Professor De Boer's list of
TEXTBOOK ERRATA
(last update 1/28/2017)
Alexander & Sadiku,
Fundamentals of Electric Circuits, 4th Ed.
ISBN 978-0-07-352955-4, McGraw-Hill, 2009.
(link to errata for the 5th edition)
If you are considering purchasing this textbook and worrying that it is a poor choice due to the length of this list of errata, please don't worry about that. Competing textbooks have about as many errata, but perhaps no list like this. "Better the devil you know than the devil you don't." Professor De Boer likes this book enough to find it worthwhile to publish this list of errata.
First, see the author's list of errata for this textbook.
(The authors' list of errata was once hosted on McGraw-Hill's
textbook companion web pages but was taken down some years ago.
A copy of that file is hosted here for Prof. De Boer's students
in his EGR 220/PHYS 206 course. If you get a login screen, log
in, then back up to this link and click the link again. After
you are logged in the link will work. If you are not a student
in Prof. De Boer's EGR 220 course but would like this file,
please try your favorite web search engine. E.g.search on "Errata 4th edition Alexander Sadiku." There usually are
various copies of it available but the URL's are not stable
enough to be worth linking. Prof. De Boer believes the most
recent version is dated July 17, 2009.)
Secondly, check Professor Reeder's list of errata for this textbook.
The following additional errata have been found to date:When specific corrections are illustrated, additions are in blue text.
Deletions are instrikeout blue text. Commentary is in green text.
Pages 5, About 2/3 down the page
Edit the text to replace "principal" with "base," improve the grammar, and identify the derived unit by changing the paragraph as follows:
there are
ical quantities can be derived. Table 1.1 shows
derived unit (the coulomb) that are related to this text.
"Principal" is the wrong word here. "Principle" was intended by the authors but the standards actually use the word "base." Also there are seven base units, not six. Finally, "The SI units. . ." is a weak way to state the intention of this sentence. Most students would not comprehend the intent. (The SI base unit not mentioned in Table 1.1 is the mole.) For reference see:Page 7, The highlighted text boxes and nearby text on the lower third of the page.
Bureau International des Poids et Mesures
National Institute of Standards and Technology
(Posted 8/17/2012.)
Both of these definitions are overly restrictive. Additionally, the text around them conflates the concept of a constant current versus a time-varying current with the concepts of dc versus ac. Correct the surrounding text and the definitions, starting with the paragraph that begins with, "If the current does not change. . ."
If the current does not change with time,
Current can also be classified as direct or alternating.
A direct current (dc) is a current that never reverses direction.
Direct current could be time-varying. For example, i(t) = –|3cos(377t)| A
is a direct current because it never changes sign, indicating that it never reverses
direction.
A time-varying curer is represented by the symbol i. A common
form of time-varying current is the sinusoidal current or alternating current (ac).
An alternating current (ac) is a current that changes direction from time to time.
Such current is used in your household, to run the air conditioner,
refrigerator, washing machine, and other electric appliances. Figure 1.4
There are various definitions of dc and ac in the literature. The definitions created by these corrections represent the more commonly found definitions. A statement such as, "A rectifier converts ac to dc." is consistent with the definitions given by the corrections above, but not the uncorrected definitions in the textbook. (The "dc" current that flows out of a rectifier is usually not constant with respect to time. The waveform of the "ac" current that flows into a rectifier is usually not sinusoidal.) A direct current does not have to be a constant amount. It may pulsate or otherwise vary with time so long as it does not reverse direction.Page 9, The line above Eq. 1.3
It is philosophically satisfing that we should be able to classify a current as either dc or ac. The definitions in the textbook allow for currents that are neither dc or ac, for example, a triangle wave that oscillates between 1 A and –1 A. By allowing alternating current to have any waveform and any period or no period (aperiodic) so long as it changes direction from time to time, all currents can then be classified as dc or ac.
(Posted 10/06/2012, updated 10/09/2012)
Change, "from a to b," to "from b to a."
a unit charge from
Voltage is measured between two specific points and it has a defined polarity. It is conventional that the second subscript in a voltage variable is the reference point. In Equation 1.3 the second subscript is b. Thus b is the reference point, or the the starting point where the charge is initially located. See also the comment for the errata on page 23.Page 9, The highlighted text box
For reference see:
Halladay, Resnik, and Walker, Fundamental of Physics, Part 3, 8e, John Wiley & Sons, Inc., 2008, see Chapter 24 and especially Equation 24-6.
(Posted 9/30/2011, explanation updated 8/17/2012.)
Voltage (or potential difference) is the energy required to move a unit
charge
Voltage is defined between two specific points (no circuit element is needed) and it has a defined polarity. For reference see:Page 23. Summary, point 2
Halladay, Resnik, and Walker, Fundamental of Physics, Part 3, 8e, John Wiley & Sons, Inc., 2008, see Chapter 24 and especially Equation 24-6.
(Posted 8/17/2012.)
2. The International System of Units (SI) is the international mea-
surment language, which enables engineeers to communicate their
results. From the
quantities can be derived.
Table 1.1 on page 5 is not wrong, but it is incomplete. It omits one basic SI unit,
the mole. (The coulomb, for measuring electric charge, is the one derived unit in
this table.)
References:
Bureau International des Poids et Mesures
National Institute of Standards and Technology
(Posted 9/30/2011. Updated 8/17/2012.)
Page 23. Summary, point 3
3. Current is the rate of charge flow past a given point in a given direction.
Current has location and direction. From a definitional perspective this is important.
(Posted 9/30/2011, updated 8/17/2012.)
Page 23. Summary, point 4
4. Voltage is the energy required to move 1 C of charge
Also, in the equation that follows this sentence replace "v" with "vab."
Voltage is defined with respect to two points. The pathway, "through an element," is irrelevant. Any path from b to a results in the same voltage.Page 27. Problem 1.34 part (b)
See also the comment for the errata on page 9.
(Posted 9/30/2011, updated 8/17/2012.)
Delete, the phrase, "per hour." Add, "over the total 24 hour period."
(b) the average power
"Power per hour" is a the rate of change of power. It may be used to characterize the ramp rate of an electric power generating station, but it has nothing to do with this problem. Just plain "average power" is the authors' intent. Reference: "Confusion of watts, watt-hours, and watts per hour" section of the Wikipedia article "Watt" (Posted 9/20/2010, updated 8,16,2012.)Page 59. Figure 2.55 part (b)
Near the symbol for the plug an extranious line (creating a short circuit) needs to be removed. A corrected illustration is shown below.
Figure 2.55 (b) Corrected.
(Posted 4/06/2012.)
Page 60. Practice Problem 2.16
Change the problem statement to read as follows:
Refer to Fig. 2.55 and assume there are 10 light bulbs that can be con-
nected in parallel and 10 different light bulbs that can be connected in series
In either case, each light bulb is to operate at
If the voltage at the plug is 110 V for the parallel and series connections,
calculate the current through and the voltage accress each bulb for both cases.
Answer: 364 mA, 110 V (parallel); 3.64 A, 11.0 V (series)
As written, the problem is ambiguous. Most students assume the situation of a familiar high-school physics experiment in which a few light bulbs are connected in parallel and then the same light bulbs are reconnected in series. This is not the author's intent for this practice problem.Page 75. Problem 2.61, the last line
(Posted 4/06/2012.)
