FUNDAMENTALS OF ELECTRICAL ENGINEERING, 2/e

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p. 46, Fig. P1.14: reverse the polarity of v2
p. 46, Fig. P1.17: label the voltage source vs
p. 52, Problem 1.54: (c) Find v2 when v1 =
p. 64, Fig. 2.9: insert current i1 (directed to the right) through the 2-ohm resistor
p. 71, line 3: . . . in terms of the mesh currents.
p. 154, Equation (3.35): i(0) = -A1 e -8 (0) - (1/4)A2 e -2 (0) ==>
p. 194, 2nd equation from the bottom: cos theta = a/M    and    sin theta = b/M
p. 415, Fig. P6.75(b): change -10 to -15
p. 459, line 1: Q1, (2) connect the emitter of Q2 to the base of Q, . . .
p. 479, 4th equation: iC5 is approximately equal to iB5 + iC5 . . .
p. 503, Introduction, line 3: FETs not FETS
p. 538, first equation: change ID to iD
p. 545, line 19: Fig. 8.23b).
p. 627, 5th equation: change pC to p
p. 755, Example 11.3: (261.64)8 = (177.8125)10
p. 780, line 2: change (Fig. 11.9) to (Fig. 11.8)
p. 782, first equation:

p. 786, 2nd equation:

p. 804, Problem 11.33:

p. 814, first equation:

p. 1128, answer to Problem 1.43: this is the answer to Problem 1.42
p. 1128, answer to Problem 2.28: (a) -(RL + R1 + R1RL/R2)is (b) (1 + R1/R2)is
p. 1128, answer to Problem 2.40: (b) v = -2.4 V
p. 1131, answer to Problem 4.67: (a) 31.9 + j19.8 ohm, (b) 37.3 µF
p. 1133, answer to Problem 8.63: 5V, 4 mA, -4 mA, 5 V

The following errors have been corrected in the latest printing of the book:

p. 21, lines 6 and 7: For example, a 4-ohm resistance is the same thing as a ¼-mho conductance.
p. 37, Example 1.12, line 3: 0.5i V. In this case the constant 0.5 has . . .
p. 45, Fig. P1.7: change 6-ohm resistor to 1-ohm resistor, change i5 to i1
p. 50, Problems 1.39 and 1.40: Req = Vs/is
p. 59, line 1: v1 = 3 V. (add the period)
p. 71, line 6: . . . constraint . . .
p. 90, line 20: . . . (e.g., see Problem 2.43).
p. 90, line 24: lem 2.48).
p. 100, Problem 2.11: RL = 10 kohms
p. 105, Problem 2.41: v1 = 20 V
p. 107, Problem 2.46: Hint: To find Ro, apply . . .
p. 107, Fig. 2.47: change v1 to vs
p. 113, Drill Exercise 3.1: 0 A for t less than or equal to 0 s
p. 119, Drill Exercise 3.4: ANSWER -2 V for 0 < t . . .
p. 128, next to last line: change 3.12 to 3.13a
p. 128, last line: change 3.13 to 3.13b
p. 179, Fig. P3.28: change the 10-V battery to a 12-V battery
p. 182, Problem 3.41: . . . find the unit step responses . . .
p. 182, Problem 3.42: For the circuit shown in Fig. P3.42, find the step responses . . .
p. 184, Problem 3.70: is(t) = 4u(t) A
p. 187, line 16: function goes through a complete cycle in 2 pi/omega seconds.
p. 197, Drill Exercise 4.2: Find the exponential and rectangular forms . . .
p. 198, Section 4.3, line 6: response to KA sin(omega t + theta) . . .
p. 198, Section 4.3, line 11: B cos(omega t + Ø) + jB sin(omega t + Ø) = Bej(omega t + Ø)
p. 200, line 14: . . . ang(j omega CV) = ang(j omega C) + . . .
p. 255, Problem 4.3(b): A2 = 4e-j30
p. 288, line 19: Besigma t cos(omega t + phi) + jBesigma t sin(omega t + phi) = . . .
p. 299, line 8: . . . Eq. 5.9,
p. 300, Fig. 5.32: (2/3s2) X
p. 335, Fig. P5.23: replace Z --> with Y -->
p. 392, line 6: By symmetry, i1 = i2 = (7.1 x 10-3)/2 = . . .
p. 402, Fig. 6.59: Output voltage should be labeled "(b)"
p. 411, Prob. 6.34: v1 = 1 V
p. 429, Fig. 7.10(b): resistor connected between the base and the collector should be 330 kohms
p. 446, Drill Exercise 7.8, line 2: of the BJT and a second . . .
p. 480, 2nd equation: vo = -vBE5 . . .
p. 499, Fig. P7.62: change the 12-kohm resistor to 13 kohms
p. 503, Introduction, line 5: transistor (MOSFET).
p. 503, Introduction, line 3 and line 10: FETs not FETS
p. 522, Drill Exercise 8.4: Vp = -4 V
p. 526, 2nd equation: 9 x 10-3
p. 526, last equation: vDS = -250(16 . . .
p. 535, line 9: . . . --this is the load curve and is done as follows.
p. 539, line 3: b, respectively.
p. 543, Fig. 8.32: insert a terminal symbol at v2
p. 543: Case 1:
p. 550, line 22: . . . 4 - 10 = -6 V.
p. 593, line 3: iD = -vGS/RS
p. 676, fifth line from the bottom: . . . (0.55 x 10-3) . . .
p. 726, last paragraph: (suppressed)
p. 786, line 1: Taking the complement . . .
p. 791, fourth line from bottom: . . . m5, m6, m7,
p. 813, Fig. 12.7: Rightmost block should be labeled Half-Adder
p. 841, line 10: (making Q = 1) before . . .
p. 846, last paragraph, line 2: and the present state. The first three columns . . .
p. 922, line 11: where, of course . . .
p. 936, line 1: Thus the impedance Vg/I1
p. 936, 3rd equation: Z in the denominator should be Z
p. 975, Example 15.8, line 1: . . . an ohmmeter . . .
p. 1037, line 6: indicates . . .
p. 1045, line 7: . . . the default value . . .
p. 1128, answer to Problem 1.11(d): -10.5 V
p. 1128, answer to Problem 2.30(a): (1 + R2/R1) vs
p. 1128, answer to Problem 2.40(a): -4 V
p. 1128, answer to Problem 2.45: (1 + R2/R1) vs
p. 1130, answer to Problem 3.76: replace e-3t with e-t
p. 1130, answer to Problem 3.80 is instead answer to Problem 3.81
p. 1130: delete answer to Problem 3.84
p. 1130, answer to Problem 4.1: Answer given for (c) is the answer to (b).
p. 1130, answer to Problem 4.4(d): -1.41 + j1.41
p. 1132, answer to Problem 6.28(b): 623 ohms
p. 1136, answer to Problem 14.25: Answer given for (c) is the answer to (b).

Updated: 10-22-99

Please report any new errors to:

bobrow@ecs.umass.edu



For reporting new errors, thanks to:
Margreet Mol, TNO Institute of Appled Physics, Delft, The Netherlands