Electrical engineering principles and applications 5th solutions ISM

Đầu sách dành cho các kỹ sư điện điện tử, sinh viên các trường đào tạo Điện Tử Viễn Thông hoặc Cơ Điện Tử.Mình thấy sách rất hay và có ích, hi vọng mọi người sẽ nghiên cứu quyển sách nàyElectrical Engineering: Principles Applications (5th Edition)

SOLUTION MANUAL CONTENTS Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174 Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221 Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 329 Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 286 Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347 Chapter 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 359 Chapter 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 407 Chapter 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 458 Chapter 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 487 Chapter 14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 520 Chapter 15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 572 Chapter 16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 608 Chapter 17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 646 Appendix A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 685 Appendix C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 689 Complete solutions to the in-chapter exercises, answers to the end-ofchapter problems marked by an asterisk *, and complete solutions to the Practice Tests are available to students at www.pearsonhighered.com/hambley CHAPTER 1 Exercises E1.1 Charge = Current × Time = (2 A) × (10 s) = 20 C E1.2 i (t ) = E1.3 Because i2 has a positive value, positive charge moves in the same direction as the reference. Thus, positive charge moves downward in element C. dq (t ) d = (0.01sin(200t) = 0.01 × 200cos(200t ) = 2cos(200t ) A dt dt Because i3 has a negative value, positive charge moves in the opposite direction to the reference. Thus positive charge moves upward in element E. E1.4 Energy = Charge × Voltage = (2 C) × (20 V) = 40 J Because vab is positive, the positive terminal is a and the negative terminal is b. Thus the charge moves from the negative terminal to the positive terminal, and energy is removed from the circuit element. E1.5 E1.6 iab enters terminal a. Furthermore, vab is positive at terminal a. Thus the current enters the positive reference, and we have the passive reference configuration. (a) pa (t ) = v a (t )ia (t ) = 20t 2 10 10 20t 3 w a = ∫ pa (t )dt = ∫ 20t dt = 3 0 0 10 2 = 0 20t 3 = 6667 J 3 (b) Notice that the references are opposite to the passive sign convention. Thus we have: pb (t ) = −v b (t )ib (t ) = 20t − 200 10 10 0 0 w b = ∫ pb (t )dt = ∫ (20t − 200)dt = 10t 2 − 200t 1 10 0 = −1000 J E1.7 (a) Sum of currents leaving = Sum of currents entering ia = 1 + 3 = 4 A (b) 2 = 1 + 3 + ib ib = -2 A ⇒ (c) 0 = 1 + ic + 4 + 3 ⇒ ic = -8 A E1.8 Elements A and B are in series. Also, elements E, F, and G are in series. E1.9 Go clockwise around the loop consisting of elements A, B, and C: -3 - 5 +vc = 0 ⇒ vc = 8 V Then go clockwise around the loop composed of elements C, D and E: - vc - (-10) + ve = 0 ⇒ ve = -2 V E1.10 Elements E and F are in parallel; elements A and B are in series. E1.11 The resistance of a wire is given by R = substituting values, we have: 9. 6 = 1.12 × 10 −6 × L π (1.6 × 10 − 3 )2 / 4 ρL A . Using A = πd 2 / 4 and ⇒ L = 17.2 m E1.12 P =V 2 R ⇒ R =V 2 / P = 144 Ω E1.13 P =V 2 R ⇒ V = PR = 0.25 × 1000 = 15.8 V ⇒ I = V / R = 120 / 144 = 0.833 A I = V / R = 15.8 / 1000 = 15.8 mA E1.14 Using KCL at the top node of the circuit, we have i1 = i2. Then, using KVL going clockwise, we have -v1 - v2 = 0; but v1 = 25 V, so we have v2 = -25 V. Next we have i1 = i2 = v2/R = -1 A. Finally, we have PR = v 2i2 = ( −25) × ( −1) = 25 W and Ps = v 1i1 = (25) × ( −1) = −25 W. E1.15 At the top node we have iR = is = 2A. By Ohm’s law we have vR = RiR = 80 V. By KVL we have vs = vR = 80 V. Then ps = -vsis = -160 W (the minus sign is due to the fact that the references for vs and is are opposite to the passive sign configuration). Also we have PR = v R iR = 160 W. 2 Problems P1.1 Four reasons that non-electrical engineering majors need to learn the fundamentals of EE are: 1. To pass the Fundamentals of Engineering Exam. 2. To be able to lead in the design of systems that contain electrical/electronic elements. 3. To be able to operate and maintain systems that contain electrical/electronic functional blocks. 4. To be able to communicate effectively with electrical engineers. P1.2 Broadly, the two objectives of electrical systems are: 1. To gather, store, process, transport, and display information. 2. To distribute, store, and convert energy between various forms. P1.3 Eight subdivisions of EE are: 1. 2. 3. 4. 5. 6. 7. 8. Communication systems. Computer systems. Control systems. Electromagnetics. Electronics. Photonics. Power systems. Signal Processing. P1.4 Responses to this question are varied. P1.5 (a) Electrical current is the time rate of flow of net charge through a conductor or circuit element. Its units are amperes, which are equivalent to coulombs per second. (b) The voltage between two points in a circuit is the amount of energy transferred per unit of charge moving between the points. Voltage has units of volts, which are equivalent to joules per coulomb. (c) The current through an open switch is zero. The voltage across the switch can be any value depending on the circuit. 3 (d) The voltage across a closed switch is zero. The current through the switch can be any value depending of the circuit. (e) Direct current is constant in magnitude and direction with respect to time. (f) Alternating current varies either in magnitude or direction with time. P1.6 (a) A conductor is analogous to a frictionless pipe. (b) An open switch is analogous to a closed valve. (c) A resistance is analogous to a constriction in a pipe or to a pipe with friction. (d) A battery is analogous to a pump. P1.7* The reference direction for iab points from a to b. Because iab has a negative value, the current is equivalent to positive charge moving opposite to the reference direction. Finally, since electrons have negative charge, they are moving in the reference direction (i.e., from a to b). For a constant (dc) current, charge equals current times the time interval. Thus, Q = (3 A) × (3 s) = 9 C. 2 coulomb/s = 12.5 × 1018 1.60 × 10 − 19 coulomb/electron P1.8 Electrons per second = P1.9* i (t ) = P1.10 The positive reference for v is at the head of the arrow, which is terminal b. The positive reference for vba is terminal b. Thus, we have v ba = v = −10 V. Also, i is the current entering terminal a, and iba is the dq (t ) d (2t + t 2 ) = 2 + 2t A = dt dt current leaving terminal a. Thus, we have i = −iba = −3 A. The true polarity is positive at terminal a, and the true current direction is entering terminal a. Thus, current enters the positive reference and energy is being delivered to the device. P1.11 To cause current to flow, we make contact between the conducting parts of the switch, and we say that the switch is closed. The corresponding fluid analogy is a valve that allows fluid to pass through. This corresponds to an open valve. Thus, an open valve is analogous to a closed 4 switch. P1.12* P1.13 ∞ ∞ 0 0 Q = ∫ i (t )dt = ∫ 2e −t dt = −2e −t | ∞0 = 2 coulombs (a) The sine function completes one cycle for each 2π radian increase in the angle. Because the angle is 200πt , one cycle is completed for each time interval of 0.01 s. The sketch is: (b) Q = 0.01 0.01 0 0 ∫ i (t )dt = ∫ 10 sin(200πt )dt = (10 / 200π ) cos(200πt ) =0 C (b) Q = 0.015 0.015 0 0 0 ∫ i (t )dt = ∫ 10 sin(200πt )dt = (10 / 200π ) cos(200πt ) = 0.0318 C P1.14* 0.01 0.015 0 The charge flowing through the battery is Q = (5 amperes) × (24 × 3600 seconds) = 432 × 10 3 coulombs The stored energy is Energy = QV = ( 432 × 10 3 ) × (12) = 5.184 × 10 6 joules (a) Equating gravitational potential energy, which is mass times height times the acceleration due to gravity, to the energy stored in the battery and solving for the height, we have Energy 5.184 × 10 6 h= = = 17.6 km mg 30 × 9.8 (b) Equating kinetic energy to stored energy and solving for velocity, we have 5 v = 2 × Energy m = 587.9 m/s (c) The energy density of the battery is 5.184 × 10 6 = 172.8 × 10 3 J/kg 30 which is about 0.384% of the energy density of gasoline. dq (t ) d (2t + e −2t ) = 2 − 2e −2t A = dt dt P1.15 i (t ) = P1.16 The number of electrons passing through a cross section of the wire per second is 5 N = = 3.125 × 1019 electrons/second − 19 1.6 × 10 The volume of copper containing this number of electrons is volume = 3.125 × 1019 = 3.125 × 10 −10 m3 1029 The cross sectional area of the wire is A= πd 2 4 = 3.301 × 10 − 6 m2 Finally, the average velocity of the electrons is volume u = = 94.67 µm/s A P1.17* Q = current × time = (10 amperes) × (36,000 seconds) = 3.6 × 10 5 coulombs Energy = QV = (3.6 × 10 5 ) × (12.6) = 4.536 × 10 6 joules P1.18 Q = current × time = (2 amperes) × (10 seconds) = 20 coulombs Energy = QV = (20) × (5) = 100 joules Notice that iab is positive. If the current were carried by positive charge, it would be entering terminal a. Thus, electrons enter terminal b. The energy is delivered to the element. 6 P1.19 The electron gains 1.6 × 10 −19 × 120 = 19.2 × 10 −18 joules P1.20* (a) P = -vaia = 30 W Energy is being absorbed by the element. (b) P = vbib = 30 W Energy is being absorbed by the element. (c) P = -vDEiED = -60 W Energy is being supplied by the element. P1.21 If the current is referenced to flow into the positive reference for the voltage, we say that we have the passive reference configuration. Using double subscript notation, if the order of the subscripts are the same for the current and voltage, either ab or ba, we have a passive reference configuration. P1.22* Q = w V = (600 J) (12 V) = 50 C . To increase the chemical energy stored in the battery, positive charge should move through the battery from the positive terminal to the negative terminal, in other words from a to b. Electrons move from b to a. P1.23 The amount of energy is W = QV = (4 C) × (15 V) = 60 J. Because the reference polarity for vab is positive at terminal a and the voltage value is negative, terminal b is actually the positive terminal. Because the charge moves from the negative terminal to the positive terminal, energy is removed from the device. P1.24* Energy = P = Cost $60 = = 500 kWh Rate 0.12 $/kWh Energy 500 kWh = = 694.4 W Time 30 × 24 h Reduction = I = 60 × 100% = 8.64% 694.4 7 P 694.4 = = 5.787 A 120 V P1.25 Notice that the references are opposite to the passive configuration. p (t ) = −v (t )i (t ) = −30e −t W ∞ Energy = ∫ p (t )dt = 30e −t | 0∞ = − 30 joules 0 Because the energy is negative, the element delivers the energy. P1.26 ( a) p (t ) = v ab iab = 50 sin(200πt ) W (b) w = 0.0025 ∫ p (t )dt = 0.0025 ∫ 50 sin(200πt )dt 0 0 = 79.58 mJ (c) w = 0.01 ∫ p (t )dt 0 =0 J = 0.01 ∫ 50 sin(200πt )dt 0 0.0025 = − (50 / 200π ) cos(200πt ) 0 0.01 = −(50 / 200π ) cos(200πt ) 0 *P1.27 (a) P = 50 W taken from element A. (b) P = 50 W taken from element A. (c) P = 50 W delivered to element A. P1.28 (a) P = 50 W delivered to element A. (b) P = 50 W delivered to element A. (c) P = 50 W taken from element A. P1.29 The current supplied to the electronics is i = p /v = 25 / 12.6 = 1.984 A. The ampere-hour rating of the battery is the operating time to discharge the battery multiplied by the current. Thus, the operating time is T = 80 / 1.984 = 40.3 h. The energy delivered by the battery is W = pT = 25(40.3) = 1008 Wh = 1.008 kWh. Neglecting the cost of recharging, the cost of energy for 250 discharge cycles is Cost = 85 /(250 × 1.008) = 0.337 $/kWh. 8 P1.30 The power that can be delivered by the cell is p = vi = 0.45 W. In 10 hours, the energy delivered is W = pT = 4.5 Whr = 0.0045 kWhr. Thus, the unit cost of the energy is Cost = (1.95) /(0.0045) = 433.33 $/kWhr which is 3611 times the typical cost of energy from electric utilities. P1.31 The sum of the currents entering a node equals the sum of the currents leaving. It is true because charge cannot collect at a node in an electrical circuit. P1.32 A node is a point that joins two or more circuit elements. All points joined by ideal conductors are electrically equivalent. Thus, there are five nodes in the circuit at hand: P1.33 The currents in series-connected elements are equal. P1.34* Elements E and F are in series. P1.35 For a proper fluid analogy to electric circuits, the fluid must be incompressible. Otherwise the fluid flow rate out of an element could be more or less than the inward flow. Similarly the pipes must be inelastic so the flow rate is the same at all points along each pipe. P1.36* At the node joining elements A and B, we have ia + ib = 0. Thus, ia = −2 A. For the node at the top end of element C, we have ib + ic = 3 . Thus, 9 ic = 1 A . Finally, at the top right-hand corner node, we have 3 + ie = id . Thus, id = 4 A . Elements A and B are in series. P1.37* We are given ia = 2 A, ib = 3 A, id = −5 A, and ih = 4 A. Applying KCL, we find ic = ib − ia = 1 A ie = ic + ih = 5 A if = ia + id = −3 A i g = if − ih = −7 A P1.38 (a) Elements C and D are in series. (b) Because elements C and D are in series, the currents are equal in magnitude. However, because the reference directions are opposite, the algebraic signs of the current values are opposite. Thus, we have ic = −id . (c) At the node joining elements A, B, and C, we can write the KCL equation i b = i a + ic = 4 − 1 = 3 A . Also, we found earlier that i d = −ic = 1 A. P1.39 We are given ia = 1 A, ic = −2 A, i g = 7 A, and ih = 2 A. Applying KCL, we find ib = ic + ia = −1 A id = if − ia = 8 A ie = ic + ih = 0 A if = i g + ih = 9 A P1.40 If one travels around a closed path adding the voltages for which one enters the positive reference and subtracting the voltages for which one enters the negative reference, the total is zero. KCL must be true for the law of conservation of energy to hold. P1.41* Summing voltages for the lower left-hand loop, we have − 5 + v a + 10 = 0, which yields v a = −5 V. Then for the top-most loop, we have v c − 15 − v a = 0, which yields v c = 10 V. Finally, writing KCL around the outside loop, we have − 5 + v c + v b = 0, which yields v b = −5 V. P1.42* Applying KCL and KVL, we have ic = ia − id = 1 A v b = v d − v a = −6 V ib = −ia = −2 A vc = vd = 4 V The power for each element is PA = −v a ia = −20 W PC = v c ic = 4 W Thus, PA + PB + PC + PD = 0 PB = v b ib = 12 W PD = v d id = 4 W 10 P1.43 (a) Elements A and B are in parallel. (b) Because elements A and B are in parallel, the voltages are equal in magnitude. However because the reference polarities are opposite, the algebraic signs of the voltage values are opposite. Thus, we have v a = −v b . (c) Writing a KVL equation while going clockwise around the loop composed of elements A, C and D, we obtain v a − v d − v c = 0. Solving for v d and substituting values, we find v d = 5 V. Also, we have v b = −v a = −12 V. P1.44 There are two nodes, one at the center of the diagram and the outer periphery of the circuit comprises the other. Elements A, B, C, and D are in parallel. No elements are in series. P1.45 We are given v a = 15 V, v b = −7 V, vf = 10 V, and v h = 4 V. Applying KVL, we find P1.46 vd = v a + v b = 8 V ve = −v a − vc + vd = 22 V vc = −v a − vf − v h = −29 V v g = ve − v h = 18 V The points and the voltages specified in the problem statement are: Applying KVL to the loop abca, substituting values and solving, we obtain: v ab − v cb − v ac = 0 15 + 7 − v ac = 0 v ac = 22 V Similiarly, applying KVL to the loop abcda, substituting values and solving, we obtain: v ab − v cb + v cd + v da = 0 15 + 7 + vcd + 10 = 0 vcd = −32 V 11 P1.47 (a) In Figure P1.36, elements C, D, and E are in parallel. (b) In Figure P1.42, elements C and D are in parallel. (c) In Figure P1.45, no element is in parallel with another element. P1.48 Six batteries are needed and they need to be connected in series. A typical configuration looking down on the tops of the batteries is shown: P1.49 (a) The voltage between any two points of an ideal conductor is zero regardless of the current flowing. (b) An ideal voltage source maintains a specified voltage across its terminals. (c) An ideal current source maintains a specified current through itself. (d) The voltage across a short circuit is zero regardless of the current flowing. When an ideal conductor is connected between two points, we say that the points are shorted together. (e) The current through an open circuit is zero regardless of the voltage. P1.50 Provided that the current reference points into the positive voltage reference, the voltage across a resistance equals the current through the resistance times the resistance. On the other hand, if the current reference points into the negative voltage reference, the voltage equals the negative of the product of the current and the resistance. P1.51 Four types of controlled sources and the units for their gain constants are: 1. Voltage-controlled voltage sources. V/V or unitless. 2. Voltage-controlled current sources. A/V or siemens. 3. Current-controlled voltage sources. V/A or ohms. 4. Current-controlled current sources. A/A or unitless. 12 P1.52* P1.53 P1.54 The resistance of the copper wire is given by RCu = ρCu L A , and the resistance of the tungsten wire is RW = ρW L A . Taking the ratios of the respective sides of these equations yields RW RCu = ρW ρCu . Solving for RW and substituting values, we have RW = RCu ρW ρ Cu = (1.5) × (5.44 × 10 -8 ) (1.72 × 10 −8 ) = 4.74 Ω P1.55 Equation 1.10 gives the resistance as R = ρL A (a) Thus, if the length of the wire is doubled, the resistance doubles to 20 Ω . (b) If the diameter of the wire is doubled, the cross sectional area A is increased by a factor of four. Thus, the resistance is decreased by a factor of four to 2.5 Ω . 13 P1.56 P1.57 P1.58* P1.59 R ( V1 )2 = P2 (V ) = 2 = P1 R 1002 = 100 Ω 100 2 = 902 = 81 W for a 19% reduction in power 100 The power delivered to the resistor is p (t ) = i 2 (t ) R = 10 exp( − 6t ) and the energy delivered is ∞ ∞  10  ∞ 10 w = ∫ p (t )dt = ∫ 10 exp( −6t )dt =  exp( −6t ) = = 1.667 J − 6 6   0 0 0 P1.60 The power delivered to the resistor is p (t ) = v 2 (t ) / R = 5 sin 2 (2πt ) = 2.5 − 2.5 cos(4πt ) and the energy delivered is 2 w = ∫ p (t )dt = 0 2 ∫ [2.5 − 2.5 cos(4πt )]dt 0 2 2. 5   = 2.5t − sin(4πt ) = 5 J 4π  0 14 P1.61 (a) Ohm's law gives iab = vab/2. (b) The current source has iab = 2 independent of vab, which plots as a horizontal line in the iab versus vab plane. (c) The voltage across the voltage source is 5 V independent of the current. Thus, we have vab = 5 which plots as a vertical line in the iab versus vab plane. (d) Applying KCL and Ohm's law, we obtain iab = v ab / 2 + 1 . (e) Applying Ohm's law and KVL, we obtain v ab = iab + 2 which is equivalent to iab = v ab − 2. The plots for all five parts are shown. P1.62* (a) Not contradictory. (b) A 2-A current source in series with a 3-A current source is contradictory because the currents in series elements must be equal. (c) Not contradictory. (d) A 2-A current source in series with an open circuit is contradictory because the current through a short circuit is zero by definition and currents in series elements must be equal. (e) A 5-V voltage source in parallel with a short circuit is contradictory because the voltages across parallel elements must be equal and the voltage across a short circuit is zero by definition. 15 P1.63* As shown above, the 2 A current circulates clockwise through all three elements in the circuit. Applying KVL, we have v c = v R + 10 = 5iR + 10 = 20 V Pcurrent − source = −v c iR = −40 W. Thus, the current source delivers power. PR = (iR ) 2 R = 22 × 5 = 20 W. The resistor absorbs power. Pvoltage − source = 10 × iR = 20 W. The voltage source absorbs power. P1.64* Applying Ohm's law, we have v 2 = (5 Ω ) × (1 A) = 5 V . However,v 2 is the voltage across all three resistors that are in parallel. Thus, i3 = v2 = 1 A , and i2 = v2 = 0.5 A . Applying KCL, we have 10 i1 = i2 + i3 + 1 = 2.5 A . By Ohm's law: v 1 = 5i1 = 12.5 V . Finally using KVL, 5 we havev x = v 1 + v 2 = 17.5 V . P1.65 The power for each element is 30 W. The voltage source delivers power and the current source absorbs it. 16 P1.66 This is a parallel circuit, and the voltage across each element is 15 V positive at the top end. Thus, the current flowing through the resistor is 15 V iR = = 3A 5Ω Applying KCL, we find that the current through the voltage source is 6 A flowing upward. Computing power for each element, we find Pcurrent − source = 45 W Thus, the current source absorbs power. PR = (iR )2 R = 45 W Pvoltage − source = −90 W The voltage source delivers power. P1.67 Ohm’s law for the right-hand 5-Ω resistor yields: v1 = 5 V. Then, we have i1 = v1 / 5 = 1 A. Next, KCL yields i2 = i1 + 1 = 2 A. Then for the 10-Ω resistor, we have v 2 = 10i2 = 20 V. Using KVL, we have v 3 = v1 + v 2 = 25 V. Next, applying Ohms law, we obtain i3 = v3 / 10 = 2.5 A. Finally applying KCL, we have I x = i2 + i3 = 4.5 A. P1.68 (a) The 3-Ω resistance, the 2-Ω resistance, and the voltage source Vx are in series. (b) The 6-Ω resistance and the 12-Ω resistance are in parallel. (c) Refer to the sketch of the circuit. Applying Ohm's law to the 6-Ω resistance, we determine that v1 = 12 V. Then, applying Ohm's law to the 12-Ω resistance, we have i1 = 1 A. Next, KCL yields i2 = 3 A. Continuing, 17 we use Ohm's law to find that v 2 = 6 V and v 3 = 9 V. Finally, applying KVL, we have Vx = v 3 + v1 + v2 = 27 V. P1.69* (a) Applying KVL, we have 10 = v x + 5v x , which yields v x = 10 / 6 = 1.667 V (b) ix = v x / 3 = 0.5556 A (c) Pvoltage − source = −10ix = −5.556 W. (This represents power delivered by the voltage source.) PR = 3(ix ) 2 = 0.926 W (absorbed) Pcontrolled − source = 5v x ix = 4.63 W (absorbed) P1.70* We have a voltage-controlled current source in this circuit. v x = (4 Ω) × (1 A) = 4 V is = v x / 2 + 1 = 3 A Applying KVL around the outside of the circuit, we have: v s = 3is + 4 + 2 = 15 V P1.71 This is a current-controlled current source. First, we have vo = Po 8 = 8 V. Then, we have i1 = vo / 32 = 0.25 A and io = vo / 8 = 1 A. KCL gives 2000iin = i1 + io = 1.25 A. Thus we have iin = 1.25 / 2000 = 625 µA. Then, vin = 2iin = 1.25 mV, and finally we have I x = iin + vin / 5 = 875 µA. 18 P1.72 (a) No elements are in series. (b) Rx and the 8-Ω resistor are in parallel. Also, the 3-Ω resistor and the 6-Ω resistor are in parallel. Thus, the voltages across the parallel elements are the same, as labeled in the figure. (c) P1.73 vy = 3 × 2 = 6 V i6 = v y / 6 = 1 A v x = 10 − v y = 4 V i8 = v x / 8 = 0. 5 i s = i 6 − i8 = 0. 5 A i x = i s + 2 = 2. 5 A Rx = v x / ix = 1.6 Ω i2 = v / 2 v =6 V i6 = v / 6 i2 = 3 A i2 + i6 = v / 2 + v / 6 = 4 i6 = 1 A 19 P1.74 This circuit contains a voltage-controlled voltage source. Applying KVL around the periphery of the circuit, we have − 16 + v x + 3v x = 0, which yields v x = 4 V. Then, we have v 12 = 3v x = 12 V. Using Ohm’s law we obtain i12 = v 12 / 12 = 1 A and ix = v x / 2 = 2 A. Then KCL applied to the node at the top of the 12-Ω resistor gives i x = i12 + i y which yields i y = 1 A. P1.75 Consider the series combination shown below on the left. Because the current for series elements must be the same and the current for the current source is 2 A by definition, the current flowing from a to b is 2 A. Notice that the current is not affected by the 10-V source in series. Thus, the series combination is equivalent to a simple current source as far as anything connected to terminals a and b is concerned. P1.76 Consider the parallel combination shown below. Because the voltage for parallel elements must be the same, the voltage vab must be 10 V. Notice that vab is not affected by the current source. Thus, the parallel combination is equivalent to a simple voltage source as far as anything connected to terminals a and b is concerned. 20 P1.77 (a) 20 = v 1 + v 2 (b) v 1 = 4i v 2 = 6i (c) 20 = 4i + 6i i =2 A (d) Pvoltage − source = −20i = −40 W. (Power delivered by the source.) P1 = 4i 2 = 16 W (Power absorbed by R1 .) P1 = 6i 2 = 24 W (Power absorbed by R2 .) P1.78 (a) 3 = i1 + i2 (b) i1 = v / 5 i2 = v / 10 (c) 3 = v / 5 + v / 10 v = 10 V (d) Pcurrentsource = −I sv = −30 W (Power is supplied by the source.) P1 = v 2 / R1 = 20 W (Power is absorbed by R1.) P2 = v 2 / R2 = 10 W (Power is absorbed by R2.) P1.79 The source labeled is is an independent current source. The source labeled aix is a current-controlled current source. Applying ohm's law to the 5-Ω resistance gives: i x = −20 V/5 Ω = −4 A Applying KCL for the node at the top end of the controlled current source: is = 0.5ix − ix = −0.5ix = 2 A P1.80 . The source labeled 10 V is an independent voltage source. The source labeled aix is a current-controlled voltage source. Applying Ohm's law and KVL, we have − 10 + 7i x + 3i x = 0. Solving, we obtain i x = 1 A. 21 Practice Test T1.1 (a) 4; (b) 7; (c) 16; (d) 18; (e) 1; (f) 2; (g) 8; (h) 3; (i) 5; (j) 15; (k) 6; (l) 11; (m) 13; (n) 9; (o) 14. T1.2 (a) The current Is = 3 A circulates clockwise through the elements entering the resistance at the negative reference for vR. Thus, we have vR = −IsR = −6 V. (b) Because Is enters the negative reference for Vs, we have PV = −VsIs = −30 W. Because the result is negative, the voltage source is delivering energy. (c) The circuit has three nodes, one on each of the top corners and one along the bottom of the circuit. (d) First, we must find the voltage vI across the current source. We choose the reference shown: Then, going around the circuit counterclockwise, we have − v I +Vs + v R = 0 , which yields v I =Vs + v R = 10 − 6 = 4 V. Next, the power for the current source is PI = I sv I = 12 W. Because the result is positive, the current source is absorbing energy. Alternatively, we could compute the power delivered to the resistor as PR = I s2R = 18 W. Then, because we must have a total power of zero for the entire circuit, we have PI = −PV − PR = 30 − 18 = 12 W. T1.3 (a) The currents flowing downward through the resistances are vab/R1 and vab/R2. Then, the KCL equation for node a (or node b) is I2 = I1 + v ab v ab + R1 R2 Substituting the values given in the question and solving yields vab = −8 V. 22 (b) The power for current source I1 is PI 1 = v ab I1 = −8 × 3 = −24 W . Because the result is negative we know that energy is supplied by this current source. The power for current source I2 is PI 2 = −v ab I 2 = 8 × 1 = 8 W . Because the result is positive, we know that energy is absorbed by this current source. 2 (c) The power absorbed by R1 is PR 1 = v ab / R1 = ( −8)2 / 12 = 5.33 W. The 2 power absorbed by R2 is PR 2 = v ab / R2 = ( −8) 2 / 6 = 10.67 W. T1.4 (a) Applying KVL, we have −Vs + v 1 + v 2 = 0. Substituting values given in the problem and solving we find v1 = 8 V. (b) Then applying Ohm's law, we have i = v1 / R1 = 8 / 4 = 2 A. (c) Again applying Ohm's law, we have R2 = v 2 / i = 4 / 2 = 2 Ω. T1.5 Applying KVL, we have −Vs + v x = 0. Thus, v x = Vs = 15 V. Next Ohm's law gives ix = v x / R = 15 / 10 = 1.5 A. Finally, KCL yields i sc = i x − av x = 1.5 − 0.3 × 15 = −3 A. 23 CHAPTER 2 Exercises E2.1 (a) R2, R3, and R4 are in parallel. Furthermore R1 is in series with the combination of the other resistors. Thus we have: 1 Req = R1 + = 3Ω 1 / R2 + 1 / R3 + 1 / R4 (b) R3 and R4 are in parallel. Furthermore, R2 is in series with the combination of R3, and R4. Finally R1 is in parallel with the combination of the other resistors. Thus we have: 1 =5Ω Req = 1 / R1 + 1 /[R2 + 1 /(1 / R3 + 1 / R4 )] (c) R1 and R2 are in parallel. Furthermore, R3, and R4 are in parallel. Finally, the two parallel combinations are in series. 