ECE 22X: Solutions Errata
Fundamentals of Electric Circuits, 2nd ed, 1st printing
Alexander and Sadiku

See the official errata too. This list does not repeat the official errata.

Chapter 1

• Problem 1.13: The solution uses the wrong identity for cos^2 x. The correct identity is listed on page 860 of the text. This problem results in the wrong final answer listed in the solution.
• Problem 1.15: There are at least two errors in the solution to part a: the sign of the exponent at the end of the first line of expressions should be negative and the exponent should be -4 in the second line. The correct answer is 1.4725 C. They also leave out the negative sign in the exponent in part of the solution to part c.
• Problem 1.25: The solution divides 4 minutes by 6 instead of 60 to convert minutes to hours.
• Problem 1.31: The solution is completely wrong. It should be (120 W x 4 hrs/day + 60 W x 8 hrs/day) x 365 days x 1 kW/1000 W * 0.12 $/kWh = $42.05
• Problem 1.37: The final answer should be -961.0 J, not -901.2 as listed in the solution.

Chapter 2

• Problem 2.11: The solution contains an arithmitic error. The final answer for V1 should be -4.0 V, not 4 V as listed in the solution.
• Problem 2.21: The solution is for a different circuit than that shown in the text. The solution has a 10 resistor in place of the 1 resistor.
• Problem 2.35: There is a typo in the equation solving for the equivalent parallel resistance of 20 and 5 . It lists 20||15, it should be 20||5. This did not affect the final answer.
• Problem 2.51: The answer to part a should be 9.23 , not 9.31 as listed.

Chapter 3

• Problem 3.12: There are a few error in equation (1) of the solution to this problem. The first equation should be (V1-V0)/2 + 5 = (40-V1)/1. The solution lists V0 for this last term rather than V1. This equation then simplifies to V1-V0 = 70.
• Problem 3.28: The circuit in the solutions differs from that in the text. The voltage source on the right side of the circuit should be 12 V to be consistent with the text. The negative sign is also missing from the 6 V source at the top of the circuit.
• Problem 3.31: The second term on the left side of the equation for node 3 should be v3/4, not v2/4 as stated.
• Problem 3.37: The solution for ix should be 2.105 A, not 4.105 A as listed.
• Problem 3.38: The second sentence labeled (2) on the right hand side should read, "so that (1) becomes i1 = (16/6) i2" instead of "so that i1 = (7/12) i2". This error is propagated throughout the rest of the solution. The correct answer is v0 = 1.175 V and i0 = 0.1961 A.
• Problem 3.40: G₁₁ = 1 + 3 + 5 = 9 S, not 8 S as listed in the solution.
• Problem 3.55: The figure in the solutions manual has the 10 k resistor replaced with a 100 k resistor, though the solution is worked out with the 10 k resistor.
• Problem 3.64: The second equation in (2) should be i1=(16/6)i2.
• Problem 3.74: The last element of the vector on the right side of the last equation in the solution should be V2-V4, not just V4.

Chapter 4

• Problem 4.3: In part c, vo is 0.5 V when R = 10 , not 5 V as listed. In the same part, the equation for the current Io is alsow wrong.
• Problem 4.27: The resistor in series with the 40 V voltage source was wrongly converted to 10 resistor in the solution for vx3. However, even though the picture is wrong, the solution and equations are correct.
• Problem 4.18: The solution has the answer in units of mA (milliamps). It should be in units of V (volts).
• Problem 4.19: The solution solves for vx, but the problem asks you to solve for ix.
• Problem 4.25: The solution applies the source transformation incorrectly to the 3A source (wrong polarity) and the equation for KVL has the wrong signs on most of the voltage sources. The final solution should be -6.6 V.
• Problem 4.31: The algebra used to solve this problem is odd. The numerical answer has two digits swapped. The correct answer is 84/23 = 3.652, not 3.625 as stated.
• Problem 4.42: The solution should read "to find Vth we transform the 20V and 5A source" instead of "the 20V and 5V source".
• Problem 4.57: Equation 2 has the first term as (1-vo)/2. It should be (1-vx)/2.
• Problem 4.64: The equivalent resistance is -2 , not -1 as the solution states.
• Problem 4.66: The solution contains a typo when they solve for the mesh current. The solution should be -6, not -5. The correct value for i was used in the remainder of the solution.
• Problem 4.67: i2 is listed as 1.5 A, it should be -1.5 A.
• Problem 4.69: The thevenin equivalent voltage is -220.0 V, the short circuit current is 176 A, and the equivalent resistance is-1.25 .