An "I" is missing. Change the last line of the problem statement to read as follows:
such that I lies within the range I = 1.2 A ±5 percent.
(Posted 10/29/2013.)
Page 109. Figure 3.40(b)
At the dependent current source, change "bIB," to "βIB"
(Posted 9/24/2010, updated 9/14/2011)
Page 119. Problem 3.34
As printed, what is being requested is not precisely clear, especially for students who are learning the concept of a planar circuit. Change the problem statement so that it reads as follows:
For each
If it is planar
(Posted 10/01/2014.)
Page 164. Problem 4.20
Replace the problem with the following:
Without the changes shown there is no reasonable way to apply the concept of equivalency in order to reduce the circuit in a meaningful way. The changes to the figure are the addition of the dashed box, and the addition of the terminals labeled "a" and "b."Page 173. Problem 4.93, in Figure 4.149
(posted 3/19/2012)
At the dependent current source, change "bix," to "βix" (Posted 9/24/2010.)
Page 180. About two inches from the top of the page
In the line that starts, "characteristics of the. . ." change the word "characteristics" to "properties." The line then reads as follows:
properties
(Posted 10/29/2013.)
Page 218. Equations 6.5 and 6.6.
*Change the variable of integration from t to λ. Do not change the limit. This is the first of a number of similar errors in this textbook. See the note marked by an asterisk below on this list of errata. In this case, the correct equations are:
(Posted 10/20/2010.)
Page 249. Problem 6.67.
An initial condidtion is needed to solve this problem. Add it to the end of the problem statement. The problem is then as follows:
6.67 An op amp integrator has R = 100 kΩ and C =
0.01 μF. If the input voltage is vi = 10sin50t mV,
optain the output voltage. Assume that at t = 0
the output voltage is zero.
Also see the errata on the answer to this problem.
(Alternatively, one could pick a non-zero initial condition. For example, if the initial condition is chosen as 200 mV then the correct answer is 200cos(50t) mV. However an initial condition of zero is realistic and more instructive for studnets.)(Posted 9/26/2012.)
Page 270. Practice Problem 7.6
Add units to the answers
10[u(t) – 2u(t – 2) + u(t – 4)] A, 10[r(t) – 2r(t – 2) +
r(t – 4)] A·s
(Posted 10/29/2013.)
Page 321. Five lines below Equation 8.11
second (rad/s); and α is the neper frequency
Also see related errata on pages 323, 361, and I-3.
The phrase, neper frequency is preferable for naming the quantity denoted by α in this textbook. In addition to the definition of damping factor implied in this textbook, the phrase damping factor has a plurality of other definitions applying in other contexts. For three unique examples see here and here (in "Variations and hybrid methods" section) and this textbook (See page 506 of the textbook). Furthermore, Prof. De Boer has not so far found any place where the phrase damping factor is treated synonymously with neper frequency except in undergraduate circuits and linear systems textbooks or World Wide Web pages intended for that same audience. This limited context is too narrow to justify introducing the phrase damping factor here.Page 323. The lines below Equation 8.22b
A nearly similar term, damping ratio, (not "factor") is uniformly well defined in many contexts, including RLC circuits, but the neper frequency (α in our textbook) and the damping ratio are not the same thing. Usually the damping ratio is represented by the variable ζ (zeta). Using our textbook's notation, the damping ratio would be defined as ζ = α/ωo. Since most engineering students will encounter the term damping ratio in their careers, the confusingly similar name, damping factor (also known as damping attenuation factor, attenuation factor, and damping coefficient, etc.) should not be introduced in the context of RLC circuits. The name neper frequency for α is more memorable, more universal in the literature, and thus more helpful.
(Posted 10/28/2010, updated 9/28/2011)
and also about ten lines from the bottom of the page
On the line below Equation 8.22b change the phrase,
"damping frequency" to "damped frequency"
where j = √ –1 and ωd = √ ωo2 – α2, which is called the damped
Also, about ten lines from the bottom of the page, change
"The damping factor. . ." to "The neper frequency. . ."
. . .due to the presence of resistance R. The
α determines the rate at which. . .
Also see related errata on pages 321, 361, and I-3.
The name "damping frequency" is incorrect—a grammatic error. "Damped frequency" is one of several names commonly used for this quantity.Page 327. Fifth line from the bottom of the pageReferences:(Posted 10/28/2010, updated 8/17/2012)
Katsuhiko Ogata, Modern Control Engineering, fifth edition,
Prentice Hall, 2010, page 166.
Norman S. Nise, Control Systems Engineering, fourth edition,
Wiley, 2004, page 197.
Richard C. Dorf, Modern Control Engineering, twelfth edition,
Prentice Hall, 2011, Page 567.
James W. Nielson, Electric Circuits, fourth edition,
Addison Wesley, 1994, page 332.
Wikipedia article on "Damping"
Voltage is confused for current. Change the line to read as follows:
iR = v/R and the capacitor current
(Posted 10/29/2013.)
Page 350. About 2/3 down the page
Change these lines:
Eq. (8.11). We must keep in mind that the
ited to planar circuits. Nonplanar circuits have no duals, as they can-
not be described by a system of mesh equations.
method described here for
finding a dual is limited to planar circuits. Finding a dual for a
nonplanar circuit is beyond the scope of this textbook because nonplanar
circuits cannot be described by a system of mesh equations.
Methods for finding a dual for a non-planar circuit can be found in the literature. For one example, A. Bloch, "On Methods for the Construction of Networks Dual to Non-Planar Networks," Proceedings of the Physical Society, Volume 58, Number 6, November 1, 1946.(Posted 11/01/2010.)
The basis for duality is found in the duality of electric and magnetic fields implied by the Lorentz transformation. The Lorentz transformation has its roots in the theory of special relativity. Thus the theory of duality found in electric circuits is not restricted to planar circuits, only the method shown in this textbook is so restricted. However the utility of the method shown in this textbook merits its mention.
Page 360. Problem 8.14
Add this sentence to the problem statement:
Assume the voltage across the capacitor,
vc(t) = 0 for all t < 0.
(Posted 10/20/2010, updated 10/27/2010.)
Page 361. Problem 8.23
Change "damping factor" to "neper frequency."
Also see related errata on pages 321, 323, and I-3.
(Posted 11/03/2010.)
Page 363. Problem 8.45, Figure 8.92
The Figure is inconsistent with the given assumptions. In particular, i(0) would be zero based on Figure 8.45 as it is printed. To correct this error, add a 1 A source as shown below and change "4u(t) A" to "3u(t) A."
(An interesting variation on the above problem is to change the value of R from 2 Ω to 1 Ω. In this case, the answers are v(t) = 6e-tsin(t) V andPage 394. Practice Problem 9.10
i(t) = 4 – 3e-t[cos(t) + sin(t)] A
(Posted 10/29/2010, updated 11/05/2010, 11/11/2011, and 10/26/2012.)
The answer needs a unit.