1 1 + = 52.1 Ω Req = 1 / R1 + 1 / R2 1 / R3 + 1 / R4 (d) R1 and R2 are in series. Furthermore, R3 is in parallel with the series combination of R1 and R2. 1 = 1.5 kΩ Req = 1 / R3 + 1 /(R1 + R2 ) E2.2 (a) First we combine R2, R3, and R4 in parallel. Then R1 is in series with the parallel combination. Req = 1 = 9.231 Ω 1 / R2 + 1 / R3 + 1 / R4 i1 = 20 V 20 = = 1.04 A R1 + Req 10 + 9.231 v eq = Req i1 = 9.600 V i2 = v eq / R2 = 0.480 A i4 = v eq / R4 = 0.240 A 24 i3 = v eq / R3 = 0.320 A (b) R1 and R2 are in series. Furthermore, R3, and R4 are in series. Finally, the two series combinations are in parallel. Req 1 = R1 + R2 = 20 Ω v eq = 2 × Req = 20 V Req 2 = R3 + R4 = 20 Ω Req = i1 = v eq / Req 1 = 1 A 1 / Req 1 1 = 10 Ω + 1 / Req 2 i2 = v eq / Req 2 = 1 A (c) R3, and R4 are in series. The combination of R3 and R4 is in parallel with R2. Finally the combination of R2, R3, and R4 is in series with R1. Req 1 = R3 + R4 = 40 Ω Req 2 = v 2 = i1Req 2 = 20 V E2.3 (a) v 1 = v s 1 = 20 Ω 1 / Req 1 + 1 / R2 i1 = vs =1A R1 + Req 2 i2 = v 2 / R2 = 0.5 A i3 = v 2 / Req 1 = 0.5 A R1 = 10 V . v 2 = v s R2 R1 + R2 + R3 + R4 R1 + R2 + R3 + R4 Similarly, we find v 3 = 30 V and v 4 = 60 V . 25 = 20 V . (b) First combine R2 and R3 in parallel: Req = 1 (1 / R2 + 1 R3 ) = 2.917 Ω. R1 = 6.05 V . Similarly, we find R1 + Req + R4 Then we have v 1 = v s v2 = vs E2.4 Req = 5.88 V and v 4 = 8.07 V . R1 + Req + R4 (a) First combine R1 and R2 in series: Req = R1 + R2 = 30 Ω. Then we have Req R3 15 30 i1 = is = = 1 A and i3 = is = = 2 A. R3 + Req 15 + 30 R3 + Req 15 + 30 (b) The current division principle applies to two resistances in parallel. Therefore, to determine i1, first combine R2 and R3 in parallel: Req = Req 5 = = 1A. 1/(1/R2 + 1/R3) = 5 Ω. Then we have i1 = is R1 + Req 10 + 5 Similarly, i2 = 1 A and i3 = 1 A. E2.5 Write KVL for the loop consisting of v1, vy , and v2. The result is -v1 - vy + v2 = 0 from which we obtain vy = v2 - v1. Similarly we obtain vz = v3 - v1. E2.6 Node 1: E2.7 Following the step-by-step method in the book, we obtain v1 − v3 v1 − v2 v − v1 v2 v2 − v3 + = ia Node 2: 2 + + =0 R1 R2 R3 R4 R2 v v − v2 v3 − v1 Node 3: 3 + 3 + + ib = 0 R5 R4 R1 1 1  +  R1 R2  − 1  R2   0  E2.8 − 1 R2 + 1 R2 1 R3 − 1 R4    v 1   − is  1    − v = 0  R4   2   1 1  v 3   is  +  R4 R5  0 + 1 R4 Instructions for various calculators vary. The MATLAB solution is given in the book following this exercise. 26 E2.9 (a) Writing the node equations we obtain: v − v3 v1 v1 − v2 Node 1: 1 + + =0 20 5 10 v −v3 v − v1 Node 2: 2 + 10 + 2 =0 10 5 v − v1 v3 v3 − v2 Node 3: 3 + + =0 20 10 5 (b) Simplifying the equations we obtain: 0.35v 1 − 0.10v 2 − 0.05v 3 = 0 − 0.10v 1 + 0.30v 2 − 0.20v 3 = −10 − 0.05v 1 − 0.20v 2 + 0.35v 3 = 0 (c) and (d) Solving using Matlab: >>clear >>G = [0.35 -0.1 -0.05; -0.10 0.30 -0.20; -0.05 -0.20 0.35]; >>I = [0; -10; 0]; >>V = G\I V= -27.2727 -72.7273 -45.4545 >>Ix = (V(1) - V(3))/20 Ix = 0.9091 E2.10 Using determinants we can solve for the unknown voltages as follows: 6 − 0.2 1 0. 5 3 + 0.2 v1 = = = 10.32 V 0.7 − 0.2 0.35 − 0.04 − 0.2 0.5 0.7 6 − 0.2 1 0 . 7 + 1. 2 v2 = = = 6.129 V 0.7 − 0.2 0.35 − 0.04 − 0.2 0.5 Many other methods exist for solving linear equations. 27 E2.11 First write KCL equations at nodes 1 and 2: Node 1: v 1 − 10 v 1 v1 − v2 =0 2 5 10 v − 10 v 2 v 2 − v 1 Node 2: 2 + + =0 10 5 10 + + Then, simplify the equations to obtain: 8v 1 − v 2 = 50 and − v 1 + 4v 2 = 10 Solving manually or with a calculator, we find v1 = 6.77 V and v2 = 4.19 V. The MATLAB session using the symbolic approach is: >> clear [V1,V2] = solve('(V1-10)/2+(V1)/5 +(V1 - V2)/10 = 0' , ... '(V2-10)/10 +V2/5 +(V2-V1)/10 = 0') V1 = 210/31 V2 = 130/31 Next, we solve using the numerical approach. >> clear G = [8 -1; -1 4]; I = [50; 10]; V = G\I V= 6.7742 4.1935 E2.12 The equation for the supernode enclosing the 15-V source is: v3 − v2 v3 − v1 v1 v2 + = + R3 R1 R2 R4 This equation can be readily shown to be equivalent to Equation 2.37 in the book. (Keep in mind that v3 = -15 V.) 28 E2.13 Write KVL from the reference to node 1 then through the 10-V source to node 2 then back to the reference node: − v 1 + 10 + v 2 = 0 Then write KCL equations. First for a supernode enclosing the 10-V source, we have: v1 v1 − v3 v2 − v3 + + =1 R1 R2 R3 Node 3: v3 v3 − v1 v3 − v2 + + =0 R4 R2 R3 Reference node: v1 v3 + =1 R1 R4 An independent set consists of the KVL equation and any two of the KCL equations. E2.14 (a) Select the reference node at the left-hand end of the voltage source as shown at right. Then write a KCL equation at node 1. v 1 v 1 − 10 + +1 = 0 R1 R2 Substituting values for the resistances and solving, we find v1 = 3.33 V. 10 − v 1 = 1.333 A. Then we have ia = R2 (b) Select the reference node and assign node voltages as shown. Then write KCL equations at nodes 1 and 2. 29 v 1 − 25 v 1 v 1 − v 2 + + =0 R2 R4 R3 v 2 − 25 v 2 − v 1 v 2 + + =0 R1 R3 R5 Substituting values for the resistances and solving, we find v1 = 13.79 V v −v2 = -0.259 A. and v 2 = 18.97 V. Then we have ib = 1 R3 E2.15 (a) Select the reference node and node voltage as shown. Then write a KCL equation at node 1, resulting in v 1 v 1 − 10 + − 2ix = 0 5 5 Then use ix = (10 − v 1 ) / 5 to substitute and solve. We find v1 = 7.5 V. 10 − v 1 = 0.5 A. Then we have ix = 5 (b) Choose the reference node and node voltages shown: Then write KCL equations at nodes 1 and 2: v1 5 + v 1 − 2i y 2 +3= 0 v2 5 + v 2 − 2i y 10 30 =3 Finally use i y = v 2 / 5 to substitute and solve. This yields v 2 = 11.54 V and i y = 2.31 A. E2.16 >> clear >> [V1 V2 V3] = solve('V3/R4 + (V3 - V2)/R3 + (V3 - V1)/R1 = 0', ... 'V1/R2 + V3/R4 = Is', ... 'V1 = (1/2)*(V3 - V1) + V2' ,'V1','V2','V3'); >> pretty(V1), pretty(V2), pretty(V3) R2 Is (2 R3 R1 + 3 R4 R1 + 2 R4 R3) ----------------------------------------------2 R3 R1 + 3 R4 R1 + 3 R1 R2 + 2 R4 R3 + 2 R3 R2 R2 Is (3 R3 R1 + 3 R4 R1 + 2 R4 R3) ----------------------------------------------2 R3 R1 + 3 R4 R1 + 3 R1 R2 + 2 R4 R3 + 2 R3 R2 Is R2 R4 (3 R1 + 2 R3) ----------------------------------------------2 R3 R1 + 3 R4 R1 + 3 R1 R2 + 2 R4 R3 + 2 R3 R2 E2.17 Refer to Figure 2.33b in the book. (a) Two mesh currents flow through R2: i1 flows downward and i4 flows upward. Thus the current flowing in R2 referenced upward is i4 - i1. (b) Similarly, mesh current i1 flows to the left through R4 and mesh current i2 flows to the right, so the total current referenced to the right is i2 - i1. (c) Mesh current i3 flows downward through R8 and mesh current i4 flows upward, so the total current referenced downward is i3 - i4. (d) Finally, the total current referenced upward through R8 is i4 - i3. E2.18 Refer to Figure 2.33b in the book. Following each mesh current in turn, we have R1i1 + R2 (i1 − i4 ) + R4 (i1 − i2 ) − v A = 0 R5i2 + R4 (i2 − i1 ) + R6 (i2 − i3 ) = 0 R7i3 + R6 (i3 − i2 ) + R8 (i3 − i4 ) = 0 R3i4 + R2 (i4 − i1 ) + R8 (i4 − i3 ) = 0 31 In matrix form, these equations become − R4 0 − R2  i1  v A  (R1 + R2 + R4 )  i   0   − R4 (R4 + R5 + R6 ) − R6 0  2  =     i3   0   0 − R6 (R6 + R7 + R8 ) − R8      − R2 0 − R8 (R2 + R3 + R8 ) i 4   0   E2.19 We choose the mesh currents as shown: Then, the mesh equations are: 5i 1 + 10(i1 − i2 ) = 100 and 10(i2 − i1 ) + 7i2 + 3i2 = 0 Simplifying and solving these equations, we find that i1 = 10 A and i2 = 5 A. The net current flowing downward through the 10-Ω resistance is i1 − i2 = 5 A. To solve by node voltages, we select the reference node and node voltage shown. (We do not need to assign a node voltage to the connection between the 7-Ω resistance and the 3-Ω resistance because we can treat the series combination as a single 10-Ω resistance.) 32 The node equation is (v 1 − 10) / 5 + v 1 / 10 + v 1 / 10 = 0 . Solving we find that v1 = 50 V. Thus we again find that the current through the 10-Ω resistance is i = v 1 / 10 = 5 A. Combining resistances in series and parallel, we find that the resistance “seen” by the voltage source is 10 Ω. Thus the current through the source and 5-Ω resistance is (100 V)/(10 Ω) = 10 A. This current splits equally between the 10-Ω resistance and the series combination of 7 Ω and 3 Ω. E2.20 First, we assign the mesh currents as shown. Then we write KVL equations following each mesh current: 2(i1 − i3 ) + 5(i1 − i2 ) = 10 5i2 + 5(i2 − i1 ) + 10(i2 − i3 ) = 0 10i3 + 10(i3 − i2 ) + 2(i3 − i1 ) = 0 Simplifying and solving, we find that i1 = 2.194 A, i2 = 0.839 A, and i3 = 0.581 A. Thus the current in the 2-Ω resistance referenced to the right is i1 - i3 = 2.194 - 0.581 = 1.613 A. E2.21 Following the step-by-step process, we obtain − R3 − R2  i1   v A  (R2 + R3 )  i  =  − v   −R (R3 + R4 ) 0 3  2   B    − R2 0 (R1 + R2 ) i3   v B  33 E2.22 Refer to Figure 2.39 in the book. In terms of the mesh currents, the current directed to the right in the 5-A current source is i1, however by the definition of the current source, the current is 5 A directed to the left. Thus, we conclude that i1 = -5 A. Then we write a KVL equation following i2, which results in 10(i2 − i1 ) + 5i2 = 100. E2.23 Refer to Figure 2.40 in the book. First, for the current source, we have i2 − i1 = 1 Then, we write a KVL equation going around the perimeter of the entire circuit: 5i1 + 10i2 + 20 − 10 = 0 Simplifying and solving these equations we obtain i1 = -4/3 A and i2 = -1/3 A. E2.24 (a) As usual, we select the mesh currents flowing clockwise around the meshes as shown. Then for the current source, we have i2 = -1 A. This is because we defined the mesh current i2 as the current referenced downward through the current source. However, we know that the current through this source is 1 A flowing upward. Next we write a KVL equation around mesh 1: 10i1 − 10 + 5(i1 − i2 ) = 0. Solving, we find that i1 = 1/3 A. Referring to Figure 2.30a in the book we see that the value of the current ia referenced downward through the 5 Ω resistance is to be found. In terms of the mesh currents, we have ia = i1 − i2 = 4 / 3 A . 34 (b) As usual, we select the mesh currents flowing clockwise around the meshes as shown. Then we write a KVL equation for each mesh. − 25 + 10(i1 − i3 ) + 10(i1 − i2 ) = 0 10(i2 − i1 ) + 20(i2 − i3 ) + 20i2 = 0 10(i3 − i1 ) + 5i3 + 20(i3 − i2 ) = 0 Simplifying and solving, we find i1 = 2.3276 A, i2 = 0.9483 A, and i3 = 1.2069 A. Finally, we have ib = i2 - i3 = -0.2586 A. E2.25 (a) KVL mesh 1: − 10 + 5i1 + 5(i1 − i2 ) = 0 For the current source: i2 = −2ix However, ix and i1 are the same current, so we also have i1 = ix. Simplifying and solving, we find ix = i1 = 0.5 A. (b) First for the current source, we have: i1 = 3 A Writing KVL around meshes 2 and 3, we have: 2(i2 − i1 ) + 2i y + 5i2 = 0 10(i3 − i1 ) + 5i3 − 2i y = 0 However i3 and iy are the same current: i y = i3 . Simplifying and solving, we find that i3 = i y = 2.31 A. 35 E2.26 Under open-circuit conditions, 5 A circulates clockwise through the current source and the 10-Ω resistance. The voltage across the 10-Ω resistance is 50 V. No current flows through the 40-Ω resistance so the open circuit voltage is Vt = 50 V. With the output shorted, the 5 A divides between the two resistances in parallel. The short-circuit current is the current through the 40-Ω 10 = 1 A. Then, the Thévenin resistance is resistance, which is isc = 5 10 + 40 Rt = v oc / isc = 50 Ω. E2.27 Choose the reference node at the bottom of the circuit as shown: Notice that the node voltage is the open-circuit voltage. Then write a KCL equation: v oc − 20 v oc + =2 5 20 Solving we find that voc = 24 V which agrees with the value found in Example 2.17. E2.28 To zero the sources, the voltage sources become short circuits and the current sources become open circuits. The resulting circuits are : 36 (a) Rt = 10 + (c) Rt = E2.29 1 = 14 Ω 1 / 5 + 1 / 20 1 + 10 6 + 1 1 (b) Rt = 10 + 20 = 30 Ω = 5Ω 1 (1 / 5 + 1 / 20) (a) Zero sources to determine Thévenin resistance. Thus 1 Rt = = 9.375 Ω. 1 / 15 + 1 / 25 Then find short-circuit current: I n = isc = 10 / 15 + 1 = 1.67 A 37 (b) We cannot find the Thévenin resistance by zeroing the sources, because we have a controlled source. Thus, we find the open-circuit voltage and the short-circuit current. v oc − 2v x v oc =2 v oc = 3v x 10 30 Solving, we find Vt = v oc = 30 V. + Now, we find the short-circuit current: 2v x + v x = 0 ⇒ vx = 0 Therefore isc = 2 A. Then we have Rt = v oc / isc = 15 Ω. E2.30 First, we transform the 2-A source and the 5-Ω resistance into a voltage source and a series resistance: 38 10 + 10 = 1.333 A. 15 From the original circuit, we have i1 = i2 − 2, from which we find i1 = −0.667 A. Then we have i2 = The other approach is to start from the original circuit and transform the 10-Ω resistance and the 10-V voltage source into a current source and parallel resistance: 1 = 3.333 Ω . 1 / 5 + 1 / 10 The current flowing upward through this resistance is 1 A. Thus the voltage across Req referenced positive at the bottom is 3.333 V and i1 = −3.333 / 5 = −0.667 A. Then from the original circuit we Then we combine the resistances in parallel. Req = have i2 = 2 + i1 = 1.333 A, as before. E2.31 Refer to Figure 2.62b. We have i1 = 15 / 15 = 1 A. Refer to Figure 2.62c. Using the current division principle, we have 5 i2 = −2 × = −0.667 A. (The minus sign is because of the reference 5 + 10 direction of i2.) Finally, by superposition we have iT = i1 + i2 = 0.333 A. E2.32 With only the first source active we have: Then we combine resistances in series and parallel: 1 Req = 10 + = 13.75 Ω 1 / 5 + 1 / 15 Thus, i1 = 20 / 13.75 = 1.455 A, and v 1 = 3.75i1 = 5.45 V. 39 With only the second source active, we have: Then we combine resistances in series and parallel: 1 Req 2 = 15 + = 18.33 Ω 1 / 5 + 1 / 10 Thus, is = 10 / 18.33 = 0.546 A, and v 2 = 3.33is = 1.818 V. Then, we have i2 = ( −v 2 ) / 10 = −0.1818 A Finally we have vT = v 1 + v 2 = 5.45 + 1.818 = 7.27 V and iT = i1 + i2 = 1.455 − 0.1818 = 1.27 A. Problems P2.1* (a) Req = 20 Ω (b) Req = 23 Ω P2.2* We have 4 + P2.3* The 12- Ω and 6- Ω resistances are in parallel having an equivalent resistance of 4 Ω . Similarly, the 18-Ω and 9-Ω resistances are in parallel and have an equivalent resistance of 6 Ω. Finally, the two parallel combinations are in series, and we have 1 = 8 which yields Rx = 5 Ω. 1 / 20 + 1 / Rx Rab = 4 + 6 = 10 Ω 40 P2.4* P2.5* The 20-Ω and 30-Ω resistances are in parallel and have an equivalent resistance of Req1 = 12 Ω. Also the 40-Ω and 60-Ω resistances are in parallel with an equivalent resistance of Req2 = 24 Ω. Next we see that Req1 and the 4-Ω resistor are in series and have an equivalent resistance of Req3 = 4 + Req1 = 16 Ω. Finally Req3 and Req2 are in parallel and the overall equivalent resistance is 1 Rab = = 9.6 Ω 1 / Req 1 + 1 / Req 2 P2.6 (a) Req = 22 Ω (b) Req = 20 Ω (c) Notice that the points labeled c are the same node and that the points labeled d are another node. Thus, the 30-Ω, 24-Ω, and 20-Ω resistors are in parallel because they are each connected between nodes c and d. The equivalent resistance is 18 Ω. P2.7 We have 1 = 48 which yields Rx = 80 Ω. 1 / 120 + 1 / Rx 41 P2.8 (a) Req = 18 Ω P2.9 We have Req = (b) Req = 10 Ω R (2R ) 2R = . Clearly, for Req to be an integer, R must be R + 2R 3 an integer multiple of 3. P2.10 Rab = 20 Ω P2.11 Because the resistances are in parallel, the same voltage v appears across both of them. The current through R1 is i1 = v/90. The current through R2 is i2 = 3i1 = 3v/90. Finally, we have R2 = v/i2 = v/(3v/90) = 30 Ω. P2.12 Combining the resistances shown in Figure P2.12b, we have 2Req 1 Req = 1 + +1 =2+ 1 / 2 + 1 Req 2 + Req Req (2 + Req ) = 2(2 + Req ) + 2Req (R ) eq 2 − 2Req − 4 = 0 Req = 3.236 Ω (Req = −1.236 Ω is another root, but is not physically reasonable.) P2.13 P2.14 Req = 1 1 1 1 + + + ... 1000 1000 1000 = 1 = n 1000 1000 n In the lowest power mode, the power is Plowest = 1202 = 83.33 W. R1 + R2 For the highest power mode, the two elements should be in parallel with an applied voltage of 240 V. The resulting power is 42 Phighest = 240 2 R1 + 240 2 R2 = 1000 + 500 = 1500 W. Some other modes and resulting powers are: R1 operated separately from 240 V yielding 1000 W R2 operated separately from 240 V yielding 500 W R1 in series with R2 operated from 240 V yielding 333.3 W R1 operated separately from 120 V yielding 250 W P2.15 For operation at the lowest power, we have 1202 P = 240 = R1 + R2 At the high power setting, we have 1202 1202 P = 1280 = + R1 R2 Solving these equations, we find R1 = 15 Ω and R2 = 45 Ω . (The second solution simply has the values of R1 and R2 interchanged.) The intermediate power settings are obtained by operating one of the elements from 120 V resulting in powers of 320 W and 960 W. P2.16 By symmetry, we find the currents in the resistors as shown below: Then, the voltage between terminals a and b is v ab = Req = 1 3 + 1 6 + 1 3 = 5 6 P2.17 R = 16 Ω. 43 P2.18 (a) For a series combination Geq = 1 1 / G1 + 1 / G2 + 1 / G3 (b) For a parallel combination of conductances Geq = G1 + G2 + G3 P2.19 To supply the loads in such a way that turning one load on or off does not affect the other loads, we must connect the loads in series with a switch in parallel with each load: To turn a load on, we open the corresponding switch, and to turn a load off, we close the switch. P2.20 We have Ra + Rb = Rab = 50, Rb + Rc = Rbc = 100 and Ra + Rc = Rca = 70 These equations can be solved to find that Ra = 10 Ω, Rb = 40 Ω, and Rc = 60 Ω. After shorting terminals b and c, the equivalent resistance between terminal a and the shorted terminals is 1 Req = Ra + = 34 Ω 1 / Rb + 1 / Rc P2.21 The equations for the conductances are 1 1 1 1 Gb + Gc = Ga + Gc = = = Ras 12 Rbs 20 Gb + Ga = 1 Rcs = 1 15 Adding respective sides of the first two equations and subtracting the respective sides of the third equation yields 1 1 1 8 1 from which we obtain Gc = S. Then we have 2Gc = + − = 12 20 15 120 30 Rc = 30 Ω. Similarly, we find Ra = 60 Ω and Rb = 20 Ω. P2.22 The steps in solving a circuit by network reduction are: 1. Find a series or parallel combination of resistances. 2. Combine them. 44 3. Repeat until the network is reduced to a single resistance and a single source (if possible). 4. Solve for the currents and voltages in the final circuit. Transfer results back along the chain of equivalent circuits, solving for more currents and voltages along the way. 5. Check to see that KVL and KCL are satisfied in the original network. The method does not always work because some networks cannot be reduced sufficiently. Then, another method such as node voltages or mesh currents must be used. P2.23* i1 = 10 Req = 10 = 1A 10 vx = 4 V v i2 = x = 0.5 A 8 P2.24* We combine resistances in series and parallel until the circuit becomes an equivalent resistance across the voltage source. Then, we solve the simplified circuit and transfer information back along the chain of equivalents until we have found the desired results. P2.25* Combining resistors in series and parallel, we find that the equivalent resistance seen by the current source is Req = 17.5 Ω. Thus, 45 v = 8 × 17.5 = 140 V. Also, i = 1 A. P2.26 Using Ohm's and Kirchhoff's laws, we work from right to left resulting in P2.27 The equivalent resistance seen by the current source is 1 1 Req = 8 + + = 24 Ω . Then, we have v = 3Req = 72 V, 1 6 + 1 12 1 20 + 1 30 i2 = 1 A, and i1 = 1.2 A. P2.28 The equivalent resistance seen by the current source is 1 Req = 6 + = 14 Ω 1 / 10 + 1 /(30 + 10) 1 v2 = 2 = 16 V Then, we have v s = 2Req = 28 V 1 / 10 + 1 /(30 + 10) i2 = P2.29 v2 10 = 1.6 A i1 = v2 10 + 30 = 0.4 A v 1 = −10i1 = −4 V. The equivalent resistance seen by the voltage source is 1 Req = + 4 = 10 Ω 1 / 18 + 1 /(7 + 2) Then, we have 46 i1 = P2.30 P2.31 P2.32 20 V Req = 2A 1 = 12 V 1 / 18 + 1 /(7 + 2) i2 = v2 18 = 0.667 A The currents through the 3-Ω resistance and the 4-Ω resistance are zero because they are series with an open circuit. Similiarly, the 6-Ω resistance is also in series with the open circuit, and its current is zero. Thus, we can consider the 8-Ω and the 7-Ω resistances to be in series. The current circulating clockwise in the left-hand loop is given by 15 i1 = = 1 A, and we have v 1 = 7i1 = 7 V. The current circulating 7+8 counterclockwise in the right hand loop is 2 A. By Ohm's law, we have v 2 = 4 V. Then, using KVL we have v ab = v 1 − v 2 = 3 V. 20 V = 1A i1 = i2 − 3 = −2 A 20 Ω Notice that current is referenced into the negative reference for the voltage for both sources, thus P = −vi for both sources. Pcurrent − source = −3 A × 20 V = −60 W Pvoltage − source = −20i1 = 40 W i2 = Power is delivered by the current source and absorbed by the voltage source. P 36 mW 1 4 i = = Req = R + = R = 3 mA v 1 R +1 R +1 R 3 12 V i = 3 × 10 − 3 = P2.33 v 2 = i1 12 Req = 12 4R / 3 R = 3000 Ω With the switch open, the current flowing clockwise in the circuit is given 12R2 12 , and we have v 2 = R2i = by i = = 8. Solving, we find R2 = 12 6 + R2 6 + R2 Ω. With the switch closed, R2 and RL are in parallel with an equivalent 1 1 = . The current resistance given by Req = 1 / R2 + 1 / RL 1 / 12 + 1 / RL 12Req 12 through Req is given by i = = 6. and we have v 2 = Req i = 6 + Req 6 + Req 47 Solving, we find Req = 6 Ω. Then, we can write Req = Solving, we find RL = 12 Ω. P2.34* Req = 1 = 3.75 Ω 1 5 + 1 15 i1 = v x 5 = 1.5 A 1 = 6. 1 / 12 + 1 / RL v x = 2 A × Req = 7.5 V i2 = v x 15 = 0.5 A P4A = 4 × 7.5 = 30 W delivering P2A = 2 × 7.5 = 15 W absorbing P5 Ω = 7.52 5 = 11.25 W absorbing P15 Ω = (7.5)2 15 = 3.75 W absorbing P2.35* Req = 1 =4Ω 1 6 + 1 12 v 1 = v 2 = Req i1 = 10 V i4 = 10 12 = 0.8333 A P2.36* R1 ×vs = 5 V R1 + R2 + R3 R3 v3 = × v s = 13 V R1 + R2 + R3 v1 = i1 = 20 V = 2. 5 A 2Req i3 = 10 6 = 1.667 A i2 = i3 − i4 = 0.8333 A v2 = R2 ×vs = 7 V R1 + R2 + R3 48 R2 i1 = P2.38* Combining R2 and R3 , we have an equivalent resistance 1 Req = = 10 Ω . Then, using the voltage-division principle, we 1 R2 + 1 R3 R1 + R2 have v = is = 1 A i2 = R1 P2.37* R1 + R2 is = 2 A Req 10 ×v s = × 10 = 3.333 V . R1 + Req 20 + 10 R2 P2.39 i3 = P2.40 (a) R1 + R2 = R2 + R3 × is = 25 × 20 = 5 mA 25 + 75 15 V = 75 Ω 0.2 A R2 R1 + R2 × 15 = 5 Solving, we find R2 = 25 Ω and R1 = 50 Ω . (b) The equivalent resistance for the parallel combination of R2 and the load is 1 Req = = 22.22 Ω 1 25 + 1 200 Then, using the voltage division principle, we have vo = P2.41 Req × 15 V = 4.615 V R1 + Req We have 120 5 = 20, which yields Rx = 15 Ω. 10 + 5 + Rx 49 P2.42 First, we combine the 60 Ω and 20 Ω resistances in parallel yielding an equivalent resistance of 15 Ω, which is in parallel with Rx. Then, applying the current division principle, we have 15 15 = 10 15 + Rx which yields Rx = 7.5 Ω . P2.43* v = 0.1 mA × Rw = 50 mV Rg = P2.44 50 mV = 25 mΩ 2 A − 0.1 mA The circuit diagram is: With iL = 0 and v L = 5 V , we must have R2 R1 + R2 × 15 = 5 V . Rearranging, this gives R1 =2 (1) R2 With iL = 50 mA and v L = 4.7 V , we have 15 − R1 (4.7 R2 + 50 mA) = 4.7 . Rearranging, this gives 4. 7 R1 + R1 × 0.05 = 10.3 . R2 (2) Using Equation (1) to substitute into Equation (2) and solving, we obtain R1 = 18 Ω and R2 = 9 Ω. Maximum power is dissipated in R1 for iL = 50 mA , for which the voltage 10.32 = 5.89 W . Thus, R1 must be rated 18 for at least 5.89 W of power dissipation. across R1 is 10.3 V. Thus, Pmax R 1 = 50 Maximum power is dissipated in R2 for iL = 0 , in which case the voltage across R2 is 5 V. Thus, Pmax R 2 = P2.45 52 = 2.78 W . 9 We need to place a resistor in series with the load and the voltage source as shown: Applying the voltage-division principle, we have 12.6 we find R = 76 Ω. P2.46 50 = 5. Solving, 50 + R We have P = 25 × 10 −3 = I L2RL = I L2 1000. Solving, we find that the current through the load is I L = 5 mA. Thus, we must place a resistor in parallel with the current source and the load. Then, we have 20 P2.47 R R + RL = 5 from which we find R = 333.3 Ω. In a similar fashion to the solution for Problem P2.12, we can write the following expression for the resistance seen by the 16-V source. 1 Req = 2 + kΩ 1 / Req + 1 / 4 The solutions to this equation are Req = 4 kΩ and Req = −2 kΩ . However, we reason that the resistance must be positive and discard the negative 51 root. Then, we have i1 = i3 = i18 = P2.48* i1 2 i1 16 V Req = 2 mA. Similarly, i4 = 29 = 4 mA, i2 = i1 i3 2 = i1 22 Req i = 1 = 2 mA, and 4 + Req 2 = 1 mA. Clearly, in + 2 = in / 2. Thus, = 7.8125 µA. At node 1 we have: v1 v1 − v2 =1 10 v v − v1 =2 At node 2 we have: 2 + 2 5 10 In standard form, the equations become 0.15v 1 − 0.1v 2 = 1 − 0.1v 1 + 0.3v 2 = 2 20 + Solving, we find v 1 = 14.29 V and v 2 = 11.43 V . v − v2 Then we have i1 = 1 = 0.2857 A. 10 P2.49* Writing a KVL equation, we have v 1 − v 2 = 10 . At the reference node, we write a KCL equation: v1 Solving, we find v 1 = 6.667 and v 2 = −3.333 . Then, writing KCL at node 1, we have is = P2.50 v2 − v1 5 − 5 v1 5 + v2 10 =1. = −3.333 A . Writing KCL equations, we have v1 v1 − v2 v1 − v3 + + =0 21 6 9 v2 − v1 v2 + =3 6 28 v3 v3 − v1 + + = −3 6 9 In standard form, we have: 0.3254v 1 − 0.1667v 2 − 0.1111v 3 = 0 − 0.1667v 1 + 0.2024v 2 = 3 − 0.1111v 1 + 0.2778v 3 = −3 G = [0.3254 -0.1667 -0.1111; -0.1667 0.2024 0; -0.1111 0 0.2778] I = [0; 3; -3] V = G\I 52 Solving, we find v 1 = 8.847 V, v 2 = 22.11 V , and v 3 = −7.261 V. If the source is reversed, the algebraic signs are reversed in the I matrix and consequently, the node voltages are reversed in sign. P2.51 Writing KCL equations at nodes 1, 2, and 3, we have v1 v1 − v2 v1 − v3 + + =0 R4 R2 R1 v2 − v1 v2 − v3 + = Is R2 R3 v3 v3 − v2 v3 − v1 + + =0 R5 R3 R1 In standard form, we have: 0.55v 1 − 0.20v 2 − 0.25v 3 = 0 − 0.20v 1 + 0.325v 2 − 0.125v 3 = 2 − 0.25v 1 − 0.125v 2 + 0.875v 3 = 0 Using Matlab, we have >> G = [0.55 -0.20 -0.25; -0.20 0.325 -0.125; -0.25 -0.125 0.875]; >> I = [0; 2; 0]; >> V = G\I V= 5.1563 10.4688 2.9688 P2.52 To minimize the number of unknowns, we select the reference node at one end of the voltage source. Then, we define the node voltages and write a KCL equation at each node. 53 v 1 − 15 + v1 − v2 5 2 In Matlab, we have =1 v2 − v1 2 + v 2 − 15 10 = −3 G = [0.7 -0.5; -0.5 0.6] I = [4; -1.5] V = G\I I1 = (15 - V(1))/5 Then, we have i1 = 1.0588 A . The 20-Ω resistance does not appear in the network equations and has no effect on the answer. The voltage at the top end of the 10-Ω resistance is 15 V regardless of the value of the 20-Ω resistance. Thus, any nonzero value could be substituted for the 20-Ω resistance without affecting the answer. P2.53 Writing KCL equations at nodes 1, 2, and 3, we have v1 v1 − v2 + + Is = 0 R3 R4 v2 − v1 v2 − v3 v2 + + =0 R4 R6 R5 v3 v − v2 + 3 = Is R1 + R2 R6 In standard form, we have: 0.15v 1 − 0.10v 2 = −5 − 0.10v 1 + 0.475v 2 − 0.25v 3 = 0 − 0.25v 2 + 0.30v 3 = 5 Solving using Matlab, we have G = [ 0.15 -0.10 0; -0.10 0.475 -0.25; 0 -0.25 0.30] I = [-5; 0; 5] V = G\I v 1 = −30.56 V v 2 = 4.167 V v 3 = 20.14 V P2.54 We must not use all of the nodes (including those that are inside supernodes) in writing KCL equations. Otherwise, dependent equations result. 