bus being served by two single-phase transformers. The individual transformer specifications are: Transformer #1: 5.25%, 2,000 kVA Transformer #2: 4.75%, 3,000 kVA (Both have the same turns ratio) How much power (most nearly) is being provided by each transformer?

Pa g e 331

A.

Transformer #1 = 1,000 kVA, Transformer #2 = 3,000 kVA

B.

Transformer #1 = 2,500 kVA, Transformer #2 = 2,500 kVA

C.

Transformer #1 = 1,000 kVA, Transformer #2 = 3,000 kVA

D.

Transformer #1 = 1,500 kVA, Transformer #2 = 2,500 kVA

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N otes / W orkspace

The question is describing a synchronous motor, so the answer is (A). A synchronous motor must be excited by an external source, so it’s not a self-starting motor. The external DC source provides the necessary power for the rotor, and it is sent through slip rings and brushes. As a synchronous motor is capable of running at unity power factor (1.0) it is often used to improve power factor correction.

The total capacity of the transformers is greater than the load. Each transformer will be loaded according to their impedances and the total will obviously be the load total only. So when calculating the contributions from each transformer, the total that is being split is the load total and not the transformer capacity total. kVAL = 4000 kVA

/ kVA \

kVA = 2000 kVA kVA„ = 3000 kVA

P

•kVA,

l kVA \

kVA \

Z1 = 5.25% 7 - 4.75%

P =1505 kVA P = 2495 kVA

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PAGE 332

M orning S olutions 17

A capacitor can provide power factor correction for what type of load? A.

resistive loads only

B.

leading loads only lagging loads only

D.

18

leading or lagging loads

There are two wye-connected generators on a single 3-phase system bus. One of the generators has nameplate ratings of 855 VA, 975 V, 0.92 lagging power factor, while the other has nameplate ratings of 925 VA, 975 V, 0.88 lagging power factor. The real power available at the bus is most nearly: A. —

19

B.

1,600 W

C.

1,725 W

D.

1,780 W

A machine is rated for 3-phase, 6.6 kV, 10 MW and has a power factor of 0.75. In order to increase the power factor to 0.90, the design engineer is going to add a wyeconnected capacitor bank. How many kvars are added per phase?

►

P a g e 333

1,500 W

A.

950 kvar/phase

B.

1,325 kvar/phase

C.

2,796 kvar/phase

D.

3,980 kvar/phase

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N otes / W orkspace A capacitor can only correct lagging power factors. The capacitor’s reactive polarity is opposite of a typical inductive load like an induction motor, which it is usually trying to correct.

To find the total power available on the bus, add the individual power output of the generators together. The answer is (B), see below: S1 = 855 VA S2 = 925 VA pf = 0.92 pf2 = 0.88

Pr - P , + P 2- ( S , ' P f , ) + (S2 ' P t ) = 1601 W

0.75 First, find the difference in reactive power with the improved power factor. Then divide that by 3 to get the per-phase value. The answer is (B), calculations are shown:

Pf2 0.90 P

Qi

10 MW Ptan ^cos"1(pf,)] 8819 kvar

Q2

Ptan cos" (pfT 4843 kvar

AQ

Q1 - Q

2

3976 kvar —

= 1325

kvar / phase

3 COPYRIGHT 2017 BY COMPLEX IMAGINARY, LLC

PAGE 334

M orning S olutions 20

21

A3.6kV, 60 hp induction motor has full load-current of 170 A. If the 3,200 ft-lb startup torque is 120% of the full-load torque, what is the rated speed of the motor? A.

60 rpm

B.

98 rpm

C.

118 rpm

D.

493 rpm

What is the current in the neutral of a 3-phase system with the following phase currents? Ia= 137 Z1127.50 A l“ = 1 0 0 /1 0 ° A I = 1 1 2 Z .2 5 0 .8 0 A

22

P a g e 335

A.

20.4Z1-8.30 A

B.

49.1/10° A

C.

49.1/18.3° A

D.

104/1-126° A

If an induction m otor is running at synchronous speed, what does this indicate about the system it is powering? A.

Full-load conditions are met

B.

No-load conditions are met

C.

Full torque is developed

D.

No torque is developed

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N otes / W orkspace For this problem, use the startup torque to calculate the full-load torque. Then, use the torque/hp/ speed formula to find the rated speed of the motor - which is the same as the full-load speed. T = 3200 ^ 2667 ft-lb FL 1.2 T=

5252 • hp rpm

5252(60) rpm = ------- -— - = 118 rpm 2667

The answer is (A). For unbalanced loads, the sum of all phase and neutral currents must be zero. See the calculations below:

1,-137^127.5" ls -100^0lc =112^250.8°

lA+ l . + ' c + ' „ - ° l „ - ( l . + l B+'c) = 20.2 - j2.9 A = 20.4Z-8.3° A

The answer is (D). No torque is produced if there is no difference between synchronous speed and actual speed. This is because the difference is necessary to induce the magnetic field that causes torque.

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PAGE 336

M orning S olutions 23

Calculate the apparent power of the load shown below.

AV = 355 + y'320 V I = 52 + y'31 A

24

25

Pa g e 337

C.

28.6 + y’5.6 kVA

D.

28.6 - y‘5.6 kVA

An 18% impedance is the rating for a 13 MVA transformer. What MVA base (most nearly) would result in a per-unit impedance of 0.27 pu? (Assume base voltage remains the same) A. 11 MVA B.

20 MVA

C.

34 MVA

D.

35 MVA

According to the NEC®, a single-phase, 2-wire, AC circuit requires what method of grounding for the conductors? A.

Only one conductor should be grounded.

B.

Both conductors should be grounded.

C.

Only the neutral conductor should be grounded.

D.

No grounding is required.

COPYRIGHT 2017 BY COMPLEX IMAGINARY, LLC

N otes / W orkspace The voltage difference shown in the circuit is the same as the voltage across the load. It is given as a complex number which means that when you want to calculate the complex power the conjugate of the current will be required. AV = 355 +

j 320 V = 478A 42° V S = A V I *

I = 52 + /31 A = 61Z.310 A

= (4 7 8 A 42 °)(6 1Z -31°)

I* = 5 2 -/3 1 A = 6 1 4.-3 1° A

= 28.6 + y'5.6 kVA

For this problem you are going to use the impedance base conversion formula for per-unit problems to find the missing variable. In this case the missing variable is the new MVA base. The answer is (B), see below: Use conversion formula:

A Zo 0.27

K So U S1

0.18 " 13 19.5 MVA

The answer is (A). NEC® section 250.26(1) states that only one conductor should be grounded in a single-phase, 2-wire system. Choice (C) is wrong because there is no neutral conductor, and (D) is incorrect.

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PAGE 338

M orning S olutions

Pa g e 339

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N otes / W orkspace Since the SCR’s fire at angle zero, there are no special integrals you need to perform. Use the standard waveform formulas to find the average of half of a sine wave. The answer is (C), see below.

P = VI V

VL -N

V

= 7.22 kV

3 Because it’s wye connected, phase and line currents are equal. iL l = 37.6 A = IL -N So: P = (7.22 kv)(37.6 a ) = 271 kW purely resistive, so: S = P = 271 kVA

In this speed regulation problem, you are given the full-load speed and the actual speed regulation value. So plug those values into the speed regulation formula and do the algebra to find the no-load value. The answer is (D), see the calculation below: FL = 945 rpm SR = 4.2% SR = NL =

N L -F L

100

(SR)(FL)+ FL 100

NL = 985 rpm COPYRIGHT 2017 BY COMPLEX IMAGINARY, LLC

PAGE 340

M orning S olutions 29

What kind of common transformer test accounts for the losses due to hysteresis and eddy currents? —

30

31

Page 341

►

A.

open-circuit test

B.

short-circuit test

C.

oil test

D.

power factor test

A wye-wye, 230 kV, 3-phase transmission line is being stepped down to 200 kV using three autotransformers connected between the neutral and the line. For each autotransformer, if the high-side line current is 330 A, what is the current output on the low-side? A. 825 A B.

380 A

C.

150 A

D.

132 A

Two single-phase motors are connected to a transformer on a single branch. These motors comply with the stipulations set out in NEC® 430.6(A). The branch’s THHW copper conductor must have an ampacity of not less than what?

A.

100% of the continuous non-motor load

B.

100% of the noncontinuous non-motor load

C.

100% of the full-load current rating of the highest rated m otor

D.

125% of the full-load ratings of all other motors in the group

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N otes / W orkspace An open-circuit test will account for the hysteresis and eddy current losses. A short-circuit test will not. The oil test tests for certain gases for other kinds of analysis. The power factor test is made up. The answer is (A).

First convert the two voltages to their phase values, because the autotransformer is connected to the phase. Then, use the voltage ratios to get the current ratios, and calculate the low-side current. The answer is (B), see: lH = 330 A w 230 kV V =— H /q

.„ n , w 132.8 kV

i

v „ 200 kV = 115,5 kV 3 V N„ I, NS + NC _L_

v

V

VH

ns + nc

I. “

= 380 A

N

NEC® 430.24 outlines the ampacity requirements for conductors serving multiple motors. It says in (3) that 100% of the noncontinuous non-m otor load is a minimum requirement for this situation. This means (B) is the correct answer.

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M orning S olutions 32

33

34

PAGE 343

A 3-phase, 150 kVA, wye-connected synchronous generator has a stator resistance of 0.4 Q and a synchronous reactance of 2.1 Q. If the internal generated phase voltage is 524 L 16° V and the output voltage is 460 Z. 0° V, what is the output phase armature current of the generator? A.

40 L -3° A

B.

68 4.-133° A

C.

68 ^ 160° A

D.

71 L -6° A

Most grounded systems employ methods of grounding for the system neutral at one or more points. What are the two general categories of grounding methods? A.

Electrode Grounding and System Capacitance Grounding

B.

Arc-Fault Grounding and Conductor Grounding

C.

Relay Grounding and Non-Relay Grounding

D.

Solid Grounding and Impedance Grounding

A 3 synchronous generator with per-unit reactances of Xd = 1.1, X’ = 0.35, and X” = 0.22 has a steady-state current of 0.97 pu. Assume base values of 250 kV and 100 MVA. The internal generated phase voltage is most nearly: A.

464 kV

B.

268 kV

C.

248 kV

D.

85 kV

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N otes / W orkspace You can calculate the armature current by calculating the current flowing through the generator’s internal impedances which are given as the stator resistance and synchronous reactance. All the givens and calculations involve phase values so no ^ factors are necessary. The answer is (D), see: RA =0.4 Q Xc =2.1 Q S EA = 5 2 4 Z 1 6 °V = 71Z - 6° A

V = 460Z00 V

The two main ways of grounding the neutral are solid grounding (i.e. a simple conductor connecting the neutral to ground) and impedance grounding (inserting resistance/reactance in between the neutral and ground). There are a variety of ways to do impedance grounding, but at a high level, these are the tw o main categories.

The internal voltage for the generator is the open-circuit armature voltage. Steady-state formula: V = L X„

= 1.07 pu 1.07(250 kV) = 268 kV

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P a g e 344

M orning S olutions 35

36

37

An engineer is using the zonal cavity method to design the lighting for a room 10 feet wide by 30 feet long. If the height of the ceiling cavity is 4 feet, what is the ceiling cavity ratio? A.

0.33

B.

0.5

C.

0.66

D.

2.67

The NEC® defines the interior of a raceway installed underground as: A.

a “wet location”

B.

a “ non-conductive electrode”

C.

a “shallow groove installation”

D.

an “ intermediate conduit”

In the circuit shown, what is the average voltage for N^? - — 277 VAC

P a g e 345

A.

125 V

B.

176 V

C.

277 V

D.