Answer (129.52 – j295) Ω
(Posted 10/29/2013, updated 1/28/2017)
Page 405. Problem 9.29
Add, "Given that v(0) = 2cos(155°) V" to the beginning of the problem statement. The problem statement then reads as follows:
9.29 Given that v(0) = 2cos(155°) V,
instantaneous voltage across a 2-μF capacitor
when the current through it is
i = 4sin(106t + 25°) A?
(Posted 11/10/2010.)
Page 405. Problem 9.35
Add "the steady-state" between the word "Find" and "current."
The problem then reads:
9.35 Find the steady-state current i in . . .
(Posted 11/10/2010.)
Page 464. Seventh line from the top
Replace "10.58" with "105.8." The line then reads as follows:
(Posted 10/29/2013.)
Page 471. Definition of Power Factor
Change the definition as follows
The power factor is the ratio of the real power to the apparent power.
A definition must be the most broadly useful statement possible, in this case to include non-sinusoidal voltages and currents too. One can then note that for sinusoidal steady state conditions, Equations 11.36 and 11.39 imply that the power factor can be found from the cosine of the angle between voltage and current or the cosine of the angle of the load impedance.
(Posted 8/29/2013.)
Page 475. Equation 11.51, 2nd and 3rd lines
The second and third lines of Equation 11.51 should show Vrms and Irms in italics rather than boldface. Change them to read as follows:
=
Apparent Power = S = |S| =
Alternatively, use more magnitude symbols as follows:
= |Vrms||Irms| /θv – θi
Apparent Power = S = |S| = |Vrms||Irms| = √P2 + Q2
Apparent power is always a non-negative real number—a magnitude.
(Thanks to Adam Van Hal for calling this one. Posted 11/21/2014)
Page 490. Review Question 11. 3
The abberviation "rms" is missing from the second line of the problem.
Change it to read as follows:
11.3 The amplitude of the voltage available in the 60 Hz,
120 V rms power outlet in your home is:
(a) 110 V (b) 120 V
(c) 170 V (d) 210 V
This is the first of a number of similar omissions that start at Problem 11.51 and continue through the end of the Chapter. Also see the note marked by a double asterisk at the end of this errata for more examples of this type of error.
(Posted 11/20/2012)
Page 492. Problem 11.13
Part (a) of the question is missing. Entirely replace the problem statement with this statement:
11.13 The Thevenin impedance of a source is
ZTh = (120 +j60) Ω, while the peak Thevenin
voltage is VTh = (110 + j0) V.
(a.) What load impedance will draw the maximum
power possible from this source?
(b.) Determine the maximum avalable power from
the source.
(Posted 11/17/2010.)
Page 495. Problem 11.48
Two minus signs are missing.
In part (a) change "Q = 150 VAR," to "Q = –150 VAR."
In part (b) change "Q = 2000 VAR," to "Q = –2000 VAR."
Also, the words "capacitive" and "inductive" should be deleted from parts (a) and (c). The problem then reads as:
11.48 Determine the complex power for the following
cases:
(a) P = 269 W, Q = –150 VAR
(b) Q = –2000 VAR, pf = 0.9 (leading)
(c) S = 600 VA, Q = 450 VAR
(d) Vrms = 220 V, P = 1 kW
| Z | = 40 Ω (inductive)
The words "capacitive" and "inductive" can be applied to an impedance, especially an impedance magnitude as in part (d). They can also be applied to the power factor in lieu of "leading" and "lagging" respectively, but these words are not correct with respect to reactive power. "Capacitively generated" reactive power is always negative. Similarly, "Inductively generated" reactive power is always positive. Stating a positive reactive power and calling it "capacitive" is confusingly incorrect. Similarly, the terms "leading" and "lagging" should not be used with respect to a (numerically) quantified reactive power either in lieu of or to augment an algebraic sign.Page 496. Problem 11.51, Figure 11.70 and most
(Posted 3/19/2012, explanation updated 11/21/2012)
following problems and figures in this chapter
Many "rms" labels are missing from units. Voltages and currents from problem 11.51 to the end of this chapter are RMS unless the context shows it is obviously not RMS. For example, in Figure 11.70 the voltage source should be "120/45° V rms." This is one of the first of a number of similar omissions. See the note marked by a double asterisk at the end of this errata for more examples of this type of error.
(Posted 11/20/2012)
Page 499. Problem 11.80, Figure 11.95
The label for the impedance,
Boldface indicates a complex quantity, but only the magnitude is given here. (Posted 11/20/2012)Page 500. Problem 11.84
Replace the word "energy" with "electricity" in part (a).
(a) Determine the annual cost of
The cost of "energy" would exclude the demand cost, but that is not the author's intention.Page 506. Figure 12.5
(Posted 11/20/2012)
The labels at the top of the figure, Van, Vbn, and Vcn should be replaced by van(t), vbn(t), and vcn(t) respectively.
This is a time-domain figure. Capital letters like Van with no "(t)" time dependencies are wrong because they indicate phasors. This is incongruent with the domain of the figure and the notation elsewhere in this textbook. (Posted 3/19/2012)Page 511. First line of Example 12.2
The word "phasor" is missing from the first line of the example.
Calculate the phasor line currents in the three-wire Y-Y system of Fig 12.13.
If a "phasor" amount is not specifically requested then by convention the answer should correspond to that which a VOM or DMM would read, which is only the magnitude of the phasor amount. In this case the complete phasor amount is the desired answer, thus it should be specifically requested.Page 512. Sixth line of Practice Problem 12.2
(Posted 9/12/2013.)
The word "phasor" is missing in two places on the last line of the problem statement.
120/30° V, find: (a) the phasor line voltages, (b) the phasor line currents.
If a "phasor" amount is not specifically requested then by convention the answer should correspond to that which a VOM or DMM would read, which is only the magnitude of the phasor amount. In this case the complete phasor amount is the desired answer, thus it should be specifically requested.Page 512. In Section 12.4, near the middle of the page
(Posted 9/12/2013.)
In the line that starts with, "positive sequence. . ." change the word "phase" to the phrase, "line-to-neutral." The line then reads as follows:
positive sequence, the line-to-neutral
In a delta connection the phrase "phase voltage" referrs to the line-to-line voltage since the phases are each connected from line-to-line. However in a wye connection the phrase "phase voltage" refers to the line-to-neutral voltage since the phases are each connected to neutral. Since the context here is a wye-delta connection, the phrase "phase voltage" is ambiguous. (Posted 10/29/2013.)Page 512. Equation 12.20, last line
The angle of Vca should be positive. Delete the negative sign.
Vca = √3Vp /150° = VCA
(Posted 10/29/2013.)Page 513. Figure 12.15
The arrow indicating the 30° angle between IAB and Ia should point in the clockwise direction, the same as the other two such arrows in this figure.
(Posted 12/01/2010.)
Page 513. The last line on the page, in Example 12.3
The word "phasor" is missing in two places in the last line of the problem statement.
culate the phasor phase and phasor line currents.
If a "phasor" amount is not specifically requested then by convention the answer should correspond to that which a VOM or DMM would read, which is only the magnitude of the phasor amount. In this case the complete phasor amount is the desired answer, thus it should be specifically requested.Page 514. Third line of Practice Problem 12.3
(Posted 9/12/2013.)