54 P2.55 The circuit with a 1-A source connected is: v1 − v2 v1 − v3 + =1 R2 R1 v2 v2 − v1 v2 − v3 + + =0 R4 R2 R3 v3 v3 − v1 v3 − v2 + + =0 R5 R1 R3 In Matlab, we use the commands [V1,V2,V3] = solve('(V1 - V2)/R2 + (V1 - V3)/R1 = 1' , ... ' V2/R4 + (V2 - V1)/R2 + (V2 - V3)/R3 = 0' , ... ' V3/R5 + (V3 - V1)/R1 + (V3 - V2)/R3 = 0'); PRETTY(V1) After some clean up, this produces Req = R4R5R2 + R2R3R5 + R2R3R1 + R5R2R1 + R4R2R1 + R4R3R1 + R4R1R5 + R4R3R5 R5R3 + R5R1 + R2R3 + R4R3 + R4R1 + R1R3 + R5R2 + R4R2 Then, the command subs(V1, {'R1','R2','R3','R4','R5'}, {15,5,20,10,8}) yields Req = 8.7946 Ω. P2.56* First, we can write: ix = v1 − v2 . 5 Then, writing KCL equations at nodes 1 and 2, we have: v1 + ix = 1 and v2 + 0.5ix − ix = 0 10 20 Substituting for ix and simplifying, we have 55 0.3v 1 − 0.2v 2 = 1 − 0.1v 1 + 0.15v 2 = 0 Solving, we have v 1 = 6 and v 2 = 4 . v − v2 Then, we have ix = 1 = 0. 4 A . 5 P2.57* v x = v2 − v1 Writing KCL at nodes 1 and 2: v 1 v 1 − 2v x v 1 − v 2 + + =1 5 15 10 v2 v 2 − 2v x v2 − v1 =2 5 10 10 Substituting and simplifying, we have + 15v 1 − 7v 2 = 30 + and v 1 + 2v 2 = 20 . Solving, we find v 1 = 5.405 and v 2 = 7.297 . P2.58 First, we can write ix = − v1 10 . Then writing KVL, we have v 1 − 5ix − v 2 = 0 . v2 = ix + 8 . Using the first 20 equation to substitute for ix and simplifying, we have 1.5v 1 − v 2 = 0 2v 1 + v 2 = 160 Writing KCL at the reference node, we have Solving, we find v 1 = 45.71 V, v 2 = 68.57 V, and ix = − Finally, the power delivered to the 8-Ω resistance is (v − v ) 2 P = 1 2 = 65.31 W. 8 P2.59 First, we can write: 5i − v 2 ix = x 10 Simplifying, we find i x = −0.2v 2 . Then write KCL at nodes 1 and 2: v 1 − 5ix v2 =3 − i x = −1 5 10 56 v1 10 = 4.571 A . Substituting for ix and simplifying, we have v 1 − v 2 = 15 and 0.3v 2 = −3 which yield v 1 = 25 V and v 2 = −10 V . P2.60 The circuit with a 1-A current source connected is: v x = v1 − v2 v1 v1 − v2 + =1 20 10 v2 − v1 v2 + − 0.1v x = 0 10 5 Using the first equation to substitute for vx and simplifying, we have 0 . 15 v 1 − 0 . 1v 2 = 1 − 0.2v 1 + 0.4v 2 = 0 Solving we find v 1 = 10 . However, the equivalent resistance is equal in value to v1 so we have Req = 10 Ω. P2.61 The circuit with a 1-A current source connected is ix = v1 9 + v1 − v2 6 v1 − v2 6 =1 57 v2 − v1 v2 + 0.5ix = 0 6 24 Using the first equation to substitute for ix and simplifying, we have 5v 1 − 3v 2 = 18 − 2v 1 + 3v 2 = 0 + Solving we find v 1 = 6 V. However, the equivalent resistance is equal in value to v1, so we have Req = 6 Ω. P2.62 The Matlab commands: SV = solve('(V1 - Vin)/R1 + (V1 - Vout)/R1 + (V1 - V2)/R1 = 0' , ... '(V2 - V1)/R1 + V2/R1 + (V2 - Vout)/R1 = 0', ... '(Vout - V1)/R1 + (Vout - V2)/R1 + Vout/R2 = 0' , 'V1','V2','Vout'); pretty(SV.Vout) result in R2 Vin 1/2 ------R2 + R1 Thus, we have P2.63 R /2 Vout = 2 Vin R2 + R1 We write equations in which voltages are in volts, resistances are in kΩ, and currents are in mA. (v − v 1 ) (v 2 − v 3 ) v 2 KCL node 2: 2 + + =0 4 3 2 (v − v 2 ) (v 3 − v 1 ) (v 3 − v 4 ) KCL node 3: 3 + + =2 3 1 2 KCL ref node: KVL: v2 + v4 2 5 v 1 − v 4 = 10 =2 Using Matlab: >> G = [-1/4 (1/2 + 1/3 + 1/4) -1/3 0; ... -1 -1/3 (1 + 1/2 + 1/3) -1/2; 0 1/2 0 1/5; 1 0 0 -1] >> I = [0; 2; 2; 10] >> V = G\I V= 58 9.3659 4.2537 6.8000 -0.6341 P2.64 Elements on the diagonal of G equal the sum of the conductances connected to any node, which is 3 S. Element gjk off the diagonal is zero if no resistance is connected between nodes j and k and equal to −1 if there is a resistance connected between the nodes. G is the same for all three parts of the problem, only the node to which the current source is attached changes. We used the MATLAB Array Editor to enter the elements of G. >> G G= 3 -1 -1 0 0 0 -1 3 0 -1 0 0 -1 0 3 -1 -1 0 0 -1 -1 3 0 -1 0 0 -1 0 3 -1 0 0 0 -1 -1 3 0 -1 0 0 0 -1 >> Ia = [1; 0; 0; 0; 0; 0; 0]; >> Ib = [0; 1; 0; 0; 0; 0; 0]; >> Ic = [0; 0; 0; 1; 0; 0; 0]; >> Va = G\Ia; >> Vb = G\Ib; >> Vc = G\Ic; >> % Here are the answers: >> Ra = Va(1) Ra = 0.5833 >> Rb = Vb(2) Rb = 0.7500 >> Rc = Vc(4) Rc = 0.8333 0 -1 0 0 0 -1 3 59 By symmetry, shorting nodes with equal node voltages, and series parallel combination, we can obtain Ra = 7 / 12 Ω, Rb = 3 / 4 Ω, and Rc = 5 / 6 Ω. P2.65* Writing KVL equations around each mesh, we have 5i1 + 15(i1 − i2 ) = 20 and 15(i2 − i1 ) + 10i2 = 10 Putting the equations into standard from we have 20i1 − 15i2 = 20 and − 15i1 + 25i2 = 10 Solving, we obtain i1 = 2.364 A and i2 = 1.818 A. Then, the power delivered to the 15-Ω resistor is P = (i1 − i2 ) 2 15 = 4.471 W. P2.66* Writing and simplifying the mesh-current equations, we have: 28i1 − 10i2 = 12 − 10i1 + 40i2 − 30i3 = 0 − 30i2 + 60i3 = 0 Solving, we obtain i1 = 0.500 i2 = 0.200 i3 = 0.100 Thus, v 2 = 5i3 = 0.500 V and the power delivered by the source is P = 12i1 = 6 W. 60 P2.67* Because of the current sources, two of the mesh currents are known. Writing a KVL equation around the middle loop we have 20(i1 − 1) + 10i1 + 5(i1 + 2) = 0 Solving, we find i1 = 0.2857 A. P2.68 Writing KVL equations around each mesh, we have 5i1 + 7(i1 − i3 ) + 31 = 0 11(i2 − i3 ) + 3i2 − 31 = 0 i3 + 11(i3 − i2 ) + 7(i3 − i1 ) = 0 Putting the equations into standard from, we have 12i1 − 7i3 = −31 14i2 − 11i3 = 31 − 7i1 − 11i2 + 19i3 = 0 Using Matlab to solve, we have >> R = [12 0 -7; 0 14 -11; -7 -11 19]; >> V = [-31; 31; 0]; >> I = R\V I= -2.0000 3.0000 1.0000 Then, the power delivered by the source is P = −31(i1 − i2 ) = 155 W. P2.69 Writing and simplifying the mesh equations, we obtain: 40i1 − 20i2 = 10 − 20i1 + 40i2 = 0 61 Solving, we find i1 = 0.3333 and i2 = 0.1667 . Thus, v = 20(i1 − i2 ) = 3.333 V . P2.70 The mesh currents and corresponding equations are: i1 = 20 mA 25(i2 − i1 ) + 75i2 = 0 Solving, we find i2 = 5 mA . However, i3 shown in Figure P2.39 is the same as i2, so the answer is i3 = 5 mA. P2.71 First, we select the mesh currents and then write three equations. Mesh 1: 30i1 + 20(i1 − i3 ) = 0 Mesh 2: 12i2 + 6(i2 − i3 ) = 0 However by inspection, we have i3 = 3 . Solving, we obtain i1 = 1.2 A and i2 = 1.0 A. 62 P2.72 Writing and simplifying the mesh equations yields: 14i1 − 8i2 = 10 − 8i1 + 16i2 = 0 Solving, we find i1 = 1.000 and i2 = 0.500 . Finally, the power delivered by the source is P = 10i1 = 10 W. P2.73 4iA + 18(iA − iB ) = 20 18(iB − iA ) + 7iB + 2iB = 0 Solving we find iA = 2 A and iB = 1.333 A. Then we have i1 = iA = 2 A and i2 = iA − iB = 0.667 A. P2.74 Mesh A: 10iA + 30iA + 10(iA − iB ) = 0 By inspection: iB = 2 Solving, we find iA = 0.4 A . Then we have i1 = iA = 0.4 A and i2 = iB − iA = 1.6 A. 63 P2.75 (a) First, we select mesh-current variables as shown. Then, we can write (Rw + Rn + R1 )i1 − Rn i2 − R1i3 = 120 − Rn i1 + (Rw + Rn + R2 )i2 − R2i3 = 120 − R1i1 − R2i2 + (R1 + R2 + R3 )i3 = 0 Alternatively, because the network consists of independent voltage sources and resistances, and all of the mesh currents flow clockwise, we can enter the matrices directly into MATLAB. >> Rw = 0.1; Rn=0.1; R1 = 15; R2 = 5; R3 = 8; >> R = [Rw+Rn+R1 -Rn -R1; -Rn Rw+Rn+R2 -R2; -R1 -R2 R1+R2+R3]; >> V = [120; 120; 0]; >> I = R\V; >> % Finally, we compute the voltages across the loads. >> Vr1 = R1*(I(1) - I(3)), Vr2 = R2*(I(2) - I(3)), Vr3 = R3*I(3) Vr1 = 117.8069 Vr2 = 113.3613 Vr3 = 231.1682 (b) Next, we change Rn to a very high value such as 109 which for practical calculations is equivalent to an open circuit, and again compute the voltages resulting in: Vr1 = 173.9130 Vr2 = 57.9710 64 Vr3 = 231.8841 The voltage across R1 is certainly high enough to damage most loads designed to operate at 110 V. P2.76 Current source in terms of mesh currents: − i1 + i2 = I s KVL for mesh 3: − R2i1 − R3i2 + (R1 + R2 + R3 )i3 = 0 KVL around outside of network: R4i1 + R5i2 + R1i3 = 0 Then using MATLAB: >> R1 = 4; R2 = 5; R3 = 8; R4 = 10; R5 = 2; Is = 2; >> R = [-1 1 0; -R2 -R3 (R1+R2+R3); R4 R5 R1]; >> V = [Is; 0; 0]; >> I = R\V; >> V3 = R5*I(2) V3 = 2.9688 P2.77 (R2 + R4 )i1 − R4i2 − R2i3 = 1 − R4i1 + (R3 + R4 + R5 )i2 − R3i3 = 0 65 − R2i1 − R3i2 + (R1 + R2 + R3 )i3 = 0 Now using MATLAB: R1 = 6; R2 = 5; R3 = 4; R4 = 8; R5 = 2; R = [(R2+R4) -R4 -R2; -R4 (R3+R4+R5) -R3; -R2 -R3 (R1+R2+R3)]; V = [1; 0; 0]; I = R\V; Req = 1/I(1) % Gives answer in ohms. Req = 4.5979 P2.78 Mesh 1: 3i1 + 7i1 + 30(i1 − i2 ) = 1 Mesh 2: 3i2 + 12(i2 − i3 ) + 4i2 + 30(i2 − i1 ) = 0 Mesh 3: 24i3 + 12(i3 − i2 ) = 0 Solving, we find i1 = 0.05 A. Then Req = 1 / i1 = 20 Ω. P2.79 We write equations in which voltages are in volts, resistances are in kΩ, and currents are in mA. KVL mesh 1: 4i1 + 1(i1 − i2 ) + 3(i1 − i3 ) = 0 KVL mesh 2: 1(i2 − i1 ) + 2(i2 − i4 ) = −10 KVL supermesh: 2i3 + 3(i3 − i1 ) + 2(i4 − i2 ) + 5i4 = 0 Current source: i4 − i3 = 2 Now, we proceed in Matlab. >> R = [8 -1 -3 0; -1 3 0 -2; -3 -2 5 7; 0 0 -1 1]; >> V = [0; -10; 0; 2]; >> I = R\V % This yields the mesh currents in mA. I= 0.4780 4.6439 -0.2732 1.7268 66 P2.80* First, we write a node voltage equation to solve for the open-circuit voltage: v oc − 10 v oc =1 10 5 Solving, we find v oc = 6.667 V . + Then zeroing the sources, we have this circuit: Thus, Rt = P2.81* 1 = 3.333 Ω . The Thévenin and Norton equivalents are: 1 10 + 1 5 The equivalent circuit of the battery with the resistance connected is i = 6 100 = 0.06 A Rt = 9−6 = 50 Ω 0.06 67 P2.82 With open-circuit conditions: Solving, we find v ab = −12 V . With the source zeroed: Rt = 1 = 5Ω 1 6 + 1 (6 + 24 ) The equivalent circuits are: Notice the source polarity relative to terminals a and b. P2.83 First, we solve the network with a short circuit: Req = 10 + 1 = 16 Ω 1 10 + 1 15 i10 = 32 Req = 2 A i15 = i10 10 = 0.8 A 10 + 15 isc = i15 = 0.8 A 68 Zeroing the source, we have: Combining resistances in series and parallel we find Rt = 12 Ω . Then the Thévenin voltage is vt = isc Rt = 9.6 V . The Thévenin and Norton equivalents are: P2.84 The 7-Ω resistor has no effect on the equivalent circuits because the voltage across the 12-V source is independent of the resistor value. 69 P2.85 The Thévenin voltage is equal to the open-circuit voltage which is 12.6 V. The equivalent circuit with the 0.1-Ω load connected is: We have 12.6 /(Rt + 0.1) = 100 from which we find Rt = 0.026 Ω. The Thévenin and Norton equivalent circuits are: Because no energy is converted from chemical form to heat in a battery under open-circuit conditions, the Thévenin equivalent seems more realistic from an energy conversion standpoint. P2.86 The Thévenin voltage is equal to the open-circuit voltage which is 15 V. The circuit with the load attached is: 10 = 5 mA and v x = Vt − 10 = 5 V. Thus, the Thévenin 2000 5V = 1 kΩ. resistance is Rt = 5 mA We have iL = P2.87 The equivalent circuit with a load attached is: 70 For a load of 2.2 kΩ, we have iL = 4.4 / 2200 = 2 mA , and we can write v L = Vt − Rt iL . Substituting values this becomes 4.4 = Vt − 0.002Rt (1) Similarly, for the 10-kΩ load we obtain 5 = Vt − 0.0005Rt (2) Solving Equations (1) and (2), we find Vt = 5.2 V and Rt = 400 Ω. P2.88 Open-circuit conditions: ix = 15 − v x 5 vx 10 + 10 then we have Vt = v oc = v x − ix + 0.5ix = 0 10 = 5 V. 10 + 10 Short-circuit conditions: 71 Solving, we find v x = 10 V and ix = 15 − v x 5 we have isc = vx vx 10 equivalents are: P2.89 10 − ix + 0.5ix = 0 Solving, we find v x = 7.5 V and then = 0.75 A. Then, we have Rt = v oc isc = 6.67 Ω . Thus the As is Problem P2.80, we find the Thévenin equivalent: Then maximum power is obtained for a load resistance equal to the Thévenin resistance. (v 2)2 = 3.333 W Pmax = t Rt P2.90 As in Problem P2.82, we find the Thévenin equivalent: Then, maximum power is obtained for a load resistance equal to the Thévenin resistance. (V 2)2 = 7.2 W Pmax = t Rt 72 P2.91* To maximize the power to RL , we must maximize the voltage across it. Thus, we need to have Rt = 0 . The maximum power is Pmax = P2.92 202 = 80 W 5 The circuit is By the current division principle: iL = I n Rt RL + Rt The power delivered to the load is PL = (iL ) RL = (I n ) 2 2 (Rt )2 RL (RL + Rt )2 Taking the derivative and setting it equal to zero, we have 2 2 2 dPL 2 (Rt ) (Rt + RL ) − 2(Rt ) RL (Rt + RL ) = 0 = (I n ) dRL (Rt + RL )4 which yields RL = Rt . The maximum power is PL max = (I n )2 Rt 4 . P2.93 For maximum power conditions, we have RL = Rt . The power taken from the voltage source is Ps = (Vt )2 Rt + RL = (Vt )2 2Rt Then, half of Vt appears across the load and the power delivered to the load is PL = (0.5Vt )2 Rt Thus, the percentage of the power taken from the source that is delivered to the load is 73 PL × 100% = 50% Ps On the other hand, for RL = 9Rt , we have η= Ps = PL = η= (Vt )2 Rt + RL = (Vt )2 10Rt (0.9Vt )2 9Rt PL × 100% = 90% Ps Design for maximum power transfer is relatively inefficient. Thus, systems in which power efficiency is important are almost never designed for maximum power transfer. P2.94* First, we zero the current source and find the current due to the voltage source. iv = 30 15 = 2 A Then, we zero the voltage source and use the current-division principle to find the current due to the current source. 10 = 2A 5 + 10 Finally, the total current is the sum of the contributions from each source. i = iv + ic = 4 A ic = 3 74 P2.95* The circuits with only one source active at a time are: 1 = 3.75 Ω 1 5 + 1 15 10 V =− = −2.667 A Req = is ,v is ,c = −1 10 = −0.667 A 10 + 5 Req Then the total current due to both sources is is = is ,v + is ,c = −3.333 A . P2.96 Zero the 2-A source and use the current-division principle: i1,1A = 1 20 = 0.5714 A 15 + 20 Then zero the 1 A source and use the current-division principle: i1,2A = −2 Finally, 5 = −0.2857 A 5 + 30 i1 = i1,1A + i1,2A = 0.2857 A 75 P2.97 The circuits with only one source active at a time are: i1.4A = 4 × = 3A 15 15 + 5 i1,2A = −2 × = −1.5 15 15 + 5 Finally, we add the components to find the current with both sources active. i1 = i1, 4A + i1,2A = 1.5 A P2.98 The circuit, assuming that v 2 = 1 V is: i2 = (v 2 5) = 0.2 A v1 = 30i2 = 6 V i10 = v 1 10 = 0.6 A i30 = v 1 30 = 0.2 A is = i2 + i10 + i30 = 1 A v s = 12is + v 1 + 6is = 24 V We have established that for v s = 24 V , we have v 2 = 1 V . Thus, for v s = 12 V , we have: v2 = 1 × 12 = 0. 5 V 24 76 P2.99 We start by assuming i2 = 1 A and work back through the circuit to determine the value of vs. The results are shown on the circuit diagram. However, the circuit actually has vs = 10 V, so the actual value ofi2 is 10 × (1 A) = 0.5 A. 20 P2.100 We start by assuming i6 = 1 A and work back through the circuit to determine the value of Is. This results in Is = -4.5 A. However, the circuit actually has Is = 10 A, so the actual value ofi6 is 10 × (1 A) = − 2.222 A. − 4. 5 P2.101 (a) With only the 2-A source activated, we have i2 = 2 and v 2 = 3(i2 )2 = 12 V (b) With only the 1-A source activated, we have i1 = 1 A and v 1 = 3(i1 )2 = 3 V (c) With both sources activated, we have i = 3 A and v = 3(i )2 = 27 V 77 Notice that i ≠ i1 + i2 . Superposition does not apply because device A has a nonlinear relationship between v and i. P2.102 From Equation 2.90, we have R2 1 kΩ R3 = × 3419 = 341.9 Ω R1 10 kΩ R 100 kΩ (b) Rx = 2 R3 = × 3419 = 34.19 kΩ R1 10 kΩ (a) Rx = P2.103* (a) Rearranging Equation 2.90, we have R3 = R1 10 4 Rx = 4 × 5932 = 5932 Ω R2 10 (b) The circuit is: The Thévenin resistance is Rt = 1 1 + = 7447 Ω 1 R3 + 1 R1 1 R2 + 1 Rx The Thévenin voltage is vt = v s R3 R1 + R3 − vs Rx Rx + R2 = 0.3939 mV Thus, the equivalent circuit is: 78 idetector = Vt = 31.65 × 10 −9 A Rt + Rdetector Thus, the detector must be sensitive to very small currents if the bridge is to be accurately balanced. P2.104 If R1 and R3 are too small, large currents are drawn from the source. If the source were a battery, it would need to be replaced frequently. Large power dissipation could occur, leading to heating of the components and inaccuracy due to changes in resistance values with temperature. If R1 and R3 are too large, we would have very small detector current when the bridge is not balanced, and it would be difficult to balance the bridge accurately. P2.105 With the source replaced by a short circuit and the detector removed, the Wheatstone bridge circuit becomes The Thévenin resistance seen looking back into the detector terminals is 1 1 Rt = + 1 R3 + 1 R1 1 R2 + 1 Rx The Thévenin voltage is zero when the bridge is balanced. 79 Practice Test T2.1 (a) 6, (b) 10, (c) 2, (d) 7, (e) 10 or 13 (perhaps 13 is the better answer), (f) 1 or 4 (perhaps 4 is the better answer), (g) 11, (h) 3, (i) 8, (j) 15, (k) 17, (l) 14. T2.2 The equivalent resistance seen by the voltage source is: 1 = 16 Ω Req = R1 + 1 / R2 + 1 / R3 + 1 / R4 is = vs = 6A Req Then, using the current division principle, we have G4 1 / 60 i4 = is = 6=1 A G2 + G 3 + G 4 1 / 48 + 1 / 16 + 1 / 60 T2.3 Writing KCL equations at each node gives v1 v1 − v2 v1 − v3 + + =0 4 5 2 v2 − v1 v2 + =2 5 10 v3 v3 − v1 + = −2 1 2 In standard form, we have: 0.95v 1 − 0.20v 2 − 0.50v 3 = 0 − 0.20v 1 + 0.30v 2 = 2 − 0.50v 1 + 1.50v 3 = −2 In matrix form, we have GV = I  0.95 − 0.20 − 0.50 v 1   0   − 0.20 0.30 0  v 2  =  2        − 0.50 0 1.50  v 3   − 2 The MATLAB commands needed to obtain the column vector of the node voltages are G = [0.95 -0.20 -0.50; -0.20 0.30 0; -0.50 0 1.50] I = [0; 2; -2] V = G\I % As an alternative we could use V = inv(G)*I 80 Actually, because the circuit contains only resistances and independent current sources, we could have used the short-cut method to obtain the G and I matrices. T2.4 We can write the following equations: KVL mesh 1: R1i1 −Vs + R3 (i1 − i3 ) + R2 (i1 − i2 ) = 0 KVL for the supermesh obtained by combining meshes 2 and 3: R4i2 + R2 (i2 − i1 ) + R3 (i3 − i1 ) + R5i3 = 0 KVL around the periphery of the circuit: R1i1 −Vs + R4i2 + R5i3 = 0 Current source: i2 − i3 = I s A set of equations for solving the network must include the current source equation plus two of the mesh equations. The three mesh equations are dependent and will not provide a solution by themselves. T2.5 Under short-circuit conditions, the circuit becomes Thus, the short-circuit current is 1 A flowing out of b and into a. Zeroing the sources, we have Thus, the Thévenin resistance is 1 Rt = = 24 Ω 1 / 40 + 1 /(30 + 30) and the Thévenin voltage is Vt = I sc Rt = 24 V . The equivalent circuits are: 81 Because the short-circuit current flows out of terminal b, we have oriented the voltage polarity positive toward b and pointed the current source reference toward b. T2.6 With one source active at a time, we have Then, with both sources active, we have 82 We see that the 5-V source produces 25% of the total current through the 5-Ω resistance. However, the power produced by the 5-V source with both sources active is zero. Thus, the 5-V source produces 0% of the power delivered to the 5-Ω resistance. Strange, but true! Because power is a nonlinear function of current (i.e., P = Ri 2 ), the superposition principle does not apply to power. 83 CHAPTER 3 Exercises E3.1 v (t ) = q (t ) / C = 10 −6 sin(10 5t ) /(2 × 10 −6 ) = 0.5 sin(10 5t ) V dv i (t ) = C = (2 × 10 −6 )(0.5 × 10 5 ) cos(10 5t ) = 0.1 cos(10 5t ) A dt E3.2 Because the capacitor voltage is zero at t = 0, the charge on the capacitor is zero at t = 0. t q (t ) = ∫ i (x )dx + 0 0 t = ∫ 10 −3 dx = 10 −3t for 0 ≤ t ≤ 2 ms 0 = 2E −3 ∫ 10 −3 dx + 0 t ∫ − 10 −3 dx = 4 × 10 -6 − 10 −3t for 2 ms ≤ t ≤ 4 ms 2E −3 v (t ) = q (t ) / C = 10 4t for 0 ≤ t ≤ 2 ms = 40 − 10 4t for 2 ms ≤ t ≤ 4 ms p (t ) = i (t )v (t ) = 10t for 0 ≤ t ≤ 2 ms = −40 × 10 −3 + 10t for 2 ms ≤ t ≤ 4 ms w (t ) = Cv 2 (t ) / 2 = 5t 2 for 0 ≤ t ≤ 2 ms = 0.5 × 10 −7 (40 − 10 4t ) 2 for 2 ms ≤ t ≤ 4 ms in which the units of charge, electrical potential, power, and energy are coulombs, volts, watts and joules, respectively. Plots of these quantities are shown in Figure 3.8 in the book. E3.3 Refer to Figure 3.10 in the book. Applying KVL, we have v = v1 + v2 + v3 Then using Equation 3.8 to substitute for the voltages we have 84 v (t ) = 1 C1 t 1 t ∫ i (t )dt + v (0) + C ∫ i (t )dt + v 1 0 2 0 2 (0) + 1 C3 t ∫ i (t )dt + v This can be written as t  1 1 1   ∫ i (t )dt + v 1 ( 0 ) + v 2 ( 0 ) + v 3 ( 0 ) v (t ) =  + +  C1 C2 C3 0 Now if we define  1 1 1 1   and v (0) = v 1 (0) + v 2 (0) + v 3 (0) =  + + C eq  C 1 C 2 C 3  we can write Equation (1) as t 1 v (t ) = ∫ i (t )dt + v (0) C eq 3 (0) 0 (1) 0 Thus the three capacitances in series have an equivalent capacitance given by Equation 3.25 in the book. E3.4 (a) For series capacitances: 1 1 = = 2 / 3 µF C eq = 1 / C1 + 1 / C2 1 / 2 + 1 / 1 (b) For parallel capacitances: C eq = C 1 + C 2 = 1 + 2 = 3 µF E3.5 From Table 3.1 we find that the relative dielectric constant of polyester is 3.4. We solve Equation 3.26 for the area of each sheet: 10 −6 × 15 × 10 −6 Cd Cd = = = 0.4985 m 2 A= ε ε r ε 0 3.4 × 8.85 × 10 −12 Then the length of the strip is L = A /W = 0.4985 /(2 × 10 −2 ) = 24.93 m E3.6 v (t ) = L di (t ) d [ = (10 × 10 −3 ) 0.1 cos(10 4t )] = −10 sin(10 4t ) V dt dt w (t ) = 21 Li 2 (t ) = 5 × 10 −3 × [0.1 cos(10 4t )] = 50 × 10 −6 cos2 (10 4t ) J 2 85 E3.7 1 t t 1 i (t ) = ∫ v (x )dx + i (0) = v (x )dx L0 150 × 10 −6 ∫0 t = 6667 ∫ 7.5 × 10 6 xdx = 25 × 10 9t 2 V for 0 ≤ t ≤ 2 µs 0 = 6667 2E-6 ∫ 7.5 × 10 6 xdx = 0.1 V for 2µs ≤ t ≤ 4 µs 0 2E-6 t   = 6667  ∫ 7.5 × 10 6 xdx + ∫ (− 15)dx  = 0.5 − 10 5t V for 4 µs ≤ t ≤ 5 µs 4 E-6  0  A plot of i(t) versus t is shown in Figure 3.19b in the book. E3.8 Refer to Figure 3.20a in the book. Using KVL we can write: v (t ) = v 1 (t ) + v 2 (t ) + v 3 (t ) Using Equation 3.28 to substitute, this becomes di (t ) di (t ) di (t ) v (t ) = L1 + L2 + L3 (1) dt dt dt Then if we define Leq = L1 + L2 + L3 , Equation (1) becomes: v (t ) = Leq di (t ) dt which shows that the series combination of the three inductances has the same terminal equation as the equivalent inductance. E3.9 Refer to Figure 3.20b in the book. Using KCL we can write: i (t ) = i1 (t ) + i2 (t ) + i3 (t ) Using Equation 3.32 to substitute, this becomes t t t 1 1 1 i (t ) = ∫ v (t )dt + i1 (0) + ∫ v (t )dt + i2 (0) + ∫ v (t )dt + i3 (0) L1 0 L2 L3 0 0 This can be written as t 1 1 1   v (t ) =  + +  ∫ v (t )dt + i1 (0) + i2 (0) + i3 (0)  L1 L2 L3  0 Now if we define 1 1 1 1 =  + +  and i (0) = i1 (0) + i2 (0) + i3 (0) Leq  L1 L2 L3  we can write Equation (1) as 86 (1) i (t ) = 1 Leq t ∫ v (t )dt + i (0) 0 Thus, the three inductances in parallel have the equivalent inductance shown in Figure 3.20b in the book. E3.10 Refer to Figure 3.21 in the book. (a) The 2-H and 3-H inductances are in series and are equivalent to a 5H inductance, which in turn is in parallel with the other 5-H inductance. This combination has an equivalent inductance of 1/(1/5 + 1/5) = 2.5 H. Finally the 1-H inductance is in series with the combination of the other inductances so the equivalent inductance is 1 + 2.5 = 3.5 H. (b) The 2-H and 3-H inductances are in series and have an equivalent inductance of 5 H. This equivalent inductance is in parallel with both the 5-H and 4-H inductances. The equivalent inductance of the parallel combination is 1/(1/5 + 1/4 + 1/5) = 1.538 H. This combination is in series with the 1-H and 6-H inductances so the overall equivalent inductance is 1.538 + 1 + 6 = 8.538 H. E3.11 The MATLAB commands including some explanatory comments are: % We avoid using i alone as a symbol for current because % we reserve i for the square root of -1 in MATLAB. Thus, we % will use iC for the capacitor current. syms t iC qC vC % Define t, iC, qC and vC as symbolic objects. iC = 0.5*sin((1e4)*t); ezplot(iC, [0 3*pi*1e-4]) qC=int(iC,t,0,t); % qC equals the integral of iC. figure % Plot the charge in a new window. ezplot(qC, [0 3*pi*1e-4]) vC = 1e7*qC; figure % Plot the voltage in a new window. ezplot(vC, [0 3*pi*1e-4]) The plots are very similar to those of Figure 3.5 in the book. An m-file (named Exercise_3_11) containing these commands can be found in the MATLAB folder on the OrCAD disk. 87 E.12 The MATLAB commands including some explanatory comments are: % We avoid using i by itself as a symbol for current because % we reserve i for the square root of -1 in MATLAB. Thus, we % will use iC for the capacitor current. syms t vC iC pC wC % Define t, vC, iC, pC and wC as symbolic objects. vC = 1000*t*(heaviside(t)- heaviside(t-1)) + ... 1000*(heaviside(t-1) - heaviside(t-3)) + ... 500*(5-t)*(heaviside(t-3) - heaviside(t-5)); ezplot(vC, [0 6]) iC = (10e-6)*diff(vC, 't'); % iC equals C times the derivative of vC. figure % Plot the current in a new window. ezplot(iC, [0 6]) pC = vC*iC; figure % Plot the power in a new window. ezplot(pC, [0 6]) wC = (1/2)*(10e-6)*vC^2; figure % Plot the energy in a new window. ezplot(wC, [0 6]) The plots are very similar to those of Figure 3.6 in the book. An m-file (named Exercise_3_12) containing these commands can be found in the MATLAB folder on the OrCAD disk. Problems P3.1 A dielectric material is an electrical insulator through which virtually no current flows, assuming normal operating voltages. Some examples of dielectrics mentioned in the text are air, Mylar, polyester, polypropylene, and mica. Some others are porcelain, glass, and certain types of oil. P3.2 Charges (usually in the form of electrons) flow in and accumulate on one plate. Meanwhile, an equal amount of charge flows out of the other plate. Thus, current seems to flow through a capacitor. P3.3 Capacitors consist of two conductors separated by an insulating material. Frequently, the conductors are sheets of metal that are separated by a thin layer of the insulating material. 