  ## Errata

### October 2017 Printing:

There is a page numbering error:

All of the pages are there. There was an error which for some reason did not allow the new page numbers on those pages to be seen. The beginning of chapter three goes back to 61 instead of picking up at 67. Sorry for Any confusion this may have caused. There is also a page number issue around page 116. It had to do with a font not being recognized during printing. Notice that the example / information continues as normal.

Page 5 – Should say “for a completely resistive load, θ = 0°, cos 0° = 1, therefore the power is all real.”

Page 15 – Should say “ Vbn = Vp angle - 120°”

Page 25 – P should be 217.6 W instead of 219 W

Page 38 – Diagram 1.36 should say 240 KV instead of 230 KV

Page 159 & 160 – The text should refer to figure 6.32 instead of figure 109

Page 161 – The text should refer to figure 6.34 instead of figure 112

### July 2017 Printing:

p. 2 In the capacitance section it should say – As capacitance (C) increases, Xc decreases

p. 5 The instantaneous power formula should not have an = sign after (VMAX IMAX)/2.

p. 7 Figure 1.6 should be labeled as Power Triangle

p. 17 Figure 1.17 - Delta Phasor Diagram.

The equation indicates:
Ia=Iab*sqrt(3) (angle 30 degrees)

This should say minus 30 degrees.

p.19 In Ex 5, the denominator in the calculation for line current should be 23 + j25Ω.

p. 27 The angle noted “83.1” in the vector diagram needs to be “77.2”.

The vector named “Ib” should be “Ic”.

p. 40 Ex. 22 was supposed to say “If the transformer ratings are chosen as the voltage base values"

p. 40 & 41 Ex. 22, Sbase = 100 MVA.

p. 143 / 154 / 226 I only solved for Ib. For a 2LG fault we need to also solve for Ic (same magnitude, supplementary angle) and add them together. If = Ib + Ic

p. 145 Ex. 59, should Z1's units on figure 6.19 and be labeled as PU and not ohms

p. 229 C = Q/V

p. 307 Q 80. Qsend should be 0.525

On problem 74 in the practice exam theta = -60 deg, w2=0, pf=0.5 lagging, it should be LEADING.

### March 2017 Printing:

p. 2 In the capacitance section it should say – As capacitance (C) increases, Xc decreases

p. 4 The instantaneous power formula should not have an = sign after (VMAX IMAX)/2.

p. 7 Figure 1.6 should be labeled as Power Triangle

p.19 In Ex 5, the denominator in the calculation for line current should be 23 + j25Ω.

p. 36 Ex. 18 was supposed to say “If the transformer ratings are chosen as the voltage base values"

p. 125 / 136 / 203 I only solved for Ib. For a 2LG fault we need to also solve for Ic (same magnitude, supplementary angle) and add them together. If = Ib + Ic

p. 127 Ex. 50, should Z1's units on figure 98 and be labeled as PU and not ohms

p. 206 C = Q/V

### 2016 Printings:

On Pg 34 Figure 33 should say:

“Figure 33 – Circuit for Example 17 with values”

On pages 37&38, (Vbaseold/Vbasenew) should be written in units of KV.

Ex 18 should say: If the low-side transformer ratings are chosen as the base values in the generator circuit, find the base values and per unit quantities for the system in the diagram.

Ex. 29 should say:

What is the synchronous speed of an induction motor with 2 poles per phase and a frequency of 60 hz?

Ex. 31  On ex. 32: Page 99 – There are two definitions that say Static Var Compensators. The first should say Synchronous Condenser.

### Errata - Jan 2015 Printing:

Important!

There is an error in the PT section of the book that will be corrected in the next printing:

There is a Copy & Paste error. Where it says:

“Just as in CTs, never open the secondary of a PT while current is flowing in the primary circuit, as large periodic secondary voltages develop during periods of flux change. Because of this, if a meter or relay secondary circuit of a PT must be disconnected, you must first short-circuit the secondary winding, and then remove the component. The short across the secondary winding may be removed after the secondary circuit is closed”

This is incorrect information.

While the secondary winding may seem to be isolated from the primary winding, There may be large voltages built up due to the distributed capacitance. Never short circuit the secondary winding. As this may lead to a fatal shock.

P 4 The RMS formula should be: P 8 ex. 2, The concepts shown in this problem are correct. However for this size load & generator, we should be using a three phase solution:  P 9, Ex 2.

Vp should be as follows: P 19 On page 21, I should have been more specific on my formulas: P 23 334.49 should be in VA, not W

Calculation should say 87.34 angle -11.3 V (not Amps)

P 26 ex. 12 Power should be in KW not KVA

P45 The dot on the secondary coil for the subtractive polarity diagram should be on the lower right. P 49 - Ex. 22 Should say 4% - 0.3 = 3.7%, So the range will be 3.7% to 4.3%

P 54 It should say: P 60 P 65 Ex 30.

Coil pitch = 20 deg X 8 = 160 deg P 69 Ex. 31 The final answers are correct but the lagging pf angle should be +35.9 deg, the leading pf angle should be -35.9deg.

P 71 Ex. 32b is incomplete: P 81. Use THHN CU wire. It should be 28A X 125% (for cont motor loads) = 35A

P 127 I missed the generator impedance on this one. The positive sequence impedance should be 0.2 + j0.546 instead of 0.2 + 0.466