250 V

©

- N ----------VH

•14Q

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N otes / W orkspace The zonal cavity method uses the cavity ratios of the three zones of a room (ceiling, room, and floor) and the reflectances within the room as part of its analysis. Determining the cavity ratios is a first step with this method. The answer is (D), see equation below: 5 •h | Room Length + Room W id th ) — 7 ^ ----------------------------------- r— - = CCR (Room Length xRoom Width) (5 x 4)(30 + 10) -- ---------------------------------------- r ----------

(3 0 x 1 0 )

800 =

----------------

-

Z ' O f

300

Section 300.5 (B) defines interior raceways installed underground as “wet locations” . Wet locations are technically defined in Article 100. All other answer choices do not appear in the NEC®.

The diode makes this a half-rectified waveform as shown below. First, calculate the peak value of the normal sine wave, since it’s the same as the peak value for the half-rectified wave. Then, use the average-to-peak relationship of a half-rectified wave to calculate the average value. First, find the peak value: Vpeak = 277^2 = 391.7 V Now, find the average value: Vavg 2 V peak

vavg COPYRIGHT 2017 BY COMPLEX IMAGINARY, LLC

jt = 124.7 V

PAGE 346

M orning S olutions 38

If a 3-phase full-wave-rectification was performed on this input waveform, what would the resulting waveform look like assuming no smoothing or filtering com ponents on the circuit?

P a g e 347

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N otes / W orkspace This is a fully-rectified waveform involving all three phases. This means the peaks for both polarities of each phase should now be represented as positive peaks. Other than this flipping, no other modifications are being made to the wave shapes. As such, This eliminates (D) because it is a square wave and has been modified beyond flipping a negative polarity peak. (A) is incorrect because it shows a half-rectified waveform. (C) is incorrect because the frequency doesn’t match to question’s frequency.

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P a g e 348

M orning S olutions 39

40

All of the following elements are commonly used to complete a typical power flow study, EXCEPT: A One-Line Diagram

B.

Calculation of all bus voltages and angles

C.

Calculation of all transformer magnetic flux contributions

D.

Calculation of all real and reactive power in the transmission lines

A “ lux” is analogous to what other standard unit? — ..... ►

Pa g e 349

A.

A.

foot-candle

B.

lumen

C.

candela

D.

watt

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N otes / W orkspace This question tests your understanding of power flow studies. A One-Line Diagram is used to see the configuration of all circuit elements. The purpose of a power flow study is to calculate all bus voltages and angles, and the total power of lines in the circuit. Magnetic flux calculations are not used in power flow studies, so the correct answer is (C).

The answer is (A), foot-candle. Lux and foot-candle are both a unit for illuminance, So what is the difference between illuminance (lux, foot-candle) and luminous intensity (candela)? Illuminance is luminous intensity per area (e.g. lumens per square foot).

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A fternoon S olutions

P a g e 351

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N otes / W orkspace The phase voltage is 450 V and it is wye-connected, so the line voltage is: 450 -•

= 779.4 V

So the real power is: „

V*2

(779 ) 312 = 1947 W

The answer is (D). Backup relay protection should be delayed such that the primary protection has enough time to trigger first. So in a step curve, as shown here, the two additional steps are the backup protection because they take longer to trigger than the Z1 unit, giving Z1 a chance to trigger first.

The current \Ais given in the problem. Use the current division formulas to find I2. The answer is (A), see below:

R2 + JyX,2 = 18.1Z98° A

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P a g e 352

A fternoon S olutions 44

What is the type of relay shown in the diagram below? Protected Equipment

m

45

P a g e 353

Impedance relay

B.

Differential relay

C.

Overcurrent relay

D.

Voltage relay

V\AA-

-y\AA-

A 15 MVA, 3-phase, 12.5/6.6 kV transformer for a substation has a rated impedance of 6.25%. What is the actual ohmic impedance as seen from the transformer’s primary (assume rated conditions)? ■

46

A.

nn

»

A.

0.65 Q

B.

2.9 Q

C.

6.3 O

D.

12.6 Q

Which of the following instances will NOT result in unequal loading conditions when two transformers are connected in parallel? A.

Different impedances; Same turns-ratio; Same kVA

B.

Different impedances; Same turns-ratio; Different kVA

c.

Different impedances; Different turns-ratio; Same kVA

D.

Different impedances; Different turns-ratio; Different kVA

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N otes / W orkspace

The diagram is showing a differential relay, so the answer is (B). A way to tell this is a differential relay is there are two tap points; one before and after the load being protected. A relay that straddles the load in this manner is a differential relay. The relay is comparing the incoming and outgoing current, and will operate if there is a difference.

This problem is asking to convert the given per-unit impedance to an actual ohmic impedance. When the transformer impedance is given as a percentage, the base values used are the transformers rated MVA and the high-side voltage. Use these base values to calculate the base impedance values and use that to convert the transformer’s impedance to ohms. The answer is (A) see below: % ZT = 6.25% S b =15 MVA V =12500 V

calculate base impedance: V2 (l250o)2 Z = - ^ = 4--------- = 10.4 Q b Sh 15x10® b ZT =0.0625(10.4) = 0.65 Q

In this situation, the first thing to consider is the kVA rating. Transformers will be unequally loaded if their kVAs are different. That leaves tw o answer possibilities in (A) and (C). In answer (A), having the same turns-ratios will prevent the load from being split equally between the two transformers because the impedances are different. On the other hand, in answer (C), it will be possible for the load to be split equally between the tw o transformers. This is because having different turns-ratios can offset the effect of having differing impedances. For example, let’s say the turns-ratio for each transformer can be adjusted independently. Then the tapped position can be adjusted and the engineer can create a situation where the loads are balanced. The answer is (C).

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P a g e 354

A fternoon Solutions 47

Which of the following switchgear topics are NOT covered by the National Electrical Safety Code®? Switchgear Inversions B.

Dead-Front Power Switchboards

C.

Motor Control Centers

D.

Control Switchboards

The standard means a Pulse Width Modulator Variable Frequency Drive converts DC to 3-phase AC is:

49

A.

Reversing the phase current’s polarity.

B.

Alternating turning transistors on and off to create square waves.

C.

Shifting the phase currents to match the desired voltage.

D.

Augmenting the DC rms value with auxiliary AC currents.

Inform your answer in accordance with the 2017 National Electrical Code®. An AC generator supplies voltage to a 70 A wye-connected, 3-phase load protected by a relay. Provided the load is balanced and operating at a pf of 0.96 lagging, the voltage at the load is most nearly: 70 ft 208/120 V 30

P a g e 355

A.

113.3 V

B.

114.5 V

C.

116.1V

D.

118.0V

Load Phase Conductor and Neutral Specs: Made of Aluminum Size 3 AWG In Aluminum Conduit 75° C (167°F)_______ ________

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N otes / W orkspace The NESC® covers switchgear topics in section 18, this section makes no reference to switchgear inversions.

The pulse width modulation technique uses transistors to shape DC waveforms into pseudosinusoidal waveforms. These are then shaped into the familiar AC sinusoidal waveforms using additional steps beyond the scope of this question. The answer is (B).

First go to Table 9 of the NEC® and get the impedances for the conductors specified in the question. The answer is (D), see below: From Table 9 of the NEC®:

= 0.0287 + 0.00329 I = 70Z. - 16.3° A

V = V o-IZ = 118.0 V

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Ra g e 356

A fternoon S olutions 50

51

52

When a capacitor is connected to the load side of a 460 V motor circuit’s overload protection device, the setting of this device is based on what according to the National Electrical Code®? A.

The kVA rating of the capacitor

B.

125% the capacitor’s rated ampacity

C.

The improved power factor of the circuit

D.

The m otor’s rated reactance

During which type of fault will the amount of available power in the affected system be the greatest (without losing synchronism)? A.

Three-phase fault

B.

Line-line fault

C.

Line-ground fault

D.

Double line-ground fault

Given the information in the system shown below, what is x : y (turns ratios)? Z = (10 + y'150) Q

[ ^ ) ---------5 €----------V W

©

30 kVA 1000 V

P a g e 357

-

A.

9:1 6

B.

16 : 9

C.

2 :9

D.

9 :2

600 : x

-------- 5 €-------------1 m y : 2700

10TVA 8000 V

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N otes / W orkspace NEC® 460.9 states: For a capacitor connected on the load side of the motor overload device, the rating or setting of the motor overload device shall be based on the improved power factor of the motor circuit.

This is a trickier way of asking “ which fault is least severe?” The answer is (C), line-ground fault. The other faults are all more severe, and will result in more loss of available power in the system.

This is s turns ratio problem with a little twist. Instead of calculating one turns ratio, you are asked to calculate one part of tw o different turns ratios. First, start with the given voltages then algebraically solve for the missing x and y variables shown in the diagram.

V 1000

600

V

x

8000

x

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n

2700

16:9

P a g e 358

A fternoon S olutions 53

54

55

To decrease charging current in an energized high-voltage transmission line requires a(n): increase in line length

B.

decrease in line reactance

C.

increase in line voltage

D.

decrease in line capacitance

Using the MVA method, find the 3-phase short-circuit fault current for a transform er’s secondary terminal. The low-side of the transformer is served by a generator. The generator’s nameplate data indicates ratings of: 20 MVA, 23% reactance. The transform er’s ratings are: 35 MVA, 13% impedance, 12.5/34.5 kV. Assume no shortcircuit contributions downstream of the transformer. A.

10,300 A

B.

5,970 A

C.

1,910 A

Which of the following would increase the synchronous speed of a squirrel-cage motor operating at 60-Hz?

mammsa^t*

P a g e 359

A.

A.

Remove the motor winding

B.

Decrease the size of poles

q_

Decrease the number of poles

D.

Increase the applied voltage to strengthen the m otor’s magnetic field

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N otes / W orkspace The charging current formula is shown below. There is a direct relationship between capacitance and charging current, therefore decreasing capacitance will lead to a decrease in charging current. The answer is (D), see below: Charging current equation: lc = ytoCV w here: a) = frequency V = voltage C = capacitan ce

First, draw the system:

Transformer:

Generator: Convert values to MVA_ values: SC

MVA sc

20 MVA 0.23

= 87 MVA

MVA

sc

= 35 MVA = 270 MVA 0.13

Since they are in series: MVA

1'2

=

a/3V

87 • 9 70

= 65.8 MVA 87 + 270 65.8 MVA = 1100 A a/3(34.5

kv)

The formula for synchronous speed shows that the synchronous speed can be increased in tw o ways: either by increasing frequency, or decreasing the number of poles. The answer is (C), see below: s rpm

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120 -f P

P a g e 360

A fternoon S olutions 56

57

The National Electrical Code® horsepower rating for a 3-phase, 208 V, squirrel-cage induction motor with a 960 A maximum locked rotor current is most nearly:

Page 361

50 hp

B.

75 hp

C.

120 hp

D.

60 hp

A private utility company is considering purchasing portable solar equipment that costs $85,000 for a conceptual project. The life of the project will be 5 years. The project manager estimates that the equipment will generate an annual cash flow of $32,000 to be paid at the end of each year of the project. If the com pany’s MARR is 22% , what is the net present value of the equipment purchase?

E

58

A.

A.

-$16,500

B.

$6,636

C.

$75,000

D.

$91,640

What is the power factor for a 3-phase inductive load rated at 22 MVA, 5 kV with a power output of 15 MW? A.

0.59

B.

0,68

C.

0.78

D.

0.89

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N otes / W orkspace This question is only answerable by looking up NEC® code. In table 430.251 (B), you’ll find the answer to be 60 hp.