88 P3.4 Because we have i = Cdv / dt for a capacitance, the current is zero if the voltage is constant. Thus, we say that capacitances act as open circuits for constant dc voltages. P3.5* i =C P3.6* i (t ) = C dv dt 100 × 10 −6 dv i = = = 0.05 V/s dt C 2000 × 10 −6 ∆v 100 ∆t = = = 2000 s dv dt 0.05 dv d (100 sin 1000t ) = cos(1000t ) = 10 −5 dt dt p (t ) = v (t )i (t ) = 100 cos(1000t ) sin(1000t ) = 50 sin(2000t ) w (t ) = 1 C [v (t )]2 = 0.05 sin2 (1000t ) 2 89 v (t ) = P3.7* 1 C t ∫ i (t )dt + v (0) 0 t v (t ) = 2 × 10 4 ∫ 3 × 10 −3 dt − 20 0 v (t ) = 60t − 20 V p (t ) = i (t )v (t ) = 3 × 10 −3 (60t − 20 ) W Evaluating at t = 0 , we have p (0 ) = −60 mW . Because the power has a negative value, the capacitor is delivering energy. At t = 1 s , we have p (1) = 120 mW . Because the power is positive, we know that the capacitor is absorbing energy. P3.8* W = power × time = 5 hp × 746 W / hp × 3600 s = 13.4 × 10 J 6 2W 2 × 13.4 × 10 6 = 51.8 kV 0.01 C It turns out that a 0.01-F capacitor rated for this voltage would be much too large and massive for powering an automobile. Besides, to have reasonable performance, an automobile would need much more than 5 hp for an hour. V = P3.9 = The net charge on each plate is Q = CV = (1 × 10 −6 ) × 200 = 200 µC. One plate has a net positive charge and the other has a net negative charge so the net charge for both plates is zero. P3.10 dv C dt d ( = 10 −5 10e −500t ) dt = −0.05e −500t A iC (t ) = C pC (t ) = v C (t )iC (t ) = −0.5e −1000t W 1 C [v C (t )]2 2 = 0.5 × 10 −3 × e −1000t J w C (t ) = 90 The sketches should resemble the following plots. P3.11 Q = Cv = 15 × 10 −6 × 500 = 7.5 mC W = P = 3.12 1 1 Cv 2 = × 15 × 10 −6 × (500 )2 = 1.875 J 2 2 ∆W 1.875 = = 468.75 kW ∆t 4 × 10 −6 The sketches should resemble the following plots. The units for the quantities in these plots are A, V, W, J and s. 91 P3.13 v C (t ) = 1 C t ∫ iC (t )dt + v C (0) 0 t v C (t ) = 0.5 × 10 6 ∫ iC (t )dt 0 pC (t ) = v C (t )iC (t ) w C (t ) = 1 Cv C2 (t ) 10 −6 × v C2 (t ) 2 The sketches should be similar to the following plots. The units for the quantities in these plots are A, V, W, J and s. 92 P3.14 Because the switch is closed prior to t = 0 , the initial voltage is zero, and we have t t 1 v (t ) = ∫ i (t )dt + v (0 ) = 10 5 ∫ (5 × 10 −3 )dt + 0 = 500t C 0 −3 v (20 × 10 ) = 10 V p = vi = 2.5t p (20 × 10 −3 ) = 50 mW 0 1 Cv 2 (t ) = 1.25t 2 J 2 w (20 × 10 −3 ) = 500 µJ w (t ) = P3.15 v (t ) = 1 t 1 t ∫ i (t )dt + v (0) = C ∫ I m cos(ωt )dt C 0 I = m sin(ωt ) ωC 0 93 = Im [sin(ωt ) − sin(0)] ωC Clearly for ω → ∞ , the voltage becomes zero, so the capacitance becomes the equivalent of a short circuit. P3.16 v C (t ) = 1 C t ∫ iC (t )dt + v C (0) 0 t vC (t ) = 0.2 × 10 ∫ (100t )dt + 0 = 107t 2 for 0 ≤ t ≤ 1 ms 6 0 v C (t ) = 0.2 × 10 6 t ∫ (100t − 0.2)dt + 10 = 10 7 (t − 2 × 10 −3 ) 2 for 1 ms ≤ t ≤ 2 ms 10 −3 pC (t ) = v C (t )iC (t ) 1 Cv C2 (t ) = 2.5 × 10 −6 × v C2 (t ) 2 The sketches should be similar to the following plots. The units for the quantities in these plots are A, V, W, J and s. w C (t ) = 94 P3.17 We can write 1 w = Cv 2 = 2 × 10 −5v 2 = 80 2 Solving, we find v = 2000 V. Because the stored energy is decreasing, the power is negative. Thus, assuming that the current reference points into the positive voltage reference, we have p − 200 i = = = − 100 mA 2000 v Because the stored energy, voltage, and stored charge are decreasing, current actually leaves the positive terminal of the capacitor. P3.18 i (t ) = P3.19 A capacitance initially charged to 10 V has t 1 v (t ) = ∫ i (t )dt + 10 dq (t ) d [C (t )v (t )] = dt dt d [50 × 200 × 10 −12 + 50 × 50 × 10 −12 sin(5000t )] i (t ) = dt i (t ) = 12.5 cos(5000t ) µA C 0 However, if the capacitance is infinite, this becomes v (t ) = 10 V which describes a 10-V voltage source. Thus, a very large capacitance initially charged to 10 V is an approximate 10-V voltage source. P3.20 We can write v (t 0 ) = Vf = 1 C t0 + ∆t ∫t i (t )dt . The integral represents the area of 0 the current pulse, which has units of ampere seconds or coulombs and must equal Vf C . The pulse area represents the net charge transferred by the current pulse. Because a constant-amplitude pulse has the largest area for a given peak amplitude, we can say that the peak amplitude of the current pulse must be at least as large as Vf C / ∆t . We conclude that the area of the pulse remains constant and that the peak amplitude approaches infinity as ∆t approaches zero. In the limit, this type of pulse is called an impulse. P3.21 By definition, the voltage across a short circuit must be zero. Since we have v = Ri for a resistor, zero resistance corresponds to a short circuit. 95 For an initially uncharged capacitance, we have t 1 v (t ) = ∫ i (t )dt C 0 For the voltage to be zero for all values of current and time, the capacitance must be infinite. Thus, an infinite initially uncharged capacitance is equivalent to a short circuit. For an open circuit, the current must be zero. This requires infinite resistance. However for a capacitance, we have i (t ) = C dv (t ) dt Thus, a capacitance of zero is equivalent to an open circuit. P3.22 We assume that i (t ) is referenced into the positive reference for v (t ) , and we have i (t ) = C dv (t ) dt d [10 − 10 exp(−2000t )] = 0.4 exp(−2000t ) A dt p (t ) = v (t )i (t ) = [10 − 10 exp( −2000t )][0.4 exp( −2000t )] W = 20 × 10 −6 Evaluating at t = 0 , we have p (0 ) = 0 W and the capacitor is neither absorbing nor delivering energy. At t = 0.5 ms, we find p (0.5) = 930 mW. Because the power is positive, we know that the capacitor is absorbing energy. P3.23 Capacitances in parallel are combined by adding their values. Thus, capacitances in parallel are combined as resistances in series are. Capacitances in series are combined by taking the reciprocal of the sum of the reciprocals of the individual capacitances. Thus, capacitances in series are combined as resistances in parallel are. P3.24* (a) C eq = 1 + 1 = 2 µF 1 2+1 2 (b) The two 4- µ F capacitances are in series and have an equivalent 1 = 2 µ F . This combination is a parallel with the capacitance of 1 4+1 4 2- µ F capacitance, giving an equivalent of 4 µ F. Then the 12 µ F is in 96 series, giving a capacitance of 1 = 3 µF . Finally, the 5 µ F is in 1 12 + 1 4 parallel, giving an equivalent capacitance of C eq = 3 + 5 = 8 µF . P3.25* As shown below, the two capacitors are placed in series with the heart to produce the output pulse. While the capacitors are connected, the average voltage supplied to the heart is 4.95 V. Thus, the average current is I pulse = 4.95 500 = 9.9 mA . The charge removed from each capacitor during the pulse is ∆Q = 9.9 mA × 1 ms = 9.9 µC . This results in a 0.l V change in voltage, so C ∆Q 9.9 × 10 −6 we have C eq = = = = 99 µF . Thus, each capacitor has a 2 0.1 ∆v capacitance of C = 198 µF . Then as shown below, the capacitors are placed in parallel with the 2.5-V battery to recharge them. The battery must supply 9.9 µ C to each battery. Thus, the average 2 × 9.9 µC = 19.8 µA . The current supplied by the battery is I battery = 1s ampere-hour rating of the battery is 19.8 × 10 −6 × 5 × 365 × 24 = 0.867 Ampere hours . P3.26 (a) C eq = 3 + (b) C eq = 1 1 + = 11 µF 1 / 6 + 1 / 3 1 / 18 + 1 (5 + 4 ) 1 1 + = 9 µF 1 12 + 1 4 1 8 + 1 24 97 P3.27 C eq = 1 = 4 µF 1 C1 + 1 C2 The charges stored on each capacitor and on the equivalent capacitance are equal because the current through each is the same. Q = C eq × 20 V = 80 µC Q = 16 V C1 Q v2 = =4V C2 v1 = As a check, we verify that v 1 + v 2 = 20 V . P3.28 The equivalent capacitance is C eq = 100 µF and its initial voltage is 50 V. The energies are w 1 = 21 CV1 2 = 21 200 × 10 −6 × 50 2 = 0.25 J and w 2 = 1.0 J. Thus, the total energy stored in the two capacitors is w total = 1.25 J. On the other hand, the energy stored in the equivalent capacitance is w eq = 21 C eqVeq2 = 21 100 × 10 −6 × 50 2 = 0.125 J. The reason for the discrepancy is that all of the energy stored in the original capacitances cannot be accessed as long as they are connected in series. Net charge is trapped on the plates that are connected together. C eq = 1 + P3.29 P3.30 1 = 3 µF 1/6 + 1/3 We obtain the maximum capacitance of 3 µ F by connecting all three 1µ F capacitors in parallel. We obtain the minimum capacitance of 1/3 µ F by connecting all three 1- µ F capacitors in series. P3.31* C = εr ε 0A d = 15 × 8.85 × 10 −12 × 10 × 10 −2 × 30 × 10 −2 = 0.398 µF 0.01 × 10 −3 98 P3.32* The charge Q remains constant because the terminals of the capacitor are open-circuited. Q = C 1V1 = 1000 × 10 −12 × 1000 = 1 µC W1 = (1 2)C 1 (V1 )2 = 500 µJ After the distance between the plates is doubled, the capacitance becomes C 2 = 500 pF . The voltage increases to V2 = 10 −6 Q = = 2000 V and the stored C 2 500 × 10 − 12 energy is W2 = (1 2)C 2 (V2 ) = 1000 µJ . The force needed to pull the plates 2 apart supplies the additional energy. P3.33 C = εr ε 0A d = εr ε 0WL d (a) Thus if W and L are both halved, the capacitance is decreased by a factor of four resulting in C = 50 pF. (b) If d is halved, the capacitance is doubled resulting in C = 400 pF. (c) The relative dielectric constant of air is approximately unity. Thus, replacing air with oil increases εr by a factor of 35 increasing the capacitance to 7000 pF. P3.34 Using C = εr ε 0A d = εr ε 0WL d and Vmax = Kd to substitute into 1 1 εr ε 0WL 2 2 1 2 CVmax , we have Wmax = K d = εr ε 0K 2WLd . 2 2 d 2 However, the volume of the dielectric is Vol = WLd, so we have 1 Wmax = εr ε 0K 2 (Vol) 2 Thus, we conclude that the maximum energy stored is independent of W, L, and d if the volume is constant and if both W and L are much larger than d. To achieve large energy storage per unit volume, we should look for a dielectric having a large value for εr K 2 . The dielectric should have Wmax = high relative dielectric constant and high breakdown strength. P3.35 The capacitance of the microphone is ε A 8.85 × 10 −12 × 10 × 10 −4 C = 0 = d 100[1 + 0.005 cos(5000t )]10 −6 ≅ 8.85 × 10 −11 [1 − 0.005 cos(5000t )] 99 The current flowing through the microphone is dq (t ) d [Cv ] i (t ) = = ≅ 442.5 × 10 −9 sin(5000t ) dt P3.36 dt A Referring to Figure P3.36 in the book, we see that the transducer consists of two capacitors in parallel: one above the surface of the liquid and one below. Furthermore, the capacitance of each portion is proportional to its length and the relative dielectric constant of the material between the plates. Thus for the portion above the liquid, the capacitance in pF is 100 − x pF 100 in which x is the height of the liquid in cm. For the portion of the plates below the surface of the liquid: C above = 100 C below = 100(35) x 100 Then the total capacitance is: C = C above + C below C = 100 + 34x P3.37 pF With the tank full, we have 2 w full = 21 C fullVfull = 21 × 2500 × 10 −12 × 1000 2 = 1.25 mJ Q = C fullVfull = 2500 × 10 −12 × 1000 = 2.5 µC The charge cannot change when the tank is drained, so we have 2.5 × 10 −6 Q Vempty = = = 25000 V C empty 100 × 10 −12 2 w empty = 21 C emptyVempty = 21 × 100 × 10 −12 × 25000 2 = 31.25 mJ The added energy is supplied from the gravitational potential energy of the insulating fluid. When there is liquid between the plates, the charge separation of the dielectric partly cancels the electrical forces of the charges on the plates. When the dielectric fluid drains, this cancellation effect is lost, which is why the voltage increases. The charge on the plates creates a small force of attraction on the fluid, and it is the force of gravity acting against this force of attraction as the fluid drains that accounts for the added energy. 100 P3.38 dv c (t ) d [10 cos(100t )] = −10 −3 sin(100t ) = 10 −6 dt dt −3 v r (t ) = Ric (t ) = −2 × 10 sin (100t ) v (t ) = vc (t ) + v r (t ) ic (t ) = C v (t ) = 10 cos (100t ) − 2 × 10 −3 sin (100t ) Thus, v (t ) = v c (t ) to within 1% accuracy, and the resistance can be neglected. Repeating for v c (t ) = 0.1 cos(10 7t ) , we find ic (t ) = − sin(10 7t ) v r (t ) = −2 sin(10 7 ) v (t ) = v c (t ) + v r (t ) = 0.1 cos(10 7t ) − 2 sin(10 7t ) In this case, the voltage across the parasitic resistance is larger in magnitude than the voltage across the capacitance and the parasitic resistance cannot be neglected. P3.39 The required volume is 2Wmax 2 × 132 × 10 6 = Vol = Ad = = 2.9 × 10 6 m 3 2 2 12 5 − ε 0ε r K 8.85 × 10 (32 × 10 ) Clearly an air dielectric capacitor is not a practical energy storage device for an electric car! The thickness of the dielectric is V 10 4 d min = max = = 3.125 mm 32 × 10 5 K P3.40 Before the switch closes, the energies are W1 = (1 2)C 1 (V1 )2 = 12.5 mJ W2 = (1 2)C 2 (V2 )2 = 12.5 mJ Thus, the total stored energy is 25 mJ. The charge on the top plate of C 1 is Q1 = C 1V1 = +500 µC . The charge on the top plate of C 2 is Q2 = C 2V2 = −500 µC . Thus, the total charge on the top plates is zero. When the switch closes, the charges cancel, the voltage becomes zero, and the stored energy becomes zero. Where did the energy go? Usually, the resistance of the wires absorbs it. If the superconductors are used so that the resistance is zero, the energy can be accounted for by considering the inductance of the circuit. 101 (It is not possible to have a real circuit that is precisely modeled by Figure P3.40; there is always resistance and inductance associated with the wires that connect the capacitances.) P3.41 Inductors consist of coils of wire wound on coil forms such as toriods. P3.42 A fluid flow analogy for an inductor consists of an incompressible fluid flowing through a frictionless pipe of constant diameter. The pressure differential between the ends of the pipe is analogous to the voltage across the inductor and the flow rate of the liquid is proportional to the current. (If the pipe had friction, the electrical analog would have series resistance. If the ends of the pipe had different diameters, a pressure differential would exist for constant flow rate, whereas an inductance has zero voltage for constant flow rate.) P3.43* L = 2H v L (t ) = L diL (t ) dt p (t ) = v L (t )iL (t ) w (t ) = 1 L[iL (t )] 2 2 102 P3.44* iL (t ) = 1 t L ∫0 t v L (t )dt + iL (0 ) = 20 × 10 ∫ 10dt − 0.1 = 2 × 105t − 0.1 A 3 0 Solving for the time that the current reaches + 100 mA , we have iL (tx ) = 0.1 = 2 × 10 5t0 − 0.1 tx = 1 µs di 4 = 0.5 = 10 V 0. 2 dt P3.45* vL = L P3.46 iL (t1 ) = 1 L t1 ∫ v L (t )dt + iL (0 ) 0 p (1) = v L (1)iL (1) = 72 W P3.47 Because we have v L (t ) = L iL (1) = 1 1 2 ∫ 12dt + 0 = 6 A 0 w (1) = LiL2 (1) = 36 J 1 2 diL (t ) , the voltage is zero when the current is dt constant. Thus, we say that inductors act as short circuits for steady dc currents. P3.48 P3.49 1 2 Li , the energy stored 2 in the inductor increases when the current magnitude increases. Therefore, energy is flowing into the inductor. Because the energy stored in an inductor is w = L = 0. 2 H i L (t ) = 2 cos(2000πt ) A di (t ) v L (t ) = L L dt = −800π sin(2000πt ) V p (t ) = v L (t )i L (t ) = −1600π cos(2000πt ) sin(2000πt ) = −800π sin(4000πt ) W 1 L[i L (t )] 2 2 = 0.4 cos 2 (2000πt ) J w (t ) = 103 The sketches should be similar to the following plots. The units for the quantities in these plots are A, V, W, J and s. P3.50 L = 0.3 H v L (t ) = L i L (t ) = 5e −200t di L (t ) = −300e −200t V dt p (t ) = v L (t )i L (t ) = (− 300e −200t )(5e −200t ) = −1500e −400t W 1 L[i L (t )] 2 = 3.75e − 400t J 2 The sketches should be similar to the following plots. The units for the quantities in these plots are A, V, W, J and s. w (t ) = 104 P3.51 L = 0. 5 H i L (t ) = 1 t t ∫v L (t )dt + iL (0) = 2∫v L (t )dt L0 i L (t ) = 20t = 40 0 for 0 ≤ t ≤ 2 s for 2 s ≤ t ≤ 4 s = 20(6 − t ) for 4 s ≤ t ≤ 6 s =0 otherwise p (t ) = v L (t )iL (t ) w (t ) = 1 L[i (t )]2 2 L = [iL (t )] 2 The sketches should be similar to the following plots. The units for the quantities in these plots are A, V, W, J and s 105 P3.52 For an inductor with an initial current of 10 A, we have t 1 i (t ) = ∫ v (t )dt + 10 . However for an infinite inductance, this becomes Lt 0 i (t ) = +10 which is the specification for a 10-A current source. Thus, a very large inductance with an initial current of 10 A is an approximation to a 10-A current source. P3.53 L = 50 µH v L (t ) = 5 cos(2π 10 6t ) iL (t ) = 1 L t t 4 6 ∫ v L (t )dt + iL (0 ) = 2 × 10 ∫ 5 cos(2π 10 t )dt 0 = (1 / 20π ) sin(2π 10 t ) A 0 6 106 p (t ) = v L (t )iL (t ) = (1 / 4π ) sin(2π 10 6t )cos(2π 10 6t ) = (1 / 8π ) sin(4π × 10 6t ) w (t ) = 1 L[iL (t )]2 = (62.5 × 10 −9 / π 2 ) cos2 (10 6t ) µJ 2 The sketches should be similar to the following plots. The units for the quantities in these plots are V, A, W, J and s 107 P3.54 P3.55 i (t ) = 1 t 1 t v (t )dt + i (0 ) = ∫Vm cos(ωt )dt L∫ L 0 0 = Vm [sin(ωt ) − sin(0)] ωL V = m sin(ωt ) ωL Clearly for ω → ∞ , the current becomes zero, so the inductance becomes the equivalent of an open circuit. P3.56 We can write i L (t 0 + ∆t ) = If = 1 L t0 + ∆t ∫t v L (t )dt . The integral represents the 0 area of the voltage pulse, which has units of volt seconds and must equal If L . Because a constant-amplitude pulse has the largest area for a given peak amplitude, we can say that the peak amplitude of the voltage pulse must be at least as large as If L / ∆t . We conclude that the area of the pulse remains constant and that the peak amplitude approaches infinity as ∆t approaches zero. In the limit, this type of pulse is called an impulse. P3.57 w (4) = 21 LiL2 (4) = 400 J ⇒ iL (4) = 20 A Since a reference is not specified, we can choose iL (4) = +20 A. Also, because the stored energy is decreasing, the power for the inductor carries a minus sign. Thus p (4) = −200 = v L (4)iL (4) and we have v L (4) = −10 V. Finally, because the current and voltage have opposite algebraic signs, the current flows into the negative polarity. 108 P3.58 Because the current through an open circuit is zero by definition and we t 1 have i (t ) = ∫ v (t )dt (assuming zero initial current), infinite inductance Lt 0 corresponds to an open circuit. Because the voltage across a short circuit is zero by definition and di (t ) v L = L L , we see that L = 0 corresponds to a short circuit. dt P3.59 Inductances are combined in the same way as resistances. Inductances in series are added. Inductances in parallel are combined by taking the reciprocal of the sum of the reciprocals of the several inductances. P3.60* (a) Leq = 1 + 1 = 3H 1 6 + 1 (1 + 2) (b) 9 H in parallel with 18 H is equivalent to 6H. Also, 20 H in parallel 1 = 6H with 5 H is equivalent to 4 H. Finally, we have Leq = 1 15 + 1 (6 + 4 ) P3.61* i (t ) = i1 (t ) = 1 Leq 1 L1 t ∫ v (t )dt = 0 L1 + L2 t v (t )dt L1 L2 ∫0 t ∫ v (t )dt 0 Thus, we can write i1 (t ) = L2 i (t ) = 2 i (t ) . 3 L1 + L2 L1 1 i (t ) = i (t ) . Similarly, we have i2 (t ) = L1 + L2 3 This is similar to the current-division principle for resistances. Keep in mind that these formulas assume that the initial currents are zero. P3.62 (a) Leq = 5 H . (b) The 6 H inductor and 3 H inductor have no effect because they are in parallel with a short circuit. Thus, Leq = 7 H . 109 P3.63 We need to place L = 3 H in series with the original 4-H inductance. P3.64 If all three inductors are connected in series, we obtain the maximum inductance: Lmax = 18 H By connecting all three inductors in parallel, we obtain the minimum inductance: 1 Lmin = = 2H 1 6+1 6+1 6 P3.65 Ordinarilly negative inductance is not practical. Thus, adding inductance in series always increases the equivalent inductance. However, placing inductance in parallel results in smaller inductance. Thus, we need to consider a parallel inductance such that 1 =3 1/L + 1/4 Solving, we find that L = 12 H. P3.66 P3.67 (a) For i (t ) = 0.1 sin(20t ) , we have v R (t ) = 0.05 sin(20t ) v L (t ) = 0.02 cos(20t ) v (t ) = 0.05 sin(20t ) + 0.02 cos(20t ) In this case, the parasitic resistance cannot be neglected. (b) For i (t ) = 0.1 sin(10 6t ) , we have 110 vR (t ) = Ri (t ) = 0.05 sin(106t ) di (t ) v L (t ) = L = 1000 cos(106t ) dt v (t ) = vR (t ) + v L (t ) = 0.05 sin(106t ) + 1000 cos(106t ) In this case to obtain 1% accuracy, the resistance can be neglected. P3.68 diL (t ) = 40 × 10 −3 × 5000[ − sin(5000t )] = −200 sin(5000t ) V dt dv (t ) iC (t ) = C = − cos(5000t ) A dt i (t ) = iC (t ) + i L (t ) = 0 v (t ) = L w C (t ) = 21 Cv 2 (t ) = 20 sin2 (5000t ) = 10 − 10 cos(10000t ) mJ w L (t ) = 21 LiL2 (t ) = 20 cos 2 (5000t ) = 10 + 10 cos(10000t ) mJ w (t ) = w C (t ) + w L (t ) = 20 mJ The values in this circuit have been carefully selected so the source current is zero. Because the source current is zero and there are no resistances, there is no source or sink for energy in the circuit. Thus, we would expect the total energy to be constant, as the equations show. The total energy surges back and forth between the capacitance and the inductance. In a real circuit, the parasitic resistances would eventually absorb the energy. The circuit is analogous to a swinging pendulum or a ringing bell. P3.69 Because v L = L di L (t ) = 0 for currents that are constant in time, we dt conclude that the inductance behaves as a short-circuit for dc currents. Thus, the circuit simplifies to the single resistance Rs , which is given 400 mV by Rs = = 2 Ω. 200 mA P3.70 dv C (t ) = 250 × 10 −6 × 10 × 1000[ − sin(1000t )] = −2.5 sin(1000t ) A dt di (t ) v L (t ) = L = −10 cos(1000t ) V dt v (t ) = v C (t ) + v L (t ) = 0 i (t ) = C w C (t ) = 21 Cv C2 (t ) = 12.5 cos 2 (1000t ) = 6.25 + 6.25 cos(2000t ) mJ w L (t ) = 21 Li 2 (t ) = 12.5 sin2 (1000t ) = 6.25 − 6.25 cos(2000t ) mJ w (t ) = w C (t ) + w L (t ) = 12.5 mJ 111 The values in this circuit have been carefully selected so the source voltage is zero. Because the source voltage is zero and there are no resistances, there is no source or sink for energy in the circuit. Thus, we would expect the total energy to be constant, as the equations show. The total energy surges back and forth between the capacitance and the inductance. In a real circuit, the parasitic resistances would eventually absorb the energy. The circuit is analogous to a swinging pendulum or a ringing bell. P3.71 When a time-varying current flows in a coil, a time-varying magnetic field is produced. If some of this field links a second coil, voltage is induced in it. Thus time-varying current in one coil results in a contribution to the voltage across a second coil. P3.72* (a) As in Figure 3.23a, we can write di1 (t ) di (t ) +M 2 dt dt di (t ) di (t ) v 2 (t ) = M 1 + L2 2 dt dt However, for the circuit at hand, we have i (t ) = i1 (t ) = i2 (t ) . v 1 (t ) = L1 Thus, di (t ) dt di (t ) v 2 (t ) = (L2 + M ) dt Also, we have v (t ) = v 1 (t ) + v 2 (t ) . di (t ) Substituting, we obtain v (t ) = (L1 + 2M + L2 ) . dt di (t ) Thus, we can write v (t ) = Leq , in which dt v 1 (t ) = (L1 + M ) 112 Leq = L1 + 2M + L2 . (b) Similarly, for the dot at the bottom end of L2 , we have Leq = L1 − 2M + L2 P3.73 (a) Refer to Figures 3.23 and P3.73. For the dots as shown in Figure P3.73, we have di1 (t ) di (t ) +M 2 = 20 cos(20t ) + 15 cos(30t ) V dt dt di (t ) di (t ) v 2 (t ) = M 1 + L2 2 = 20 cos(20t ) + 30 cos(30t ) V dt dt v 1 (t ) = L1 (b) With the dot moved to the bottom of L2 , we have di1 (t ) di (t ) −M 2 = 20 cos(20t ) − 15 cos(30t ) V dt dt di (t ) di (t ) v 2 (t ) = −M 1 + L2 2 = −20 cos(20t ) + 30 cos(30t ) dt dt v 1 (t ) = L1 P3.74 In general, we have di di1 ±M 2 dt dt di di v 2 (t ) = ±M 1 + L2 2 dt dt v 1 (t ) = L1 Substituting the given information, we have v 1 (t ) = −2 × 10 4 sin(1000t ) and − 5000 sin(1000t ) = mM 10 4 sin(1000t ) We deduce that M = 0.5 H. Furthermore, because the upper of the two algebraic signs applies, we know that the currents are referenced into like (i.e., both dotted or both undotted) terminals. P3.75 With a short circuit across the terminals of the second coil, we have di1 (t ) di (t ) +M 2 dt dt di (t ) di (t ) v 2 (t ) = 0 = M 1 + L2 2 dt dt Solving the second equation for di2 (t ) / dt and substituting into the first v 1 (t ) = L1 equation, we have 113 di1 (t ) M 2 di1 (t ) L1 L2 − M 2 di1 (t ) v 1 (t ) = L1 − = dt L2 dt L2 dt from which we can see that LL − M2 Leq = 1 2 L2 P3.76 Because of the parallel connection, we have v 1 (t ) = v 2 (t ) = v (t ) and the equations for the mutually coupled inductors become di1 (t ) di (t ) +M 2 dt dt di (t ) di (t ) v (t ) = M 1 + L2 2 dt dt v (t ) = L1 Using determinants to solve for the derivatives, we have v (t ) M L1 v (t ) L −M di1 (t ) v (t ) L2 L −M di2 (t ) M v (t ) v ( t ) v (t ) = = 2 = = 1 2 L1 M L1 M L1 L2 − M L1L2 − M 2 dt dt M M L2 L2 Then, we have i (t ) = i1 (t ) + i2 (t ) di (t ) di1 (t ) di1 (t ) L1 + L2 − 2M v (t ) = + = L1L2 − M 2 dt dt dt from which we conclude that LL − M 2 Leq = 1 2 L1 + L2 − 2M P3.77 diL d [exp(−2t ) sin(4πt )] = 0.2 dt dt v L = 0.8π exp( −2t ) cos(4πt ) − 0.4 exp( −2t ) sin(4πt ) vL = L A sequence of MATLAB commands that produces the desired plots is syms iL vL t iL = exp(-2*t)*sin(4*pi*t); vL = 0.2*diff(iL,t); ezplot(iL, [0 2]) hold on ezplot(vL, [0 2]) The result is: 114 P3.78 Using either tables or integration by parts, we have i L (t ) = 1 L t ∫ v L (t )dt + iL (0) = 0 1 t t exp( −t )dt L∫ 0 = 1 L [− exp(−t )(t + 1)]0 t i L (t ) = 1 − t exp( −t ) − exp( −t ) A sequence of MATLAB commands to verify our result for iL(t) and obtain the desired plots is syms iL vL t vL = t*exp(-t); iL = int(vL,t,0,t) ezplot(vL, [0 10]) hold on ezplot(iL, [0 10]) The result is: 115 P3.79 The MATLAB commands are: syms vC iC pC wC t iC=10e-3*(heaviside(t)-heaviside(t-2e-3)) - ... 20e-3*(heaviside(t-4e-3)-heaviside(t-5e-3)); vC=(0.5e6)*int(iC,t,0,t); pC=vC*iC; wC=(1/2)*(2e-6)*vC^2; subplot(2,2,1) ezplot(iC, [0 6e-3]), title('\iti_C\rm versus \itt') subplot(2,2,2) ezplot(vC, [0 6e-3]), title('\itv_C \rmversus \itt') subplot(2,2,3) ezplot(pC, [0 6e-3]), title('\itp_C \rmversus \itt') subplot(2,2,4) ezplot(wC, [0 6e-3]), title('\itw_C \rmversus \itt') The resulting plots are shown in the solution to problem P3.13. P3.80 The MATLAB commands are: syms vC iC pC wC t 116 iC = 100*t*(heaviside(t)-heaviside(t-1e-3)) - ... (0.2-100*t)*(heaviside(t-1e-3)-heaviside(t-2e-3)); vC = (0.2e6)*int(iC,t,0,t); pC = vC*iC; wC = (1/2)*(5e-6)*vC^2; subplot(2,2,1) ezplot(iC, [0 3e-3]), title('\iti_C\rm versus \itt') subplot(2,2,2) ezplot(vC, [0 3e-3]), title('\itv_C \rmversus \itt') subplot(2,2,3) ezplot(pC, [0 3e-3]), title('\itp_C \rmversus \itt') subplot(2,2,4) ezplot(wC, [0 3e-3]), title('\itw_C \rmversus \itt') The resulting plots are shown in the solution to problem P3.16. P3.81 The MATLAB commands are: syms vL iL pL wL t vL = 10*(heaviside(t)-heaviside(t-2)) - 10*(heaviside(t-4)-heaviside(t-6)); iL = (2)*int(vL,t,0,t); pL = vL*iL; wL = (1/2)*(0.5)*iL^2; subplot(2,2,1) ezplot(vL, [0 8]), title('\itv_L \rmversus \itt') subplot(2,2,2) ezplot(iL, [0 8]), title('\iti_L\rm versus \itt') subplot(2,2,3) ezplot(pL, [0 8]), title('\itp_L \rmversus \itt') subplot(2,2,4) ezplot(wL, [0 8]), title('\itw_L \rmversus \itt') The resulting plots are shown in the solution to problem P3.51. 