ELEMENTARY LINEAR CIRCUIT ANALYSIS, 2/e

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p. 2, line 3: . . . and we say . . .
p. 9, Fig. DE1.3: the current through R is 1 mA directed down
p. 9, 2nd papragraph: . . . Fig. 1.20 . . .
p. 9, 2nd paragraph, second line: . . . a positive resistance, of course) . . .
p. 48, Prob. 1.46: . . . change the 5-ohm resistor to a 3-ohm resistor. . .
p. 54, next to last equation: i1 + i3 - 17 = 0
p. 54, last equation: -i2 - i3 + 21 = 0
p. 55:

                  |  17      -3  | 



           del1 = |              | 



                  |  -21      5  | 



p. 64, line 23: v2 = -1.5 V
p. 66, line 4: to the supernode . . .
p. 69, line 6: v1 = 1.5 V
p. 99, Fig. P2.22: the voltage source should be labeled "v"
p. 106, numerator for second equation: 5.0 x 10-5 + 10.0
p. 106, last equation: . . . (10 x 103)(1.0 x 10-3)2
p. 110, 1st line: Operational-amplifier . . .
p. 116, 1st equation: v = Rgig = . . .
p. 123, Fig. 3.31: insert node
p. 124, last line: . . . bargain . . .
p. 128, Equation (3.11): 9v1 + . . .
p. 129, 1st equation: v1 = -(5/3)v'
p. 138, 1st equation: v1 = 6i1 = . . .
p. 150, Fig. P3.4: the op amp should be in color
p. 155, Problem 3.24: . . . Fig. P3.24 . . .
p. 157, Fig. P3.39: remove the dot directly above the voltage source
p. 162, last line: interchange the answers to (b) and (c)
p. 168, answer to Drill Exercise 4.4: (c) 8 V
p. 182, Fig. 4.30(b): label the origin "0"
p. 202, line 15: . . . cannot change instantaneously . . .
p. 202, Equation (4.3): iC(t) = i1(t) - i2(t) = -vC(t)/1 - iL(t)
p. 202, line 22: iC(0) = -vC(0) - iL(0) = -4 - 2 = -6 A
p. 202, last line of text: . . . capacitor . . .
p. 203, 5th equation: dvC(0)/dt = 4iC(0) = 4(-6) = -24 V/s
p. 203, 6th equationn: diC(t)/dt = -dvC(t)/dt - diL(t)/dt
p. 203, 7th equation: diC(0)/dt = -dvC(0)/dt - diL(0)/dt = 24 - 0 = 24 A/s
p. 203, 9th equation: ... = -2(0) - 24 = -24 V/s
p. 206, 4th equation: iC = -(5/6)iL - (1/6)vC
p. 206, 5th equation: iC(0)= -(5/6)iL(0) - (1/6)vC(0) = -(5/6)(3) - (1/6)(0) = -2.5 A
p. 207, 1st equation:vL(0) = . . . = -5(-2.5) - 5(3) = -2.5 V
p. 207, 3rd equation: diL(0)/dt = (1/3)vL(0) = (1/3)(-2.5) = -0.83 A/s
p. 207, 5th equation: dvC(0)/dt = (1/2)iC(0) = (1/2)(-2.5) = -1.25 V/s
p. 207, Drill Exercise 4.19: . . . initial conditions . . .
p. 207, Drill Exercise 4.19: answers out of sequence
p. 217, Problem 4.78: replace colon with semicolon after part (b)
p. 221, line 16: . . . initially . . .
p. 224, answer to Drill Exercise 5.1: (b) -2e-4t u(t) A
p. 225, line 10: . . . resistance.
p. 256, Example 5.9, second line: . . . input voltage