The word "phasor" is missing in two places on the last line of the problem statement.
20/40° Ω, find the phasor phase and phasor line currents. Assume the abc sequence.
If a "phasor" amount is not specifically requested then by convention the answer should correspond to that which a VOM or DMM would read, which is only the magnitude of the phasor amount. In this case the complete phasor amount is the desired answer, thus it should be specifically requested.Page 515. Last two lines on the page, in Example 12.4
(Posted 9/12/2013.)
The word "phasor" is missing in two places on the last lines of the problem statement.
Vab = 330/0° V. Calculate the phasor phase currents of the load and the phasor
line currents.
If a "phasor" amount is not specifically requested then by convention the answer should correspond to that which a VOM or DMM would read, which is only the magnitude of the phasor amount. In this case the complete phasor amount is the desired answer, thus it should be specifically requested.Page 519. Second line of Practice Problem 12.5. (About 2/3 down the page)
(Posted 9/12/2013.)
The word "phasor" is missing on the last line of the problem statement.
Calculate the phasor line currents.
If a "phasor" amount is not specifically requested then by convention the answer should correspond to that which a VOM or DMM would read, which is only the magnitude of the phasor amount. In this case the complete phasor amount is the desired answer, thus it should be specifically requested.Page 526. The second line of Example 12.9
(Posted 9/12/2013.)
The word "phasor" is missing in two places in the second line of the problem statement.
and the abc sequence. Caculate the phasor line currents and the phasor neutral cur-
If a "phasor" amount is not specifically requested then by convention the answer should correspond to that which a VOM or DMM would read, which is only the magnitude of the phasor amount. In this case the complete phasor amount is the desired answer, thus it should be specifically requested.Page 526. The second line of Practice Problem 12.9
(Posted 9/12/2013.)
The word "phasor" is missing in the second line of the problem statement.
voltages of 240 V in the postitive sequence. Find the phasor line currents. Take
If a "phasor" amount is not specifically requested then by convention the answer should correspond to that which a VOM or DMM would read, which is only the magnitude of the phasor amount. In this case the complete phasor amount is the desired answer, thus it should be specifically requested.Page 527. The first line on the page, in Example 12.10
(Posted 9/12/2013.)
The word "phasor" is missing in the first line of the problem statement.
For the unbalanced circuit in Fig. 12.25, find: (a) the phasor line currents, (b) the
If a "phasor" amount is not specifically requested then by convention the answer should correspond to that which a VOM or DMM would read, which is only the magnitude of the phasor amount. In this case the complete phasor amount is the desired answer, thus it should be specifically requested.Page 528. The first line of Practice Problem 12.10
(Posted 9/12/2013.)
The word "phasor" is missing in the first line of the problem statement.
Find the phasor line currents. . .
If a "phasor" amount is not specifically requested then by convention the answer should correspond to that which a VOM or DMM would read, which is only the magnitude of the phasor amount. In this case the complete phasor amount is the desired answer, thus it should be specifically requested.Pages 534–535 First paragraph in Sec. 12.10.
(Posted 9/12/2013.)
This paragraph contains poor grammar and some substantive errors. It also touches on some complicated issues. For a more complete discussion of this paragraph see the entry in the "Extensions and Clarifications" section of this document. Only corrections are shown here.
Both the wye and delta connections have important practical appl-
ications. The wye source connection is used for long distance trans-
mission of electric power, where resistive losses (I2R) should be
______(page break)_______
voltage that is √3 greater than the delta connection, hence for the
same power, the line current is √3 smaller. The
nection is also used when three single-phase circuits are desired from a
three-phase source. A single-phase load may be connected
from any individual phase line of a wye to the neutral
of a wye. This conversion from three-phase to single-phase
is required in
wiring, because
appliances use single-phase power. Three-phase power is used
industrial wiring
tions, it is immaterial whether the load is wye- or delta-connected.
For example, both connections are satisfactory with induction motors.
in a wye for 440 V so that one line of motors can be readily adapted
to two different voltages.
undesirable due to the potential of having large circulating currents
that can overheat or even destroy equipment.
References:Page 537. The paragraph just above Example 12.13
Roland E. Thomas, Albert J. Rosa, The Analysis and Design
of Linear Circuits, pp. 811–812, Prentice Hall, 1994.
Richard C. Dorf, James A. Svoboda, Introduction to Electric
Circuits, 8th Ed., p. 571, Wiley, 2010.
Hadi Saadat, Power Systems Analysis 2nd ed., page 31,
McGraw-Hill, 2002.
IEEE Recommended Practice for Electric Power Distribution
for Industrial Plants, 2nd ed., pp. 437-440, IEEE, New
York, 1986.
IEEE Recommended Practice for Electric Power Systems in
Commercial Buildings , pp. 127-130, IEEE, New York, 1983.
E. Lakervi, E.J. Holmes, Electricity Distribution Network
Design, pp. 110–112, Peter Peregrinus Ltd., London, 1989.
Eng-tips discussion forums, "advantages and disadvantages. . ."
Wikipedia, "Three-phase electric power."
(Posted 3/19/2012. updated 8/17/2012)
The paragraph does not comprehensively cover the topic. Problem 12.71 is an example of a situation not properly covered by the paragraph as printed. Correct the paragraph as follows. . .
Although these results were derived from a balanced wye-connected load, they are equally valid for a balanced delta-connected load. The two-wattmeter method (Eq. 12.67) can also be used to find the total power of an unbalanced load. However The two-wattmeter method (Eq. 12.69) does not work for calculating the reactive power of an unbalanced load.
See also the Alternative development of Equation 12.67.
(Posted 12/12/2012.)
Page 541. Figure 12.38
On the schematic, the middle conductor leading upwards from node "W" to the label "To other houses" should have an arch in it where it crosses over node "B." Similarly, the conductor from "R" leading upward, "To other houses" should have two arches in it to indicate no connection to nodes "W" and "B."
(For more details on schmatic style conventions, see the related article on Circuit Diagrams in Wikipedia. Our textbook uses "non-CAD" styles as defined in Wikipedia.)Page 544. Problem 12.7(Posted 12/12/2012, updated 11/20/2014.)
The word "phasor" is missing.
Obtain the phasor line currents in the. . .
If a "phasor" amount is not specifically requested then by convention the answer should correspond to that which a VOM or DMM would read, which is only the magnitude of the phasor amount. In this case the complete phasor amount is the desired answer, thus it should be specifically requested.Page 544. Problem 12.8
(Posted 12/01/2010.)
The word "phasor" is missing from the second-to-last line of the problem statement.
. . .10 + j14 Ω. Calculate the phasor line currents and the. . .
If a "phasor" amount is not specifically requested then by convention the answer should correspond to that which a VOM or DMM would read, which is only the magnitude of the phasor amount. In this case the complete phasor amount is the desired answer, thus it should be specifically requested.Page 544. Problem 12.9
(Posted 9/12/2013.)
The word "phasor" is missing from the last line of the problem statement.
. . .the phasor line currents and neutral current.