117 Practice Test v ab (t ) = T3.1 1 C t t ∫ iab (t )dt + vC (0) = 10 ∫ 0.3 exp(−2000t )dt 5 0 0 t v ab (t ) = − 15 exp( −2000t ) 0 v ab (t ) = 15 − 15 exp(−2000t ) V w C (∞ ) = T3.2 1 1 Cv C2 (∞ ) = 10 − 5 (15) 2 = 1.125 mJ 2 2 The 6-µF and 3-µF capacitances are in series and have an equivalent capacitance of 1 C eq1 = = 2 µF 1/ 6 + 1/3 Ceq1 is in parallel with the 4-µF capacitance, and the combination has an equivalent capacitance of C eq 2 = C eq 1 + 4 = 6 µF Ceq2 is in series with the 12-µF and the combination, has an equivalent capacitance of 1 = 4 µF 1 / 12 + 1 / 6 Finally, Ceq3 is in parallel with the 1-µF capacitance, and the equivalent capacitance is C eq = C eq 3 + 1 = 5 µF C eq3 = ε r ε 0A 80 × 8.85 × 10 −12 × 2 × 10 −2 × 3 × 10 −2 = = 4248 pF d 0.1 × 10 − 3 T3.3 C = T3.4 v ab (t ) = L diab = 2 × 10 − 3 × 0.3 × 2000 cos(2000t ) = 1.2 cos(2000t ) V dt The maximum value of sin(2000t) is unity. Thus the peak current is 0.3 A, and the peak energy stored is 1 2 1 w peak = Li peak = × 2 × 10 − 3 (0.3) 2 = 90 µJ 2 2 T3.5 The 2-H and 4-H inductances are in parallel and the combination has an equivalent inductance of 118 1 = 1.333 H 1/2 + 1/ 4 Also, the 3-H and 5-H inductances are in parallel, and the combination has an equivalent inductance of 1 Leq 2 = = 1.875 H 1/3 + 1/5 Finally, Leq1 and Leq2 are in series. The equivalent inductance between terminals a and b is Leq = Leq 1 + Leq 2 = 3.208 H Leq 1 = T3.6 For these mutually coupled inductances, we have di1 (t ) di (t ) −M 2 dt dt di (t ) di (t ) v 2 (t ) = −M 1 + L2 2 dt dt v 1 (t ) = L1 in which the currents are referenced into the positive polarities. Thus the currents are i1 (t ) = 2 cos(500t ) and i2 (t ) = −2 exp(−400t ) Substituting the inductance values and the current expressions we have v 1 (t ) = −40 × 10 −3 × 1000 sin(500t ) − 20 × 10 −3 × 800 exp( −400t ) v 1 (t ) = −40 sin(500t ) − 16 exp(−400t ) V v 2 (t ) = 20 × 10 −3 × 1000 sin(500t ) − 30 × 10 −3 × 800 exp( −400t ) v 2 (t ) = 20 sin(500t ) − 24 exp( −400t ) V T3.7 One set of commands is syms vab iab t iab = 3*(10^5)*(t^2)*exp(-2000*t); vab = (1/20e-6)*int(iab,t,0,t) subplot(2,1,1) ezplot(iab, [0 5e-3]), title('\iti_a_b\rm (A) versus \itt\rm (s)') subplot(2,1,2) ezplot(vab, [0 5e-3]), title('\itv_a_b\rm (V) versus \itt\rm (s)') The results are 119 v ab = 15 15 − exp( −2000t ) − 7500 exp( −2000t ) − 7.5 × 10 6t 2 exp( −2000t ) 4 4 120 CHAPTER 4 Exercises E4.1 The voltage across the circuit is given by Equation 4.8: v C (t ) = Vi exp( −t / RC ) in which Vi is the initial voltage. At the time t1% for which the voltage reaches 1% of the initial value, we have 0.01 = exp( −t1% / RC ) Taking the natural logarithm of both sides of the equation, we obtain ln(0.01) = −4.605 = −t1% / RC Solving and substituting values, we find t1% = 4.605RC = 23.03 ms. E4.2 The exponential transient shown in Figure 4.4 is given by v C (t ) = Vs −Vs exp( −t / τ ) Taking the derivative with respect to time, we have dv C (t ) Vs = exp( −t / τ ) dt τ Evaluating at t = 0, we find that the initial slope is VS / τ . Because this matches the slope of the straight line shown in Figure 4.4, we have shown that a line tangent to the exponential transient at the origin reaches the final value in one time constant. E4.3 (a) In dc steady state, the capacitances act as open circuits and the inductances act as short circuits. Thus the steady-state (i.e., t approaching infinity) equivalent circuit is: From this circuit, we see that ia = 2 A. Then ohm’s law gives the voltage as v a = Ria = 50 V. 121 (b) The dc steady-state equivalent circuit is: Here the two 10-Ω resistances are in parallel with an equivalent resistance of 1/(1/10 + 1/10) = 5 Ω. This equivalent resistance is in series with the 5-Ω resistance. Thus the equivalent resistance seen by the source is 10 Ω, and i1 = 20 / 10 = 2 A. Using the current division principle, this current splits equally between the two 10-Ω resistances, so we have i2 = i3 = 1 A. E4.4 (a) τ = L / R2 = 0.1 / 100 = 1 ms (b) Just before the switch opens, the circuit is in dc steady state with an inductor current of Vs / R1 = 1.5 A. This current continues to flow in the inductor immediately after the switch opens so we have i (0+) = 1.5 A . This current must flow (upward) through R2 so the initial value of the voltage is v (0 +) = −R2i (0 +) = −150 V. (c) We see that the initial magnitude of v(t) is ten times larger than the source voltage. (d) The voltage is given by v (t ) = − Vs L exp( −t / τ ) = −150 exp( −1000t ) R1τ Let us denote the time at which the voltage reaches half of its initial magnitude as tH. Then we have 0.5 = exp( −1000tH ) Solving and substituting values we obtain tH = −10 −3 ln(0.5) = 10 −3 ln(2) = 0.6931 ms 122 E4.5 First we write a KCL equation for t ≥ 0. t v (t ) 1 + ∫ v (x )dx + 0 = 2 R L 0 Taking the derivative of each term of this equation with respect to time and multiplying each term by R, we obtain: dv (t ) R + v (t ) = 0 dt L The solution to this equation is of the form: v (t ) = K exp( −t / τ ) in which τ = L / R = 0.2 s is the time constant and K is a constant that must be chosen to fit the initial conditions in the circuit. Since the initial (t = 0+) inductor current is specified to be zero, the initial current in the resistor must be 2 A and the initial voltage is 20 V: v (0+) = 20 = K Thus, we have v (t ) = 20 exp( −t / τ ) 1 iR = v / R = 2 exp( −t / τ ) t t 1 iL (t ) = ∫ v (x )dx = [− 20 τ exp( −x / τ )] = 2 − 2 exp( −t / τ ) L0 2 0 E4.6 Prior to t = 0, the circuit is in DC steady state and the equivalent circuit is Thus we have i(0-) = 1 A. However the current through the inductor cannot change instantaneously so we also have i(0+) = 1 A. With the switch open, we can write the KVL equation: di (t ) + 200i (t ) = 100 dt The solution to this equation is of the form i (t ) = K 1 + K 2 exp( −t / τ ) in which the time constant is τ = 1 / 200 = 5 ms. In steady state with the switch open, we have i (∞) = K 1 = 100 / 200 = 0.5 A. Then using the initial 123 current, we have i (0+) = 1 = K 1 + K 2 , from which we determine that K 2 = 0.5. Thus we have i (t ) = 1.0 A for t < 0 = 0.5 + 0.5 exp( −t / τ ) for t > 0. v (t ) = L di (t ) dt = 0 V for t < 0 = −100 exp( −t / τ ) for t > 0. E4.7 As in Example 4.4, the KVL equation is t 1 Ri (t ) + ∫ i (x )dx + v C (0+) − 2 cos(200t ) = 0 C 0 Taking the derivative and multiplying by C, we obtain di (t ) RC + i (t ) + 400C sin(200t ) = 0 dt Substituting values and rearranging the equation becomes di (t ) 5 × 10 −3 + i (t ) = −400 × 10 −6 sin(200t ) dt The particular solution is of the form i p (t ) = A cos(200t ) + B sin(200t ) Substituting this into the differential equation and rearranging terms results in 5 × 10 −3 [− 200A sin(200t ) + 200B cos(200t )] + A cos(200t ) + B sin(200t ) = −400 × 10 −6 sin(200t ) Equating the coefficients of the cos and sin terms gives the following equations: − A + B = −400 × 10 −6 and B + A = 0 from which we determine that A = 200 × 10 −6 and B = −200 × 10 −6 . Furthermore, the complementary solution is iC (t ) = K exp( −t / τ ) , and the complete solution is of the form i (t ) = 200 cos(200t ) − 200 sin(200t ) + K exp( −t / τ ) µA At t = 0+, the equivalent circuit is 124 from which we determine that i (0+) = 2 / 5000 = 400 µA. Then evaluating our solution at t = 0+, we have i (0 +) = 400 = 200 + K , from which we determine that K = 200 µA. Thus the complete solution is i (t ) = 200 cos(200t ) − 200 sin(200t ) + 200 exp(−t / τ ) µA E4.8 The KVL equation is t 1 Ri (t ) + ∫ i (x )dx + v C (0+) − 10 exp(−t ) = 0 C 0 Taking the derivative and multiplying by C, we obtain di (t ) RC + i (t ) + 10C exp( −t ) = 0 dt Substituting values and rearranging, the equation becomes di (t ) 2 + i (t ) = −20 × 10 −6 exp( −t ) dt The particular solution is of the form i p (t ) = A exp( −t ) Substituting this into the differential equation and rearranging terms results in − 2A exp(−t ) + A exp( −t ) = −20 × 10 −6 exp( −t ) Equating the coefficients gives A = 20 × 10 −6. Furthermore, the complementary solution is iC (t ) = K exp( −t / 2) , and the complete solution is of the form i (t ) = 20 exp(−t ) + K exp( −t / 2) µA At t = 0+, the equivalent circuit is 125 from which we determine that i (0+) = 5 / 10 6 = 5 µA. Then evaluating our solution at t = 0+, we have i (0 +) = 5 = 20 + K , from which we determine that K = −15 µA. Thus the complete solution is i (t ) = 20 exp( −t ) − 15 exp( −t / 2) µA E4.9 (a) ω 0 = 1 LC = 1 10 −3 × 10 −7 = 10 5 α= 1 2RC = 2 × 10 5 ζ = α =2 ω0 (b) At t = 0+, the KCL equation for the circuit is v ( 0 +) 0.1 = + iL (0 +) + Cv ′(0 +) (1) R However, v (0 +) = v (0 −) = 0 , because the voltage across the capacitor cannot change instantaneously. Furthermore, iL (0 +) = iL (0−) = 0 , because the current through the inductance cannot change value instantaneously. Solving Equation (1) for v ′(0 +) and substituting values, we find that v ′(0+) = 10 6 V/s. (c) To find the particular solution or forced response, we can solve the circuit in steady-state conditions. For a dc source, we treat the capacitance as an open and the inductance as a short. Because the inductance acts as a short v p (t ) = 0. (d) Because the circuit is overdamped (ζ > 1), the homogeneous solution is the sum of two exponentials. The roots of the characteristic solution are given by Equations 4.72 and 4.73: s 1 = −α − α 2 − ω 02 = −373.2 × 10 3 s 2 = −α + α 2 − ω 02 = −26.79 × 103 Adding the particular solution to the homogeneous solution gives the general solution: 126 v (t ) = K 1 exp(s1t ) + K 2 exp(s 2t ) Now using the initial conditions, we have v ( 0 +) = 0 = K 1 + K 2 v ′(0+) = 10 6 = K 1s1 + K 2s 2 Solving we find K 1 = −2.887 and K 2 = 2.887. Thus the solution is: v (t ) = 2.887[exp(s 2t ) − exp(s1t )] E4.10 (a) ω 0 = 1 LC = 1 10 −3 × 10 −7 = 10 5 α= 1 2RC = 10 5 ζ = α =1 ω0 (b) The solution for this part is the same as that for Exercise 4.9b in which we found thatv ′(0 +) = 10 6 V/s. (c) The solution for this part is the same as that for Exercise 4.9c in which we found v p (t ) = 0. (d) The roots of the characteristic solution are given by Equations 4.72 and 4.73: s1 = −α − α 2 − ω 02 = −10 5 s 2 = −α + α 2 − ω 02 = −10 5 Because the circuit is critically damped (ζ = 1), the roots are equal and the homogeneous solution is of the form: v (t ) = K 1 exp(s1t ) + K 2t exp(s1t ) Adding the particular solution to the homogeneous solution gives the general solution: v (t ) = K 1 exp(s1t ) + K 2t exp(s1t ) Now using the initial conditions we have v ( 0 +) = 0 = K 1 v ′(0+) = 10 6 = K 1s1 + K 2 Solving we find K 2 = 10 6 Thus the solution is: v (t ) = 10 6t exp( −10 5t ) E4.11 (a) ω 0 = 1 LC = 1 10 −3 × 10 −7 = 10 5 α= 1 2RC = 2 × 10 4 ζ = α = 0.2 ω0 (b) The solution for this part is the same as that for Exercise 4.9b in which we found that v ′(0+) = 10 6 V/s. (c) The solution for this part is the same as that for Exercise 4.9c in 127 which we found v p (t ) = 0. (d) Because we have (ζ < 1), this is the underdamped case and we have ω n = ω 02 − α 2 = 97.98 × 10 3 Adding the particular solution to the homogeneous solution gives the general solution: v (t ) = K 1 exp( −αt ) cos(ω nt ) + K 2 exp( −αt ) sin(ω nt ) Now using the initial conditions we have v ( 0 +) = 0 = K 1 v ′(0+) = 10 6 = −αK 1 + ω n K 2 Solving we find K 2 = 10.21 Thus the solution is: v (t ) = 10.21 exp( −2 × 10 4t ) sin(97.98 × 103t ) V E4.12 The commands are: syms ix t R C vCinitial w ix = dsolve('(R*C)*Dix + ix = (w*C)*2*cos(w*t)', 'ix(0)=-vCinitial/R'); ians =subs(ix,[R C vCinitial w],[5000 1e-6 1 200]); pretty(vpa(ians, 4)) ezplot(ians,[0 80e-3]) An m-file named Exercise_4_12 containing these commands can be found in the MATLAB folder on the OrCAD disk. E4.13 The commands are: syms vc t % Case I R = 300: vc = dsolve('(1e-8)*D2vc + (1e-6)*300*Dvc+ vc =10', ... 'vc(0) = 0','Dvc(0)=0'); vpa(vc,4) ezplot(vc, [0 1e-3]) hold on % Turn hold on so all plots are on the same axes % Case II R = 200: vc = dsolve('(1e-8)*D2vc + (1e-6)*200*Dvc+ vc =10',... 'vc(0) = 0','Dvc(0)=0'); vpa(vc,4) ezplot(vc, [0 1e-3]) % Case III R = 100: vc = dsolve('(1e-8)*D2vc + (1e-6)*100*Dvc+ vc =10',... 'vc(0) = 0','Dvc(0)=0'); vpa(vc,4) 128 ezplot(vc, [0 1e-3]) An m-file named Exercise_4_13 containing these commands can be found in the MATLAB folder on the OrCAD disk. Problems P4.1 The time constant τ is the interval required for the voltage to fall to exp(-1) ≅ 0.368 times its initial value. The time constant is given by τ = RC . Thus to attain a long time constant, we need large values for both R and C. P4.2* The voltage across the capacitor is given by Equation 4.8. v (t ) = Vi exp( −t / RC ) in which Vi = 100 V is the initial voltage, C = 100 µF is the capacitance, and R is the leakage resistance. The energy stored in the capacitance is 1 w = Cv 2 (t ) = 0.5 × 10 −4 × 100 2 exp( −2t / RC ) 2 Since we require the energy to be 90% of the initial value after one minute, we can write 0.9 × 0.5 × 10 −4 × 100 2 ≤ 0.5 × 10 −4 × 100 2 exp( −120 / RC ) Solving we determine that RC must be greater than 1139 s. Thus the leakage resistance must be greater than 11.39 MΩ. P4.3* The solution is of the form of Equation 4.17: v C (t ) = Vs + K 2 exp(− t RC ) = 10 + K 2 exp(− t (2 × 10 −3 ) ) in which K 2 is a constant to be determined. At t = 0 + , we have v C (0 + ) = −10 = 10 + K 2 Solving, we find that K 2 = −20 and the solution is v C (t ) = 10 − 20 exp(− t (2 × 10 −3 ) ) V Setting the voltage equal to zero at time t0 and solving we obtain: 0 = 10 − 20 exp(− t0 (2 × 10 −3 ) ) 1 / 2 = exp(− t0 (2 × 10 −3 ) ) − ln(2) = −t0 /(2 × 10 −3 ) t0 = 2 ln(2) = 1.386 ms 129 P4.4* The initial energy is W1 = 1 1 C (Vi )2 = 100 × 10 −6 × 1000 2 = 50 J 2 2 1 C [v (t2 )]2 , which 2 yields v (t2 ) = 707.1 V . The voltage across the capacitance is given by At t = t2 , half of the energy remains, and we have 25 = v C (t ) =Vi exp(− t RC ) = 1000 exp(− 10t ) for t > 0 Substituting, we have 707.1 = 1000 exp(− 10t2 ) . Solving, we obtain ln( 0.7071) = −10t2 t2 = 0.03466 seconds P4.5* This is a case of a capacitance discharging through a resistance. The voltage is given by Equation 4.8: v C (t ) = Vi exp(− t RC ) At t = 0 , we have v C (0 ) = 50 = Vi . At t = 30 , we have v C (30 ) = 25 . Thus, we can write 25 = 50 exp(− 30 RC ) . Dividing by 50 and taking the natural logarithm of both sides, we obtain − ln(2) = − 30 RC . Rearranging, we 30 = 4.328 MΩ . have R = C ln(2) P4.6* v (t ) = V1 exp[−(t − t1 ) / RC ] for t ≥ t1 P4.7 We have v (t ) = Vi exp( −t / τ ) and w (t ) = 1 2 voltage. The initial stored energy is Wi = Cv 2 (t ) in which Vi is the initial 1 2 CVi 2 . After two time constants, we have v (t ) = Vi exp(−τ / τ ) ≅ 0.135Vi so that 13.5 percent of the initial voltage remains. The energy is w (τ ) = 1 2 Cv 2 (τ ) = 1 2 CVi 2 exp(−4) = Wi exp( −4) ≅ 0.0183Wi so only 1.83% of the initial energy remains. P4.8 The solution is of the form given in Equation 4.19: v C (t ) = Vs −Vs exp(− t RC ) RC = 50 × 103 × 0.04 × 10 − 6 = 2 ms 130 Thus, we have v C (t ) = 10 − 10 exp(− t / 2 × 10 −3 ) V P4.9 Equation 4.8 gives the expression for the voltage across a capacitance discharging through a resistance: v C (t ) =Vi exp(− t RC ) After one-half-life, we have v C (thalf ) = Vi Vi = Vi exp(− thalf RC ) . 2 2 Dividing by Vi and taking the natural logarithm of both sides, we have and − ln(2) = − thalf RC Solving, we obtain thalf = RC ln(2) = 0.6931RC = 0.6931τ P4.10 The voltage across the capacitor is given by Equation 4.8. v (t ) = Vi exp( −t / RC ) in which Vi is the initial voltage. Substituting values, we have v (0.1 ) = 5 = Vi exp( −0.1 / 0.15) = Vi exp( −0.6667) Vi = 9.74 V P4.11 (a) RC = 10 ms vC (t ) = 10 for t < 0 = 10 exp(− t 0.01) = 10 exp(− 100t ) for t > 0 v R (t ) = 0 for t < 0 = 10 exp(− t 0.01) = 10 exp(− 100t ) for t > 0 (b) pR (t ) = [v R (t )]2 R = exp(− 200t ) W for t > 0 pR (t ) = 0 for t < 0 131 (c) ∞ W = ∫ pR (t )dt 0 ∞ = ∫ exp(− 200t )dt 0 = − (1 / 200) exp(− 200t ) 0 ∞ = 5 mJ (d) The initial energy stored in the capacitance is 1 W = C [v C (0 )]2 2 1 = × 100 × 10 − 6 × 10 2 2 = 5 mJ P4.12 We have P = Pi (0.97)t = Pi exp( −t / τ ) in which t is time in years, Pi is the initial purchasing power and τ is the time constant in years. Solving we have (0.97)t = exp( −t / τ ) t ln(0.97) = −t / τ τ = −1 / ln(0.97) = 32.83 years P4.13 Prior to t = 0 , we have v C (t ) = 0 because the switch is closed. After t = 0 , we can write the following KCL equation at the top node of the circuit: v C (t ) dv C (t ) +C = 10 mA R dt Multiplying both sides by R and substituting values, we have dv (t ) 0.02 C + v C (t ) = 20 dt The solution is of the form v C (t ) = K 1 + K 2 exp(− t RC ) = K 1 + K 2 exp(− 50t ) (1) (2) Using Equation (2) to substitute into Equation (1), we eventually obtain K 1 = 20 The voltage across the capacitance cannot change instantaneously, so we have v C (0 + ) = v C (0 − ) = 0 v C (0 + ) = 0 = K 1 + K 2 132 Thus, K 2 = −K 1 = −20 , and the solution is v C (t ) = 20 − 20 exp(− 50t ) for t > 0 The sketch should resemble the following plot: P4.14 During the charging interval, the time constant is τ 1 = R1C = 20 s, and the voltage across the capacitor is given by v C (t ) = 2500[1 − exp( −t / τ 1 )] 0 ≤ t ≤ 40 At the end of the charging interval (t = 40 s), this yields v C (40) = 2500[1 − exp( −2)] = 2162 V The time constant during the discharge interval is τ 2 = R2C = 50 s. Working in terms of the time variable t ′ = t − 40, the voltage during the discharge interval is v C (t ′) = 2162 exp( −t ′ / τ 2 ) 0 ≤ t ′ At t ′ = 100 − 40 = 60, this yields 2162 exp( −1.2) = 651 V P4.15 The voltage across the resistance and capacitance is v C (t ) =Vi exp(− t RC ) The initial charge stored on the capacitance is Qi = CVi The current through the resistance is 133 iR (t ) = Vi exp(− t RC ) R The total charge passing through the resistance is ∞ Q = ∫ iR (t )dt 0 = ∞ V ∫ Ri exp(− t RC )dt 0 V ∞ = i [− RC exp(− t RC )] 0 R = CVi P4.16 The initial current is Vi / R = 20000 / 100 = 200 A. No wonder one jumps!!! The time constant is τ = RC = 0.01 µs . P4.17 The final voltage for each 2 s interval is the initial voltage for the succeeding interval. We have τ = RC = 1 s. For 0 ≤ t ≤ 2, we have v (t ) = 10 − 10 exp( −t ) which yields v (2) = 8.647 V. For 2 ≤ t ≤ 4, we have v (t ) = 8.647 exp[ −(t − 2)] which yields v (4) = 1.170 V. For 4 ≤ t ≤ 6, we have v (t ) = 10 − (10 − 1.170) exp[ −(t − 4)] which yields v (6) = 8.805 V. Finally, for 6 ≤ t ≤ 8, we have v (t ) = 7.176 exp[ −(t − 6)] which yields v (8) = 1.192 V. P4.18 (a) The voltages across the capacitors cannot change instantaneously. Thus, v 1 (0 + ) = v 1 (0 − ) = 100 V and v 2 (0 + ) = v 2 (0 − ) = 0 . Then, we can write v (0 + ) − v 2 (0 + ) 100 − 0 i (0 + ) = 1 = = 0.5 mA 200 × 103 R (b) Applying KVL, we have 1 C1 t − v 1 (t ) + Ri (t ) + v 2 (t ) = 0 1 t ∫ i (t )dt − 100 + Ri (t ) + C ∫ i (t )dt + 0 = 0 0 2 0 Taking a derivative with respect to time and rearranging, we obtain 134 di (t ) 1 + dt R  1 1  i (t ) = 0  +  C1 C2  (c) The time constant is τ = R (1) C 1C 2 = 800 ms . C1 + C2 (d) The solution to Equation (1) is of the form i (t ) = K 1 exp(− t τ ) However, i (0 + ) = 0.5 mA , so we have K 1 = 0.5 mA and i (t ) = 0.5 exp(− 1.25t ) mA . (e) The final value of v 2 (t ) is v 2 (∞ ) = 1 C2 ∞ ∫ i (t )dt + v (0 + ) 2 0 = 0.2 × 10 6 t ∫ 0.5 × 10 −3 exp(− 1.25t )dt + 0 0 = −80 exp(− 1.25t ) ∞0 = 80 V P4.19 For a dc steady-state analysis: 1. Replace capacitances with open circuits. 2. Replace inductances with short circuits. 3. Solve the resulting circuit, which consists of dc sources and resistances. P4.20 In dc steady state conditions, the voltages across the capacitors are constant. Therefore, the currents through the capacitances, which are given by iC = C dv dt , are zero. Open circuits also have zero current. Similarly, the currents through the inductances are constant. Therefore, the voltages across inductances, which are given by v = L di P4.21* dt , are zero. Short circuits also have zero voltage. In steady state, the equivalent circuit is: 135 Thus, we have i1 = 0 i3 = i2 = 2 A P4.22* After the switch opens and the circuit reaches steady state, the 10-mA current flows through the 1-k Ω resistance, and the voltage is 10 V. The initial voltage is v C (0 + ) = 0 . The time constant of the circuit is RC = 10 ms . As a function of time, we have v C (t ) = 10 − 10 exp(− t RC ) = 10 − 10 exp(− 100t ) Let t99 denote the time at which the voltage reaches 99% is its final value. Then we have v C (t99 ) = 9.9 = 10 − 10 exp(− 100t99 ) Solving, we find t99 = 46.05 ms P4.23* With the switch in position A and the circuit in steady state prior to t = 0, the capacitor behaves as an open circuit, v R (0−) = 0, and 10 kΩ v C (0−) = 30 = 10 V. Because there cannot be infinite 10 kΩ + 20 Ωk current in this circuit, we have v C (0 +) = v R (0+) = 10 V. After switching and the circuit reaches steady state, we have v R (∞) = 0 V. For t ≥ 0, the resistance across the capacitor is R = 200 kΩ and the time constant is τ = RC = 2 s. The general form of the solution for t ≥ 0 is v R (t ) = K 1 + K 2 exp( −t / τ). However, we know that v R (0+) = 10 = K 1 + K 2 and v R (∞) = 0 = K 1 . Solving we find that K 2 = 10. Thus, we have 136 v R (t ) = 0 t < 0 = 10 exp( −t / τ ) P4.24 V for t ≥ 0 Prior to t = 0 , the steady-state equivalent circuit is: and we see that v c = 12 V . A long time after t = 0 , the steady-state equivalent circuit is: and we have v c = 12 P4.25 40 = 8V. 40 + 20 In steady state with a dc source, the inductance acts as a short circuit and the capacitance acts as an open circuit. The equivalent circuit is: 137 i 4 = (20 V ) (2 kΩ ) = 10 mA i3 = 0 i2 = (20 V ) (5 kΩ ) = 4 mA i1 = i2 + i3 + i 4 = 14 mA v C = 20 V P4.26 iL = 15 mA v x = 45 V v C = −20 V P4.27 iR = 4 mA v C = 20 V P4.28 With the circuit in steady state prior to t = 0, the capacitor behaves as an open circuit, the two 10-kΩ resistors are in parallel, and v C (0−) = (5 mA) × (5 kΩ) = 25 V. Because there cannot be infinite current in this circuit, we have v C (0 +) = 25 V. After the switch opens and the circuit reaches steady state, we have v C (∞) = (5 mA) × (10 kΩ) = 50 V. For t ≥ 0, the Thévenin resistance seen by the capacitor is Rt = 10 kΩ and the time constant is τ = Rt C = 0.1 s. The general form of the solution for t ≥ 0 is v C (t ) = K 1 + K 2 exp( −t / τ ). However, we know that v C (0+) = 25 = K 1 + K 2 and v C (∞) = 50 = K 1 . Solving, we find that K 2 = −25. Thus, we have v C (t ) = 25 t ≤ 0 = 50 − 25 exp( −t / τ ) t ≥ 0 138 P4.29 With the switch closed and the circuit in steady state prior to t = 0, the capacitor behaves as an open circuit, and v C (0−) = 9 V. Because there cannot be infinite current in this circuit when the switch opens, we have v C (0+) = 9 V. After switching and the circuit reaches steady state, we have 3 kΩ v C (∞) = 9 = 3 V. For t ≥ 0, the Thévenin resistance seen by 3 kΩ + 6 Ωk 1 the capacitor is Rt = = 2 kΩ, and the time constant 1/6 + 1/3 is τ = Rt C = 40 ms. The general form of the solution for t ≥ 0 is v C (t ) = K 1 + K 2 exp( −t / τ ). However, we know that v C (0+) = 9 = K 1 + K 2 and v C (∞) = 3 = K 1 . Solving, we find that K 2 = 6. Thus, we have v C (t ) = 9 t ≤ 0 = 3 + 6 exp( −t / τ ) t ≥0 139 P4.30 With the switch closed and the circuit in steady state prior to t = 0, the capacitor behaves as an open circuit, and v C (0−) = 10 V. Because there cannot be infinite currents in this circuit, we have v C (0+) = 10 V. After the switch opens and the circuit reaches steady state, we have v C (∞) = 30 V. For t ≥ 0, the Thévenin resistance seen by the capacitor is Rt = 2 MΩ and the time constant is τ = Rt C = 4 s. The general form of the solution for t ≥ 0 is v C (t ) = K 1 + K 2 exp( −t / τ ). However, we know that v C (0+) = 10 = K 1 + K 2 and v C (∞) = 30 = K 1 . Solving we find that K 2 = −20. Thus, we have v C (t ) = 10 t ≤ 0 = 30 − 20 exp( −t / τ ) t ≥ 0 A plot of the capacitor voltage is: 140 P4.31 The time constant τ is the interval required for the current to fall to exp(−1) ≅ 0.368 times its initial value. The time constant is given by τ = L / R . Thus to attain a long time constant, we need a large value for L and a small value for R. P4.32 The general form of the solution is x (t ) = A + B exp( −t / τ ) . A is the steady-state solution for t >> 0. We determine the value of the desired current or voltage immediately after t = 0, denoted by x (0+). Then, we solve x (0+) = A + B for the value of B. Finally, we determine the Thévenin resistance Rt from the perspective of the energy storage element (i.e., the resistance seen looking back into the circuit with the energy storage element removed) and compute the time constant: τ = Rt C for a capacitance or τ = L / Rt for an inductance. P4.33* In steady state with the switch closed, we have i (t ) = 0 for t < 0 because the closed switch shorts the source. In steady state with the switch open, the inductance acts as a short circuit and the current becomes i (∞ ) = 1 A . The current is of the form i (t ) = K 1 + K 2 exp(− Rt L ) for t ≥ 0 141 in which R = 20 Ω , because that is the Thévenin resistance seen looking back from the terminals of the inductance with the switch open. Also, we have i (0 + ) = i (0 − ) = 0 = K 1 + K 2 i (∞ ) = 1 = K 1 Thus, K 2 = −1 and the current (in amperes) is given by for t < 0 i (t ) = 0 = 1 − exp(− 20t ) P4.34* for t ≥ 0 The general form of the solution is iL (t ) = K 1 + K 2 exp(− R t L ) At t = 0 + , we have iL (0 + ) = iL ((0 − ) = −0.2 = K 1 + K 2 At t = ∞ , the inductance behaves as a short circuit, and we have iL (∞ ) = 0.3 = K 1 Thus, the solution for the current is iL (t ) = −0.2 for t < 0 = 0.3 - 0.5exp (- 2 × 10 5t ) for t > 0 The voltage is di (t ) dt = 0 for t < 0 v (t ) = L = 1000 exp(− 2 × 10 5t ) for t > 0 142 P4.35 The solution is similar to that for Problem P4.34. iL (t ) = 0.3 − 0.3 exp(− 2 × 10 5t ) for t > 0 v (t ) = 600 exp(− 2 × 10t t ) for t > 0 P4.36* The current in a circuit consisting of an inductance L and series resistance R is given by i (t ) = I i exp( −Rt / L ) in which Ii is the initial current. The initial energy stored in the inductance is w i = (1 / 2)LI i 2 and the energy stored as a function of time is w (t ) = (1 / 2)LI i 2 exp( −2Rt / L) Thus we require 0.75 × (1 / 2)LI i 2 ≤ (1 / 2)LI i 2 exp[ −2R (3600) / 10] Solving we determine that we require R ≤ 399.6 µΩ. In practice the only practical way to attain such a small resistance for a 10-H inductance is to use superconductors. P4.37 Before the switch closes, the inductor current is zero and 2 A of current circulates through the source and the two 3-Ω resistors. Immediately after the switch closes, the inductor current remains 0 A, because infinite voltage is not possible in this circuit. (Because the inductor current is zero, we can consider the inductor to be an open circuit at t = 0+.) Therefore, the current through the resistors is unchanged, and 143 is (0+) = 2 A. In steady state, the inductor acts as a short circuit, and we have is (∞) = 4 A. The Thévenin resistance seen by the inductor is 1.5 Ω because the two 3-Ω resistors are in parallel when we zero the source and look back into the circuit from the inductor terminals. Thus, the time constant is τ = L / R = 500 ms. The general form of the solution is is (t ) = K 1 + K 2 exp( −t / τ ). Using the initial and final values, we have is (0+) = 2 = K 1 + K 2 and is (∞) = 4 = K 1 which yields K 2 = −2. Thus, the current is i (t ) = 2 t ≤ 0 = 4 − 2 exp( −t / τ ) t ≥ 0 P4.38 The solution is of the form i (t ) = K 1 + K 2 exp(− Rt L ) At t = 0 + , we have i (0 + ) = 0 = K 1 + K 2 and at t = ∞ , we have i (∞ ) = (20 V ) (10 Ω ) = 2. 0 = K 1 The time constant is τ = L / R = 200 ms. Thus, the solution is i (t ) = 2 − 2 exp(− 5t ) The voltage across the inductor is v L (t ) = L di (t ) = 20 exp(− 5t ) dt 144 P4.39 In steady state, the inductor acts as a short circuit. With the switch closed, the steady-state current is (20 V ) (5 Ω ) = 4 A . With the switch opened, the current eventually approaches iL (∞ ) = (20 V ) (20 Ω ) = 1 A . For t > 0 , the current has the form iL (t ) = K 1 + K 2 exp(− Rt L ) where R = 20 Ω , because that is the resistance with the switch open. Now, we have i L (0 + ) = i L (0 − ) = 4 = K 1 + K 2 i L (∞ ) = 1 = K 1 Thus, we have K 2 = 3 . The current is iL (t ) = 4 t < 0 (switch closed) = 1 + 3 exp(− 10t ) t ≥ 0 (switch open) 145 P4.40 In steady state with the switch closed, the current is i (t ) = 2 A for t < 0. The resistance of a voltmeter is very high -- ideally infinite. Thus, there is no path for the current in the inductance when the switch opens, and di dt is very large in magnitude at t = 0 . Consequently, the voltage induced in the inductance is very large in magnitude and an arc occurs across the switch. With an ideal meter and switch, the voltage would be infinite. The voltmeter could be damaged in this circuit. P4.41 (a) iL (t ) = I i exp(− Rt L ) for t ≥ 0 (b) pR (t ) = RiL2 (t ) = R (I i )2 exp(− 2Rt L ) for t ≥ 0 ∞ (c) W = ∫ pR (t )dt 0 ∞ = ∫ R (I i )2 exp (− 2Rt L )dt 0 t =∞  L  = R (I i )2  − exp (− 2Rt L )  2R t = 0 1 = L (I i )2 2 which is precisely the expression for the energy stored in the inductance at t = 0 . P4.42 Prior to t = 0, the current source is shorted, so we have i L (t ) = 0 for t < 0 After the switch opens at t = 0, the current i L (t ) increases from zero headed for 2 A, which it would eventually reach if the switch stayed open. The inductance sees a Thévenin resistance of 18 Ω, and the time constant is τ = L / 18 = 0.3333 s, so we have iL (t ) = 2 − 2 exp( −3t ) for 0 < t < 1 At t = 1, the current reaches 1.900 A. Then, the switch closes, the source is shorted, and the current decays toward zero. Because the inductance sees a resistance of 6 Ω, the time constant is τ = L / 6 = 1 s. iL (t ) = 1.900 exp(−t ) for 1 < t 146 P4.43 With the circuit in steady state before the switch opens, the inductor acts as a short circuit, the current through the inductor is iL (t ) = 4 A, and v R (t ) = 0. Immediately after the switch opens, the inductor current remains 4 A because infinite voltage is not possible in this circuit. Then the return path for the inductor current is through the 1 kΩ resistor so v R (0+) = −4000 V. After the switch opens, the current and voltage decay exponentially with a time constant τ = L / R = 1 ms. Thus, we have v R (t ) = 0 t < 0 = −4000 exp( −1000t ) t > 0 P4.44 1. Write the circuit equation and, if it includes an integral, reduce the equation to a differential equation by differentiating. 