The MARR is the minimum interest percentage the company requires before investing in something. In this problem the net present value is being calculated using the MARR as the interest rate. If the net present value is positive, that means the company considers the project economically viable. The answer is (B) which is a positive net present value, see below: Net present value: C = 85000

NPV = -C + A(P/A, 22%, 5)

n=5 A = 32000 i = 0.22

/

l

\ p P/A, 22%, 5 —

’

)

a

(l + i)n -1

Ml

NPV = -85000 + 32000 (P/A, 22%, 5 ) = 6636

The real and apparent power are given in the problem. Divide the two and you’ll get the power factor. The answer is (B), see below:

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MW

15

MVA

22

=

0.68

P a g e 362

A fternoon S olutions

Pa g e 363

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N otes / W orkspace With lighting problems, there are a lot of variables and numbers. Lighting design engineers are very specialized, and the PE exam will probably not ask too detailed of a question about this subject. So a lot of the information given here is misdirect information. The answer is (D), see the calculations below: 90,000 lumens A = L • W = 60 -20 = 1,200 ft2 cu = 0.82

fc MM A = 61.5 fc

The bipolar transmission circuit is depicted in (C), which is the correct answer. The distinguishing characteristic of a bipolar transmission circuit is that there are two conductors transmitting the DC power. The two lines have the same magnitude of voltage, but opposite polarity.

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P a g e 364

A fternoon S olutions

61

62

W hat is the line current of a delta-connected, 3-phase, 1,325 kVA, 2.4 kV inductive load? A.

220 A

B.

319 A

C.

452 A

D.

552 A

An electrician is perform ing a test on a single-phase 480/208 V transform er. follow ing values are measured:

The

Core losses = 57 W C opper losses = 223 W S econdary voltage = 208 V S econdary current = 58 A Load pow er fa c to r = 0.85 W hat is the efficiency of the transform er?

Page 365

A.

95.3%

B.

96.3%

D.

97.7%

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N otes / W orkspace

The m otor is connected delta, so the phase and the line voltages are the same, but the phase and line currents are different so the J 3 must be used. The answer is (B), see below: S

1325 kVA

>/i v

V3-2.4 kV

= 318.7 A

The problem gives the losses which leaves the calculation of power output as the remaining step. Use the given voltage, current, and power factor to calculate the power output. Then plug these values into the efficiency formula. The answer is (D) see below: C loss = 5 7 W CU loss = 223 W VS = 208 V ls = 5 8

Pout

- V s ' s M

poutt =10254W eff Pout+ C , loss + CU loss = 97.3%

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PAGE 366

A fternoon S olutions 63

64

65

In a rectifier circuit, the resulting DC output voltage level is controlled by what? A.

Inducing harmonic effects on the AC input voltage

B.

Changing the firing times of the devices in the rectifier circuit

C.

Applying or limiting external gain to the devices in the rectifier circuit

D.

Reversing the polarity of the DC input voltage

A 3-phase, 3-wire, ABC-delta source is connected to a balanced wye load. The voltage on leg “AB” of the source is 460 V, and the current in the same leg is 37 A. W hat is the current in each leg of the wye load? A.

12 A

B.

37 A

D.

111 A

In rectification applications what device is commonly used as a voltage regulator due to its ability to produce a constant voltage under large current fluctuations? A.

Capacitor Bank Zener Diode

P a g e 367

C.

LED inverter

D.

Split-Capacitor Motor

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N otes / W orkspace Rectifier circuits can change the output DC voltage level by controlling the firing times of the devices such as the thyristors. The firing times affect the time-average of the output voltage, which results in the ability to increase or decrease the average level, so the answer is (B). These kinds of circuits are sometimes referred to as “chopper” circuits.

It is a good idea to first visualize the circuit that is being described. Once you properly connect the wye and the delta nodes together it will help you see what is happening to the currents and voltages. The answer is (C), see below:

The main attribute of a zener diode is its reverse voltage breakdown characteristics. Zener diodes have the ability to produce a relatively constant voltage for a large reverse current range. This is called the zener breakdown region. For this reason they are often used as voltage regulators in rectification applications.

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PAGE 368

A fternoon S olutions 66

67

Symmetrical components analysis is a method which converts three unbalanced current phasors into: A.

three phasors of differing magnitudes

B.

a single current phasor

C.

three sets of balanced currents

D.

three equal currents differing only in phase angles

In the ferromagnetic core shown below, the flux density of the center leg B is 1.5 Teslas (T). What is the flux in each of the tw o outer legs (A and C)?

20 cm

—

P a g e 369

►

A.

0.0053 Wb

B.

0.011 Wb

C.

0.015 Wb

D.

0.15 Wb

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N otes / W orkspace The purpose of the symmetrical component method is to take three unbalanced currents (which is mathematically difficult to deal with) and change them to three sets of balanced currents, thereby simplifying the mathematics.

First calculate the flux in the center “ B” leg. Then divide this flux proportionally through the other two legs. Since both sides are symmetrical, the flux is halved between the two. The answer is (A), see below: Leg B AREA b = (0 .l)(0 .0 7 ) = 0.007m2

Legs A & C 4>b = 4>a +c

ctbA =cb = -2 B= 0.0053 Wb “C

4>b - B • AREAe B = 1.5 T B =0.011 Wb total flux

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P a g e 370

A fternoon S olutions

P a g e 371

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N otes / W orkspace Most of the diagram shown is misdirect information, but it does help you visualize the situation. The given numbers on the right are all you need to perform the calculations. The power output value is not given, so it has to be derived using the other values. Afterwards, plug each number into the efficiency formula and the answer is (A), see below: Pout = Vs Is cos 0

r|

Vs =2800 V

Pout pCO + p1 c + p

•100

is =6.6 A 9=

-

Pco =75 W

22 °

p

Pout =17.1 kW

1 cu

=241 W

r] = 98.2%

To calculate the kVA value, you must first have both the voltage and current on either side of the transformer. Since only one of each side is given, you must use the turns-ratio to find the other. The answer is (B), see:

200

a = ----- = 4 50 i2 = 75A V1 = 250 V

V V = - L - 62.5 V 2 a S - V , i2 - (75)(62.5) = 4688 VA = 4.7 kVA

The answer to this question is in Table 9 of the NEC®. Find the resistance/reactance for the indicated cable, and multiply it by the distance, and the answer is (C). This is single-phase, so make sure to multiply the distance values by two. See below: NEC per 1000' Rl = 0.31 XL = 0.048

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L = 100 R - ° '31 x 1 0 0 (2 )-0 .0 6 2 Q 1000 1 ’ 0.048 . _ _. _ X = --------x 100 2 = 0.01 Q 1000 v ’ Z = 0.062 + y'0.01 Q PAGE 372

A fternoon S olutions

Time In Seconds

N otes / W orkspace

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PAGE 374

A fternoon S olutions 72

73

74

Pa g e 375

A 3-phase motor needs to have its power factor improved by 1,008 kvars. If 48 kvar capacitor units are available, which of the per-phase capacitor bank arrangements shown below will provide the necessary result? (Assume balanced load conditions)

Where should the conductor be positioned when using a donut-type currenttransformer? A.

At the primary polarity mark

B.

Through the center opening

C.

On the transform er’s bushings

D.

On the transformer’s contact nodes

The presence of hysteresis in a ferromagnetic transformer is indicative of:

A.

A ground fault in the system

B.

Losses due to winding resistance

C.

No-load losses

D.

Secondary feeder line losses COPYRIGHT 2017 BY COMPLEX IMAGINARY, LLC

N otes / W orkspace The problem gives the total kvar input necessary for the correction. This total amount of reactive power will be split equally between the phases. After doing the calculations shown below, the correct number of capacitor units per-phase is 7, which is answer (D). 1008 x , . -------= 21 total units 48 — = 7 units / phase 3

The donut-type current transformer employs the method of running the conductor through the hole in the donut (also called the “w indow ” of the transformer). As the conductor passes through the opening in the center of this sort of transformer, it constitutes one primary turn. The ratio of the transformer can be altered by putting multiple wraps of the conductor through the donut hole.

Hysteresis in a transformer means that the magnetic actions/reactions going on in the transformer between the windings and core are not ideal. An undesired magnetic “ memory” (because of the current alternating all the time) is taking place in the core that results in losses. Since it is a type of core loss, that means it is classified as a “ no-load” loss, which makes the answer (C).

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P a g e 376

A fternoon S olutions 75

76

A station engineer is considering the purchase of a $125,000 motor for a project. For the two year life of the project, the m otor will generate a cash benefit of $75,000 the first year, and $85,000 the second. The company’s minimum acceptable rate of return for investments is 12%. Based on this information, what would be the present worth of this investment should the engineer decide to purchase the motor? A.

-$2,294

B.

$5,267

C.

$9,726

D.

$35,000

The diagram below shows an equivalent circuit for part of a transmission system:

How much power is lost in the transmission line (per-unit)?

»

77

P a g e 377

A.

0.0019 pu

B.

0.0029 pu

C.

0.0031 pu

D.

0.0039 pu

3.2 L 23° pu

A 3-phase motor bank consists of 4 motors: 2 of them are induction motors, and the other 2 are synchronous motors. The synchronous motors are running in an overexcited state, while the induction motors operate with their usual lagging power factors. Compared to a situation where all motors running with lagging power factors, what does this indicate for the overall system? ___________ A.

The overall system efficiency is increased.

B.

The overall system efficiency is decreased.

C.

The transmission supply line losses are increased.

D.

The transmission supply line currents are increased. COPYRIGHT 2017 BY COMPLEX IMAGINARY, LLC

N otes / W orkspace In this problem, there are tw o (P/F) factors that need to be calculated: one for the positive cash flow from the first year, and another for the second year. Then, add these to the cost of the m otor to get the present worth of the purchase. The answer is (C), see below: First Year

C = -125000 F = 75000

(P/F, 12%, 2)

+

II 1

U-T

i = 0.12

( P /F , 12%, Cl

F2 =85000

Second Year

1\ / = 66964

P ,- F .( n ir = 67761

PW = C +

+ P2

= 9726

This is a basic electrical engineering problem, made slightly more challenging because of the perunit values. The trickiest part of this problem is knowing which parts of the impedance values apply for the formulas. When calculating the current, you must use all the impedance values of the circuit because the current is the same for the entire loop shown. However, when calculating the power lost in the line, only choose the line impedance (not the load) and, furthermore, only the resistive value because that is the only one that causes heat loss, or power loss.

V = IZ V _________ 1.QZQ0_______

Power Lost P = I2R

” Z " (0.021+ j0.35) + 3.2Z23°

= (0.30)2 (0.021)

= 0.30Z. - 28°

= 0.0019 pu

The answer is (A). The synchronous motors act as generators when they have a leading power factor. This extra power is used by the overall system and the system efficiency is increased. Line losses are decreased, which means that the line currents are also decreased.

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P a g e 378

A fternoon S olutions 78

Below is an example of a simplified transmission circuit including the line impedance. Calculate the apparent power output of the generator. 12:1

79

PAGE 379

233 kVA

B.

403 kVA

c.

698 kVA

D.

1,209 kVA

A capacitor on a 3-phase system corrects the power factor produced by an induction motor to 0.96 by adding 10.8 kvar. The rated voltage for the motor is 230 V, and the power used is 23.9 kW. What was the uncorrected power factor of this inductive load? »

80

A.

3:20

A.

0.80

B.

0.67

C.

0.87

D.

0.75

A delta-wye transformer is rated for: 34.5 kV/480 V, 125 A primary (high-side) line current. What is the transformer’s power rating (kVA) and the secondary current? A.

4,300 kVA, 8,980 A

B.

4,300 kVA, 15,600 A

C.

7,500 kVA, 8,980 A

D.