Chapter 5

• Problem 5.7: The op amp shown has positive feedback and would therefore not work as analyzed in the solution. Also, it is not clear that the independent voltage source has the same reference as the power supply for the op amp. There should be a ground symbol on the negative terminal to make this clear.
• Problem 5.8: The equation va = va - 2 should be vo = va - 2.
• Problem 5.13: The load resistor should be 10 k, not 4 k as shown. Note the polarity of the op amp shown in the text is also wrong, as listed in the textbook errata.
• Problem 5.17: The first sentence should read, "We convert the voltage source to a current source and then back to a voltage source (two source transformations)". The next to last line should be 4 + 4/3, not 4 x 4/3.
• Problem 5.21: The equation at the bottom left of the solution has an extra negative sign, but the final answer is still correct.
• Problem 5.23: The equation, "vo = 20/(20+20) vo" should be "vo = 20/(20+20) vx". The power dissipated by the 60 k resistor is 2.4 nW, not 204 nW.
• Problem 5.46: The first line has a resistor labeled "Rx" that should be labeled "Rf".
• Problem 5.60: Equation 1 is wrong: it should be 3Vo = -10 Is. Equation 2 is also wrong. It should be vo = v1 (3 + 3)/3 = 2 v1.
• Problem 5.61: In the last line the solution should be for Vo, not V1 .
• Problem 5.62: The subscript "f" was replaced by "4" in the solutions.
• Problem 5.65: This circuit has positive feedback in the right-most op amp and cannot be analyzed as shown in the solution.
• Problem 5.66: The first term in the equation should be -100/25 not -110/25.
• Problem 5.70: The first term in the equation should be multiplied by 1 not 10. The equation for the voltage labeled vb should have 10 in the numerator, not 60. The answer of -2 V is correct.
• Problem 5.71: The bottom-most op amp has positive feedback and the circuit cannot be analyzed as shown in the solution.
• Problem 5.73: The solution to this problem is completely wrong. The correct answer is vL = 1.8 + 9 = 10.8 V.

Chapter 6

• Problem 6.7: The answer should be 0.04t^2 + 10 V.
• Problem 6.9: The calculation of the voltage at t=2 is incorrect and the final answer is wrong.
• Problem 6.13: The solution uses a resistor of 30 (left side of circuit), rather than 3 . Using the 30 resistor the equation for v1 should be "v1 = 30 i1", not "v1 = 30 i2".
• Problem 6.17: Part c, 3 F in series with 6 F = (3 x 6)/9 = 2 F, not 6 F as listed.
• Problem 6.28: Cc is in parallel with 20 uF, not series. The solution also has a typo in the 5th equation from the top of page 244. The capacitor labeled C₆ should be labeled C_b.
• Problem 6.31: On page 246 near the end of the solution it states i1 = c2 (dv/dt). It should be i2 = c2 (dv/dt).
• Problem 6.46: The expression for Wc should be 1/2(2)(0), instead of, 1/2(2)(V).
• Problem 6.47: Last equation, (R x 2) in the denominator should be (R + 2).
• Problem 6.61: The solution in part B uses the current division principle incorrectly. This can only be used when there are zero initial conditions, which is not true for this problem (see problem 6.52). The solution in part C for W10mH should be (1/2 X 10), not (1/2 X 30).
• Problem 6.65: The units of the answers to parts (a) and (b) should be in units of Joules, not Watts. The answer to (c) is completely wrong (wrong defining equation, missed a negative sign due to the passive sign convention, etc.)
• Problem 6.71: The last term in the equation for Vo should be in terms of V3 and R3 instead of R2 and V2.