2. Form the particular solution. Often this can be accomplished by adding terms like those found in the forcing function and its derivatives, including an unknown coefficient in each term. Next, solve for the unknown coefficients by substituting the trial solution into the differential equation and requiring the two sides of the equation to be identical. 3. Form the complete solution by adding the complementary solution x c (t ) = K exp(−t / τ ) to the particular solution. 4. Use initial conditions to determine the value of K. 147 P4.45* Applying KVL, we obtain the differential equation: L diL (t ) + RiL (t ) = 5 exp(− t ) for t > 0 dt We try a particular solution of the form: iLp (t ) = A exp(− t ) (1) (2) in which A is a constant to be determined. Substituting Equation (2) into Equation (1), we have − LA exp(− t ) + RA exp(− t ) = 5 exp(− t ) which yields 5 = −1 R −L The complementary solution is of the form iLc (t ) = K 1 exp(− Rt L ) A= The complete solution is iL (t ) = iLp (t ) + iLc (t ) = − exp(− t ) + K 1 exp(− Rt L ) Before the switch closes, the current must be zero. Furthermore, the current cannot change instantaneously, so we have iL (0 + ) = 0 . Therefore, we have iL (0 + ) = 0 = −1 + K 1 which yields K 1 = 1 . Finally, the current is given by iL (t ) = − exp(− t ) + exp(− Rt L ) for t ≥ 0 . P4.46* Write a current equation at the top node: v (t ) dv C (t ) 2 exp( −3t ) = C +C R dt Substitute the particular solution suggested in the hint: 2 exp( −3t ) ≡ A exp( −3t ) − 3AC exp( −3t ) R Solving for A and substituting values of the circuit parameters, we find A = −10 6. The time constant is τ = RC = 1 s, and the general form of the solution is v C (t ) = K 1 exp( −t ) + v p (t ) = K 1 exp( −t ) − 10 6 exp( −3t ) However because of the closed switch, we have v C (0+) = 0. Substituting this into the general solution we find K 1 = 10 6 . Thus v C (t ) = 10 6 exp( −t ) − 10 6 exp( −3t ) t > 0 P4.47* Write a current equation at the top node: v (t ) 1 t 5 cos(10t ) = + ∫ v (t )dt + i L (0) R L 0 148 Differentiate each term with respect to time to obtain a differential equation: 1 dv (t ) v (t ) − 50 sin(10t ) = + R dt L Substitute the particular solution suggested in the hint: 1 1 − 50 sin(10t ) ≡ [−10A sin(10t ) + 10B cos(10t )] + [A cos(10t ) + B sin(10t )] R L Equating coefficients of sine and cosine terms, we have − 10A B − 50 = + L A R L Solving for A and B and substituting values of the circuit parameters, we find A = 25 and B = −25. The time constant is τ = L / R = 0.1 s, and the 0= 10B R + general form of the solution is v (t ) = K 1 exp( −t / τ) + v p (t ) = K 1 exp( −t / τ) + 25 cos(10t ) − 25 sin(10t ) However, because the current in the inductor is zero at t = 0+, the 5 A supplied by the source must flow through the 10-Ω resistor and we have v (0+) = 50. Substituting this into the general solution we find K 1 = 25 . Thus P4.48. v (t ) = 25 exp( −t / τ ) + 25 cos(10t ) − 25 sin(10t ) t ≥ 0 Applying KVL, we obtain the differential equation: diL (t ) + 2iL (t ) = 5 exp(− 2t ) cos(3t ) for t > 0 dt (1) Because the derivative of the forcing function is − 10 exp(− 2t ) cos(3t ) − 15 exp( −2t ) sin(3t ) we try a particular solution of the form: iLp (t ) = A exp(− 2t ) sin(3t ) + B exp( −2t ) cos(3t ) (2) in which A and B are constants to be determined. Substituting Equation (2) into Equation (1), we have − 2A exp(− 2t ) sin(3t ) − 2B exp( −2t ) cos(3t ) + 3A exp( −2t ) cos(3t ) − 3B exp( −2t ) sin(3t ) + 2A exp( −2t ) sin(3t ) + 2B exp( −2t ) cos(3t ) ≡ 5 exp(− 2t ) cos(3t ) which yields and − 2B + 3A + 2B = 5 − 2A − 3B + 2A = 0 Solving we find A = 5 / 3 and B = 0. The complementary solution is of the form 149 iLc (t ) = K 1 exp(− Rt L ) = K 1 exp( −2t ) The complete solution is iL (t ) = iLp (t ) + iLc (t ) = (5 / 3) exp( −2t ) sin(3t ) + K 1 exp( −2t ) Before the switch closes, the current must be zero. Furthermore, the current cannot change instantaneously, so we have iL (0 + ) = 0 . Therefore, we have iL (0 + ) = 0 = +K 1 which yields K 1 = 0 . Thus, the current is given by iL (t ) = (5 / 3) exp(−2t ) sin(3t ) for t ≥ 0 P4.49 The differential equation is obtained by applying KVL for the node at the top end of the capacitance: v C (t ) − v (t ) dv C (t ) +C =0 R dt Rearranging this equation and substituting v (t ) = t , we have RC dv C (t ) + v C (t ) = t for t > 0 dt We try a particular solution of the form v Cp (t ) = A + Bt (1) (2) in which A and B are constants to be determined. Substituting Equation (2) into Equation (1), we have RCB + A + Bt = t Solving, we have B =1 A = −RC Thus, the particular solution is v Cp (t ) = −RC + t The complementary solution (to the homogeneous equation) is of the form v Cc (t ) = K 1 exp(− t RC ) Thus, the complete solution is v C (t ) = v Cp (t ) + v Cc (t ) = −RC + t + K 1 exp(− t RC ) However, the solution must meet the given initial condition: v C (0 ) = 0 = −RC + K 1 Thus, K 1 = RC and we have v C (t ) = v Cp (t ) + v Cc (t ) = −RC + t + RC exp(− t RC ) 150 P4.50 Using KVL, we obtain the differential equation dis (t ) + Ris (t ) = v (t ) dt di (t ) 2 s + 300is (t ) = 15 cos(300t ) dt L We try a particular solution of the form isp (t ) = A cos(300t ) + B sin(300t ) in which A and B are constants to be determined. Substituting the proposed solution into the differential equation yields: − 600A sin(300t ) + 600B cos(300t ) + 300A cos(300t ) + 300B sin(300t ) = 15 cos(300t ) Equating coefficients of cosines yields 600B + 300A = 15 Equating coefficients of sines yields − 600A + 300B = 0 From these equations, we find that B = 1 50 and A = 1 100 . The complementary solution is isc (t ) = K 1 exp(− Rt L ) = K 1 exp(− 150t ) and the complete solution is is (t ) = isc (t ) + isp (t ) is (t ) = K 1 exp(− 150t ) + (1 100 ) cos(300t ) + (1 50 ) sin(300t ) Finally, we use the given initial condition is (0 ) = 0 = K 1 + 1 100 to determine that K 1 = − 1 100 . Thus, the solution for the current in amperes is is (t ) = −(1 100 ) exp(− 150t ) + (1 100 ) cos(300t ) + (1 50 ) sin(300t ) 151 P4.51 Applying KVL to the circuit, we have diL (t ) + RiL (t ) = v (t ) dt di (t ) 2 L + 5iL (t ) = 4t dt L We try a particular solution of the form iLp (t ) = A + Bt in which A and B are constants to be determined. Substituting the proposed solution into the differential equation yields 2B + 5A + 5Bt = 4t . From this, we find that B = 0.8 and A = −0.32 . The complementary solution is iLc (t ) = K 1 exp(− Rt L ) = K 1 exp(− 2.5t ) and the complete solution is iL (t ) = iLc (t ) + iLp (t ) = K 1 exp(− 2.5t ) − 0.32 + 0.8t Finally, we use the given initial condition: iL (0 ) = 0 = K 1 − 0.32 to determine that K 1 = 0.32 . Thus, the solution is iL (t ) = 0.32 exp(− 2.5t ) − 0.32 + 0.8t P4.52 Usually, the particular solution includes terms with the same functional forms as the terms found in the forcing function and its derivatives. In this case, there are four different types of terms in the forcing function and its derivatives, namely t sin(t ), t cos(t ), sin(t ), and cos(t ). Thus, we we are led to try a particular solution of the form v p (t ) = At sin(t ) + Bt cos(t ) + C sin(t ) + D cos(t ) Substituting into the differential equation, we have 2[A sin(t ) + At cos(t ) + B cos(t ) − Bt sin(t ) + C cos(t ) − D sin(t )] + At sin(t ) + Bt cos(t ) + C sin(t ) + D cos(t ) ≡ 5t sin(t ) We require the two sides of the equation to be identical. Equating coefficients of like terms, we have 2A − 2D + C = 0 2B + 2C + D = 0 2A + B = 0 A − 2B = 5 Solving these equations, we obtain A = 1 , B = −2 , C = 6 / 5, and D = 8 / 5. Thus, the particular solution is v p (t ) = t sin(t ) − 2t cos(t ) + 6 8 sin(t ) + cos(t ) 5 5 152 P4.53 The particular solution includes terms with the same functional forms as the terms found in the forcing function and its derivatives. In this case, there are three different types of terms in the forcing function and its derivatives, namely t 2 exp( −t ), t exp( −t ), and exp( −t ). Thus, we we are led to try a particular solution of the form v p (t ) = At 2 exp( −t ) + Bt exp( −t ) + C exp( −t ) Substituting into the differential equation, we have 2At exp( −t ) − At 2 exp(−t ) + B exp(−t ) − Bt exp( −t ) − C exp(−t ) + 3At 2 exp( −t ) + 3Bt exp( −t ) + 3C exp(−t ) ≡ t 2 exp(−t ) We require the two sides of the equation to be identical. Equating coefficients of like terms, we have 2A = 1 2A + 2B = 0 B + 2C = 0 Solving these equations, we obtain A = 1 / 2 , B = −1 / 2 , and C = 1 / 4. Thus, the particular solution is v p (t ) = (1 / 2)t 2 exp( −t ) − (1 / 2)t exp( −t ) + (1 / 4) exp( −t ) P4.54 di (t ) + 6i = 12 exp( −3t ) dt (b) τ = L / R = 1 / 3 s (a) 2 ic (t ) = A exp( −3t ) (c) A particular solution of the form i p (t ) = K exp(−3t ) does not work because the left-hand side of the differential equation is identically zero for this choice. (d) Substituting i p (t ) = Kt exp(−3t ) into the differential equation produces − 6Kt exp(−3t ) + 2K exp(−3t ) + 6Kt exp(−3t ) ≡ 12 exp( −3t ) from which we have K = 6. (e) Adding the particular solution and the complementary solution, we have i (t ) = A exp( −3t ) + 6t exp( −3t ) However, the current must be zero in the inductor prior to t = 0 because of the open switch, and the current cannot change instantaneously in this circuit, so we have i (0 +) = 0. This yields A = 0. Thus, the solution is i (t ) = 6t exp(−3t ) 153 P4.55 dv (t ) v (t ) + = 5 × 10 − 6 exp( −10t ) 3 dt 50 × 10 v c (t ) = A exp( −10t ) (b) τ = RC = 0.1 s (a) 2 × 10 − 6 (c) A particular solution of the form v p (t ) = K exp(−10t ) does not work because the left-hand side of the differential equation is identically zero for this choice. (d) Substituting v p (t ) = Kt exp( −10t ) into the differential equation produces 2 × 10 − 6 [ −10Kt exp( −10t ) + K exp(−10t )] + Kt exp(−10t ) = 5 × 10 − 6 exp(−10t ) 50 × 103 from which we have K = 2.5. (e) Adding the particular solution and the complementary solution we have v (t ) = A exp( −10t ) + 2.5t exp(−10t ) However, the voltage across the capacitance must be zero prior to t = 0 because of the closed switch, and the voltage cannot change instantaneously in this circuit, so we have v (0 +) = 0. This yields A = 0. Thus, the solution is v (t ) = 2.5t exp( −10t ) V P4.56 We look at a circuit diagram and combine all of the inductors that are in series or parallel. Then, we combine all of the capacitances that are in series or parallel. Next, we count the number of energy storage elements (inductances and capacitances) in the reduced circuit. If there is only one energy-storage element, we have a first-order circuit. If there are two, we have a second-order circuit. P4.57 First, we write the differential equation for the system and put it in the form d 2 x (t ) dx (t ) + 2α + ω02 (t ) = f (t ) 2 dt dt Then compute the damping ratio ζ = α / ω0 . If we have ζ < 1, the system is underdamped, and the complementary solution is of the form x c (t ) = K 1 exp(− αt ) cos(ωnt ) + K 2 exp(− αt ) sin(ωnt ) in which ωn = ω02 − α 2 . 154 If we have ζ = 1, the system is critically damped, and the complementary solution is of the form x c (t ) = K 1 exp(s 1t ) + K 2t exp(s 1t ) in which s1 is the root of the characteristic equation s 2 + 2αs + ω02 = 0 . If we have ζ > 1, the system is overdamped, and the complementary solution is of the form x c (t ) = K 1 exp(s 1t ) + K 2 exp(s 2t ) in which s1 and s2 are the roots of the characteristic equation s 2 + 2αs + ω02 = 0 . P4.58 The unit step function is defined by u (t ) = 0 for t < 0 = 1 for t ≥ 0 The unit step function is illustrated in Figure 4.27 in the book. P4.59 One way to determine the particular solution is to assume that it is a constant [ x p (t ) = K ], substitute into the differential equation, and solve for K. A second method is to replace the inductors by short circuits the capacitances by open circuits and solve for the steady-state dc response. P4.60 The sketch should resemble the response shown in Figure 4.29 for ζ = 0.1. Second-order circuits that are severely underdamped ( ζ ω0 , this is the overdamped case. The roots of the characteristic equation are found from Equations 4.72 and 4.73 in the text. s1 = −α + α 2 − ω02 = −0.2679 × 10 4 s 2 = − α − α 2 − ω02 = −3.732 × 10 4 The complementary solution is v Cc (t ) = K 1 exp(s1t ) + K 2 exp(s 2t ) and the complete solution is v C (t ) = 50 + K 1 exp(s 1t ) + K 2 exp(s 2t ) The initial conditions are v C (0 ) = 0 and i (0 ) = 0 = C Thus, we have v C (0 ) = 0 = 50 + K 1 + K 2 dv C (t ) dt t =0 dv C (t ) dt t =0 = 0 = s1K 1 + s 2K 2 Solving, we find K 1 = −53.87 and K 2 = 3.867 . Finally, the solution is v C (t ) = 50 − 53.87 exp(s1t ) + 3.867 exp(s 2t ) 156 P4.62* As in the solution to P4.61, we have v Cp (t ) = 50 α= R = 10 4 2L 1 ω0 = LC = 10 4 Since we have α = ω0 ,this is the critically damped case. The roots of the characteristic equation are equal: s1 = −α + α 2 − ω02 = −10 4 s 2 = −α − α 2 − ω02 = −10 4 The complementary solution is given in Equation 4.75 in the text: v Cc (t ) = K 1 exp(s 1t ) + K 2t exp(s 1t ) and the complete solution is v C (t ) = 50 + K 1 exp(s 1t ) + K 2t exp(s 1t ) The initial conditions are v C (0 ) = 0 and i(0 ) = 0 = C Thus, we have dv C (t ) dt t =0 v C (0 ) = 0 = 50 + K 1 dv C (t ) = 0 = s1K 1 + K 2 dt t =0 Solving, we find K 1 = −50 and K 2 = −50 × 10 4 . Finally, the solution is v C (t ) = 50 − 50 exp(s1t ) − (50 × 10 4 )t exp(s1t ) P4.63* As in the solution to P4.61, we have v Cp (t ) = 50 α= ω0 = R = 0.5 × 10 4 2L 1 LC = 10 4 Since we have α < ω0 , this is the under-damped case. The natural frequency is given by Equation 4.76 in the text: ωn = ω02 − α 2 = 8.660 × 10 3 The complementary solution is given in Equation 4.77 in the text: v Cc (t ) = K 1 exp(− αt ) cos(ωnt ) + K 2 exp(− αt ) sin(ωnt ) and the complete solution is 157 v C (t ) = 50 + K 1 exp(− αt ) cos(ωnt ) + K 2 exp(− αt ) sin(ωnt ) The initial conditions are v C (0 ) = 0 and Thus, we have i (0 ) = 0 = C dv C (t ) dt t =0 v C (0 ) = 0 = 50 + K 1 dv C (t ) = 0 = − αK 1 + ωn K 2 dt t =0 Solving, we find K 1 = −50 and K 2 = −28.86 . Finally, the solution is v C (t ) = 50 − 50 exp(− αt ) cos(ωnt ) − (28.86) exp(− αt ) sin(ωnt ) P4.64 (a) Using Equation 4.103 from the text, the damping coefficient is 1 α= = 20 × 10 6 2RC Equation 4.104 gives the undamped resonant frequency 1 ω0 = = 10 × 10 6 LC Equation 4.71 gives the damping ratio ζ = α ω0 = 2 Thus, we have an overdamped circuit. (b) Because the switch has been open for a long time the inductor acts as a short circuit and we have iL (0−) = 1 A. Because the current in the inductance cannot change instantaneously in this circuit, we have iL (0+) = 1 A. Writing a current equation at t = 0 + , we have v (0 + ) + iL (0 + ) + Cv ′(0 + ) = 1 R Substituting R = 25 Ω, v (0 + ) = 25 V and iL (0 + ) = 1 A, yields v ′(0 + ) = −10 9 V/s (c) Under steady-state conditions, the inductance acts as a short circuit. Therefore, the particular solution for v (t ) is: v p (t ) = 0 (d) The roots of the characteristic equation are found from Equations 4.72 and 4.73 in the text. 158 s1 = −α + α 2 − ω02 = −2.679 × 10 6 s 2 = −α − α 2 − ω02 = −37.32 × 10 6 The complementary solution is v c (t ) = K 1 exp(s1t ) + K 2 exp(s 2t ) and (since the particular solution is zero) the complete solution is v (t ) = K 1 exp(s 1t ) + K 2 (s 2t ) The initial conditions are v (0 + ) = 0 and v ′(0 + ) = −10 9 Thus, we have v (0 + ) = 25 = K 1 + K 2 v ′(0 + ) = −10 9 = s1K 1 + s2K 2 Solving, we find K 1 = −1.934 and K 2 = 26.93 . Finally, the solution is v (t ) = −1.934 exp(− 2.679 × 10 6t ) + 26.93 exp(− 37.32 × 10 6t ) V P4.65 (a) Using Equation 4.103 from the text, the damping coefficient is 1 α= = 10 × 10 6 2RC Equation 4.104 gives the undamped resonant frequency: 1 ω0 = = 10 × 10 6 LC Equation 4.71 gives the damping ratio: ζ = α / ω0 = 1 Thus, we have a critically damped circuit. (b) Because the switch has been open for a long time the inductor acts as a short circuit and we have iL (0−) = 1 A. Because the current in the inductance cannot change instantaneously in this circuit, we have iL (0+) = 1 A. Writing a current equation at t = 0 + , we have v (0 + ) + iL (0 + ) + Cv ′(0 + ) = 1 R Substituting R = 50 Ω, v (0 + ) = 25 V and iL (0 + ) = 1 A, yields v ′(0 + ) = −0.5 × 10 9 V/s (c) Under steady-state conditions, the inductance acts as a short circuit. Therefore, the particular solution for v (t ) is: 159 v p (t ) = 0 (d) The roots of the characteristic equation are found from Equations 4.72 and 4.73 in the text. s1 = −α + α 2 − ω02 = −10 × 10 6 s 2 = −α − α 2 − ω02 = −10 × 10 6 The complementary solution is given in Equation 4.75 in the text: v C (t ) = K 1 exp(s1t ) + K 2t exp(s 1t ) and the complete solution is v (t ) = v C (t ) + v p (t ) = K 1 exp(s1t ) + K 2t exp(s1t ) The initial conditions are v (0 + ) = 25 and v (0 + ) = −0.5 × 109 Thus, we have v (0 + ) = 25 = K 1 v ′(0 + ) = −0.5 × 10 9 = s1K 1 + K 2 Solving, we find K 1 = 25 and K 2 = −250 × 10 6 . Finally, the solution is v C (t ) = 25 exp(− 10 7t ) − 250 × 10 6t exp(− 10 7t ) P4.66 (a) Using Equation 4.103 from the text, the damping coefficient is 1 α= = 10 6 2RC Equation 4.104 gives the undamped resonant frequency: 1 ω0 = = 10 × 106 LC Equation 4.71 gives the damping ratio: ζ = α ω0 = 0.1 Thus, we have an underdamped circuit. (b) Because the switch has been open for a long time, the inductor acts as a short circuit and we have iL (0−) = 1 A. Because the current in the inductance cannot change instantaneously in this circuit, we have iL (0+) = 1 A. Writing a current equation at t = 0 + , we have v (0 + ) + iL (0 + ) + Cv ′(0 + ) = 1 R Substituting R = 500 Ω, v (0 + ) = 25 V and iL (0 + ) = 1 A, yields 160 v ′(0 + ) = −50 × 10 6 V/s (c) Under steady-state conditions, the inductance acts as a short circuit. Therefore, the particular solution for v (t ) is: v p (t ) = 0 (d) The natural frequency is given by Equation 4.76 in the text: ωn = ω02 − α 2 = 9.950 × 10 6 The complementary solution is given in Equation 4.77 in the text: v C (t ) = K 1 exp(− αt ) cos(ωnt ) + K 2 exp(− αt ) sin(ωnt ) and the complete solution is v (t ) = K 1 exp(− αt ) cos(ωnt ) + K 2 exp(− αt ) sin(ωnt ) The initial conditions are v (0 + ) = 25 and v ′(0 + ) = −50 × 10 6 Thus, we have v (0 + ) = 25 = K 1 v ′(0 + ) = 50 × 10 6 = −αK 1 + ωn K 2 Solving, we find K 1 = 25 and K 2 = −2.513 . Finally, the solution is v (t ) = 25 exp(− αt ) cos(ωnt ) − 2.513 exp(− αt ) sin(ωnt ) P4.67 Write a KVL equation for the circuit: di (t ) 1 20 sin(100t ) = L + Ri (t ) + dt C t ∫ i (t )dt + vC (0) 0 Differentiate each term with respect to time to obtain a differential equation: d 2i (t ) di (t ) i (t ) +R + 2000 cos(100t ) = L 2 dt dt C Substitute the particular solution suggested in the hint to obtain: 2000 cos(10t ) ≡ L[ −10 4 A cos(100t ) − 10 4 B sin(100t )] + R [ −100A sin(100t ) + 100B cos(100t )] + 1 C [A cos(100t ) + B sin(100t )] Equating coefficients of sine and cosine terms, we have 161 0 = −10 4 BL − 100AR + B C 2000 = −10 4 AL + 100BR + A C Solving for A and B and substituting values of the circuit parameters, we find A = 0 and B = 0.4 Thus, the particular solution is i p (t ) = 0.4 sin(100t ) Using Equations 4.60 and 4.61 from the text, we have α= ω0 = R = 25 2L 1 LC = 100 Because we have α < ω0 , this is the under-damped case. The natural frequency is given by Equation 4.76 in the text: ωn = ω20 − α 2 = 96.82 The complementary solution is given in Equation 4.77 in the text: ic (t ) = K 1 exp(− αt ) cos(ωnt ) + K 2 exp(− αt ) sin(ωnt ) and the complete solution is i (t ) = 0.4 sin(100t ) + K 1 exp(− αt ) cos(ωnt ) + K 2 exp(− αt ) sin(ωnt ) However, because the current is zero at t = 0+, the voltage across the inductor must be v L (0+) = −v C (0+) = −20 V which implies that di (0 +) / dt = −20. Thus, we can write i (0+) = 0 = +K 1 di (0+) = −20 = 40 − αK 1 + ωn K 2 dt Solving we find K 2 = −0.6197. Finally, the solution is: i (t ) = 0.4 sin(100t ) − 0.6197 exp(− 25t ) sin(96.82t ) P4.68 As in the solution to P4.67, we have 0 = −10 4 BL − 100AR + B C 2000 = −10 4 AL + 100BR + A C Solving for A and B and substituting values of the circuit parameters, we find A = 0 and B = 0.1 . Thus, the particular solution is i p (t ) = 0.1 sin(100t ) Using Equations 4.60 and 4.61 from the text, we have 162 α= ω0 = R = 100 2L 1 LC = 100 Since we have α = ω0 , this is the critically damped case. The roots of the characteristic equation are equal: s1 = −α + α 2 − ω20 = −100 s2 = −α − α 2 − ω20 = −100 The complementary solution is given in Equation 4.75 in the text: ic (t ) = K 1 exp(s1t ) + K 2t exp(s1t ) and the complete solution is i (t ) = 0.1 sin(100t ) + K 1 exp(s1t ) + K 2t exp(s1t ) As in the solution to P4.67, The initial conditions are i ( 0 +) = 0 = K 1 di (0+) = −20 = 10 + s1K 1 + K 2 dt Solving we find K 2 = −30 . Finally, the solution is i (t ) = 0.1 sin(100t ) − 30t exp(− 100t ) P4.69 As in the solution to P4.67, we have 0 = −10 4 BL − 100AR + B C 2000 = −10 4 AL + 100BR + A C Solving for A and B and substituting values of the circuit parameters, we find A = 0.025 and B = 0. Thus, the particular solution is i p (t ) = 0.05 sin(100t ) Using Equations 4.60 and 4.61 from the text, we have α= ω0 = R = 200 2L 1 LC = 100 Since we have α > ω0 , this is the overdamped case. The roots of the characteristic equation are found from Equations 4.72 and 4.73 in the text. 163 s1 = −α + α 2 − ω20 = −26.79 s 2 = −α − α 2 − ω20 = −373.2 The complementary solution is ic (t ) = K1 exp(s1t ) + K 2 exp(s2t ) and the complete solution is i (t ) = 0.05 sin(100t ) + K 1 exp(s1t ) + K 2 exp(s2t ) As in the solution to P4.67, The initial conditions are i ( 0 +) = 0 = K 1 + K 2 di (0+) = −20 = 5 + s1K 1 + s2K 2 dt Solving, we find K 1 = −0.07127 and K 2 = 0.07127 Finally, the solution is i (t ) = 0.05 sin(100t ) − 0.07127 exp(s1t ) − 0.07127 exp(s2t ) P4.70 (a) Applying KCL, we have: t 1 dv 4 ∫v (t )dt + C = 2 sin(10 t ) L dt 0 Taking the derivative, multiplying by L, and rearranging, we have d 2v + v (t ) = 2L10 4 cos(10 4t ) 2 dt (b) This is a parallel RLC circuit having R = ∞. Using Equation 4.103 from LC the text, the damping coefficient is 1 α= =0 2RC Equation 4.104 gives the undamped resonant frequency: 1 ω0 = = 10 4 LC The natural frequency is given by Equation 4.76 in the text: ωn = ω 02 − α 2 = 10 4 The complementary solution given in Equation 4.77 becomes: v C (t ) = K 1 exp(− αt ) cos(ωnt ) + K 2 exp(− αt ) sin(ωnt ) ( ) ( = K 1 cos 10 4t + K 2 sin 10 4t ) (c)The usual form for the particular solution doesn't work because it has the same form as the complementary solution. In other words, if we substitute a trial particular solution of the form 164 v p (t ) = A cos(10 4t ) + B sin(10 4t ) , the left-hand side of the differential equation becomes zero and cannot match the forcing function. (d) When we substitute v p (t ) = At cos(10 4t ) + Bt sin(10 4t ) into the differential equation, we eventually obtain − 2A10 −4 sin(10 4t ) + 2B 10 −4 cos(10 4t ) ≡ 200 cos(10 4t ) from which we obtain B = 10 6 and A = 0. Thus, the particular solution is v p (t ) = 10 6t sin(10 4t ) (e) The complete solution is v (t ) = v p (t ) + v C (t ) = 10 6t sin(10 4t ) + K 1 cos(10 4t ) + K 2 sin(10 4t ) However, we have v (0 ) = 0, which yields K 1 = 0 . Also, by KCL, the current through the capacitance is zero at t = 0+, so we have dv (0+) = 0 = 10 4 K 2 dt which yields K 2 = 0, so the complete solution is v (t ) = v p (t ) + v C (t ) = 10 6t sin(10 4t ) 4.71 Prior to t = 0 , we have v C (t ) = 0 because the switch is closed. After t = 0 , we can write the following KCL equation at the top node of the circuit: v C (t ) dv C (t ) +C = 10 mA R dt Multiplying both sides by R and substituting values, we have 0.02 dv C (t ) + v C (t ) = 20 dt The voltage across the capacitance cannot change instantaneously, so we have v C (0 + ) = v C (0 − ) = 0 syms vC t vC = dsolve('DvC*(0.02) + vC = 20', 'vC(0) = 0') ezplot(vC,[0 100e-3]) 165 The results are: v C (t ) = 20 − 20 exp(− 50t ) for t > 0 and this plot: P4.72 The differential equation is obtained by applying KVL for the node at the top end of the capacitance: v C (t ) − v (t ) dv C (t ) +C =0 R dt Rearranging this equation and substituting v (t ) = t , we have RC dv C (t ) + v C (t ) = t for t > 0 dt Because the source is zero prior to zero time, v C (0) = 0. The MATLAB commands are syms v vC t R C v = t*heaviside(t) vC = dsolve('R*C*DvC + vC = t*heaviside(t)', 'vC(0) = 0') vC = subs(vC, [R C], [1e6 1e-6]) ezplot(vC, [-2 5]) hold on ezplot(v, [-2 5]) 166 The resulting expression is vC = -heaviside(t)+t*heaviside(t)+exp(-t)*heaviside(t) In more standard notation, we have: v C (t ) = [t − 1 + exp(t )]u (t ) P4.73 Using KVL, we obtain the differential equation dis (t ) + Ris (t ) = v (t ) dt di (t ) 2 s + 300is (t ) = 15 cos(300t ) dt L The MATLAB commands are: syms Is t % It is best to avoid variables beginning with lower case "i" Is = dsolve('2*DIs + 300*Is =15*cos(300*t)', 'Is(0) = 0') ezplot(Is, [0 80e-3]) The result is 167 P4.74 (a) Writing a KCL equation at the top node after t = 0, we have dv (t ) v (t ) 1 t C + + ∫ v (t )dt + iL (0 +) = 1 dt R L 0 Taking the derivative to eliminate the integral, and substituting component values, we have 2 dv (t ) − 9 d v (t ) 10 + 0.04 + 10 5v (t ) = 0 2 dt dt (b) Because the switch has been open for a long time, the inductor acts as a short circuit, and we have iL (0−) = 1 A. Because the current in the inductance cannot change instantaneously in this circuit, we have iL (0+) = 1 A. Writing a current equation at t = 0 + , we have v (0 + ) + iL (0 + ) + Cv ′(0 + ) = 1 R Substituting R = 25 Ω, v (0 + ) = 50 V and iL (0 + ) = 1 A, yields v ′(0 + ) = −2 × 10 9 V/s (c) The MATLAB commands are: syms v t S=dsolve('(1e-9)*D2v + 0.04*Dv + (1e5)*v = 0', 'v(0) = 50, Dv(0)= -2e9'); simple(vpa(S,6)) The result is: ans = -3.8676*exp(-2679500.*t)+53.8676*exp(-37320500.