7,500 kVA, 15,600 A

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N otes / W orkspace The apparent power is the same for all of the three “ sections” shown in the diagram. The best strategy would be the one that involves the least amount of calculations. The current and impedance are given in the right-most “ section” , so that would be the easiest one to calculate. The answer is (C), see below: 1= 73 A Z = 131Z.240 Q S = I2Z = 698 kVA

Use the corrected power factor to find the corrected reactive component. Add the additional reac­ tive power from the capacitor to this value and then calculate the original power factor. The answer is (A), see below:

CL

P

For the corrected power factor: pf = 0.96

For the original power factor: Q1+10.8 = 17.8 kvar = Qo

23.9 = 0 . 9 6 ^ )

So = ^Q 2 + P2 = 29.8 kVA

S1 = 24.9 pf = — = 0.80 29.8

Q1 = ^/s2 - P2 = 7.0 kvar

Use the given primary voltage and primary line-to-line current to calculate the kVA rating of the transformer. Then use that value to calculate the secondary line current. The answer is (C) see below: First find the kVA rating using the primary values: S = VL 1lV3 = (34.5 kv)(l25 A)V 3 = 7469 kVA Now, find the secondary line current: 7469 kVA V3 V COPYRIGHT 2017 BY COMPLEX IMAGINARY, LLC

a/3 - 48 0

= 8984 A ~ 8980 A

V PAGE 3 8 0

PRACTICE EXAM VOLUME 4 80 QUESTIONS 8 HOUR TIME LIMIT

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MORNING SESSION QUESTIONS 1-40 4 HOUR TIME LIMIT

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PAGE 384

M orning S ession 1

2

Page 385

Two single-phase transformers are paralleled together, and are serving a 5,000 kVA load. Transformer #1 is rated for 2,000 kVA, and Transformer #2 is rated for 3,000 kVA. Both of the transformers have the same impedance of 5.25%, and the same turns ratio. How much power (in kVA) is being provided by each transformer? A.

Transformer #1 = 2,000 kVA, Transformer #2 = 3,000 kVA

B.

Transformer #1 = 2,500 kVA, Transformer #2 = 2,500 kVA

C.

Transformer #1 = 3,000 kVA, Transformer #2 = 2,000 kVA

D.

Transformer #1 = 1,500 kVA, Transformer #2 = 3,500 kVA

Given the voltage waveform shown below, what is the Vrms value most nearly?

B.

0.88

C.

1.0

D.

1.2

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PAGE 386

M orning S ession 3

4

5

PAGE 387

Which of the following systems is NOT considered to be a separately derived system? A.

A transmission switching station serving a remote rural area

B.

A paint-shed powered off the secondary of a delta-wye transformer connected to the utility on the primary side

C.

An off-grid photovoltaic system installed in the desert

D.

A portable rock-crushing machine powered by a large, 3-phase generator

The most recent version of the National Electrical Code® requires the maximum locked-rotor current for selection of a squirrel-cage induction motor — rated 5 hp, 115 V, single-phase, 60 Hz, 38 A, 30 degrees out of phase at full load — disconnect to be: A.

204

B.

186

C.

480

D.

336

What would be the advantage if you were to reconnect a conventional voltage transformer as an autotransformer? A.

It can correct a lagging power factor

B.

Fewer windings are necessary

C.

It could handle more power

D.

There is no advantage

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PAGE 388

M orning S ession 6

7

8

Pa g e 389

An 11/3.3 kV single-phase transformer has an unloaded secondary (low-side) voltage of 5,009 V. Under full-load conditions, the current at the load is 50 L -25° A, and the feeder line impedance is 0.2 + y'0.4 Q. Assuming that the load’s operating voltage is 3.3 L 0° kV, what is the voltage regulation of the transformer? A.

31 %

B.

41 %

C.

51 %

D.

61 %

A surge protector with a withstand voltage of 350 V is providing a protection level of 130 V for a 120 V line with a 35 A rated load. What is the protective margin of this arrester? A.

63%

B.

68%

C.

100%

D.

169%

For a transmission line, the velocity of propagation will increase: A.

as capacitance per-unit length increases

B.

as inductance per-unit length increases

C.

as inductance per-unit length decreases

D.

as line length increases

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Pag e 390

M orning S ession 9

10

11

Pa g e 391

A 3-phase, 208 V, 75 kVA, wye-connected synchronous generator has a full-load current of 35 A with a power factor of 0.85 lagging. What is the internal generated phase-voltage if the stator resistance is 0.5 Q and the synchronous reactance is 1.2 Q? A.

2 4 6 ^ 6 .2 ° V

B.

1 5 9 /.9 .6 0 V

C.

120 Z 00 V

D.

8 7 Z .-1 7° V

How many unique voltage magnitudes are possible in a 4-wire, delta-connected system with the neutral wire center-tapped on the BC leg? A.

1

B.

2

C.

3

D.

4

A transformer has a turns ratio of 2,000:350. The secondary’s voltage while connected to the load is 350 Z 0 V. If the voltage regulation is 7.2%, what is the unloaded secondary voltage? A.

365 V

B.

375 V

C.

385 V

D.

395 V

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PAGE 392

M orning S ession 12

If a stator of more than two poles makes a single, full mechanical rotation of the magnetic field, what is produced?

;

13

14

Page 393

A.

A phase would be mitigated in the current

B.

A single, full electrical cycle is produced

C.

More than a single full electrical cycle is produced

D.

All phase sequences would be reversed

According to the National Electrical Code®, when installing a 480 V capacitor bank | to correct a m otor’s power factor, all of the following must be implemented EXCEPT: A.

A way to disconnect the capacitor bank from all voltage sources.

B.

The overcurrent protection device must be rated or set to as high as practicably possible.

C.

A permanent or automatic method to discharge the capacitor.

D.

The residual voltage of the bank shall discharge to 50 V in no more than a minute.

A 34.5 kV, 3-phase, 4-wire, wye-connected distribution panel is experiencing an unbalanced load situation. On phase A (19.9 A 0° kV) there is a 120 + y'65 kVA load, on phase C (19.9 L 240° kV) there is a 65 + 20/ kVA load. Calculate the magnitude of the neutral current. A.

6.6 A

B.

6.1 A

C.

5.9 A

D.

2.2 A

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Pag e 394

M orning S ession A 12 volt, 125 AH, lead-acid battery can supply 20 amps of current for a period of:

16

17

A.

5.25 hours

B.

2.5 hours

C.

6.25 hours

D.

3.75 hours

A single grounding electrode consisting of a pipe has a resistance of 30 ohms, what must be done to allow this circumstance to comply with the NEC®? A.

It must be driven below 60 ft of earth

B.

It must be augmented by multiple additional electrodes

C.

An uncoated and ungalvanized steel electrode must be added

D.

It must be augmented by one additional electrode

In an underground electrical duct, four conduits are irregularly spaced as shown below:

The geometric mean distance of the conductor spacing is most nearly: 1 ft 9 in A.

Pa g e 395

B.

15 in

C.

3 ft

D.

14 cm

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PAGE 396

M orning S ession 18

19

20

Pa g e 397

A voltage of 131 L 52° kV is applied across a complex load measuring 125 + /87 MVA. Calculate the current across the load. A.

6 4 0 + /1 9 6 A

B.

6 4 0 -/1 9 6 A

C.

1,110+ y'339 A

D.

1 ,1 1 0 -/3 3 9 A

The 3-phase distribution system between the Station and the Load has phase conductors made of aluminum inside steel conduit. According to the 2017 National Electrical Code®, provided these conductors are 750 kcmil, the phase voltage at the load is most nearly:

C.

575 V

D.

585V

Since the ‘80s, the proliferation of electronic equipment has brought the issue of harmonics to the forefront of the electrical engineering industry. Generally speaking, which of the following is the primary concern regarding harmonics? A.

Unwanted current phase shifts

B.

Poor power quality

C.

Unbalanced loads

D.

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PAGE 398

M orning S ession 21

A wye-connected AC generator on a 3-phase system has a full-load current output of 452 A. Provided the generator’s rated output is 1,131 kW, and has a 0.90 lagging power factor, the generator’s phase voltage rating is most nearly: A.

0.78 kV

B.

0.93 kV

C.

2.78 kV

D.

4.81 kV

When concerned with overcurrent protection, the 2017 National Electrical Code® does not permit exceeding the following size (A) overcurrent device in the protection of 14 AWG copper. A.

7A

B.

30 A

C.

20 A

D.

15 A

In the single-phase circuit shown below, when the switch is in position 1, the current \1is 77 4.-14° A. What is the circuit current when the switch changes to position 2, assuming the circuit reaches steady-state conditions?

X

rr

P a g e 399

A.

78 A

B.

278 A

C.

318 A

D.

481 A

92

V = 250 V

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PAGE 400

M orning S ession 24

25

A 480 V, 225 A, 3-phase panelboard is protected by a 200 A circuit breaker, What should be the minimum size of the copper grounding conductor for this panel? A.

8 AWG

B.

6 AWG

C.

4 AWG

D.

1/0 AWG

In the 3-phase rectifier circuit shown below, the average voltage across the load is 365.7 V. If the rms voltage across the load is 462.4 V, what is the ripple factor for the circuit?

D.

Pa g e 401

126%

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PAGE 402

M orning S ession 26

27

28

Page 403

Calculate the current at the load Z.

A.

4 1 0 Z .-1 4 2 0 A

B.

3 1 3 Z 9 9 0A

C.

136 Z l-1 4 2 0 A

D.

136Z.820 A

A continuous duty m otor is connected to a 3-phase circuit. This m otor’s full-load current is 12 A and it employs an integrated thermal overcurrent protection device. According to the 2017 National Electrical Code®, the ultimate trip current for the protector is most nearly: A.

10.5 A

B.

16.8 A

C.

18.7 A

D.

20.4 A

A relay protection diagram is shown below. The dashed lines represent Zone 1 protection regions. If a fault occurs at point F and the primary protection works successfully, which of the breakers will trip? F

A.

1 ,2

B.

1,2,5,6

C.

1,2,3,4

D.

1,2,3,5,6

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PAGE 404

M orning S ession 29

30

31

PAGE 405

The power absorbed by a complex load with an impedance of 5 L 20° Q when 66 V is applied is most nearly: A.

819 W

B.

298 W

C.

871 W

D.

503 W

A delta-connected 460 V, squirrel-cage motor is operating with a lagging power factor of 0.70. To correct this to a 0.95 lagging power factor, a capacitor bank was added. After the application of this power factor correction, the motor circuit phase current is 96 A. What is the real power used by the motor? A.

42.0 kW

B.

72.7 kW

C.

76.5 kW

D.

125.9 kW

According to the National Electrical Code®, if a nontime-delay fuse that protects a motor branch-circuit does not exceed 600 amperes it: A.

Must be replaced by a time-delay Class CC fuse

B.

Must not exceed 400 percent of the full-load current

C.

Must not be permitted to increase

D.

Must have a resistance of 1,200 Ohms

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PAGE 406

M orning S ession 32

33

34

The total short circuit current at a secondary terminal on a transformer is not only dictated by the transformer impedance (Z), but it is also m ost sig n ifica n tly determined by which of the following factors? A.

AC vs DC generator characteristics.

B.

Mutual inductance of transformer’s windings.

C.

Source contributions at transformer primary terminal.

D.

Power factor of the load.

If you decrease the low side winding of a 138 kV/12.47 kV transformer by 10%, this would affect the turns ratio in what way? A.

The turns ratio remains 11:1

B.

The turns ratio becomes 124.2 kV/12.47 kV

C.

The turns ratio becomes 138 kV/11.22 kV

D.

The high side will automatically decrease by 10% as well

What is the turns ratio for the potential transformer below O vn2)?

— W v— fosfn. 480 V

Page 407

0

A.

2:1

B.

1:2

C.

4:1

D.