If a "phasor" amount is not specifically requested then by convention the answer should correspond to that which a VOM or DMM would read, which is only the magnitude of the phasor amount. In this case the complete phasor amount is the desired answer, thus it should be specifically requested.Page 545. Problem 12.10
(Posted 9/12/2013.)
The word "phasor" is missing from the first line of the problem statement.
12.10 For the circuit in Fig. 12.43, determine the phasor current in
If a "phasor" amount is not specifically requested then by convention the answer should correspond to that which a VOM or DMM would read, which is only the magnitude of the phasor amount. In this case the complete phasor amount is the desired answer, thus it should be specifically requested.Page 545. Problem 12.11
(Posted 9/12/2013.)
In the last line of the problem statement the variables should be in italics, not boldface.
"voltage
Boldface indicates a phasor quantity but the authors intent is to ask for only the magnitude, since this is all that a DMM can measure.Page 546. Problem 12.14, Figure 12.47
(Posted 11/20/2012.)
The polarity of the voltage source having angle –120° is backwards in the textbook. The negative end should connect to the negative ends of the other two voltage sources. Additionally, there should be no connection between nodes b and C. A corrected version of the figure is shown below.
One could solve the problem exactly as it is given and claim there is no typeographic error, but then the circuit is unbalanced. This would be a silly situation. A balanced source and load makes the problem representative of a normal three-phase situation and worthy of analysis. (This textbook has adopted the convention of showing an arch to represent wires that cross without connecting. The arch is missing in the textbook's version of the figure, which is quite confusing for students. For more details on schmatic style conventions, see the related article on Circuit Diagrams in Wikipedia. Our textbook uses "non-CAD" styles as defined in Wikipedia.)Page 546. Problem 12.16, part (a)
(Posted 11/29/2012, updated 12/12/2012, 11/20/2014.)
The word "phasor" is missing.
(a) Determine the three phasor line currents. . .
If a "phasor" amount is not specifically requested then by convention the answer should correspond to that which a VOM or DMM would read, which is only the magnitude of the phasor amount. In this case the complete phasor amount is the desired answer, thus it should be specifically requested.Page 548. Problem 12.36(c)
(Posted 9/12/2013.)
Insert the phrase "line-to-neutral."
(c) the line-to-neutral voltage at the sending end.
In three-phase systems it is conventional to assume any voltage is from line-to-line unless otherwise specified.Page 551. Problem 12.66
(Posted 12/02/2010.)
More information needs to be given in order to get the author's intended answers. Change the problem statement to read as below:
12.66 A three-phase, four-wire system operating with aPage 551. Problem 12.67
208-V line voltage is shown in Figure 12.71. The
source voltages are balanced. The voltage of phase 1
is the reference phasor. The phase sequence is 1, 2, 3.
The power absorbed by the resistive wye-connected
load is measured by the three-wattmeter method.
Calculate:
(a) the line-to-neutral voltageto neutral
(b) the currents I1, I2, I3, and In,
(c) the readings of the wattmeters
(d) the total power absorbed by the load
(Posted 12/02/2010.)
The seventh line of the problem statement mentions "line c" when it should mention "line a."
The corrected line is. . .
connected as follows: 24 kW from line a to the
(Posted 9/24/2013.)
Page 553. Problem 12.72
The reference in the last line of the problem statement should be to "Problem 12.11."
The last line of the problem statement should be. . .
connected load in Problem 12.11
Page 582 Figure 13.43 and Example 13.10.
In Figure 13.43 part (a), change the "4.2 A" label on the current arrow that is about 1 cm above the "12 V" label. Change it to "4 A." Also add dots near the top of each coil (four instances) to show flux polarity. (Voltage labels do not indicate flux polarity. They indicate voltage!)
A corrected illustration is shown below.
In an autotransformer connection correct flux linkage depends importantly on flux polarities. The illustrations need to have dots added to the transformer windings to show flux polarities even though in this case all the "+" marks coincidentally correspond to places where a dot is needed. Also see the elaboration of this figure in the "Extensions and Clarifications" section below.
(Posted 3/06/2012. Thanks to Paul Chojecki for pointing out the numerical errors in the figure. Updated 8/29/2012.)
Page 588 Figure 13.54, text along the right side
In Figure 13.54 in the text along the right-hand side, the lines second from the bottom and seventh from the bottom are interchanged. Correct the text along the right-hand side to read as follows:
K K1
K_Linear
COUPLING=0.3333
L1 – L1
L2 – L2
K K2
K_Linear
COUPLING=0.433
L1 – L1
L2 – L3
K K3
K_Linear
COUPLING=0.5774
L1 – L2
L2 – L3
Make corresponding changes in the bold-face text above Figure 13.54.
(In the lines third and seventh up from the Figure.)
(Posted 9/29/2013.)
Page 598 Review Problems 13.5, 13.6, 13.7, 13.9, 13.10
In Reveiw Problem 13.5, the voltage ratio must be a ratio of phasors in order to get the intended answer. (Magnitudes alone do not contain any phase information. Figure 13.70 correctly shows phasor lablels.) Change the last line of the problem statement as follows:
N2/N1 = 10. The ratio
In Reveiw Problem 13.6, the current ratio must be a ratio of phasors in order to get the intended answer. (Magnitudes alone do not contain any phase information. Figure 13.70 correctly shows phasor lablels.) Change the last line of the problem statement as follows:
N2/N1 = 10. The ratio
Change the list of answers at the bottom of the page as follows:
Answers: 13.1b, 13.2a, 13.3b, 13.4b, 13.5d, 13.6b,
13.7
In Review Problem 13.7 when the the value of an AC voltage is requested it is by default the rms magnitude of the voltage that is desired. An rms magnitude is never negative by definition. Thus the correct answer is (b).
In Review Problem 13.9 any step-down transformer can achieve the goal of matching. A linear transformer should have a high coupling coefiecient, but given that, it can work. The lack of isolation involved in an autotransformer does not prevent it from achieving a good match. Thus (b), (d), and (e) are all correct answers. Students should give all the correct answers, not just one or two.
In Reveiw Problem 13.10 all the transformers except the autotransformer offer isolation. The correct answers are (a) and (c). Students should give both correct answers, not just one.
(Posted 10/29/2013.)
Page 636 Example 14.8
In figure 14.27 change the 10 sin(ωt) V voltage source to a 1.25 sin(ωt) mA current source.
In part (c) of the solution, delete "Y = 1/R or" and delete the entire equation that starts, "Io ="
If the circuit is driven with a voltage source the power dissipation will not be frequency dependent. To support Example 14.8, the parallel RLC circuit must be driven by a current source. (By duality, a series RLC circuit would have to be driven by a voltage source if the power dissipation is to be frequency dependent.)
(Posted 10/29/2013, updated 3/20/2015.)
Page A-75 through A- 107. Answer to Problem 1.1 and many others
Reduce the number of significant figures in parts (b) and (c) to three significant figures as follows:
1.1 (a) –0.1038 C, (b) –0.198
(d) –26.08 C
This is a systematic error in this textbook. Many of the answers shown in this appendix have the wrong number of significant digits, usually too many.
(Thanks to Alec W. for noting this. Posted 8/28/2013.)
Page A-77. Answer to Problem 3.91
Change "8.