*t) 168 P4.75 (a) Writing a KCL equation at the top node after t = 0, we have dv (t ) 1 t C + ∫ v (t )dt + iL (0 +) = 2 sin(10 4t ) dt L 0 Taking the derivative to eliminate the integral, and substituting component values, we have d 2v (t ) 10 − 6 + 100v (t ) = 2 × 10 4 cos(10 4t ) 2 dt (b) Because the switch is closed prior to t = 0, we know that v C (0−) = 0. The voltage across the capacitor cannot change instantaneously, so we have v C (0+) = 0. At t = 0, the KCL equation becomes dv (0+) =0 dt so we have v ′(0 +) = 0. C (c) The MATLAB commands are: syms v t v=dsolve('(1e-6)*D2v + 100*v = (2e4)*cos((1e4)*t)', 'v(0) = 0, Dv(0)= 0'); simple(vpa(v,6)) The result is v (t ) = 10 6t sin(10 4t ). P4.76 The KVL equations around meshes 1 and 2 are d [i1 (t ) − i2 (t )] + 2i1 (t ) = 12 dt d [i2 (t ) − i1 (t )] di (t ) + 4i2 (t ) + 2 2 =0 dt dt Because the inductor currents are zero before t = 0, and the circuit does not force the currents to change, we have i1 (0+) = 0 and i2 (0+) = 0. The MATLAB commands are: syms I1 I2 t S = dsolve('DI1 - DI2 +2*I1 = 12','3*DI2 - DI1 + 4*I2 = 0',... 'I1(0) = 0','I2(0) = 0'); I1 = S.I1 I2 = S.I2 The results are: I1 = -4*exp(-4*t) - 2*exp(-t) + 6 and I2 = -2*exp(-4*t) + 2*exp(-t) 169 P4.77 Because the circuit has been operating for a long time before t = 0 with the switch open and because the switching does not force the capacitor voltages or the inductor current to change instantaneously, the capacitor voltages and inductor current are zero at t = 0 + . The KCL equations are t d [v 1 (t ) − v 2 (t )] v 1 (t ) − 10 1 0.2 × 10 − 6 + + v (t )dt = 0 (1) ∫ 1000 0.002 0 1 dt d [v 2 (t ) − v 1 (t )] v 2 (t ) dv (t ) 0.2 × 10 − 6 + + 0.2 × 10 − 6 2 =0 (2) 2000 dt dt We differentiate the terms in the first equation to eliminate the integral. d 2 [v 1 (t ) − v 2 (t )] dv (t ) 0.2 × 10 − 6 + 10 − 3 1 + 500v 1 (t ) = 0 (3) 2 dt dt We need to solve equations 2 and 3. Because the capacitor voltages are zero at t = 0+, we know that v 1 (0+) = 0 and v 2 (0+) = 0. At t = 0+, Equations 1 and 2 become: d [v 1 (0+) − v 2 (0+)] = 0.01 dt d [v 2 (0+) − v 1 (0+)] dv (0+) 0.2 × 10 − 6 + 0.2 × 10 − 6 2 =0 dt dt 0.2 × 10 − 6 Solving we find that dv 1 (0+) dv 2 (0+) = 10 5 V/s and = 5 × 10 4 V/s. dt dt The MATLAB commands are: syms v1 v2 t S = dsolve('(2e-6)*(2*Dv2 - Dv1) + (5e-4)*v2 = 0',... %Equation 2 '(2e-6)*(D2v1 - D2v2) +(1e-3)*Dv1 +500*v1 = 0',... %Equation 3 'v1(0) = 0, v2(0) = 0, Dv1(0) = 1e5, Dv2(0) = 5e4'); v1 = vpa(S.v1,4); v2 = vpa(S.v2,4); pretty(v1) pretty(v2) The results are v1(t) = 0.02038 exp(-563.0 t) cos(22360. t) + 4.475 exp(-563.0 t) sin(22360. t) - 0.00002264 exp(-123.0 t) V v2(t) = 2.236 exp(-563.0 t) sin(22360. t) + 0.02222 exp(-563.0 t) cos(22360. t) - 0.02222 exp(-123.0 t) V 170 Practice Test T4.1 (a) Prior to the switch opening, the circuit is operating in DC steady state, so the inductor acts as a short circuit, and the capacitor acts as an open circuit. i1 (0−) = 10 / 1000 = 10 mA i2 (0−) = 10 / 2000 = 5 mA i3 (0−) = 0 v C (0−) = 10 V iL (0−) = i1 (0−) + i2 (0−) + i3 (0−) = 15 mA (b) Because infinite voltage or infinite current are not possible in this circuit, the current in the inductor and the voltage across the capacitor cannot change instantaneously. Thus, we have iL (0+) = iL (0−) = 15 mA and v C (0+) = v C (0−) = 10 V. Also, we have i1 (0+) = iL (0+) = 15 mA, i2 (0 +) = v C (0 +) / 5000 = 2 mA, and i3 (0+) = −i2 (0+) = −2 mA. (c) The current is of the form iL (t ) = A + B exp( −t / τ ). Because the inductor acts as a short circuit in steady state, we have iL (∞) = A = 10 / 1000 = 10 mA At t = 0+, we have iL (0+) = A + B = 15 mA, so we find B = 5 mA. The time constant is τ = L / R = 2 × 10 −3 / 1000 = 2 × 10 −6 s. Thus, we have iL (t ) = 10 + 5 exp( −5 × 10 5t ) mA. (d) This is a case of an initially charged capacitance discharging through a resistance. The time constant is τ = RC = 5000 × 10 −6 = 5 × 10 −3 s. Thus we have v C (t ) = Vi exp( −t / τ ) = 10 exp( −200t ) V. T4.2 di (t ) + i (t ) = 5 exp( −3t ) dt (b) The time constant is τ = L / R = 2 s and the complementary solution is of the form ic (t ) = A exp( −0.5t ) . (a) 2 (c) The particular solution is of the form i p (t ) = K exp( −3t ) . Substituting into the differential equation produces − 6K exp( −3t ) + K exp( −3t ) ≡ 5 exp( −3t ) from which we have K = −1. (d) Adding the particular solution and the complementary solution, we have 171 i (t ) = A exp(−0.5t ) − exp( −3t ) However, the current must be zero in the inductor prior to t = 0 because of the open switch, and the current cannot change instantaneously in this circuit, so we have i (0 +) = 0. This yields A = 1. Thus, the solution is i (t ) = exp( −0.5t ) − exp(−3t ) A T4.3 (a) Applying KVL to the circuit, we obtain L di (t ) + Ri (t ) + v C (t ) = 15 dt (1) For the capacitance, we have i (t ) = C dv C (t ) dt (2) Using Equation (2) to substitute into Equation (1) and rearranging, we have d 2v C (t ) dv (t ) + (R L ) C + (1 LC )v C (t ) = 15 LC 2 dt dt d 2v C (t ) dv (t ) + 2000 C + 25 × 10 6v C (t ) = 375 × 10 6 2 dt dt (3) (b) We try a particular solution of the form v Cp (t ) = A , resulting in A = 15 . Thus, v Cp (t ) = 15 . (An alternative method to find the particular solution is to solve the circuit in dc steady state. Since the capacitance acts as an open circuit, the steady-state voltage across it is 15 V.) (c) We have ω0 = 1 LC = 5000 and α = R = 1000 2L Since we have α < ω0 , this is the underdamped case. The natural frequency is given by: ωn = ω02 − α 2 = 4899 The complementary solution is given by: v Cc (t ) = K 1 exp(− 1000t ) cos(4899t ) + K 2 exp(− 1000t ) sin(4899t ) (d) The complete solution is v C (t ) = 15 + K 1 exp(− αt ) cos(ωnt ) + K 2 exp(− αt ) sin(ωnt ) The initial conditions are 172 v C (0 ) = 0 and Thus, we have i (0 ) = 0 = C dv C (t ) dt t =0 v C (0 ) = 0 = 15 + K 1 dv C (t ) = 0 = −αK 1 + ωn K 2 dt t = 0 Solving, we find K 1 = −15 and K 2 = −3.062 . Finally, the solution is v C (t ) = 15 − 15 exp(− 1000t ) cos(4899t ) − (3.062) exp(− 1000t ) sin(4899t ) V T4.4 One set of commands is syms vC t S = dsolve('D2vC + 2000*DvC + (25e6)*vC = 375e6',... 'vC(0) = 0, DvC(0) = 0'); simple(vpa(S,4)) These commands are stored in the m-file named T_4_4 on the OrCAD disk. 173 CHAPTER 5 Exercises E5.1 (a) We are given v (t ) = 150 cos(200πt − 30 o ) . The angular frequency is the coefficient of t so we have ω = 200π radian/s . Then f = ω / 2π = 100 Hz T = 1 / f = 10 ms Vrms = Vm / 2 = 150 / 2 = 106.1 V Furthermore, v(t) attains a positive peak when the argument of the cosine function is zero. Thus keeping in mind that ωt has units of radians, the positive peak occurs when ωtmax = 30 × π 180 ⇒ tmax = 0.8333 ms 2 / R = 225 W (b) Pavg = Vrms (c) A plot of v(t) is shown in Figure 5.4 in the book. E5.2 We use the trigonometric identity sin(z ) = cos(z − 90 o ). Thus 100 sin(300πt + 60 o ) = 100 cos(300πt − 30 o ) E5.3 ω = 2πf ≅ 377 radian/s T = 1 / f ≅ 16.67 ms Vm = Vrms 2 ≅ 155.6 V The period corresponds to 360 therefore 5 ms corresponds to a phase angle of (5 / 16.67) × 360 o = 108o . Thus the voltage is o v (t ) = 155.6 cos(377t − 108o ) E5.4 (a) V1 = 10∠0 o + 10∠ − 90 o = 10 − j 10 ≅ 14.14∠ − 45 o 10 cos(ωt ) + 10 sin(ωt ) = 14.14 cos(ωt − 45 o ) (b) I1 = 10∠30 o + 5∠ − 60 o ≅ 8.660 + j 5 + 2.5 − j 4.330 ≅ 11.16 + j 0.670 ≅ 11.18∠3.44 o 10 cos(ωt + 30 o ) + 5 sin(ωt + 30 o ) = 11.18 cos(ωt + 3.44 o ) (c) I2 = 20∠0 o + 15∠ − 60 o ≅ 20 + j 0 + 7.5 − j 12.99 ≅ 27.5 − j 12.99 ≅ 30.41∠ − 25.28o 20 sin(ωt + 90 o ) + 15 cos(ωt − 60 o ) = 30.41 cos(ωt − 25.28o ) 174 E5.5 The phasors are V1 = 10∠ − 30 o V2 = 10∠ + 30 o and V3 = 10∠ − 45 o v1 lags v2 by 60 o (or we could say v2 leads v1 by 60 o ) v1 leads v3 by 15 o (or we could say v3 lags v1 by 15 o ) v2 leads v3 by 75 o (or we could say v3 lags v2 by 75 o ) E5.6 (a) Z L = jωL = j 50 = 50∠90 o VL = 100∠0 o IL = VL / Z L = 100 / j 50 = 2∠ − 90 o (b) The phasor diagram is shown in Figure 5.11a in the book. E5.7 (a) Z C = 1 / jωC = − j 50 = 50∠ − 90 o VC = 100∠0 o IC = VC / Z C = 100 /( − j 50) = 2∠90 o (b) The phasor diagram is shown in Figure 5.11b in the book. E5.8 (a) Z R = R = 50 = 50∠0 o VR = 100∠0 o IR = VR / R = 100 /(50) = 2∠0 o (b) The phasor diagram is shown in Figure 5.11c in the book. E5.9 (a) The transformed network is: Vs 10∠ − 90 o = I= = 28.28∠ − 135 o mA Z 250 + j 250 175 i (t ) = 28.28 cos(500t − 135 o ) mA VR = RI = 7.07∠ − 135 o VL = jωLI = 7.07∠ − 45 o (b) The phasor diagram is shown in Figure 5.17b in the book. (c) i(t) lags vs(t) by 45 o. E5.10 The transformed network is: Z = 1 = 55.47∠ − 56.31 o Ω 1 / 100 + 1 /( − j 50) + 1 /( + j 200) V = ZI = 277.4∠ − 56.31 o V IC = V /( − j 50) = 5.547 ∠33.69 o A IL = V /( j 200) = 1.387∠ − 146.31 o A IR = V /(100) = 2.774∠ − 56.31 o A E5.11 The transformed network is: We write KVL equations for each of the meshes: j 100I1 + 100( I1 − I2 ) = 100 − j 200 I2 + j 100I2 + 100( I2 − I1 ) = 0 Simplifying, we have (100 + j 100) I1 − 100I2 = 100 − 100 I1 + (100 − j 100) I2 = 0 Solving we find I1 = 1.414∠ − 45 o A and I2 = 1∠0 o A. Thus we have i1 (t ) = 1.414 cos(1000t − 45 o ) A and i2 (t ) = cos(1000t ). 176 E5.12 (a) For a power factor of 100%, we have cos(θ ) = 1, which implies that the current and voltage are in phase and θ = 0. Thus, Q = P tan(θ ) = 0. Also I rms = P /[Vrms cos(θ )] = 5000 /[500 cos(0)] = 10 A. Thus we have I m = I rms 2 = 14.14 and I = 14.14∠40 o. (b) For a power factor of 20% lagging, we have cos(θ ) = 0.2, which implies that the current lags the voltage by θ = cos−1 (0.2) = 78.46o. Thus, Q = P tan(θ ) = 24.49 kVAR. Also, we have I rms = P /[Vrms cos(θ )] = 50.0 A. Thus we have I m = I rms 2 = 70.71 A and I = 70.71∠ − 38.46o. (c) The current ratings would need to be five times higher for the load of part (b) than for that of part (a). Wiring costs would be lower for the load of part (a). E5.13 The first load is a 10 µF capacitor for which we have Z C = 1 /( jωC ) = 265.3∠ − 90 o Ω θ C = −90 o I Crms = Vrms / Z C = 3.770 A PC = Vrms I Crms cos(θ C ) = 0 QC =Vrms I Crms sin(θ C ) = −3.770 kVAR The second load absorbs an apparent power of Vrms I rms = 10 kVA with a power factor of 80% lagging from which we have θ 2 = cos −1 (0.8) = 36.87 o. Notice that we select a positive angle for θ 2 because the load has a lagging power factor. Thus we have P2 = Vrms I 2rms cos(θ 2 ) = 8.0 kW and Q2 = Vrms I 2rms sin(θ ) = 6 kVAR . Now for the source we have: Ps = PC + P2 = 8 kW Qs = QC + Q2 = 2.23 kVAR Vrms I srms = Ps 2 + Qs2 = 8.305 kVA I srms = Vrms I srms /Vrms = 8.305 A power factor = Ps /(Vrms I srms ) × 100% = 96.33% E5.14 First, we zero the source and combine impedances in series and parallel to determine the Thévenin impedance. 177 Zt = 50 − j 25 + 1 = 50 − j 25 + 50 + j 50 1 / 100 + 1 / j 100 = 100 + j25 = 103.1∠14.04 o Then we analyze the circuit to determine the open-circuit voltage. Vt = Voc = 100 × 100 = 70.71∠ − 45 o 100 + j 100 In =Vt / Zt = 0.6858∠ − 59.04 o E5.15 (a) For a complex load, maximum power is transferred for Z L = Zt* = 100 − j 25 = RL + jX L . The Thévenin equivalent with the load attached is: The current is given by 70.71∠ − 45 o I= = 0.3536∠ − 45 o 100 + j 25 + 100 − j 25 The load power is 2 PL = RL I rms = 100(0.3536 / 2 ) 2 = 6.25 W 178 (b) For a purely resistive load, maximum power is transferred for RL = Zt = 100 2 + 252 = 103.1 Ω. The Thévenin equivalent with the load attached is: The current is given by 70.71∠ − 45 o I= = 0.3456∠ − 37.98 o 103.1 + 100 − j 25 The load power is 2 PL = RL I rms = 103.1(0.3456 / 2 ) 2 = 6.157 W E5.16 The line-to-neutral voltage is 1000 / 3 = 577.4 V. No phase angle was specified in the problem statement, so we will assume that the phase of Van is zero. Then we have Van = 577.4∠0 o Vbn = 577.4∠ − 120 o Vcn = 577.4∠120 o The circuit for the a phase is shown below. (We can consider a neutral connection to exist in a balanced Y-Y connection even if one is not physically present.) The a-phase line current is V 577.4∠0 o = 4.610∠ − 37.02o IaA = an = Z L 100 + j 75.40 The currents for phases b and c are the same except for phase. IbB = 4.610∠ − 157.02o IcC = 4.610∠82.98o V I 577.4 × 4.610 P = 3 Y L cos(θ ) = 3 cos(37.02 o ) = 3.188 kW 2 2 179 VI 577.4 × 4.610 Q = 3 Y L sin(θ ) = 3 sin(37.02o ) = 2.404 kVAR 2 E5.17 2 The a-phase line-to-neutral voltage is Van = 1000 / 3 ∠ 0 o = 577 . 4 ∠ 0 o The phase impedance of the equivalent Y is ZY = Z ∆ / 3 = 50 / 3 = 16.67 Ω. Thus the line current is V 577.4∠0 o = 34.63∠0 o A IaA = an = 16.67 ZY Similarly, IbB = 34.63∠ − 120 o A and IcC = 34.63∠120 o A. Finally, the power is P = 3(I aA / 2 ) 2 Ry = 30.00 kW E5.18 Writing KCL equations at nodes 1 and 2 we obtain V1 V − V2 + 1 = 1∠60 o 100 + j 30 50 − j 80 V2 V − V1 + 2 = 2∠30 o j 50 50 − j 80 In matrix form, these become   1 1  +  100 30 50 80 + − j j    1 −  50 − j 80      V1  = 1∠60°   1  V2  2∠30° 1   +  j 50 50 − j 80  − 1 50 − j 80 The MATLAB commands are Y = [(1/(100+j*30)+1/(50-j*80)) (-1/(50-j*80));... (-1/(50-j*80)) (1/(j*50)+1/(50-j*80))]; I = [pin(1,60); pin(2,30)]; V = inv(Y)*I; pout(V(1)) pout(V(2)) The results are V1 = 79.98∠106.21o and 180 V2 = 124.13∠116.30 o Problems P5.1 The units of angular frequency ω are radians per second. The units of frequency f are hertz, which are equivalent to inverse seconds. The relationship between them is ω = 2πf . P5.2 (a) Decreasing the peak amplitude Vm? 2. Compresses the sinusoidal curve vertically. (b) Increasing the frequency f? horizontally. (c) Increasing θ? 4. Compresses the sinusoidal curve 6. Translates the sinusoidal curve to the left. (d) Decreasing the angular frequency ω? curve horizontally. (e) Increasing the period? horizontally. 3. Stretches the sinusoidal 3. Stretches the sinusoidal curve P5.3 An angle in radians is defined to be arc length divided by radius. Thus radian measure of angle is length divided by length. In terms of physical units, radians are unitless. Thus the physical units of ω are s-1. P5.4* v (t ) = 10 sin(1000πt + 30 o ) = 10 cos(1000π − 60 o ) ω = 1000π rad/s f = 500 Hz phase angle = θ = −60 o = − π 3 radians T = 1 f = 2 ms Vrms =Vm 2 = 10 2 = 7.071 V P = (Vrms )2 R = 1 W The first positive peak occurs for 1000πt peak − π 3 = 0 t peak = 0.3333 ms 181 P5.5 v (t ) = 12 sin(400πt − 120 o ) = 12 cos(400πt − 210 o ) v (t ) = 12 cos(400πt + 150 o ) ω = 400π rad/s f = 200 Hz phase angle = θ = −210 o = 150 o = 5π 6 radians T = 1 f = 5 ms Vrms = Vm 2 = 12 2 = 8.485 V P = (Vrms ) R = 1.44 W 2 Positive peak occurs for 400πt1 + 5π 6 = 0 t1 = −2.08 ms The first positive peak after t = 0 is at t peak = t1 +T = 2.916 ms 182 P5.6* Sinusoidal voltages can be expressed in the form v (t ) =Vm cos(ωt + θ ). The peak voltage is Vm = 2Vrms = 2 × 20 = 28.28 V. The frequency is f = 1 /T = 10 kHz and the angular frequency is ω = 2πf = 2π10 4 radians/s. The phase corresponding to a time interval of ∆t = 20 µs is θ = (∆t /T ) × 360 o = 72o . Thus, we have v (t ) = 28.28 cos(2π10 4t − 72o ) V. P5.7 I m = I rms 2 = 10 2 = 14.14 A ω = 2πf = 400π rad/s i (t ) = 14.14 cos(400πt − 72o ) P5.8 f = 1 = 200 Hz T tmax θ = −360 = −72 o T The angular frequency is ω = 2πf = 2000π radians/s. We have T = 1 / f = 1 ms. The sinusoid reaches a positive peak one quarter of a cycle after crossing zero with a positive slope. Thus, the first positive peak occurs at t = 0.35 ms. The phase corresponding to a time interval of ∆t = 0.35 ms is θ = ( ∆t /T ) × 360 o = 126o . Thus, we have v (t ) = 50 cos(2000πt − 126o ) V. P5.9 i (t ) = 10 cos(2000πt ) A p (t ) = Ri 2 (t ) = 10 4 cos 2 (2000πt ) = 5000[1 + cos (4000πt )] W ( Pavg = R (I rms )2 = 100 10 2 ) 2 = 5000 W 183 P5.10 v (t ) = 1000 sin(500πt ) V p (t ) = v 2 (t ) R = 2000 sin2 (500πt ) = 1000[1 − sin(1000πt )] W Pavg = (Vrms )2 R = 1000 W P5.11 A sequence of MATLAB instructions to accomplish the desired plot for part (a) is Wx = 2*pi; Wy = 2*pi; Theta = 90*pi/180; t = 0:0.01:20; x = cos(Wx*t); y = cos(Wy*t + Theta); plot(x,y) By changing the parameters, we can obtain the plots for parts b, c, and d. The resulting plots are 184 (a) (b) (c) Vrms = P5.12* P5.13* P5.14 I rms = 1 T T i T T 2 ∫ v (t )dt = 0 2 ∫ i (t )dt = 0 (d) 1 1 15 2 dt = 10.61 V ∫ 20 2 4  1  ∫ 25dt + ∫ 4dt  = 3.808 A   4 0 2  The limits on the integral don't matter as long as they cover one period Vrms = Vrms = 1 T T /2 0.025 −T / 2 − 0.025 2 ∫ v (t )dt = 10 2 ∫ [4 cos(20πt )] dt t = 0.025 80   sin( 40πt )  =  80t + 40π  t = −0.025 185 = 10 4 =2 V 0.025 ∫ [8 + 8 cos( 40πt )]dt − 0.025 P5.15 Vrms = Vrms = 1 T T 2 ∫v (t )dt = 0 0.5 ∫ [A 1 ∫ [A cos(2πt ) + B sin(2πt )] dt 2 0 ] cos 2 (2πt ) + 2AB cos( 2πt ) sin( 2πt ) + B 2 sin 2 (2πt ) dt 2 0 t =1  A2 B2  =  t +0+ t 2 t =0  2  A2 + B 2   =  2   The MATLAB commands are syms Vrms t A B Vrms = sqrt((int((A*cos(2*pi*t)+ B*sin(2*pi*t))^2,t,0,1))) and the result is Vrms = 1/2*(2*A^2+2*B^2)^(1/2) P5.16 P5.17 Vrms = Vrms 1 T T ∫v 2 (t )dt = 0 2 1 (5t )2 dt = ∫ 40 t =2 1  t3   25  = 4.082 V 4  3 t = 0 i T 2 = v (t )dt T ∫0 Vrms = = 1 0.1 0. 1 1 0.1 0. 1 ∫ [15 + 10 cos(20πt )] 2 dt 0 ∫ [225 + 300 cos(20πt ) + 100 cos 2 (20πt )]dt 0 = 0.1 0.1 0. 1  1  + + dt t dt π 225 300 cos( 20 ) 100 cos 2 (20πt )]dt  ∫ ∫ ∫ 0.1  0 0 0  = 0.1 0.1 0. 1 0. 1  1  dt + π t dt + dt + 225 300 cos( 20 ) 50 50 cos( 40πt )]dt  ∫ ∫ ∫ ∫ 0.1  0 0 0 0  1 [22.5 + 0 + 5 + 0] 0.1 = 16.58 V = 186 P5.18 Vrms = 1 T T ∫ v (t )dt 2 = 0 1 ∫ [5 exp( −t )] dt 2 0 [− 12.5 exp( −2t )]tt ==10 = = 1 ∫ [25 exp( −2t )]dt 0 = 12.5[1 − exp( −2)] = 3.288 V P5.19 The rms values of all periodic waveforms are not equal to their peak values divided by the square root of two. However, they are for sinusoids, which are important special cases. P5.20 Vrms = 1 T T 2 ∫ v (t )dt = 0 1 ∫ [10 exp( −5t ) sin(20πt )] dt 2 0 The MATLAB commands are: syms Vrms t Vrms = sqrt((int((10*exp(-5*t)*sin(20*pi*t))^2,t,0,1))) vpa(Vrms,4) The result is Vrms = 2.228 V. P5.21 Two methods to determine the phase relationship between two sinusoids are: 1. Examine the phasor diagram and consider it to rotate counterclockwise. If phasor A points in a given direction before phasor B by an angle θ, we say that A leads B by the angle θ or that B lags A by the angle θ. 2. Examine plots of the sinusoidal waveforms versus time. If A reaches a point (such as a positive peak or a zero crossing with positive slope) by an interval ∆t before B reaches the corresponding point, we say that A leads B by the angle θ = ( ∆t /T ) × 360° or that B lags A by the angle θ. P5.22 To add sinusoids we: 1. Convert each of the sinusoids to be added into a phasor. 2. Add the phasors and convert the sum to polar form using complex arithimetic. 3. Convert the resulting phasor to a sinusoid. All of the sinusoids to be added must have the same frequency. P5.23* We are given the expression 5 cos(ωt + 75 o ) − 3 cos(ωt − 75 o ) + 4 sin(ωt ) 187 Converting to phasors we obtain 5∠75 o − 3∠ − 75 o + 4∠ − 90 o = 1.2941 + j 4.8296 − (0.7765 − j 2.8978) − j 4 = 0.5176 + j 3.7274 = 3.763∠82.09 o Thus, we have 5 cos(ωt + 75 o ) − 3 cos(ωt − 75 o ) + 4 sin(ωt ) = 3.763 cos(ωt + 82.09 o ) P5.24* v 1 (t ) = 100 cos (ωt ) v 2 (t ) = 100 sin(ωt ) = 100 cos (ωt − 90 o ) V1 = 100 ∠0 o = 100 V2 = 100 ∠ − 90 o = − j 100 Vs = V1 + V2 = 100 − j 100 = 141 .4∠ − 45 o v s (t ) = 141.4 cos (ωt − 45 o ) V2 lags V1 by 90 o Vs lags V1 by 45o Vs leads V2 by 45o P5.25* ω = 2πf = 400 π v 1 (t ) = 10 cos (400 πt + 30 o ) v 2 (t ) = 5 cos( 400 πt + 150 o ) v 3 (t ) = 10 cos (400 πt + 90 o ) v 1 (t ) lags v 2 (t ) by 120 o v 1 (t ) lags v 3 (t ) by 60 o v 2 (t ) leads v 3 (t ) by 60 o 188 P5.26 Vm = 5 V f = 1 T T = 10 ms = 1000 Hz ω = 2πf = 2000π rad/s tmax = −36o T v (t ) = 5 cos(2000πt − 36o ) V θ = −360 o V = 5∠ − 36o V 5 Vrms = = 3.536 V 2 P5.27 The magnitudes of the phasors for the two voltages are 10 2 and 7 2 V. The phase angles are not known. If the phase angles are the same, the phasor sum would have its maximum magnitude which is 17 2 . On the other hand, if the phase angles differ by 180 o , the phasor sum would have its minimum magnitude, which is 3 2 . Thus, the maximum rms value of the sum is 17 V and the minimum is 3 V. P5.28 T = 1 / f = 1 / 500 = 2 ms i1 leads i2 by the angle ∆θ = ( ∆t /T ) × 360° = (0.25 / 2) × 360° = 45°. Therefore, the angle for i2 is θ 2 = θ1 − ∆θ = 15°. P5.29 We are given the expression 15 sin(ωt − 45o ) + 5 cos(ωt − 30 o ) + 10 cos(ωt − 120 o ) Converting to phasors we obtain 15∠ − 135 o + 5∠ − 30 o + 10∠ − 120 o = − 10.61 − j 10.61 + 4.33 − j 2.5 − 5 − j 8.66 = −11.28 − j 21.77 = 24.51∠ − 117.39° Thus, we have 15 sin(ωt − 45 o ) + 5 cos(ωt − 30 o ) + 10 cos(ωt − 120 o ) = 24.51 cos(ωt − 117.39o ) P5.30 v 1 (t ) = 90 cos(ωt − 15 o ) v 2 (t ) = 50 sin(ωt − 60 o ) = 50 cos(ωt − 150 o ) V1 = 90∠ − 15o = 86.93 − j 23.29 V2 = 50∠ − 150 o = −43.30 − j 25.00 Vs = V1 + V2 = 43.63 − j 48.29 = 65.08∠ − 47.90 o v s (t ) = 65.08 cos(ωt − 47.90 o ) 189 V2 lags V1 by 135 o Vs lags V1 by 32 .90 o Vs leads V2 by 102 .1o P5.31 v 1 (t ) = 10 cos(ωt − 30 o ) V I 1m = 2 × I 1rms = 14.14 A V1 = 10∠ − 30 o V I1 = 14.14∠ − 70 o A i1 (t ) = 14.14 cos(ωt − 70 o ) A P5.32 A sequence of MATLAB commands to generate the desired plot is: t = 0:0.01:2; v = cos(19*pi*t) + cos(21*pi*t); plot(v,t) The resulting plot is Notice that the first term cos(19πt) has a frequency of 9.5 Hz while the second term cos(21πt) has a frequency of 10.5 Hz. At t = 0, the rotating vectors are in phase and add constructively. At t = 0.5, the vector for the second term has rotated one half turn more than the vector for the first term and the vectors cancel. At t = 1, the first vector has rotated 9.5 turns and the second vector has rotated 10.5 turns so they both point in the same direction and add constructively. 190 P5.33 For an inductance, we have VL = jωLIL . I For a capacitance, we have VC = C . jωC P5.34 For a pure resistance, current and voltage are in phase. For a pure inductance, current lags voltage by 90 o. For a pure capacitance, current leads voltage by 90 o. P5.35* v L (t ) = 10 cos(2000 πt ) ω = 2000 π Z L = jωL = j 200 π = 200 π∠90 o VL = 10∠0 o IL = VL Z L = (1 20 π )∠ − 90 o i L (t ) = (1 20 π ) cos (2000 πt − 90 o ) = (1 20 π ) sin(2000 πt ) iL (t ) lags v L (t ) by 90 o P5.36 V = 50∠0 o = 50 + j 0 I Because Z is pure real, the element is a resistance of 50 Ω. (a) V = 100∠30 o I = 2∠30 o Z = (b) Notice that the current is a sine rather than a cosine. V V = 100∠30 o I = 3∠ − 60 o Z = = 33.33∠90 o = j 33.33 I Because Z is pure imaginary and positive, the element is an inductance. ω = 400 L= Z = 83.33 mH ω (c) Notice that the voltage is a sine rather than a cosine. V V = 100∠ − 60 o I = 2∠30 o Z = = 50∠ − 90 o = − j 50 I 191 Because Z is pure imaginary and negative, the element is a capacitance. 1 ω = 200 = 100 µF C = Zω P5.37* v C (t ) = 10 cos(2000 πt ) ω = 2000 π −j = − j 15.92 = 15.92∠ − 90 o Ω ωC VC = 10∠0 o ZC = IC = VC Z C = 0.6283∠90 o iC (t ) = 0.6283 cos(2000 πt + 90 o ) = −0.6283 sin(2000 πt ) iC (t ) leads v C (t ) by 90 o 5.38 (a) From the plot, we see that T = 2 ms, so we have f = 1 /T = 500 Hz and ω = 1000π . Also, we see that the current leads the voltage by 0.5 ms or 90°, so we have a capacitance. Finally, 1 / ωC = Vm / I m = 200 Ω, from which we find that C = 1.592 µF. (b) From the plot, we see that T = 16 ms, so we have f = 1 /T = 62.5 Hz and ω = 125π . Also, we see that the current lags the voltage by 4 ms or 90°, so we have an inductance. Finally, ωL = Vm / I m = 400 Ω, from which we find that L = 1.018 H. P5.39 A Matlab m-file that produces the desired plots is: f=0:1:1000; ZL=2*pi*f*0.01; ZC=1./(2*pi*f*10e-6); R=50 plot(f,ZL) 192 axis([0 1000 0 100]) hold plot(f,ZC) plot(f,R) The resulting plot is: P5.40 V = 5∠0 o = 5 I Because Z is pure real, the element is a resistance of 5 Ω. (a) Z = V = 4∠90 o = j 4 I Because Z is pure imaginary and positive, the element is an inductance. (b) Z = ω = 1000 L= Z = 4 mH ω V = 20∠ − 90 o = − j 20 I Because Z is pure imaginary and negative, the element is a capacitance. 1 ω = 1000 = 50 µF C = (c) Z = Zω 193 P5.41 The steps in analysis of steady-state ac circuits are: 1. Replace sources with their phasors. 2. Replace inductances and capacitances with their complex impedances. 3. Use series/parallel, node voltages, or mesh currents to solve for the quantities of interest. All of the sources must have the same frequency. P5.42* I= = Vs R + jω L 10∠0 o 100 + j 100 = 70.71∠ − 45 o mA VR = RI = 7.071∠ − 45 o V VL = jωLI = 7.071∠45 o V I lags Vs by 45 o P5.43 I= Vs 10∠0 o = = 55.47 ∠ − 56.31o mA R + jωL 100 + j 150 VR = RI = 5.547 ∠ − 56.31o V Vs 56.31° VL = jωLI = 8.321∠33.69o V I I lags Vs by 56.31o VR P5.44* I= = Vs R − j ωC 10∠0 o 1000 − j 2000 = 4.472∠63.43o mA VR = RI = 4.472∠63.43o V VC = (− j ωC )I = 8.944 ∠ − 26.57 o V I leads Vs by 63.43o 194 VL P5.45 I= Vs = 10∠0 o = 8 + j 4 = 8.944∠26.56o mA 1000 − j 500 R − j ωC VR = RI = 8.944∠26.56o V VC = (− j ωC )I = 4.472∠ − 63.43o V I VR I leads Vs by 63.43° Vs 63.43° VC P5.46* Z = jωL + R − j ω = 500 : 1 ωC Z = j 50 + 50 − j 200 = 50 − j 150 Ω = 158.1∠ − 71.57 o ω = 1000 : Z = j 100 + 50 − j 100 = 50 Ω = 50∠0 o ω = 2000 : Z = j 200 + 50 − j 50 = 50 + j 150 Ω = 158.1∠71.57 o P5.47 Z = 1 1 = 1 Z L + 1 Z c 1 jωL + jωC + 1 / R ω = 500 : Z = 1 1 ( j 50 ) + j 0.005 + 0.001 = 4.425 + j 66.37 = 66.52∠86.19° Ω ω = 1000 : Z = 1000 + j 0 = 1000∠0° Ω ω = 2000 : Z = 4.425 − j 66.37 = 66.52∠ − 86.19° Ω P5.48 Z = 1 1 /(R + jωL) + jωC ω = 500 : Z = 1 = 1.770 + j 6.549 = 6.784∠74.88° Ω 1 /(1 + j 5) + j 0.05 195 ω = 1000 : Z = 100 − j 10 = 100.5∠ − 5.71° Ω ω = 2000 : Z = 0.1106 − j 6.659 = 6.660∠ − 89.04° Ω Is = 10∠0 o mA P5.49* V = Is 1 1 R + 1 jωL + jωC = 10 −2 1 1 1000 − j 0.005 + j 0.005 = 10∠0 o V IR = V R = 10∠0 o mA IL = V jωL = 50∠ − 90 o mA IC = V ( jωC ) = 50∠90 o mA The peak value of iL (t ) is five times larger than the source current! This is possible because current in the capacitance balances the current in the inductance (i.e., IL + IC = 0 ). 5.50 Is = 0.5∠ − 90 o A V = Is 1 1 200 + 1 j 200 = 70.71∠ − 45o V IR = V R = 0.3536∠ − 45o A IL 45° 45° IR IL = V jωL = 0.3536∠ − 135 o A V leads Is by 45 P5.51 o V Is Vs = 10∠45o V Vs 10∠45° = 0.2∠45o A I= = jωL + R − j ωC j 500 + 50 − j 500 VL = j 500 × I = 100∠135o V VR = 100 I = 10∠45o V VC = − j 500 I = 100∠ − 45 o V The peak value of v L (t ) is ten times larger than the source voltage! This is possible because the impedance of the capacitor cancels the impedance of the inductance. 196 P5.52* Is = 100∠0 o mA V = Is 1 1 100 + 1 (− j 200 ) = 8.944 ∠ − 26.56o V IR = V R = 89.44 ∠ − 26.56o mA IC = V ( jωC ) = 44.72∠63.44 o mA V lags Is by 26.56o P5.53 Ztotal = jωL + 1 1 = j 200 + = 20 + j 160 0.01 + j 0.02 1 R + jωC Ztotal = 161.25∠82.88o Ω I= 100∠0 o = 0.6202∠ − 82.88o A Ztotal − j 50 ZC IR = I = (0.6202∠ − 82.88o ) × R + ZC 100 − j 50 = 0.2774∠ − 146.31o A IC = I R 100 = (0.6202∠ − 82.88o ) × R + ZC 100 − j 50 = 0.5547 ∠ − 56.31o A P5.54 V1 = 100∠ − 90 o = − j 100 V2 = 100∠30 o − j 100 + (86.60 + j 50 ) V + V2 I= 1 = R + jωL 100 + j 100 V2 VL = 0.7071∠ − 75o A VR = 100 I = 70.71∠ − 75o V VL = j 100 I = 70.71∠15o V I leaads V1 by 15 o I I lags VL by 90 o VR V1 197 P5.55 A Matlab program that produces the desired plots is: w=0:1:2000; L=20e-3; C=50e-6; Zmaga=abs(w*L-1./(w*C)); Zmagb=abs(1./((1./(i*w*L))+i*w*C)); plot(w,Zmaga) hold on plot(w,Zmagb) axis([0 2000 0 100]) hold off The result is: P5.56 A Matlab program that produces the desired plots is: w=0:1:5000; L=20e-3; R=50; Zmaga=abs(R+i*w*L); Zmagb=abs(1./((1./(i*w*L))+1/R)); plot(w,Zmaga) hold on 198 plot(w,Zmagb) hold off The result is: P5.57 The KCL equation is V1 − j 20 V V + 10 = 0. Solving, we find + 1 + 1 j 20 − j 10 5 V1 = −12 + j 16 = 20∠126.89o V. P5.58 The KCL equation is V1 − j 10 V1 + + 2 = 0. Solving, we find 15 5 + j5 V1 = −10.58 − j 2.35 = 10.84∠ − 167.47° V. P5.59 Power factor is the cosine of the power angle. It is often expressed as a percentage. PF = cos(θ ) × 100% P5.60 The units for real power are watts (W). For reactive power, the units are volt-amperes reactive (VAR). For apparent power, the units are voltamperes (VA). 