1:4

nT fn 2

240 V

1

o

CD

Cl

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PAGE 408

M orning S ession 35

36

37

Page 409

The reflected high-side impedance of a transformer is 255 Q. What is the impedance on the low-side if the turns ratio of the transformer is 350:25 ? A.

1.3 Q

B.

10 O

C.

180

D.

25 Q

A 60-Hz induction has a rotor speed of 1,710 rpm under a 5% slip. How many poles does this motor have? A.

3

B.

2

C.

4

D.

5

Which of the following units is a measure of luminous intensity? A.

foot-candle

B.

lumens

C.

candela

D.

watt

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PAGE 410

M orning S ession 38

The primary difference between a GFCI and an AFCI is: A.

39

Page 411

GFCI is a system; AFCI is a device

B.

GFCI protects people from shocks; AFCI protects against arcing

C.

GFCI uses the Geometric mean; AFCI uses the Average

D.

GFCI is for 240V systems; AFCI is for 120V systems

There is a transformer on a 3-phase system. This transformer has a ferromagnetic core and is rated 900 kVA with a 4% impedance. The impedance under a 1 MVA base, would most nearly be: A.

3.4%

B.

4.4%

C.

5.4%

D.

5.7%

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Pag e 412

M orning S ession 40

A 3cf), wye-connected, sychronous motor draws 427 A from a 480 V source with a 0.77 lagging power factor. How many kvars would a capacitor bank need to supply to increase the power factor to 0.87? A.

47 kvar

B.

71 kvar

C.

131 kvar

D.

155 kvar

STOP

This C om pletes The M orning Session o f the Test

Page 413

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PAGE 414

AFTERNOON SESSION QUESTIONS 41-80 4 HOUR TIME LIMIT

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PAGE 416

A fternoon S ession

41

42

Page 417

A wye-wye, 230 kV, 3-phase transmission line is being stepped down to 200 kV using three autotransformers connected between the neutral and the line. For each phase, if the autotransformer is rated for 500 MVA, and the turns ratio is 5:2 (common:series) what is the apparent power in the common coil? A.

143 MVA

B.

150 MVA

C.

173 MVA

D.

250 MVA

In the circuit shown below, calculate the total low-side equivalent impedance.

A.

3 8 z l5 4 ° O

B.

45 Z l 54° O

C.

5 1 Z .6 8 0 Q

D.

53 z l 88° Q

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PAGE 418

A fternoon S ession 43

44

45

A power plant engineer plans on retiring after 30 years of service. At the tim e of his retirement, he has $175,000 in a savings account that pays an annually compounded interest rate of 4% .W hat is the maximum amount of money the engineer can wit each year if he plans for the savings to last him 15 years? A.

$11,667

B.

$15,740

C.

$18,330

D.

$19,690

A 3-phase wye-connected synchronous generator has a power factor of 0.82. If the phase voltage is 2.77 kV and the line current of 85 A, what is the total apparent power of the generator? A.

235 kVA

B.

334 kVA

C.

579 kVA

D.

706 kVA

Given the 3-phase circuit below, what is the current, l2? transformers)

PAGE 419

(T1 and T2 are potential

7 4 z 6 1 °Q

A.

7.1 A

B.

12.3 A

C.

21.3 A

D.

25.1 A

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Pag e 420

A fternoon S ession 46

47

48

Page 421

Which of the following factors does NOT affect the voltage drop calculations for feeders? A.

Load Current

B.

Feeder Source Turns Ratio

C.

Feeder Length

D.

Conductor Size

A wye-connected synchronous generator has per-unit reactances of Xd = 1.24, X’ = 0.37, and X” = 0.25. A subtransient analysis is conducted and a subtransient current of 4.8 pu is detected. What is the internal generated phase voltage in pu? Assume all analysis is performed at rated values. A.

1.0 pu

B.

1.2 pu

C.

1.8 pu

D.

6.0 pu

With the zonal cavity method, the room is considered to contain how many vertical zones/cavities?

A.

1

B.

2

C.

3

D.

4

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Pag e 422

A fternoon S ession 49

50

51

Page 423

What dictates the highest temperature a given synchronous generator can withstand during its operation? A.

The number of poles in the stator

B.

The starting torque of the machine

C.

The insulation class of the machine’s windings

D.

The machine’s terminal contact ratings

A 480 V, 3-phase bus is serving a 10 kVA motor with a power factor of 0.72 (lagging). A second 15 kVA squirrel-cage motor is attached to the bus, and the system power factor is now observed to be 0.76, lagging. What is the power factor of the new motor? A.

0.75

B.

0.77

C.

0.79

D.

0.81

A 3-phase, 12.5/7.2 kV, delta-delta transformer is rated 150 MVA. An accident occurred which caused one phase of the transformer to disconnect, leaving just two lines available on the secondary. What is the amount of power that can be delivered in this situation? A. 50 MVA B.

87 MVA

C.

100 MVA

D.

150 MVA

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PAGE 424

A fternoon S ession 52

53

54

P a g e 425

A current transformer has a 150:12 ratio and is monitoring a 38 Q load on a 6.6 kV line. Assuming nominal conditions, what is the current in the instrument? A.

0.1 A

B.

1.7 A

C.

13.9 A

D.

174 A

Once the phasor rotation reference has been chosen while performing a symmetrical components calculation, the negative sequence components will have (with respect to the reference): A.

twice the phase amplitude

B.

phasors in phase with each other

C.

the reverse phase sequence

D.

a 90° lagging phase

In a typical power-flow study, which of the following statements concerning voltagecontrolled buses is no t true? A.

The voltage magnitude is fixed

B.

The real power of the bus is calculated

C.

Most generator buses are considered to be voltage-controlled

D.

The reactive power component is calculated

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PAGE 426

A fternoon S ession 55

56

Page 427

Large disturbance rotor angle stability is concerned with the capacity of a power system to maintain synchronism when subjected to severe disturbance (e.g. short circuit on a transmission line). What is this most commonly called? A. Series Capacitance B.

Transient Stability

C.

Conductor Hysteresis Capacity

D.

First Swing Instability

There is a 3-phase, 50 kW load with a lagging power factor. After adding a 9.2 kvar capacitor the power factor gets corrected to 0.96. What was the initial pow er factor of the load? A. 0.83 B.

0.87

C.

0.91

D.

0.95

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PAGE 428

A fternoon S ession 57

58

In the circuit shown below, what is the purpose of the inductor and capacitor?

A.

To

B.

To reverse the voltage polarity

C.

To provide countervoltage for the motor

D.

To filter harmonic content

The waveforms across a thyristor are shown below. Which of the following waveforms for the gate current, lGwould achieve this situation?

c.

A.

cot 2/7

4H ” +

An

»q

2/7

4n

2/7

4n

'g

B.

D.

cot 2/7

Page 429

cot

■ 4 -+ H 4n

cot

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PAGE 430

A fternoon S ession All of the following are types of losses which occur in DC machines, EXCEPT:

60

61

Page 431

A.

Hysteresis losses

B.

Stray load losses

C.

Core losses

D.

Brush losses

An engineering firm has a surplus of $250,000 and plans to invest it at 9%, com pounded annually. How much will the investment be worth in 12 years? A.

$703,000

B.

$724,000

C.

$759,000

D.

$766,000

What is the advantage of using iron cores, rather than air cores, for electric machinery and transformers? A.

Iron’s rigidity is necessary for core composition

B.

Ferromagnetic material allows many times less flux

C.

Ferromagnetic material allows many times more flux

D.

Air produces more harmonic resonance

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PAGE 432

A fternoon S ession 62

Two induction motors have the following nameplate ratings: Motor A - 455 VA, 480 V, 0.77 lagging power factor Motor B - 365 VA, 480 V, 0.80 lagging power factor The combined real power (W) for these motors is:

63

64

P a g e 433

A.

642 W

B.

675 W

C.

735 W

D.

830 W

Given the circuit below, what is the average magnitude of the voltage, v(t)? Assume ideal AC circuit behavior.

A.

6.5 kV

B.

7.2 kV

C.

10.2 kV

D.

12.4 kV

If two 3-phase transformers are connected in parallel, but one is wye-wye, while the other is wye-delta, what will be the relationship of their secondary phase voltages? A.

They will be in phase.

B.

30° out of phase.

C.

45° out of phase.

D.

60° out of phase.

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Pag e 434

A fternoon S ession 65

66

67

PAGE 435

Which of the following does no t function as a sensor for overcurrent protective relays? A.

Zener Diode

B.

Current Transformer

C.

Potential Transformer

D.

Temperature Gage

A 25/6.6 kV, wye-wye, 3-phase transformer bank is constructed using three single­ phase transformers. What is the turns ratio for each of the individual transformers (primary:secondary)? A.

1.50:1

B.

2.18:1

C.

3.78:1

D.

6.56:1

Refer to the 2017 edition of the National Electrical Code® to answer the following question. What is the minimum required size for the single-insulated, 60° C, TW copper conductors feeding a 460 V, 3-phase, AC motor rated 150-hp operating on continuous duty? A.

4/0 AWG

B.

300 kcmil

C.

400 kcmil

D.

500 kcmil

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PAGE 436

A fternoon S ession 68

What is the configuration of this 3-phase transformer (primary-secondary)? A.

delta-wye

B.

wye-delta

C.

wye-wye

D.

delta-delta

C B A

B A

PAGE 437

Primary

Secondary

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PAGE 438

A fternoon S ession 69

Below is a circuit whose waveform from A - B looks like:

t If the capacitor is removed from the circuit, what would the resulting waveform across A - B look like?

70

71

Page 439

A 3-phase motor needs to have its power factor improved by 3360 kvars. If 140 kvar capacitor units are available, how many of them would be needed per-phase if the units are to be connected in series? (Assume balanced load conditions) A.

4

B.

6

C.

8

D.

10

What is a distinguishing characteristic of a shaded pole motor? A.

A relatively low starting torque

B.

A separate auxiliary winding for each pole

C.

Commonly used for poly-phase applications

D.

Uses a capacitor for startup

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PAGE 440

A fternoon S ession 72

In the circuit shown below, a 3-phase fault occurs at point F. If the short-circuit current is 572 A, what is the source impedance?

0

15 kV

73

74

Pa g e 441

r-F

Za= 14.75zl760 Q

A.

15.1 Q

B.

1 8 .4 Z .-5 1 0 Q

C.

26.7Z - 71° Q

D.

32.2Z. - 52° Q

* Zb= 11.25/176° Q

Which of the following will NOT cause unbalanced phase currents in a power system? A.

Abnormal System Frequencies

B.

Unbalanced Load

C.

Line-Line Faults

D.

Asymmetrical Phase Impedances

The synchronous speed of an induction motor is based on: A.

Its rating under full-load conditions

B.

The m otor’s power factor when in use

C.

The torque of the m otor operating at full-load

D.

Supply frequency and number of poles in motor winding

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PAGE 442

A fternoon S ession 75

76

77

Pa g e 443

A single-phase, 277 V conductor in a 350’ long aluminum conduit is serving a lighting load in an office building. According to the 2017 NEC®, what should be the total reactance of the conductor if the size is 2/0 AWG? A.

0.015Q

B.

0.025 Q

C.

0.03 Q

D.

0.04 Q

According to the NEC®, what is the maximum startup current for a 3-phase induction motor (code-letter H) rated for 25 hp, 480 V? A.

123 A

B.

213 A

C.

369 A

D.

450 A

In the circuit shown below, a 460 V source is feeding a balanced load. If lAB is 78 A, what is lA?

A.

26 A

B.

45 A

C.

78 A

D.

135 A

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PAGE 444

A fternoon S ession 78

79

80

A customer of a utility is asking for an additional service for a new three-phase transformer. The utility says that the short circuit current available at the building site is 7,125A at 22kV. What is the source impedance if the transformer is rated 2,500 kVA, 22/6.6kV with an impedance of 4.25%? A.