Page A-78. Answer to Problem 4.23
Change "
(Posted 9/13/2010. This erratum is also noted in the author's list)
Page A-78. Answer to Problem 4.47
Change "4.1
(Posted 9/13/2010)
Page A-78. Answer to Problem 5.5.
Change "0.99999
(Five nines, not six of them.)
(Posted 9/15/2010)
Page A-81. Answer to Problem 6.67.
Change "200cos(50t) mV" to
"200[cos(50t) – 1] mV"
Also see the errata on the problem statement.
(Posted 10/09/2012)
Page A-83. Answer to Problem 7.31(a).
Change
(The answer needs to be correctly rounded and normalized.)
(Posted 10/24/2012, updated 10/26/2102)
Page A-84. Answer to Problem 8.9.
Change "(2 + 10t)e–5t
(There is no information given to suggest any answer for t < 0.)
(Posted 10/24/2012)
Page A-86. Answer to Problem 9.1(d).
Delete "0.3 rad"
(d) 44.48 V
(Posted 11/08/2012)
Page A-86. Answer to Problem 9.9(a).
Change "50.8
(Posted 11/08/2012)
Page A-86. Answer to Problem 9.11(d).
Change "60/
(The angle should be in the fundamental range, –180° < θ ≤ 180°.)
(Posted 11/12/2010)
Page A-86. Answer to Problem 9.23(a).
Change "43.
("43.59," not "43.49" for the leading coefficient.)
(Posted 11/10/2010)
Page A-87. Answer to Problem 10.55(a).
Change "VTh =
Also change "IN = 2.236/
Magnitudes must always be expressed as non-negative amounts.
VTh = –50/30° = (–1)(50/30°) = (1/–180)(50/30°) = 50/–150°
Angles should be in the fundamental range, –180° < θ ≤ 180°.
273.4° – 360° = –86.6°
(Posted 11/12/2010, updated 11/15/2010)
Page A-88. Answer to Problem 11.5.
The answers shown for the resistors are incorrect. Correct the text as follows:
P1Ω =
P3H = P0.25F = 0 W (Posted 9/15/2011)
Page A-88. Answer to Problem 11.9.
Change "
(Posted 9/15/2011)
Page A-89. Answer to Problem 11.69(b).
Change "
(Posted 12/01/2010)
Page A-89. Answer to Problem 11.73 parts (b) and (d).
Phasor amounts were not requested. Answers in terms of rms magnitudes (what an ammeter would read) are expected. Change the answers to
(a) 12.21 kVA, (b) 50.86
(c) 4.083 kVAR, 188.0
(Posted 9/16/2013)
Page A-89. Answer to Problem 11.91.
Change "
(Posted 11/20/2012)
Page A-89. Answer to Problem 12.5.
Change the third answer from "260 cos(ωt
(The angle should be expressed in the fundamental range.)
(Posted 12/01/2010)
Page A-89. Answer to Problem 12.31 part (b).
Change the answer to part (b) from (b)
(Posted 12/17/2013)
Page A-90. Answer to Problem 12.69.
The problem asks for the line current (magnitude only). Delete the phasors from the answer and
add IL. The corrected answer is. . .
Ic = 94.32/57.95° A,
IL = 94.31 A.
(Posted 9/23/2013)
Page A-90. Answer to Problem 12.71 part (b).
The answer for part (b) is wrong. The correct answer is. . .
(b)
Since the load is unbalanced the formula Q = √3(P2 – P1) does not apply.
See also the errata for page 537 and the alternative devevelopment of Equation 12.67.
(Posted 12/13/2013)
Page I-3. About 13 lines from the bottom of the second column
Delete the index entry for "Damping factor."
Change the index entry for "Damping frequency" to "damped frequency."
Damp
Also see related errata on pages 321, 323, and 361.
(Posted 3/19/2012.)
*This textbook contains a systematic error.
In many cases where an expression is integrated with respect to time the same variable, t, is used as the variable of integration and as the upper limit of the integral. One variable should not be used for two different quantities. In these cases some other variable such as λ should be substituted for the variable of integration. As an example, the first instance of this error is in equation 6.5 on page 218. Some other instances of this error include:
Equations 6.5, 6.6, 6.8, 6.15, 6.20, 6.21, 6.23, 6.29, 6.34, 6.35, 6.15.2, 6.15.3, 7.12, 7.34, 7.39, 8.3, 8.28, 15.27, 15.28, 15.32
The error also appears in unlabeled equations on pages, 229 (three instances), 233 (four instances), 241 (four instances), 250 (Problems 6.70 and 6.71).
Some other instances of this problem likely have not yet been itemized in this errata sheet. A few instnaces of correct equations of this type also exist in this text, for example Equation 15.70 and apparently all of the remainder of Chapter 15 is correct on this point.
(The last update to the lists here was on 8/17/2012.)
**Chapter 11 contains a systematic error.
In many cases where a voltage or current is given, the units should be "V rms" or "A rms" but only "V" or "A" is shown. The first instance of this error is on page 490 in Review Question 11.3. Many other instances of this error occur in Problems 11.51 and following. These include:
Figures 11.70, 11.72, 11.73, 11.75, 11.76, 11.78, 11.80, 11.82, 11.83, 11.85, 11.91, 11.95, 11.96, and 11.97
The error also appears in some problem statements. For example in Problem 11.85,
". . .both 120-V rms and 240-V rms, 60 Hz appliances."
Similer errors occur in problems 11.88, 11.89, 11.92, and 11.96. (And just to add interest, a "V" is missing in problem 11.97.) Some other instances of this problem may not have yet been itemized in this errata sheet.
A few instnaces of correct units of this type also exist in this chapter, for example Figures 11.77, 11.79, 11.81, 11.88, and 11.90 are correct on this point.
On a related topic of subscripts for variables, see the clarification of Equation 11.38a.
(Posted 11/20/2012 The last update to the lists here was on 11/20/2012.)
Extensions and Clarifications of text material (Not errata.)
Page 459. More information on Eq. (11.5).
Equation (11.5) can be further developed to show the reactive power component. As with the development of Equation 11.5 in the textbook, this discussion and the related definitions below assume the voltage and current are sinusoidal.
Rearranging the second term in Eq. (11.5) to deliberately itemize the angle (θv – θi), the same as in the first term, gives
p(t) = (1/2)VmIm cos(θv – θi) + (1/2)VmIm cos[(2ωt + 2θv) – (θv – θi)]
Now apply the trigonometric identity cos(α – β) = cos(α)cos(β) + sin(α)sin(β)
to the second term to get
p(t) = (1/2)VmIm cos(θv – θi) +
(1/2)VmIm cos(2ωt + 2θv)cos(θv – θi) +
(1/2)VmIm sin(2ωt + 2θv)sin(θv – θi)
Associate the middle term above with the first term and factor out amounts that are constant with respect to time. The instantaneous power then is:
p(t) = (1/2)VmIm cos(θv – θi)[1 + cos(2ωt + 2θv)] +
(1/2)VmIm sin(θv – θi)sin(2ωt + 2θv)] (11.5b)
The first term above in Eq. (11.5b) pulsates due to the multiplicative term
[1 + cos(2ωt + 2θv)]. This multiplicative term has a maximum of 2, a minimum of 0, and an average of 1. Importantly, it is never negative. This means that the power represented by the first term always represents the rate of actual work. (Whether the work is absorbed or generated depends on the angle θv – θi and the sign convention associated with the labels that define the polarities of Vm and Im.) Since the average of the multiplicative term is 1, the coefficients on this multiplicative term represent the average power, P. This represents the average rate of actual work.