                        { 2 V  for   -inf < t < 0



               vs(t) =  { 



                        { 4 V  for   0 < t < inf 



p. 260, Example 5.10: . . . (see Fig. 5.26 on p. 247) . . .
p. 265, Example 5.13, line 2: . . . input current . . .
p. 268, Fig.P5.8: the current i1 (and arrow) should be in color
p. 272, Problem 5.53: and (c) vo(t).
p. 273, 1st line of Sec. 6.1: . . . initial . . .
p. 278, 7th line from bottom: . . . After . . .
p. 282, last line: di(0)/dt = . . .
p. 292, Fig. DE6.4: change inductor value to ½ H
p. 298, Fig. 6.20: units for current scale are "mA"
p. 300, last line: replace = sign with + sign
p. 303, Drill Exercise 6.9, line 2: . . . = u(t) A.
p. 304, Fig. 6.23: "v1" should be in color
p. 321, Example 7.4, line 2: vC(t) . . .
p. 321, 323, 325, Running head: . . . ZERO-INPUT CIRCUITS
p. 322, line 20: you can have the computer print out . . .
p. 326, line 00016 of program: I2K=I2+(-I2+V)*DELT
p. 339, Fig. DE7.10: label capacitor voltage "v" (plus on left) and inductor current "i" (directed down)
p. 340, line 21: . . . corresponding . . .
p. 355, 3rd equation from bottom: jtheta + . . .
p. 356, line 3: and this result is known as Euler's formula.
p. 362, 1st equation: insert underbar
p. 362, line 5 from bottom: . . . capacitor.
p. 365, answer to Drill Exercise 8.5: (b) j18.85 ohms
p. 370, next to last equation: insert underbar
p. 382, middle answer to Drill Exer. 8.12: 2.17 cos(4t + 77.47°) A
p. 383, 2nd answer to Drill Exercise 8.13: -3 cos 2t A
p. 389, Figs. P8.39 and P8.40: reverse polarity of lower voltage source
p. 392, last equation: insert underbar
p. 397, line 1: . . . when ZL is con-
p. 409, 7th equation: replace 1 in numerator with V, and insert underbar
p. 410, 6th line from bottom: the plus at the neutral wire) . . .
p. 416, Example 9.10, first line: insert underbar
p. 417, Fig. 9.25: Replace VbC with Vbc
p. 418, Example 9.11, 2nd equation: |Z| = . . .
p. 422, next to last equation: change the minus sign between the two terms to +
p. 431, 2nd line of equations: insert underbar
p. 442, equation for P1: 2 in the denom. should not be boldface
p. 446, footnote: Hendrik (not Hendrick)
p. 450, Fig. 10.10(b): ang H(jomega)
p. 459, last equation: |V| = R |I| / . . .
p. 467, 4th line from bottom: . . . 1/sqrtLC
p. 477, answer to Drill Exercise 10.8: 2.77e-3t cos(3t + 56.3°) V
p. 479, last paragraph: Again let us return. . .
p. 490, Problem 10.4: Indicate the half-power frequency.
p. 495, Problem 10.59: and angle of H(s) . . .
p. 513, 5th line from bottom: denomin.for Y(s) is s(s2 + 2s + 5)
p. 523, 525, 527, Running head: APPLICATION TO CIRCUITS
p. 529, numerator to first answer of Drill Exercise 11.20: s (not 2)
p. 556, Equation (12.3): . . . L2i22(t) . . .
p. 562, Example 12.4, first line: . . . (p. 561) . . .
p. 567, next to last equation: . . . = (-2vg)2/20 = . . .
p. 570, Example 12.6, line 6: the output (Thévenin . . .
p. 570, Drill Exercise 12.5, first line: . . . Fig. 12.16
p. 576, line 9: y22 (boldface)
p. 588, Problem 12.2: resonance frequency.
p. 608, answer to Drill Exercise 13.2: 0.667 + 0.551cos(2¶/3)t - 0.276cos(4¶/3)t + 0.138cos(8¶/3)t - 0.110cos(10¶/3)t +
p. 615, line 14: . . . the formulas for . . .
p. 650 and inside back cover: replace "dt" with "dtau" in convolution integral
p. 655, 2nd equation: . . . [6e-5tau . . .
p. 682, line 3: . . . nonindependent-source elements
p. 686, line 2: are irrelevant.
p. 687, line 13: . . . and final time, respectively, of the
p. 691, 4th line from bottom: seven characters, LYYYYYYY and LZZZZZZZ
p. 696, answer to Problem 1.20: -18 A, -6 A, -12 A, -8 A
p. 697, answer to Problem 2.17: remove the first /
p. 697, answer to Problem 2.30: 10 V, 6 V, 4 V
p. 697, answer to Problem 3.22: -2.16 A, 0.96 V
p. 698, answer to Problem 4.10, last part: 3t + 1 for 0 < t < 1
p. 700, answer to Problem 4.73: (d) 16 V/s
p. 700, answer to Problem 4.79: answers out of sequence
p. 701, answer to Problem 5.47: (a) 2r(t) - ½(1 - e-4t)u(t), (b) ¼(1 - e-4t)u(t)
p. 701, answer to Problem 6.14: answers out of sequence
p. 701, answer to Problem 6.18: variable should be v2
p. 702, answer to Problem 7.7: should be answer to Problem 7.8
p. 703, answer to Problem 9.16: (a) 0.5, (b) 0.707
p. 703, answer to Problem 9.41: third angle is 90.26°
p. 704, answer to Problem 11.10: (a) numerator is "A2s + alphaA2 + A1"
p. 704, answer to Problem 11.12: answer given to (c) is answer to (d)
p. 705, answer to Problem 11.20: (a) change "2 sin 2t" to "2 sin 3t"
p. 705, answer to Problem 11.31: add unit "V" to third answer
p. 705, answer to Problem 11.65: answer given is for (a)
p. 705, answer to Problem 12.10: numerator should be "8s(s2 - 1)"
p. 705, answer to Problem 12.14: should be answer to Problem 12.15
p. 706, answer to Problem 12.25: answer given is for (b)
p. 706, answer to Problem 12.27: 360
p. 706, answer to Problem 13.7: add terms "+ 0.424 sin 2¶t + 0.212 sin 4¶t"
p. 707, answer to Problem 13.25: 0.637 - 0.212e · · ·
p. 707, answer to Problem 14.21: (d) high-pass filter

Updated: 12-28-96