1.0%

B.

0.96%

C.

0.92%

D.

0.88%

The impedance of a 20 kV, 12 MVA, 3-phase line is 22 Q/phase. What is the per-unit value of this impedance on a 100 kV, 50 MVA base? A.

0.11 pu

B.

0.22 pu

C.

0.33 pu

D.

0.44 pu

ANSI Standard Device Numbers (ANSI/IEEE Standard C37.2) list various features and circuit elements all having the primary function of: A.

Protective Devices (e.g. Relays or Circuit Breakers)

B.

Capacitive Devices (e.g. Capacitors or Switches)

C.

Grounding Devices (e.g. Electrodes or Grounding Conductors)

D.

Amplification Devices (e.g. Thyristors or Diodes)

STOP

This Completes The A fternoon Session o f the Test P a g e 445

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PAGE 446

M orning S ession S olutions #

CORRECT SOLUTION

#

CORRECT SOLUTION

1

A

parallel transformers

21

B

generator | pow er formula | pow er factor

2

C

waveform s

22

D

NEC j overload protection

3

A

separately derived systems

23

C

circuit analysis j com plex values | o hm 's law | |

4

D

NEC | maximum locked rotor current

24

B

NEC | grounding | conductor sizing | |

5

C

transform er | autotransform er

25

C

rectification | ripple factor | waveform s

6

c

transform er | voltage regulation | ohm 's law

26

B

ohm 's law | com plex values

7

D

protective margin

27

C

m otor | NEC | overload protection

8

C

velocity of propogation

28

D

relay protection | protection zones

9

B

pow er factor | generator | ohm 's law | internal generated voltage

29

A

power form ula | com plex values

10

C

delta/wye

30

D

m otor | power facto r | power formula

11

B

transform er | turns ratio | voltage regulation

31

B

NEC | overload protection

12

C

m otor | stator

32

C

transform er | short circuit

13

B

NEC | capacitor bank | overload protection

33

C

short-circuit | fault current

14

A

delta/w ye | unbalanced load | pow er form ula | com plex values | com plex conjugate

34

A

transform er | turns ratio

15

C

battery

35

A

transform er | turns ratio

16

D

NEC | grounding

36

C

m otor | synchronous speed

17

A

geom etric mean

37

C

lighting

18

C

pow er form ula | com plex conjugate

38

B

GFCI | AFCI | standard receptacle use

19

B

NEC | voltage drop | com plex values

39

B

transform er | per-unit | base conversion

20

B

harmonics

40

B

m otor | pow er factor

Pag e 447

TOPIC(S) OF STUDY

TOPIC(S) OF STUDY

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A fternoon S ession S olutions C O RRECT SO LU TIO N

#

C O R R EC T SO LU TIO N

41

A

autotransformer 61

C

transfomrers | core | flux

42

C

transformer | turns ratio 62

A

parallel motors | power factor

43

B

A

waveforms

44

D

B

parallel transformers

45

B

circuit analysis | turns ratio

65

A

relay protection | zener diode

46

B

voltage drop

66

C

transformer | turns ratio

47

B

generator | per-unit | transients 67 | ohm's law

A

48

C

49

C

50

c

51

TO PIC (S) OF STU D Y

#

economics | uniform series | 63 capital recovery generator | power factor | 64 power formula

TO PIC(S) OF STUDY

NEC

| motor | conductor sizing

68

D

delta/wye | transformer

generator | losses 69

D

rectification | waveforms

70

C

motor | power factor

B

open delta 71

A

motor

52

C

current transformer | turns 72 ratio

C

short circuit | fault current | complex values

53

c

symmetrical components

73

A

unbalanced loads

54

B

power flow study | voltagecontrolled bus

74

D

motor | synchronous speed

55

B

transient stability

75

C

NEC

56

C

power factor

76

B

NEC | m otor

57

D

inverter | harmonics | 77 waveforms

D

delta/wye

58

B

waveforms | thyristor 78

C

short circuit | transformer | perunit

59

A

losses

79

A

per-unit

60

A

economics | future worth

80

A

ANSI device numbers

lighting | zonal cavity method

parallel motors | power factor

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PAGE 448

M orning S olutions

Page 449

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N otes / W orkspace Since the two transformers have the same impedances and turns ratios, the load will be split according to each transformer’s rated values. This is because the current division will cause each transformer to be loaded to the rated values. kVA. = 5000 kVA kVAi = 2000 kVA kVAo = 3000 kVA 2

Z. = 5.25% Zn = 5.25% (kVA ^

\Xi

P =

(kVA ^

kVA ^

•kVA.

x = 1, for transformer #1 x = 2, for transformer #2

P = 2000 kVA P2 = 3000 kVA

For triangular wave shapes (sawtooth):

.1 Y,

S

First do the calculation with the x-axis moved up to y = 0.75 So you have: V

(1 .2 -0 .7 5 )

V3

V3

0.26

Now, add the additional vertical space below the waveform to make up for the vertical shift. 0.26 + 0.75 = 1.01 Answer is (C)

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PAGE 450

M orning S olutions Which of the following systems is NOT considered to be a separately derived system? A.

A transmission switching station serving a remote rural area

B.

A paint-shed powered off the secondary of a delta-wye transformer connected to the utility on the primary side

C.

An off-grid photovoltaic system installed in the desert

D.

A portable rock-crushing machine powered by a large, 3-phase generator

The most recent version of the National Electrical Code® requires the maximum locked-rotor current for selection of a squirrel-cage induction motor — rated 5 hp, 115 V, single-phase, 60 Hz, 38 A, 30 degrees out of phase at full load — disconnect to be: A.

204

B.

186

C.

480

D.

336

What would be the advantage if you were to reconnect a conventional voltage transformer as an autotransformer?

Pa g e 451

A.

It can correct a lagging power factor

B.

Fewer windings are necessary

C.

It could handle more power

D.

There is no advantage

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N otes / W orkspace The term “separately derived system ” is sometimes difficult to be perfectly clear about. The simplest way to think about it is to ask, “ does the system in question have a physical, solid connection to a service?” If the answer is “ No” , then it is a separately derived system. The right choice is (A), since it’s the only one with a solid connection to the utility service. The paint-shed circuit is only connected off of the secondary, which means the only connection to the primary is through the transformer winding induction. The primary side doesn’t have a ground because it’s in the delta configuration, whereas the secondary side has the 4th neutral wire. The off-grid system is off-grid, so tha t’s easy. The rock crushing machine is powered by a generator, so there’s no service there, obviously.

In NEC® table 430.251 (A), you’ll find the answer to be 336 A.

Autotransformers have an apparent power rating advantage compared to a conventional transformer with the same windings. This is because autotransformers have that additional series coil that transfers power in addition to the power going through the common windings.

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Pag e 452

M orning S olutions An 11/3.3 kV single-phase transformer has an unloaded secondary (low-side) voltage of 5,009 V. Under full-load conditions, the current at the load is 50 L -25° A, and the feeder line impedance is 0.2 + y'0.4 Q. Assuming that the load’s operating voltage is 3.3 A 0° kV, what is the voltage regulation of the transformer?

7

Page 453

31 %

B.

41 %

C.

51 %

D.

61 %

A surge protector with a withstand voltage of 350 V is providing a protection level of 130 V for a 120 V line with a 35 A rated load. What is the protective margin of this arrester?

—

8

A.

►

A.

63%

B.

68%

C.

100%

D.

169%

For a transmission line, the velocity of propagation will increase: A.

as capacitance per-unit length increases

B.

as inductance per-unit length increases

c.

as inductance per-unit length decreases

D.

as line length increases

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N otes / W orkspace Since the full-load voltage for the transformer secondary is not given, it must first be calculate before being able to complete the voltage regulation formula. To doe this add the voltage loss from the line to the given load voltage and that will give you the voltage at the secondary’s terminals. FL = 3300Z0° + IZ

NL = 5009 V

= 330040° + (504 - 25°)(o.2 + 7O.4 )

VL = 3300Z0° V I = 50Z - 25° A

= 33174 0.24%

Z = 0.2 + y'0.4

wr, N L -F L ^ VR = ------------100 FL 5009-331 7 3317 - 51 %

100

The problem has three voltages given. For protective margin calculations, the two voltages necessary are the withstand voltage and the line protection level voltages (not the operating voltage). V -V PM = — ----- *■ V

Vw =350 V V =130V

3 5 0 -1 3 0 130

x 100

= 169%

The velocity of propagation is the speed of transmitted electricity. The velocity of propagation equation:

L’ l

q ,L

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per unit length

P a g e 454

M orning S olutions A 3-phase, 208 V, 75 kVA, wye-connected synchronous generator has a full-load current of 35 A with a power factor of 0.85 lagging. What is the internal generated phase-voltage if the stator resistance is 0.5 Q and the synchronous reactance is 1.2 Q? A. " " i

10

»

2 4 6 ^ 6 .2 ° V

B.

159/19.6° V

C.

1 2 0 ,/0 ° V

D.

87/1-17° V

How many unique voltage magnitudes are possible in a 4-wire, delta-connected system with the neutral wire center-tapped on the BC leg? A.

1

B.

2

C. _3 D.

11

P a g e 455

4

A transformer has a turns ratio of 2,000:350. The secondary’s voltage while connected to the load is 350 L 0 V. If the voltage regulation is 7.2%, what is the unloaded secondary voltage? A.

365 V

B.

375 V

C.

385 V

D.

395 V

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N otes / W orkspace The internal generated voltage will be higher than the output voltage. The losses are due to the stator resistance and synchronous reactance. Use these values with Ohm’s Law to add the additional voltage to the output voltage. The answer is (B). VL = 208 V V ~f= 120V V3 RA = 0.5 Q

\

=

pf = 0.85, lagging) 9 = cos"1(pf) = -^3l

=

Xs =1.2 Q IA = 35 A

IA = 35 Z. - 31.8° A EA = V(j) + jX ■* Sc IA +R Aa / \ = 120Z0° + j( l.2 j = 157 + j26.5 V = 159^9.6° V

Since the delta configuration doesn’t have a convenient center point to attach a neutral wire (like in a wye configuration) the neutral is sometimes center-tapped as described in the question. When this occurs, there are three voltages present, so the correct answer is (C). The normal line voltages and phase voltages are there, of course. The third voltage is the neutral wire connected to the line coming off the A vertex. The diagram shows this more clearly:

A

The loaded voltage and voltage regulation is given, which leaves the no-load secondary voltage to calculate. Use the voltage regulation formula to find the answer, which is (B), see below: FL = 350 V VR - 0.072

N L -F L VR = ---------FL VR(FL) = N L -F L NL = V r (f l ) + FL = 375 V

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P a g e 456

M orning S olutions 12

13

14

If a stator of more than two poles makes a single, full mechanical rotation of the magnetic field, what is produced? A phase would be mitigated in the current

B.

A single, full electrical cycle is produced

C.

More than a single full electrical cycle is produced

D.

All phase sequences would be reversed

According to the National Electrical Code®, when installing a 480 V capacitor bank to correct a m otor’s power factor, all of the following must be implemented EXCEPT: A.

A way to disconnect the capacitor bank from all voltage sources.

B.

The overcurrent protection device must be rated or set to as high as practicably possible.

C.

A permanent or automatic method to discharge the capacitor.

D.

The residual voltage of the bank shall discharge to 50 V in no more than a minute.

A 34.5 kV, 3-phase, 4-wire, wye-connected distribution panel is experiencing an unbalanced load situation. On phase A (19.9 Z 0° kV) there is a 120 + y'65 kVA load, on phase C (19.9 Z 240° kV) there is a 65 + 20/ kVA load. Calculate the magnitude of the neutral current. —

Pag e 457

A.