P = (1/2)VmIm cos(θv – θi)
The above equation is identical to Eq. (11.8) on page 460 in the text.
Turning our attention to the second term in Eq. (11.5b) notice that it's pulsation is represented by the multiplicative term, sin(2ωt + 2θv) This multiplicative term varys both positive and negative as time varys. This term has an average value of zero. It represents power that is surging back and forth as the algebraic sign of sin(2ωt + 2θv) oscillates. Thus this power does no useful work. This second term in Eq. (11.5b) is called the instantaneous reactive power. Although at any instant there is an amount of power flow related to this term, there is actually an amount of energy that surges back and forth with each cycle of the alternating current. The peak magnitude of the instantaneous reactive power is proportional to the amount of energy that oscillates back and forth. Thus the amount of reactive power is defined as the peak of the instantaneous reactive power. We use the variable Q to represent reactive power. Reactive power is given the units of "volt-amperes reactive" and is abbreviated as "VAR" or "VARs" (plural). This abbreviation is often pronounced as a word. The peak magnitude of the second term in Eq (11.5b) is
Q = (1/2)VmIm sin(θv – θi)
DEFINITION: Reactive power is the peak magnitude of power that surges equally back and forth. It has units of volt-amperes reactive (VAR).
(This definition and the related discussion here are valid for single-phase systems with sinusoidal voltages and currents. Otherwise this issue is more complicated.)
Now the instantaneous power, Eq. (11.5b), can be expressed in terms of the average power, P, and the reactive power, Q.
p(t) = P[1 + cos(2ωt + 2θv)] + Qsin(2ωt + 2θv) (11.5c)
In Section 11.4 of this textbook it is shown that
Vrms = Vm/√2 and that Irms = Im/√2
Applying these equalities gives.
P = VrmsIrms cos(θv – θi) , Q = VrmsIrms sin(θv – θi)
The above equations match Eq. (11.50) on page 474 of the text.
This illustrates the calculation of the average power and the reactive power for sinusoidal currents and voltages. The angle (θv – θi) recurs many times in discussions of AC circuits. For this reason in most textbooks and magazine articles it is simply denoted as θ. This angle is given the name power factor angle.
DEFINITION: The power factor angle or more simply, the power angle is
by definition θ = (θv – θi).
(This definition and the related discussion here are valid for single-phase systems with sinusoidal voltages and currents. With care to use phase voltages and currents or line voltages and currents and to not mix these, this definition may also be used with balanced three-phase systems that have sinusoidal voltages and currents.)
Using this definition, Eq. 11.50 on page 474 can be reduced to
P = VrmsIrms cos(θ) , Q = VrmsIrms sin(θ)
(Posted 11/12/2010, updated 11/15/2010, 11/17/2010, 9/14/2011, and 8/17/2012.)
Page 471. Eq 11.38a.
This equation subtly introduces phasors with RMS magnitudes.
In the power systems industry phasors are stated with RMS magnitudes rather than peak magnitudes. There are more implications to this than first meets the eye. In particular, the phasor transform is modified so that the phasor magnitude is in RMS units. Up to this point in this textbook, all phasor magnitudes have been peak magnitudes. In electronics work, peak magnitudes are usually used, but in power systems work RMS magnitudes are usually used.
Consider this electronics-style phasor transform having peak magnitude. This is the way the phasor transform has been used for everything in this book up to this point.
v(t) = 14.14cos(ωt + 25°) V <—> V = 14.14/25° V
Now the very same function in a power systems-style phasor transform will have an RMS magnitude. This type of phasor transform is used for everything from this point through the end of Chapter 12.
v(t) = 14.14cos(ωt + 25°) V <—> Vrms = 10/25° VRMS
Just like it is fair to say that 1 foot = 12 inches, it is also fair to say that
14.14/25° V = 10/25° VRMS because of the different units employed. And of course, the inverse transform of either style phasor is v(t) = 14.14cos(ωt + 25°) V.
However in this textbook, and in the general literature on power systems, the "RMS" subscript on the variable Vrms and on the unit "volts RMS" is usually omitted. The reader should understand that "unless otherwise stated" the units are RMS units when the context is power systems. For example consider Equation 11.41 on page 473. The equation is
S = VrmsI*rms. The very same equation appears again in Equation 11.55 on page 478, but now the subscripts are missing, S = VI*. It means exactly the same thing and the magnitudes of the voltage and current phasors are still intended to be in RMS units, even though the subscripts are missing, because it occurs in the context of a chapter on electrical power (not a chapter on radio circuits for example).
Note in passing that boldface denotes a phasor (complex in general) and normal text denotes a real number and that Vrms = |Vrms|. Similarly, S = |S|.
Note also that there are errata in Chapter 11 on the related topic of units such as V, V rms, I and I rms.
(Posted 9/15/2011. Updated 11/21/2012)
Page 534–535 First paragraph in Sec. 12.10.
The first paragraph in Section 12.10 of the textbook contains some errors of substance and some poor grammar. Several paragraphs of replacement text are offered here. This replacement corrects the errors and offers a more complete explanation of the several issues the textbook's paragraph touches on.
In modern practice, the balanced wye connection is preferred for most sources. This is due to at least three factors. First, a balanced wye connection creates a neutral (center of the wye) that can be connected to earth for safety. Providing the source with this safety connection simplifies the design of other protection such as fusing. (Providing a neutral for a delta-connected source is a much more complicated matter. See for example here and here and here.) Secondly, single-phase loads can be most easily connected to a three-phase system by connecting them between any single line and neutral, but a delta connection does not provide a simple-to-use neutral for this purpose. For this reason three-phase 208 VRMS line-to-line systems are popular in North American commercial buildings. Then 120 VRMS single-phase loads are connected from line to neutral (208/√3 = 120). Similarly in Europe and some other geographic locations a 400 VRMS line-to-line three-phase power system is used to provide 230 VRMS from any one line to neutral. Thirdly, connecting a given set of windings in a three-phase source in a wye connection gives line voltages that are a factor of √3 greater than a delta connection would give. In the case of long distance power transmission, higher line voltages reduce the line currents and consequently reduce the power losses (I2R) caused by heating of the conductors. Delta connected sources are rarely specified for modern installations because any slight imbalance between the three phases can cause problematic circulating currents, and also because protection devices (e.g. fuses) are more complicated to use and maintain effectively.(Posted 8/17/2012.)
Circulating currents are not a problem in delta-connected loads however, because each phase has appreciable impedance so that it may absorb electrical power. Delta connected loads also operate at the full line-to-line voltage which is √3 greater than the line-to-neutral voltage. This can economically reduce heating (I2R) losses in some practical situations. If for some reason a delta-connected load is slightly electrically unbalanced (say one phase draws a bit more current than the others), connecting it to a wye source distributes the unbalance over two phases, making the system as a whole more balanced. For these practical reasons delta connections are popular for loads. However, in theory it is immaterial if a load is wye or delta connected.