Chapter 7

• Problem 7.24: In part b, the unit step should start at t = 1, not t = 0. The answer should be i(t) = -10u(t-1) +20u(t-3) - 10u(t-5).
• Problem 7.30: In part b, the solution omits the 4t² term, though the answer is correct.
• Problem 7.39: In part a, the solution correctly calculates v(0) = 4, but incorrectly substitutes v(0) = 8 V in the solution for v(t).
• Problem 7.54: For t>0, the dependent voltage source has a voltage of 0 V, since it is proportional to io for t>0 which is zero. This is equivalent to a short circuit. The solutions incorrectly analyze the circuit for t>0 as though the dependent source is an open circuit.
• Problem 7.55: They drew and analyzed the wrong circuit for t>0. For t>0 i0 = 0 and the dependent voltage source acts like a short circuit.
• Problem 7.64: The solution has units of for the 10 V independent source, rather than volts. Also, the final answer should be 5/6(1 + e^-t) A, not 5/6(1 - e^-t).
• Problem 7.71: The time constant is calculated incorrectly. It should be RC = (10 k + 20 k)(100 m) = 3000 s.
• Problem 7.85: Part a) should be 659.7µs and part b) should be t = 16.63s.

Chapter 8

Chapter 9

• Problem 9.14: In part a) the denominator should be (-7+j17), not (-15+11j) as listed.
• Problem 9.34: The solution uses an inductance of 2 mH instead of 20 mH.
• Problem 9.61: There is a short circuit between the two terminals so the equivalent impedance is 0 . The solution would be correct if one of the terminals was connected at the center of the four passive elements.

Chapter 10

• Problem 10.42: The answer should list I2 = 2.23 at an angle of -51.8 degrees.
• Problem 10.47: The solution uses w = 1 rad/s, rather than 3 rad/s as stated in the text.
• Problem 10.75: The solution uses R₄ = 20 k, instead of 40 k as stated in the problem in the text.

Chapter 11

• Problem 11.2: The Thevenin impedance is 12.8 - j 49.6 , not 51.2 + j 1.6 as stated in the solution.
• Problem 11.6: I1 = 8/2^(1/2)<-25 degrees, not I1 = 10/2^(1/2)<-25 degrees as listed in the solution.
• Problem 11.41: In part b) the calculation of Z should be Z = (0.64 + j0.52) / (1.64 + j0.52) = 0.4793<21.5 degrees, not Z = (0.64 + j0.44) / (1.64 + j0.44) = 0.4793<21.5 degrees as stated.

Chapter 12

Chapter 13

• Problem 13.21: The answer to the linear set of equations should be I1 = 4.2538<-8.5169 degrees and I2 = 1.5637<27.510 degrees.
• Problem 13.26: The dot convention was applied incorrectly in the solution. In matrix form the terms j17 should be -j17. This gives I1 = 3.68 < 78.0 degrees and I2 = 1.52 < 92.0 degrees.
• Problem 13.50: For the circuit shown in Figure. b the solutions transform 10 sin (3t-30) to 10<-30 degrees, rather than converting the sine to a cosine first (10<-120 degrees). This works as long as you substitute sin for cos when you apply the inverse transform, but this is inconsistent with the definintion of the phasor transform and the examples given in the text.
• Problem 13.60: The answer for Zin should be Zin = 4 + 6/n^2 = 4 + 6/4^2 = 4 + 0.375 = 4.375 , not 100 as listed in the solution.

Chapter 15

• Problem 15.9: Part a. The second term should have s in the denominator, instead of s^2.
• Problem 15.5: Part d. The first line of the solution is wrong. 2 e^(-(t-1)) u(t) = 2 e e^(-t) u(t), not 2 e^(-t) u(t) as shown.
• Problem 15.34: Part a. The last term should be multiplied by u(t). Part b. The solution states that B = 1/2. It is actually -1/2.

Chapter 16

• Problem 16.10: The circuit diagram used to solve for the equivalent impedance has a current source labelled 1V. It should be labelled 1 A.
• Problem 16.12: This solution contains several errors. The first equation is correct, but the second equation and those that follow are wrong.
• Problem 16.17: The solution has io(t)= i2(t), but it should be io(t) = i2(t) - i3(t). Also (3) should be I1 = I2 - 4/s, since the current souce should has an s-domain equivalent of 4/s - not 4, as is used in the solution.
• Problem 16.18: The solution is missing a factor of 3 in the term e^{-s}/s of the second-to last equation in the first line of the solutions.
• Problem 16.24: The solution to this problem contains many errors. The capacitor is drawn as a resistor. In the s-domain the voltage source should have a value of 36/s, not 36. In the s-domain this voltage source should be 36/s not 36. The solution shows the coefficient of B to be -12.8 it should actually be 12.8.

Chapter 19