►

A.

6.6 A

B.

6.1 A

C.

5.9 A

D.

2.2 A

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N otes / W orkspace Since each pole pair represents one electrical cycle, having more than one pair would increase the number of electrical cycles that can be produced for each mechanical rotation. So, the answer is (C).

The answer is (B). Section 460.8 (B) states that the overcurrent protection shall be as low as possible. All of the other answers can be confirmed in Article 460.

This is a problem involving unbalanced load conditions. Use the complex values to get the tw o line currents and add them together. The resulting sum is the answer because the neutral needs to carry that total current back to the source to make a complete circuit. The answer is (A), see below: 120 + j65

=19.9 kV 0° for phase A 240° for phase C

20Z2400

t COPYRIGHT 2 017 BY COMPLEX IMAGINARY, LLC

” 6.85 Z. -28.4°

20Z0° 65 + 20j l*+l? A C

= 3.4Z. -1 3 7 °

3.5 - 5.6y = 6.6 Z -5 80 6.6 A

PAGE 458

M orning S olutions 15

16

17

A 12 volt, 125 AH, lead-acid battery can supply 20 amps of current for a period of: A.

5.25 hours

B.

2.5 hours

C.

6.25 hours

D.

3.75 hours

A single grounding electrode consisting of a pipe has a resistance of 30 ohms, what must be done to allow this circumstance to comply with the NEC®? A.

It must be driven below 60 ft of earth

B.

It must be augmented by multiple additional electrodes

C.

An uncoated and ungalvanized steel electrode must be added

D.

It must be augmented by one additional electrode

In an underground electrical duct, four conduits are irregularly spaced as shown below:

The geometric mean distance of the conductor spacing is most nearly: A. 1 ft 9 in

Page 459

B.

15 in

C.

3 ft

D.

14 cm

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N otes / W orkspace This is a very easy problem. The hardest part about this problem is getting over the thought, “ Could it really be this easy?” Yes, it’s that easy. Not all PE questions are highly difficult. You need to have the confidence in the subject matter to not turn easy problems into hard ones. The answer is: 125 7 20 = 6.25 hours. (C)

NEC® Section 250.53 (A)(2) [see also Exception note for (A)(2)] states it must be augmented by one additional electrode if over 25 ohms.

To find this geometric mean distance, you must make sure to include all the possible distances between any pair of conduits. In this problem with 4 conduits, there are six possible distances. The answer is (A), see below: V l6 x 1 4 x 1 2 x 3 0 x 2 6 x 4 2 - 2 1 .1 " - T 9 "

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P a g e 460

M orning S olutions 18

19

20

Page 461

A voltage of 131 Z. 52° kV is applied across a complex load measuring 125 + /87 MVA. Calculate the current across the load. A.

6 4 0 + /1 9 6 A

B.

6 4 0 -/1 9 6 A

C.

1,110+ y'339 A

D.

1,110 - /339 A

The 3-phase distribution system between the Station and the Load has phase conductors made of aluminum inside steel conduit. According to the 2017 National Electrical Code®, provided these conductors are 750 kcmil, the phase voltage at the load is most nearly:

B.

565 V

C.

575V

D.

585V

Since the ‘80s, the proliferation of electronic equipment has brought the issue of harmonics to the forefront of the electrical engineering industry. Generally speaking, which of the following is the primary concern regarding harmonics? A.

Unwanted current phase shifts

B.

Poor power quality

C.

Unbalanced loads

D.

Increased susceptibility to lightning strikes COPYRIGHT 2017 BY COMPLEX IMAGINARY, LLC

Notes / W orkspace The voltage and the load are given as complex values. This should tell you that you need to use the complex conjugate of the current for this calculation. S = VI*

r

S

(1 5 2 x 1 0 ^ 3 5 °

V

(l3 1 x 1 0 3)/L52°

= 1160Z. -1 7 ° A But that is the conjugate of the current, therefore: I = 1160Z17°A = 1110+ y339 A

This is a voltage drop problem using Table 9 of the NEC®. Find the appropriate impedance values using the columns there. The answer is (A), see below: From NEC® Table 9: X, = 0.048 Q \ L > p e r 1000 R, = 0.031 Q / Vo = 600 V I = 350 A pf = 0.83, lagging) 0 = cos"1(p f)= 4 3 .9 ° I = 350Z. - 33.9° A d = 2000'

= 0.062 + 0.096/ Q Vi = Vo - Z I = 563 V The main problem with harmonics is the decrease in power quality. This is due to the distortion of the voltage waveforms, which results in all sorts of problems: overheating the building wiring and transformers, nuisance tripping, overheated transformer units, and random end-user equipment failure. The answer is (B).

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Pag e 462

M orning S olutions 21

22

A wye-connected AC generator on a 3-phase system has a full-load current output of 452 A. Provided the generator’s rated output is 1,131 kW, and has a 0.90 lagging power factor, the generator’s phase voltage rating is most nearly: A.

0.78 kV

B.

0.93 kV

C.

2.78 kV

D.

4.81 kV

When concerned with overcurrent protection, the 2017 National Electrical C ode® does not permit exceeding the following size (A) overcurrent device in the protection of 14 AWG copper. A.

7A

B.

30 A 20 A

D.

23

15 A

In the single-phase circuit shown below, when the switch is in position 1, the current ^ is 77 4.-14° A. What is the circuit current when the switch changes to position 2, assuming the circuit reaches steady-state conditions?

V = 250 V

B.

278 A 318 A

D.

P a g e 463

481 A

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N otes / W orkspace This generator is wye-connected and the problem is asking for the phase voltage. That means it would be convenient to solve this using the power-per-phase formula The answer is (B), see the calculations below:

P - 3 V l( p f ) 1131x103 = 3V (452)(0.90) V =927 V

NEC® 240.4.D(3) states that the protection of a 14 AWG copper wire shall not exceed 15 amps.

First, calculate the impedance using the switch in position 1. Then, since you now know the impedances for both elements, calculate the current when the switch is in position 2. The answer is (C), see below: V - 250 V ^ -7 7 4 .- 1 4 ° A Z = — = 3.2 + ;0.79 Q X, = 0.79 Q V 1 = — = 318 A 2 /x,

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PAGE 464

M orning S olutions

P a g e 465

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N otes / W orkspace You should be able to detect that this is a code lookup question because it’s asking about sizing of equipment bonding jumpers (which are used for grounding). The answer can be found in NEC® Table 250.122, which is a commonly used table by engineers, and should probably be bookmarked. The table indicates that the size is based on the setting of the overcurrent device (the circuit breaker in this case), which is 200 A (NOT 225 A). So the answer is 6 AWG, (B).

The voltage across the load has been rectified from the incoming AC voltage. For ripple factor to be zero the waveform has to be a constant value, which would be an ideal DC current. In this situation the waveform across the load is not constant which is why the rms and average voltage are different. Use the ripple factor formula to find the answer, which is (C). See below: Vavg = 365.7 V Vrms = 462.4 V = 77.4%

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P a g e 466

M orning S olutions 26

27

28

P a g e 467

Calculate the current at the load Z.

A.

4 1 0 /1 -1 4 2 ° A

B.

3 1 3 /1 9 9 ° A

C.

1 3 6 Z -1 4 2 0 A

D.

136/182° A

A continuous duty motor is connected to a 3-phase circuit. This m otor’s full-load current is 12 A and it employs an integrated thermal overcurrent protection device. According to the 2017 National Electrical Code®, the ultimate trip current for the protector is most nearly: A.

10.5 A

B.

16.8 A

C.

18.7 A

D.

20.4 A

A relay protection diagram is shown below. The dashed lines represent Zone 1 protection regions. If a fault occurs at point F and the primary protection works successfully, which of the breakers will trip? F

A.

1 ,2

B.

1,2,5,6

C.

1,2,3,4

D.

1,2,3,5,6 COPYRIGHT 2017 BY COMPLEX IMAGINARY, LLC

N otes / W orkspace This is a voltage drop problem with complex values. The problem gives the starting and ending voltage, and the difference between the two gives the voltage drop due to the line impedance shown. Use this difference and Ohm ’s Law to calculate the line current. The answer is (B) see below: AV = IR

4740 Z 9 0 ° - 4 1 0 0 Z 0 0 — ----- ------------------------ = 313.36Z98.860 A 2 0 Z 32°

The answer for this code lookup question can be found in section 430.32(A)(2). Since the m otor’s full-load current is 12 A, the maximum current that can cause the unit to trip is 156% • 12 = 18.7 A. Answer is (C).

In this relay protection problem, the fault occurs in two overlapping primary protection zones. This means that all the breakers in those tw o zones will trip. One of the zones will trip breakers 1,2,5,6 and the other zone will trip breaker 3. So the answer is (D). F

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PAGE 468

M orning Solutions 29

The power absorbed by a complex load with an impedance of 5 L 20° Q when 66 V is applied is most nearly: ... ■■

30

31

Pa g e 469

»

A.

819 W

B.

298 W

C.

871 W

D.

503 W

A delta-connected 460 V, squirrel-cage motor is operating with a lagging power factor of 0.70. To correct this to a 0.95 lagging power factor, a capacitor bank was added. After the application of this power factor correction, the motor circuit phase current is 96 A. What is the real power used by the motor? A.

42.0 kW

B.

72.7 kW

C.

76.5 kW

D.

125.9 kW

According to the National Electrical Code®, if a nontime-delay fuse that protects a motor branch-circuit does not exceed 600 amperes it: A.

Must be replaced by a time-delay Class CC fuse

B.

Must not exceed 400 percent of the full-load current

C.

Must not be permitted to increase

D.

Must have a resistance of 1,200 Ohms

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N otes / W orkspace First calculate the apparent power for the impedance. But, the problem asks for the power absorbed which indicates only the real power component. Convert the apparent power to its rectangular form, and the real component is the answer. Z = 5 Z. 20°Q V = 66 V

V2 662 S = — ------------- 871.2Z - 20° VA Z 5Z.200 = 8 1 9 -/2 9 8 VA Power absorbed is the real component: P = 819 W

In this power factor correction problem, it is not necessary to use the original power factor, just use the corrected one. Take the given voltage, current, and corrected power factor to calculate the power output of the motor. Since the motor is delta-connected and the phase current is given, the total power is going to be three times the power-per-phase. P = 3VI, (p f) = 3 (4 6 0 )(9 6 )(0 .9 5 ) = 125.9 kW

This question is only answerable by looking up NEC® code. In section 430.52(C)(1)(ex. 2)(a), you’ll find the answer to be (B).

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P a g e 470

M orning S olutions 32

33

34

The total short circuit current at a secondary terminal on a transformer is not only dictated by the transformer impedance (Z), but it is also m ost sig n ific a n tly determined by which of the following factors? A.

AC vs DC generator characteristics.

B.

Mutual inductance of transformer’s windings.

C.

Source contributions at transformer primary terminal.

D.

Power factor of the load.

If you decrease the low side winding of a 138 kV/12.47 kV transformer by 10%, this would affect the turns ratio in what way? A.

The turns ratio remains 11:1

B.

The turns ratio becomes 124.2 kV/12.47 kV

C.

The turns ratio becomes 138 kV/11.22 kV

D.

The high side will automatically decrease by 10% as well

What is the turns ratio for the potential transformer below (n ^n j?

A /W 480 V

0

m rO Cn2

240 V 1

Pa g e 471

A.

2:1

B.

1:2

C.

4:1

D.

1:4

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N otes / W orkspace The whole idea of a short-circuit is that something went wrong with the normal power flow controls on a circuit. So, like a water dam breaking, there is going to be an inrush of a lot of current that is present in the system. The most significant contribution to this inrush of power would obviously be the source where the power is generated. The less power available at the source, the less power available for a short circuit anywhere else in the system. The answer is (C).