For example, either a wye or a delta connection is satisfactory with induction motors. Some induction motors have terminal boards with six terminals, one for each end of the three windings. Then the customer can choose a wye or delta connection. In this way a manufacturer can offer one line of motors to operate at two different voltages. The higher rated voltage option will then be √3 greater than the lower rated option. For example a motor might be rated for 277 delta/480 wye VRMS. In this example the motor phases (coils) are designed to operate at 277 VRMS. The motor may be connected to a 277 VRMS line-to-line voltage system using a delta connection which gives each phase (coil) 277 VRMS. Alternatively, the same motor may be connected to a 480 VRMS line-to-line voltage system by using a wye connection. Then each phase (coil) will again be driven with 277 VRMS. Some of these motors are designed to start with the wye connection and then, as the motor comes up to speed, a set of relays switches it to the delta connection to bring it to full power for normal running conditions. Connecting the example motor above for wye-start-delta-run would require a 277 VRMS line-to-line three-phase power source. This starting method minimizes voltage brown-outs in the power supply system when the motor starts by minimizing the surge current needed to start the motor.
Other motors achieve a dual voltage rating by using dual windings for each of the three phases (six windings in total). These motors have terminal boards with nine terminals. The dual windings of each phase may then be connected in either series or parallel. Some connections are made inside the motor by the manufacturer and are not accessible to the user. These internal connections will make the motor connections either wye or delta as prescribed by the manufacturer. The customer will only have the choice of operating voltage, not wye or delta connection. The two voltage ratings will differ by a factor of two, for example 240/480 VRMS. The dual rating can also be used on a power system corresponding to the lower voltage rating to minimize the surge current needed to start the motor. In this case the motor is stared with the pairs of windings connected in series. Then, as the motor comes up to speed, a set of relays reconnects the windings in parallel for normal running. This type of starting method generally provides more starting torque than the wye-start-delta-run method.
Finally, some motors have dual windings and twelve terminals. These allow the customer up to four different operating voltages by choosing combinations of series and parallel windings and wye or delta connections. This flexibility may also be used to achieve a wye-start delta-run configuration at either of two voltage levels or a series-start-parallel-run configuration, again at either of two voltages.References:
EASA Engineering Handbook, "Three-Phase Motors—Single
Speed."
Hamid A. Toliyat, Gerald B. Kliman, editors, Handbook of Electric Motors
2nd edition revised and expanded, Marcel Dekker, Inc., 2004,
pages 455-456.
Roland E. Thomas, Albert J. Rosa, The Analysis and Design
of Linear Circuits, pp. 811–812, Prentice Hall, 1994.
Richard C. Dorf, James A. Svoboda, Introduction to Electric
Circuits, 8th Ed., p. 571, Wiley, 2010.
Hadi Saadat, Power Systems Analysis 2nd ed., page 31,
McGraw-Hill, 2002.
IEEE Recommended Practice for Electric Power Distribution
for Industrial Plants, 2nd ed., pp. 437-440, IEEE, New
York, 1986.
IEEE Recommended Practice for Electric Power Systems in
Commercial Buildings , pp. 127-130, IEEE, New York, 1983.
E. Lakervi, E.J. Holmes, Electricity Distribution Network
Design, pp. 110–112, Peter Peregrinus Ltd., London, 1989.
Eng-tips discussion forums, "advantages and disadvantages. . ."
Wikipedia, "Three-phase electric power."
Wikipedia, "Mains Electricity."
Page 536 Alternative development of Equation 12.67.
The textbook's development of Equation 12.67 is unecessarily restricted to balanced loads. In Figure 12.35 one may take the load to be unbalanced. Identify each phase with a different load such as ZYa, ZYb, ZYc and assume that these are not necessarily equal to each other. Also consider node b (the phase b line) to be the ground reference. Now there are effectively only two voltage sources driving the load, Vab and Vcb as labeled in Figure 12.35. Each wattmeter measures the power supplied by one source. Thus it is self-evident that PT = P1 + P2, even if the load is unbalanced.
See also the errata for page 537 and the answer to Problem 12.71.
(posted 12/13/2013.)
Page 582 Example 13.10.
Although an autotransformer is customarily thought of as a transformer that employs one tapped winding as both the primary and the secondary, it is not necessarily constructed this way.
It can be helpful to think of an autotransformer as an ordinary transformer that is used in a special type of connection, the "autotransformer connection." Any transformer can be connected in an autotransformer connection, assuming it has adequate insulation which is usually the case. Below, Figure 13.43 has been augmented with part (c) to show how the transformer of part (a) can be connected to perform identically to the autotransformer of part (b).
When the voltage will be autotransformed by a small amount, that is by a ratio that is close to 1:1, then the turns ratio of the basic transformer, as in part (a) of Figure 13.43, needs to be far from 1:1. In this example the voltage transformation ratio is 20:21 and the turns ratio is 20:1. Then usually the winding with fewer turns needs to be made of larger gauge wire in order to carry the correspondingly larger current. Thus for the maximum economic benefit, the autotransformer illustrated in part (b) of the figure might actually be constructed from two windings connected as in part (c). This way the primary winding can be made from a thinner conductor since it does not need to carry the full 4 A load current. The transformer is still called an "autotransformer," as if it had only one winding, because of the presence of the connection that places the two physical windings into one electrical winding with a tap.
However, if the voltage transformation ratio is 1:2 then the turns ratio of the transformer is 1:1. This case is interesting because the current in each winding is identical in magnitude. For example if the turns ratio of the transformer in part (a) of the figure is changed to 1:1 then the secondary voltage changes to 240 V and the secondary current changes to 0.2 A. The transformer would be operating at
(240 V)(0.2 A) = 48 VA. But, in the autotransformer connection of part (c) the current flowing into the circuit from the left would be 0.4 A. This current would divide with 0.2 A going into the primary winding and 0.2 A going into the lower terminal of the secondary winding. Each winding experiences 0.2 A in this case. The net power delivered to the load would be 480 V at 0.2 A or 96 VA, which is two times greater than the transformer can handle when connected as a transformer. This is the type of situation in which a true single winding with a tap, as implied in the definition of an autotransformer, is economic in terms of a savings of copper and iron, and weight and size.
Regardless of the internal construction of an autotransformer (as truly one winding with a tap or as two physical windings of different gauge conductors connected in series) the economic advantages of an autotransformer are best realized when the voltage transformation ratio is between 2:1 and 1:2. When a wider voltage transformation ratio is needed, at some point the advantage of isolation that a standard transformer connection has becomes compelling and the economic advantage of the autotransformer dwindles in comparison.
See also the errata on this figure as published in the textbook.
(Posted 3/19/2012)
Disclaimer: This list of errata is provided by Professor De Boer for the use of his students in his courses. Professor De Boer has no connection to the book's publisher or the authors of the textbook. This list is offered as is, with no guarantee of any kind. (This list is likely to be incomplete at the least.)
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