When looking at transformer high and low sides, it’s important to know which is which (obviously). For this problem, the low side is the second number and the high side is the first number of the turns ratio. (D) cannot be correct because changes can be made to one side of the ratio which don’t automatically (as a mere consequence of function) affect the other.

The high-side voltage form the generator is 480 V and the load-side voltage is 240 V. The ratio is 2:1. More detailed calculations are provided below: V

V

n p n

480

ni

P

240 “ n2 n2

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=2

PAGE 472

M orning S olutions 35

36

37

P a g e 473

The reflected high-side impedance of a transformer is 255 Q. What is the impedance on the low-side if the turns ratio of the transformer is 350:25 ? A.

1.3 0

B.

100

C.

180

D.

25 0

A 60-Hz induction has a rotor speed of 1,710 rpm under a 5% slip. How many poles does this motor have? A.

3

B.

2

D.

5

Which of the following units is a measure of luminous intensity? A.

foot-candle

B.

lumens

C.

candela

D.

watt

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N otes / W orkspace When reflecting impedances across a transformer, it’s important to realize that the impedances are proportional to the SQUARE of the turns-ratio. The answer is (A), see below: 350 a = ----- = 14 25

Z

-2 5 5 L

a2

142

= 1.3 Q

The induction m otor will run at speeds lower than the synchronous speed. You can use the slip percentage to calculate the exact pole numbers. However, the easier way is to understand that the speed will be slightly less than synchronous speed. So whatever synchronous speed for pole pairs that is closest to 1,710 rpm at 60 Hz is the answer. In this case, it’s 1,800 rpm (95% of 1,800 is 1,710). So the answer is 4 poles, choice (C). s rpm = P=

120 f 120-f

120(60)

s

1800

=4

The intensity of a light source is measure in candelas, so the answer is (C). It’s difficult to visualize the differences in these lighting units. A foot-candle is a measure of illumination. A lumen measures the amount of light flowing through an imaginary surface (like a flux). A candela measures luminous intensity. And the watt is not specific to lighting, of course. The most confusing of the choices are foot-candle and candela. “ Foot-candle” can be described as “the illuminance cast on a surface by a one-candela source one foot away.”

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PAGE 474

M orning S olutions 38

The primary difference between a GFCI and an AFCI is: A.

39

Page 475

GFCI is a system; AFCI is a device

B.

GFCI protects people from shocks; AFCI protects against arcing

C.

GFCI uses the Geometric mean; AFCI uses the Average

D.

GFCI is for 240V systems; AFCI is for 120V systems

There is a transformer on a 3-phase system. This transformer has a ferromagnetic core and is rated 900 kVA with a 4 % impedance. The impedance under a 1 MVA base, would most nearly be: A.

3.4%

B.

4.4%

C.

5.4%

D.

5.7%

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N otes / W orkspace GFCI stands for Ground Fault Circuit Interrupter, and it’s a device used expressly to protect people from electrical shocks (commonly found in potentially wet areas like bathrooms and kitchens). AFC1 stands for Arc Fault Circuit Interrupter, and it’s a device dedicated to respond to arc faults and mitigate the risk of fire and damage caused by arcing. So the correct answer is (B).

Sometimes, when per-unit calculations are involved, it is necessary to change the impedance to a different base. This problem is asking to change a transformer’s impedance value to a different base. Use the impedance conversion formula with the original base values being that of the transformer’s ratings with the new base being 1 MVA. Use conversion formula: Z,

0.04

RfV\

0.9

Z. = 0.044 = 4.4%

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PAGE 476

M orning S olutions 40

A 3, wye-connected, sychronous m otor draws 427 A from a 480 V source with a 0.77 lagging power factor. How many kvars would a capacitor bank need to supply to increase the power factor to 0.87?

—

P a g e 477

A.

47 kvar

B.

71 kvar

C.

131 kvar

D.

155 kvar

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N otes / W orkspace For the corrected pf: pf = 0.87 Q2 = Ptan [c o s '1( p f)j First find the real power for the original pf:

= 273 x 103tan [c o s '1(o.87)

pf = 0.77 P= V i v L iL ( p f )' V 3 (4 8 0 )(4 2 7 )(0 .7 7 ) 273 kW

Q 1 = P ta n (o )

= 155 kvar Q1- Q Z = 2 2 6 -1 5 5 = 71 kvar

= P ta n |^cos"1( p f ) j = 2 7 3 x 1 0 3 tan co s"

= 226 kvar

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PAGE 478

A fternoon S olutions 41

A wye-wye, 230 kV, 3-phase transmission line is being stepped down to 200 kV using three autotransformers connected between the neutral and the line. For each phase, if the autotransformer is rated for 500 MVA, and the turns ratio is 5:2 (common:series) what is the apparent power in the common coil? ■

42

»

A.

143 MVA

B.

150 MVA

C.

173 MVA

D.

250 MVA

In the circuit shown below, calcul

A.

38 Z l 54° Q

B.

45 Z l 54° O 51 Z l 68° Q

....C. D.

P a g e 479

53z l 88° Q

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N otes / W orkspace Use the autotransformer formulas to find the apparent power in the common coil. The answer is (A), see below: _ 2

S = 500 MVA

NS + NC

N

_ -| _j___ c_

Nc s S

N

s

500 MVA

H Nc “ 1 + _c Ns

H _5 1+ 2

143 MVA

Since the low side is fixed: Z =49/171° Q Find reflection of Z2:

n

p _

ns 3

IF

__ p_

\1Z 2o |

zp

19 “ \ 151/117° 9

Z

p

361 ~ 151Z170

Since the impedances are in series in the equivalent circuit: Z =49Z71° + 3.8Z17°

Z = 9 (151Z17 P 36 v = 3.8Z170 Q

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= 19.6 + j 47.4 = 51/168° Q

P a g e 480

A fternoon S olutions 43

44

45

A power plant engineer plans on retiring after 30 years of service. At the tim e of his retirement, he has $175,000 in a savings account that pays an annually compounded interest rate of 4%. What is the maximum amount of money the engineer can withdraw each year if he plans for the savings to last him 15 years? A.

$11,667

~B.

$15,740

C.

$18,330

D.

$19,690

A 3-phase wye-connected synchronous generator has a power factor of 0.82. If the phase voltage is 2.77 kV and the line current of 85 A, what is the total apparent power of the generator? A.

235 kVA

B.

334 kVA

C.

579 kVA

D.

706 kVA

Given the 3-phase circuit below, what is the current, l2? transformers)

(T1 and T2 are potential

74^61° Q L

©

Page 481

A.

7.1 A

B.

12.3 A

C.

21.3 A

D.

25.1 A

230 kV

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N otes / W orkspace For engineering economics problems, the most important reference to have is a table of the formulas (or “factors”) for doing calculations based on the financial situation described. The formula for this question is called a “ uniform-series, capital-recovery factor” and the calculations are shown. The answer is (B). P = 175000 i = 0.04 n = 15

( a /P, i%, r ) -----1- M A _ p = 1A = 15740

To find the answer, find the apparent power for each phase, and multiply it by 3 to get the total for the generator. The use of the power factor is not necessary in the calculations because the question asks for apparent power and not real power. The answer is (D), see below: V = 2.77 kV $ I =8 5 A ♦

S = 3 - V(|) I4 = 706.35 kVA

There is some misdirect information in this problem. What it boils down to is a current conversion problem, using the turns ratio of the T2 transformer. The answer is (B), see below: I1 = 82 A

n2

20

ii = l '2 a

l2 = a x l = 12.3 A

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P a g e 482

A fternoon S olutions 46

47

48

Pag e 483

Which of the following factors does NOT affect the voltage drop calculations for feeders? A.

Load Current

B.

Feeder Source Turns Ratio

C.

Feeder Length

D.

Conductor Size

A wye-connected synchronous generator has per-unit reactances of Xd = 1.24, X’ = 0.37, and X” = 0.25. A subtransient analysis is conducted and a subtransient current of 4.8 pu is detected. What is the internal generated phase voltage in pu? Assume all analysis is performed at rated values. A.

1.0 pu

B.

1.2 pu

C.

1.8 pu

D.

6.0 pu

With the zonal cavity method, the room is considered to contain how many vertical zones/cavities?

A.

1

B.

2

D.

4

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N otes / W orkspace The voltage drop formula is (3-phase): VD-

^ R -l-L conductor size

The turns ratio has no affect on voltage drop. (B) is the correct answer. This is a standard theoretical voltage drop question. All of the other choices are directly involved in voltage drop calculations.

In this per-unit problem all the values have already been converted to per-unit and there are no base conversions necessary. So, you can go into the Ohm’s Law steps without any of the preparatory work necessary for per-unit problems. In problems involving transient nomenclature, the subtransient quantities are indicated with the double tick mark (e.g. X”). X V 0.25 = — l4.8 V = 1.2 pu

The zonal cavity method is a widely accepted methodology for lighting design. It provides more accuracy than the older lumen method. The room is divided into 3 cavities as shown below: Ceiling Cavity Luminaire Rane Room Cavity Work Plane Floor Cavity

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P a g e 484

A fternoon S olutions 49

What dictates the highest temperature a given synchronous generator can withstand during its operation? A.

The number of poles in the stator

B.

The starting torque of the machine The insulation class of the machine’s windings

D.

50

51

PAGE 485

The machine’s terminal contact ratings

A 480 V, 3-phase bus is serving a 10 kVA motor with a power factor of 0.72 (lagging). A second 15 kVA squirrel-cage motor is attached to the bus, and the system power factor is now observed to be 0.76, lagging. What is the power factor of the new motor? A.

0.75

B.

0.77

C.

0.79

D.

0.81

A 3-phase, 12.5/7.2 kV, delta-delta transformer is rated 150 MVA. An accident occurred which caused one phase of the transformer to disconnect, leaving just two lines available on the secondary. What is the amount of power that can be delivered in this situation? A. 50 MVA B.

87 MVA

C.

100 MVA

D.

150 MVA

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N otes / W orkspace The answer is (C), excitation losses. In an open-circuit, the secondary terminals are open, so there is no load current present. The test will only show the losses due to starting the transformer. Core losses are considered “ no-load” losses, so they don’t apply to an open-circuit test (that would make it a short-circuit test). Mechanical losses are caused by specific environmental conditions (vibration, screw tension, etc.) which are not measure by the standard open and short-circuit tests for transformers.

This problem is an exercise of complex mathematics. The apparent power of the bus with the two motors attached is the sum of each individual m otor’s apparent power. After adding the two together, use complex mathematics to find out the missing variable, which is the power factor of the second motor: sT= s + s 2 System Motor 1 Motor 2 10 kVA 15 kVA S2 = S T - S , = 1 1 .8 + j9 .3 kVA 10 + 15 = 25 kVA pf = 0.72 0 = cos~1(p f) - cos 1( 0 .72 ) = 43.9° S1 =10Z.43.9° kVA

pf = 0.76

= 15Z.38.20 kVA

0 = cos"1(o.76j = 40.5° ST = 25Z40.50 kVA = 19 + j16.2 kVA

Find pf of S2 pf = cos(0j = cos(38.2°) = 0.79

- 7.2 + j6.9 kVA

This was a really wordy way to describe a transformer in the open-delta configuration. In an opendelta connection, the maximum power available would be about 58% of the normal rated power if the delta was fully connected. So 58% of 150 MVA is 87 MVA, the answer is (B). Where does the 58% come from? If you do the circuit analysis of the delta vs. open delta connection, and once you finish all that Kirchoff business, you will find that the answer is 58%. We w on’t do that whole exercise, but w e’ll start you off with the diagram below. It is probably a good idea to memorize the 58% number for open-delta configurations.

Total Power Available S OD = V 3 V I