199 P5.61 Complex power is equal to one half of the phasor voltage across the element times the complex conjugate of the phasor current provided that the current reference direction points into the positive reference for the voltage. Average power P equals the real part of the complex power and reactive power Q equals the imaginary part. P5.62 A load with a leading power factor is capacitive and has negative reactive power. A load with a lagging power factor is inductive and has positive reactive power. P5.63 (a) For a pure resistance, the power is positive and the reactive power is zero. (b) For a pure inductance, the power is zero and the reactive power is positive. (c) For a pure capacitance, the power is zero, and the reactive power is negative. P5.64 Usually, power factor correction refers to adding capacitances in parallel with an inductive load to reduce the reactive power flowing from the power plant through the distribution system to the load. To minimize power system losses, we need 100% power factor for the combined load. Ultimately, an economic analysis is needed to balance the costs of power factor correction against the costs of losses and additional distribution capacity needed because of reactive power. P5.65 See Figure 5.23 in the book. P5.66 Real power represents a net flow, over time, of energy from the source to the load. This energy must be supplied to the system from hydro, fossil-fuel, or nuclear sources. Reactive power represents energy that flows back and forth from the source to the load. Aside from losses in the transmission system (lines and transformers), no net energy must be supplied to the system to create the reactive power. Reactive power is important mainly because of the increased system losses associated with it. 200 P5.67* 1000 2∠0 o 1000 2∠0 o = 14.14 + j 5.331 = 15.11∠20.66o + 100 − j 265.3 P = Vrms I rms cos θ = 10 kW I= Q = VrmsI rms sin θ = −3.770 kVAR Apparent power = Vrms I rms = 10.68 kVA Power factor = cos(20.66o ) = 0.9357 = 93.57% leading P5.68 I= 1000 2∠0 o 1000 2∠0 o + = 14.14 − j 7.502 100 j 188.5 = 16.01∠ − 27.95o P = Vrms I rms cos θ = 10 kW Q = Vrms I rms sin θ = 5.305 kVAR Apparent power =Vrms I rms = 11.32 kVA Power factor = cos(27.95o ) = 0.8834 = 88.34% lagging P5.69* This is a capacitive load because the reactance is negative. 2 P = I rms R = (15) 2 100 = 22.5 kW 2 Q = I rms X = (15) 2 ( −50) = −11.25 kVAR Q  θ = tan −1   = tan −1 ( −0.5) = 26.57 o P  power factor = cos(θ ) = 89.44% apparent power = P 2 + Q 2 = 25.16 KVA P5.70 If we flip the voltage reference so the current points into the positive voltage reference, we have V = −1500 2∠ − 120 o = 1500 2∠60 o S = 21 VI * = 21.73 − j 5.823 kVA θ = θv − θi = 60 o − 75o = −15o power factor = cos(θ ) = 96.59% leading P = Vrms I rms cos(θ ) = 21.73 kW Q = Vrms I rms sin(θ ) = −5.823 kVAR apparent power = Vrms I rms = 1500 × 15 = 22.50 KVA Z = V 1500 2∠60° = 100∠ − 15° Ω = I 15 2∠75° 201 P5.71 This is a capacitive load because the reactance is negative. Z = 40 − j 30 = 50∠ − 36.87° I = S = 21 VI * = 23.04 − j 17.28 kVA V Z = 1200 2∠30° = 24 2∠66.87° 50∠ − 36.87° 2 P = I rms R = (24) 2 40 = 23.04 kW 2 Q = I rms X = (24)2 ( −30) = −17.28 kVAR θ = −36.87 o power factor = cos(θ ) = 80% apparent power = Vrms I rms = 1200 × 24 = 28.80 KVA P5.72 S = 21 VI * = 21 (10 4 2∠75°) × (25 2∠30°)* = 176.8 + j 176.8 kVA Vrms = 10 4 V I rms = 25 A θ = θv − θi = 75o − 30 o = 45o power factor = cos θ × 100% = 70.71% lagging P = Vrms I rms cos(θ ) = 176.8 kW Q = Vrms I rms sin(θ ) = 176.8 kVAR Apparent Power = Vrms I rms = 250 KVA This is an inductive load. P5.73 I= 260 2∠50 o − 220 2∠30 o = 7.091 2∠37.32 o 5 + j 12 I rms = 7.091 A Delivered by Source A: PA = 260I rms cos(50 − 37.32) = 1.799 kW QA = 260I rms sin(50 − 37.32) = 0.4046 kVAR Absorbed by Source B: PB = 220I rms cos(30 − 37.32) = 1.547 kW QA = 220I rms sin(30 − 37.32) = −0.1989 kVAR Absorbed by resistor: 2 PR = I rms R = 0.2514 kW Absorbed by inductor: 2 QL = I rms X = 0.6013 kVAR P5.74 (a) For a resistance in series with an inductance, real power is positive and reactive power is positive. (b) For a resistance in series with a capacitance, real power is positive and reactive power is negative. 202 (c) For a pure resistance, real power is positive and reactive power is zero. P5.75 If the inductive reactance is greater than the capacitive reactance, the total impedance is inductive, real power is zero, and reactive power is positive. If the inductive reactance equals the capacitive reactance, the total impedance is zero, and the current is infinite. Real power and reactive power are indeterminate. This case is rarely, if ever, of practical importance. If the inductive reactance is less than the capacitive reactance, the total impedance is capacitive, real power is zero, and reactive power is negative. P5.76 If the inductive reactance is greater than the capacitive reactance, the total impedance is capacitive, real power is zero, and reactive power is negative. If the inductive reactance equals the capacitive reactance, the total impedance is infinite, and the current is zero. Real power and reactive power are zero. If the inductive reactance is less than the capacitive reactance, the total impedance is inductive, real power is zero, and reactive power is positive. P5.77 Apparent power = Vrms I rms ⇒ 2500 = 220I rms I rms = 11.37 A Vrms = 19.36 Ω I rms P 2000 ⇒ θ = −36.87° (We know that θ is negative cos(θ ) = = Vrms I rms 2500 Z = because the impedance is capacitive.) Z = 19.36∠ − 36.87° = 15.49 − j 11.61 = R + 1 /( jωC ) 1 R = 15.49 Ω C = = 228.5 µF 11.61(2π 60) P5.78* ⇒ Load A: PA = 10 kW θ A = cos − 1 (0.9) = 25.84 o QA = PA tan θ A = 4.843 kVAR 203 Load B: Vrms I Brms = 15 kVA θB = cos − 1 (0.8) = 36.87 o QB = Vrms I Brms sin(θB ) = 9 kVAR PB = Vrms I Brms cos(θB ) = 12 kW Source: Ps = PA + PB = 22 kW Qs = QA + QB = 13.84 kVAR Apparent power = Power factor = P5.79 (Ps )2 + (Qs )2 Ps Apparent power = 26 kVA = 0.8462 = 84.62% lagging Load A: PA = 50 kW θ A = cos − 1 (0.6) = 53.13o QA = PA tan θ A = 66.67 kVAR Load B: PB = 75 kW θB = cos − 1 (0.8) = 36.86o QB = PB tan θB = 56.25 kVAR Source: Ps = PA + PB = 125 kW Qs = QA + QB = 122.92 kVAR Apparent power = Power factor = P5.80 (Ps )2 + (Qs )2 Ps Apparent power = 175.3 kVA = 0.7130 = 71.30% lagging VA = (15 + j 10)10 2∠10 o + 240 2∠ − 20 o = 358.4 2∠6.80 o VArms = 358.4 V rms Delivered by Source A: 204 PA = 358.4(10) cos(6.80 − 10) = 3.579 kW QA = 358.4(10) sin(6.80 − 10) = −0.200 kVAR Absorbed by Source B: PB = 240(10) cos(−20 − 10) = 2.079 kW QB = 240(10) sin(−20 − 10) = −1.200 kVAR Absorbed by resistor: 2 PR = I rms R = 1.500 kW Absorbed by inductor: 2 QL = I rms X = 1.000 kVAR P5.81 P ( Vrms )2 = R 1500 2 = = 45000 W 50 Q = QL + QC = (Vrms )2 Q = 19630 VAR XL + (Vrms )2 XC = 15002 15002 + 94.25 ( −530.5) Apparent power = P 2 + Q 2 = 49095 VA Power factor = P5.82 P Apparent power = 91.65% lagging 1500 2∠0 o 1500 2∠0 o = = 3.416 2∠83.46o R + jωL − j ωC 50 + j 94.25 − j 530.5 S = 21 VI * = 583.4 − j 5090.7 I= P = Re(S) = 583.4 W Q = Im(S) = −5090.7 VAR Apparent power = S = 5124 VA Power factor = cos(θ ) = cos( −83.46°) = 11.38% leading P5.83* (a) cos θ = 0.25 θ = 75.52 o P = Vrms I rms cos(θ ) 100 kW P = = 400 A I rms = Vrms cos(θ ) 1 kV (0.25 ) I = 400 2∠ − 75.52 o 205 Qload = Vrms I rms sin θ = 387 .3 kVAR Qtotal = 0 = Qload + QC (b) QC = −387 .3 × 10 3 = X C = −2.582 = − C = 1027 µF 1 ωC (Vrms )2 XC The capacitor must be rated for at least 387.3 kVAR. With the capacitor in place, we have: P = 100 kW = Vrms I rms I rms = 100 A I = 100 ∠0o (c) P5.84 The line current is smaller by a factor of 4 with the capacitor in place, reducing I 2R losses in the line by a factor of 16. The ac steady state Thévenin equivalent circuit for a two-terminal circuit consists of a phasor voltage source Vt in series with a complex impedance Zt. The ac steady state Norton equivalent circuit for a two-terminal circuit consists of a phasor current source In in parallel with a complex impedance Zt. Vt is the open-circuit voltage of the original circuit. In is the short circuit current of the original network. The impedance can be found by zeroing the independent sources and finding the impedance looking into the terminals of the original network. Also, we have Vt = Zt It . P5.85 The load voltage is given by the voltage divider principle. VL = Vt ZL Zt + Z L The load voltage VL is larger than the Thévenin voltage in magnitude if the magnitude of Zt + Z L is less than the magnitude of Z L which can happen when the imaginary parts of Zt and Z L have opposite signs. This does not happen in purely resistive circuits. 206 P5.86 To attain maximum power, the load must equal (a ) the complex conjugate of the Thévenin impedance if the load can have any complex value; (b) the magnitude of the Thévenin impedance if the load must be a pure resistance. P5.87* (a) Zeroing the current source, we have: Thus, the Thévenin impedance is Zt = 100 + j 50 = 111.8∠26.57 o Ω Under open circuit conditions, there is zero voltage across the inductance, the current flows through the resistance, and the Thévenin voltage is Vt = Voc = 200∠0 o In = Vt Zt = 1.789∠ − 26.57 o Thus, the Thévenin and Norton equivalent circuits are: (b) For maximum power transfer, the load impedance is Z load = 100 − j 50 Iload = Vt 200 = =1 Zt + Z load 100 + j 50 + 100 − j 50 Pload = Rload (I rms −load )2 = 100(1 / 2 ) 2 = 50 W (c) In the case for which the load must be pure resistance, the load for maximum power transfer is Z load = Zt = 111.8 I load = Vt 200 = = 0.9190∠ − 13.28 o Zt + Z load 100 + j 50 + 111.8 Pload = Rload (I rms −load )2 = 47.21 W 207 P5.88 Under open-circuit conditions, we have Vt = Vab − oc = (4 + j 5)2∠ − 30° = 11.92 + j 4.66 = 12.80∠21.34° V With the source zeroed, we look back into the terminals and see Zt = 5 − j 3 + j 5 + 4 = 9 + j 2 Ω Next, the Norton current is V In = t = 1.389∠8.81° A Zt P5.89 Zeroing sources, we have: Thus, the Thévenin impedance is 1 Zt = − j 10 = 7.0711∠ − 45 o = 5 − j 5 1 10 + 1 j 10 Writing a current equation for the node at the upper end of the current source under open circuit conditions, we have Voc − 75∠30 o V + oc = 5 10 j 10 Vt = Voc = 85.50∠63.07 o In = Vt Zt = 12.09∠108.07 o The Thévenin and Norton equivalent circuits are: For the maximum power transfer, the load impedance is 208 Z load = 5 + j 5 Iload = Vt 85.50∠63.07 o = = 8.55∠63.07 o A Zt + Z load 5 − j 5 + 5 + j 5 Pload = Rload (I rms −load )2 = 182.75 W In the case for which the load must be pure resistance, the load for maximum power transfer is Z load = Zt = 7.0711 I load = Vt 85.50∠63.07 o = = 6.544∠85.57 o Zt + Z load 5 − j 5 + 7.0711 Pload = Rload (I rms −load )2 = 151.4 W P5.90 At the lower left-hand node under open-circuit conditions, KCL yields Ix + 0.5Ix = 0 from which we have Ix = 0. Then, the voltages across the 6-Ω resistance and the j8-Ω inductance are zero, and KVL yields Vt = Vba − oc = 10∠30° With short circuit conditions, we have 10∠30° Ix = = 1∠ − 23.13° 6 + j8 In = Iba − sc = 0.5Ix + Ix = 1.5∠ − 23.13° The Thévenin impedance is given by V 10∠30° Zt = t = = 6.667 ∠53.13° In 1.5∠ − 23.13° Finally, the equivalent circuits are: 209 P5.91* For maximum power transfer, the impedance of the load should be the complex conjugate of the Thévenin impedance: Z load = 10 − j 5 Yload = 1 Z load = 0.08 + j 0.04 Yload = 1 Rload + jωC load = 0.08 + j 0.04 Setting real parts equal: 1 Rload = 0.08 Rload = 12.5 Ω Setting imaginary parts equal: ωC load = 0.04 C load = 106.1 µF P5.92 For maximum power transfer, the impedance of the load should be the complex conjugate of the Thévenin impedance: Z load = 10 − j 5 = Rload − j (ωC load ) Setting real parts equal: Rload = 10 Ω Setting imaginary parts equal: − 5 = − 1 (ωC load ) C load = 1 (5ω ) = 530.5 µF P5.93 We are given: v an (t ) = 100 cos(ωt − 60 o ) v bn (t ) = 100 cos(ωt + 60 o ) v cn (t ) = −100 cos(ωt ) The phasor diagram is: For counterclockwise rotation, the sequence of phasors is acb. Thus, this is a negative sequence source. From the phasor diagram, we can determine that 210 Vab = Van − Vbn = 100 3∠ − 90 o Vbc = Vbn − Vcn = 100 3∠ + 30 o Vca = Vcn − Van = 100 3∠ + 150 o v ab (t ) = 100 3 cos(ωt − 90 o ) v bc (t ) = 100 3 cos(ωt + 30 o ) v ca (t ) = 100 3 cos(ωt + 150 o ) P5.94 We are given v an (t ) = 120 cos(100πt + 75o ) (a) By inspection, ω = 2πf = 100π and we have f = 50 Hz. P5.95* (b) v bn (t ) = 120 cos(100πt − 45o ) V (c) v bn (t ) = 120 cos(100πt − 165o ) V v cn (t ) = 120 cos(100πt − 45o ) V v cn (t ) = 120 cos(100πt − 165o ) V ZY = = 1 1 R + j ωC 1 1 50 + j 377 × 10 −4 = 10.98 − j 20.70 = 23.43∠ − 62.05 o Z ∆ = 3ZY = 70.29∠ − 62.05 o Ω P5.96* VL = 3 ×VY = 3 × 440 = 762.1 V rms V 440 IL = Y = = 14.67 A rms 30 R P = 3VYI L cos(θ ) = 3 × 440 × 14.67 × cos(0 ) = 19.36 kW 211 P5.97 Total power flow in a balanced system is constant with time. For a single phase system the power flow pulsates. Reduced vibration in generators and motors is a potential advantage for the three-phase system. In addition, less wire is needed for the same power flow in a balanced threephase system. P5.98 This is a positive sequence source. The phasor diagram is shown in Figure 5.41 in the book. Thus, we have 440 2 ∠0 o V 3 The impedance of an equivalent wye-connected load is Van = ZY = Z∆ = 4 + j1 Ω 3 The equivalent circuit for the a-phase of an equivalent wye-wye circuit is: Thus, the line current is Van IaA = 1 + j 2 + ZY = 61.61∠ − 30.96o VAn = Van − IaA (1 + j 2) = 254.03∠ − 16.93o VAB = 440∠13.07 o IAB = VAB Z∆ = 35.57 ∠ − 0.96o Pload = 3(I ABrms )2 × 12 = 22.77 kW Pline = 3(I aArms )2 × 1 = 5.69 kW 212 P5.99* This is a positive sequence source. The phasor diagram is shown in Figure 5.41 in the book. Thus, we have: 440 2 Van = ∠0 o 3 The impedance of a equivalent wye-connected load is ZY = Z∆ = 5 − j2 Ω 3 The equivalent circuit for the a-phase of an equivalent wye-wye circuit is: Thus, the line current is: Van IaA = 1 + j 2 + ZY = 59.87∠0 o VAn = Van − IaA (1 + j 2) = 322.44∠ − 21.80 o VAB = 558∠8.20 o IAB = VAB Z∆ = 34.56∠30 o Pload = 3(I ABrms )2 × 12 = 26.89 kW Pline = 3(I aArms )2 × 1 = 5.38 kW P5.100 The line-to-line voltage is 277 3 = 480 V rms. The impedance of each arm of the delta is 1 Z∆ = = 13.42∠26.57 o 1 / 15 + 1 /( j 30) The equivalent wye has impedances of ZY = Z∆ o = 4.472 ∠26.57 3 Then working with one phase of the wye-wye, we have 213 I line = 277 ZY = 61.94 A rms power factor = cos(26.57 o ) × 100% = 89.44% P = 3(277)(61.94)(0.8944) = 46.04 kW P5.101 Vab = Van − Vbn = 3 ×VY ∠ − 30 o Vbc = Vbn − Vcn = 3 ×VY ∠90 o Vca = Vcn − Van = 3 ×VY ∠ − 150 o P5.102 VY = VL 3 = 208 = 120 V rms 3 Van = 120 2∠0 o Vbn = 120 2∠ − 120 o Vcn = 120 2∠120 o Vab = 208 2∠30 o Vbc = 208 2∠ − 90 o Vca = 208 2∠150 o 214 Ia = Van = 2.4 2∠ − 53.13o 30 + j 40 Ib = 2.4 2∠ − 173.13o Ic = 2.4 2∠66.86o P = 3VYrms I Lrms cos(θ ) = 3 × 120 × 2.4 × cos(53.13) = 518.4 W Q = 3VYrms I Lrsm sin(θ ) = 3 × 120 × 2.4 × sin(53.13o ) = 691.2 VAR P5.103 As suggested in the hint given in the book, the impedances of the circuits between terminals a and b with c open must be identical. Equating the impedances, we obtain: Z Z + Z B ZC 1 Za + Zb = = A C 1 / Z C + 1 /(Z A + Z B ) Z A + Z B + Z C Similarly for the other pairs of terminals, we obtain Z Z + Z B ZC 1 Z a + Zc = = A B 1 / Z B + 1 /(Z A + Z C ) Z A + Z B + Z C Zb + Zc = Z Z + ZAZB 1 = A C 1 / Z A + 1 /(Z C + Z B ) Z A + Z B + Z C (1) (2) (3) Then adding the respective sides of Equations 1 and 2, subtacting the corresponding sides of Equation 3, and dividing boths sides of the result by 2, we have: Za = Z B ZC Z A + Z B + ZC Similarly we obtain: Zb = Z AZC Z A + Z B + ZC and 215 Zc = ZAZB Z A + ZB + ZC P5.104 As suggested in the hint, consider the circuits shown below. The admittances of the circuits between terminals must be identical. First, we will solve for the admittances of the delta in terms of the impedances of the wye. Then we will invert the results to obtain relationships between the impedances. Z b + Zc 1 (1) YC + YB = = 1 Z Z + Z Z + Z Z a c a c b b Za + 1 / Z b + 1 / Zc Similarly working with the other terminials, we obtain Za + Zb YA + YB = (2) Z a Z b + Z b Zc + Z a Zc Z a + Zc YA + YC = (3) Z a Zb + Z b Zc + Z a Zc Then adding the respective sides of Equations 2 and 3, subtacting the corresponding sides of Equation 1, and dividing boths sides of the result by 2, we have: YA = Za Z a Zb + Z b Zc + Z a Zc Inverting both sides of this result yields: ZA = Z a Z b + Z b Zc + Z a Zc Za Similarly, we obtain: ZB = Z a Z b + Z b Zc + Z a Zc Zb 216 and Z C = Z a Z b + Zb Zc + Z a Zc Zc P5.105* First we write the KVL equation: V1 − V2 = 10∠0 o Then we enclose nodes 1 and 2 in a closed surface to form a supernode and write a KCL equation: V1 V V V + 1 + 2 + 2 =0 10 j 20 15 − j 5 The MATLAB commands are: echo on Y=[1 -1;(0.1+1/(j*20)) (1/15+(1/(-j*5)))] I=[10;0] V=inv(Y)*I pout(V(1)) pout(V(2)) The answers are: V1 = 9.402∠29.58o V2 = 4.986∠111.45 o P5.106 Writing KVL equations around the meshes, we obtain 10I1 + j 20( I1 − I2 ) = 0 j 20( I2 − I1 ) + 15( I2 − I3 ) = −10 − j 5I3 + 15( I3 − I2 ) = 0 The MATLAB commands are: echo on Z=[(10+j*20) -j*20 0; -j*20 (15+j*20) -15;0 -15 (15-j*5)] V=[0;-10;0] I=inv(Z)*V pout(I) Solving, we obtain: I1 = 0.9402∠ − 150.4 o I2 = 1.051∠ − 177.0 o I3 = 0.9972∠ − 158.6o P5.107* Writing KVL equations around the meshes, we obtain 5I1 + j 20( I1 − I2 ) = j 20 − j 10 I2 + j 20( I2 − I1 ) = 10 The MATLAB commands are: Z=[(5+j*20) -j*20;-j*20 j*10] 217 V=[j*20;10] I=inv(Z)*V pout(I) Solving, we obtain: I1 = 1.372∠120.96o I2 = 1.955∠136.22o P5.108 The current through the current source is I1 − I2 = 2 Writing KVL around the perimeter of the circuit, we have 15I1 + (5 + j 5) I2 = j 10 The MATLAB commands are: Z=[1 -1; 15 (5+j*5)] V=[2;j*10] I=inv(Z)*V pout(I) Solving, we obtain: I1 = 1.0847 ∠49.40 o and I2 = 1.5339∠147.52° P5.109 Writing KCL equations at nodes 1 and 2 we obtain V1 V − V2 + 1 = 1∠0 o 20 5 + j 8 V2 V − V1 + 2 = 1∠30 o − j 20 5 + j 8 Y=[(1/20+1/(5+j*8)) -1/(5+j*8);... -1/(5+j*8) (1/(-j*20)+1/(5+j*8))] I=[1;pin(1,30)] V=inv(Y)*I pout(V) Solving, we obtain V1 = 23.75∠ − 37.26o V2 = 30.55∠ − 37.06o 218 Practice Test T5.1 I rms = 1 T T 2 ∫ i (t )dt = 0 2 2 1 (3t ) 2 dt = t 3 = 8 = 2.828 A ∫ 0 30 P = I rms R = 8(50) = 400 W 2 T5.2 T5.3 V = 5∠ − 45° + 5∠ − 30° = 3.5355 − j 3.5355 + 4.3301 − j 2.5000 V = 7.8657 − j 6.0355 = 9.9144∠ − 37.50° v (t ) = 9.914 cos(ωt − 37.50°) 15 = 10.61 V 2 (b) f = 200 Hz (c) ω = 400π radians/s (d) T = 1 / f = 5 ms (e) V1 = 15∠ − 45° and V2 = 5∠ − 30° (a) V1rms = V1 lags V2 by 15° or V2 leads V1 by 15° T5.4 I= V1 Vs 10∠0° 10∠0° = 0.7071∠ − 45° A = = R + jωL − j / ωC 10 + j 15 − j 5 14.14∠45° VR = 10 I = 7.071∠ − 45° V VC = − j 5I = 5.303∠ − 135° V T5.5 V2 VL = j 15I = 10.606∠45° V S = 21 VI * = 21 (440∠30°)(25∠10°) = 5500∠40° = 4213 + j 3535 VA P = Re(S) = 4213 W Q = Im(S) = 3535 VAR Apparent power = S = 5500 VA Power factor = cos(θv − θ I ) = cos(40°) = 76.6% lagging T5.6 We convert the delta to a wye and connect the neutral points with an ideal conductor. ZY = Z ∆ / 3 = 2 + j 8 / 3 Ztotal = Z line + ZY = 0.3 + j 0.4 + 2 + j 2.667 = 2.3 + j 3.067 219 Ztotal = 3.833∠53.13° IaA = T5.7 Van Ztotal = 208∠30° = 54.26∠ − 23.13° A 3.833∠53.13° The mesh equations are: j 10I1 + 15( I1 − I2 ) = 10∠45° − j 5I2 + 15( I2 − I1 ) = −15 In matrix form these become − 15   I1  10∠45° (15 + j 10) =  − 15 (15 − j 5)  I2   − 15   The commands are: Z = [(15+j*10) -15; -15 (15-j*5)] V = [pin(10,45); -15] I = inv(Z)*V pout(I(1)) pout(I(2)) 220 CHAPTER 6 Exercises E6.1 (a) The frequency of v in (t ) = 2 cos(2π ⋅ 2000t ) is 2000 Hz. For this frequency H (f ) = 2∠60 o. Thus, Vout = H (f )Vin = 2∠60 o × 2∠0 o = 4∠60 o and we have v out (t ) = 4 cos(2π ⋅ 2000t + 60 o ). (b) The frequency of v in (t ) = cos(2π ⋅ 3000t − 20 o ) is 3000 Hz. For this frequency H (f ) = 0. Thus, Vout = H (f )Vin = 0 × 2∠0 o = 0 and we have v out (t ) = 0. E6.2 The input signal v (t ) = 2 cos(2π ⋅ 500t + 20 o ) + 3 cos(2π ⋅ 1500t ) has two components with frequencies of 500 Hz and 1500 Hz. For the 500-Hz component we have: Vout,1 = H (500)Vin = 3.5∠15 o × 2∠20 o = 7 ∠35 o v out,1 (t ) = 7 cos(2π ⋅ 500t + 35 o ) For the 1500-Hz component: Vout,2 = H (1500)Vin = 2.5∠45 o × 3∠0 o = 7.5∠45 o v out,2 (t ) = 7.5 cos(2π ⋅ 1500t + 45 o ) Thus the output for both components is v out (t ) = 7 cos(2π ⋅ 500t + 35 o ) + 7.5 cos(2π ⋅ 1500t + 45 o ) E6.3 The input signal v (t ) = 1 + 2 cos(2π ⋅ 1000t ) + 3 cos(2π ⋅ 3000t ) has three components with frequencies of 0, 1000 Hz and 3000 Hz. For the dc component, we have v out,1 (t ) = H (0) ×v in ,1 (t ) = 4 × 1 = 4 For the 1000-Hz component, we have: Vout,2 = H (1000)Vin,2 = 3∠30 o × 2∠0 o = 6∠30 o v out,1 (t ) = 6 cos(2π ⋅ 1000t + 30 o ) For the 3000-Hz component: Vout,3 = H (3000)Vin,3 = 0 × 3∠0 o = 0 v out,3 (t ) = 0 Thus, the output for all three components is v out (t ) = 4 + 6 cos(2π ⋅ 1000t + 30 o ) 221 E6.4 Using the voltage-division principle, we have: Vout = Vin × R R + j 2πfL Then the transfer function is: V R 1 1 H (f ) = out = = = Vin R + j 2πfL 1 + j 2πfL / R 1 + jf / fB E6.5 From Equation 6.9, we have fB = 1 /(2πRC ) = 200 Hz , and from Equation V 1 . 6.9, we have H (f ) = out = 1 + jf / fB Vin For the first component of the input, the frequency is 20 Hz, H (f ) = 0.995∠ − 5.71 o , Vin = 10∠0 o , and Vout = H (f )Vin = 9.95∠ − 5.71 o Thus the first component of the output is v out,1 (t ) = 9.95 cos(40πt − 5.71 o ) For the second component of the input, the frequency is 500 Hz, H (f ) = 0.371∠ − 68.2o , Vin = 5∠0 o , and Vout = H (f )Vin = 1.86∠ − 68.2o Thus the second component of the output is v out,2 (t ) = 1.86 cos(40πt − 68.2o ) For the third component of the input, the frequency is 10 kHz, H (f ) = 0.020∠ − 88.9 o , Vin = 5∠0 o , and Vout = H (f )Vin = 0.100∠ − 88.9 o Thus the third component of the output is v out,2 (t ) = 0.100 cos(2π × 10 4t − 88.9o ) Finally, the output with for all three components is: v out (t ) = 9.95 cos(40πt − 5.71 o ) + 1.86 cos(40πt − 68.2o ) + 0.100 cos(2π × 10 4t − 88.9 o ) E6.6 H (f ) dB = 20 log H (f ) = 20 log(50) = 33.98 dB E6.7 (a) H (f ) dB = 20 log H (f ) = 15 dB log H (f ) = 15/20 = 0.75 H (f ) = 10 0.75 = 5.623 222 (b) H (f ) dB = 20 log H (f ) = 30 dB log H (f ) = 30/20 = 1.5 H (f ) = 101.5 = 31.62 1000 × 22 = 4000 Hz is two octaves higher than 1000 Hz. 1000 / 23 = 125 Hz is three octaves lower than 1000 Hz. 1000 × 10 2 = 100 kHz is two decades higher than 1000 Hz. 1000 / 10 = 100 Hz is one decade lower than 1000 Hz. E6.8 (a) (b) (c) (d) E6.9 (a) To find the frequency halfway between two frequencies on a logarithmic scale, we take the logarithm of each frequency, average the logarithms, and then take the antilogarithm. Thus f = 10[log(100) +log(1000)] / 2 = 10 2.5 = 316.2 Hz is half way between 100 Hz and 1000 Hz on a logarithmic scale. (b) To find the frequency halfway between two frequencies on a linear scale, we simply average the two frequencies. Thus (100 + 1000)/2 = 550 Hz is halfway between 100 and 1000 Hz on a linear scale. E6.10 To determine the number of decades between two frequencies we take the difference between the common (base-ten) logarithms of the two frequencies. Thus 20 Hz and 15 kHz are log(15 × 10 3 ) − log(20) = 2.875 decades apart. Similarly, to determine the number of octaves between two frequencies we take the difference between the base-two logarithms of the two frequencies. One formula for the base-two logarithm of z is log(z ) log2 (z ) = ≅ 3.322 log(z ) log(2) Thus the number of octaves between 20 Hz and 15 kHz is log(15 × 10 3 ) log(20) − = 9.551 log(2) log(2) E6.11 The transfer function for the circuit shown in Figure 6.17 in the book is 1 /( j 2πfC ) V 1 1 H (f ) = out = = = Vin R + 1 /( j 2πfC ) 1 + j 2πRCf 1 + jf / fB 223 in which fB = 1 /(2πRC ) = 1000 Hz. Thus the magnitude plot is approximated by 0 dB below 1000 Hz and by a straight line sloping downward at 20 dB/decade above 1000 Hz. This is shown in Figure 6.18a in the book. The phase plot is approximated by 0 o below 100 Hz, by − 90 o above 10 kHz and by a line sloping downward between 0 o at 100 Hz and − 90 o at 10 kHz. This is shown in Figure 6.18b in the book. E6.12 Using the voltage division principle, the transfer function for the circuit shown in Figure 6.19 in the book is j (f / fB ) j 2πRC V R H (f ) = out = = = Vin R + 1 /( j 2πfC ) 1 + j 2πRCf 1 + j (f / fB ) in which fB = 1 /(2πRC ). E6.13 Using the voltage division principle, the transfer function for the circuit shown in Figure 6.22 in the book is j (f / fB ) j 2πfL j 2πfL / R V H (f ) = out = = = Vin R + j 2πfL 1 + j 2πfL / R 1 + j (f / fB ) in which fB = R /(2πL). E6.14 E6.15 A first-order filter has a transfer characteristic that decreases by 20 dB/decade below the break frequency. To attain an attenuation of 50 dB the signal frequency must be 50/20 = 2.5 decades below the break frequency. 2.5 decades corresponds to a frequency ratio of 10 2.5 = 316.2. Thus to attenuate a 1000 Hz signal by 50 dB the highpass filter must have a break frequency of 316.2 kHz. Solving Equation 6.22 for capacitance and substituting values, we have 1 1 = = 503.3 pF C = 2πfB R 2π × 1000 × 316.2 × 10 3 1 = 2533 pF ω L (2πf0 ) L (2π 10 ) 10 × 10 −6 R = ω 0 L / Qs = 1.257 Ω B = f0 / Qs = 20 kHz fL ≅ f0 − B / 2 = 990 kHz fH ≅ f0 + B / 2 = 1010 kHz C = 1 2 0 = 1 2 = 6 2 224 E6.16 At resonance we have VR = Vs = 1∠0 o VL = jω 0 LI = jω 0 LVs / R = jQs Vs = 50∠90 o V VC = (1 / jω 0C ) I = (1 / jω 0C )Vs / R = − jQs Vs = 50∠ − 90 o V E6.17 1 L= 1 = = 1 = 2.156 µH (2π × 5 × 10 ) 470 × 10 −12 ω C (2πf0 ) C Qs = f0 / B = (5 × 10 6 ) /(200 × 10 3 ) = 25 R = 2 0 2 1 = ω 0CQs 1 6 2 1 = 2.709 Ω 2π × 5 × 10 × 470 × 10 −12 × 25 6 R = 22.36 ω 0L E6.18 f0 = E6.19 Qp = f0 / B = 50 E6.20 A second order lowpass filter with f0 = 5 kHz is needed. The circuit configuration is shown in Figure 6.34a in the book. The normalized transfer function is shown in Figure 6.34c. Usually we would want a filter without peaking and would design for Q = 1. Given that L = 5 mH, the other component values are 2πf0 L 1 R = = 157.1 Ω C = = 0.2026 µF Q (2πf0 ) 2 L 2π LC = 711.8 kHz L= Qp = R = 0.3183 µH ω 0Qp C = B = f0 / Qp = 31.83 kHz Qp ω 0R = 795.8 pF The circuit is shown in Figure 6.39 in the book. E6.21 We need a bandpass filter with fL = 45 kHz and fH = 55 kHz. Thus we have f0 ≅ R = fL + fH 2 2πf0 L Q = 50 kHz = 62.83 Ω B = fH − fL = 10 kHz C = 1 (2πf0 ) 2 L = 10.13 nF The circuit is shown in Figure 6.40 in the book. 225 Q = f0 / B = 5 E6.22 The files Example_6_8 and Example_6_9 can be found in the MATLAB folder on the OrCAD disk. The results should be similar to Figures 6.42 and 6.44. E6.23 (a) Rearranging Equation 6.56, we have 0. 9 τ a = = = 0. 9 T 1 − a 1 − 0. 9 Thus we have τ = 9T . (b) From Figure 6.49 in the book we see that the step response of the digital filter reaches 0.632 at approximately n = 9. Thus the speed of response of the RC filter and the corresponding digital filter are comparable. E6.24 Writing a current equation at the node joining the resistance and capacitance, we have y (t ) d [ y (t ) − x (t )] +C =0 R dt Multiplying both sides by R and using the fact that the time constant is τ = RC, we have dy (t ) dx (t ) y (t ) + τ −τ =0 dt dt Next we approximate the derivatives as dy (t ) ∆y y (n ) − y (n − 1) dx (t ) ∆x x (n ) − x (n − 1) ≅ = and ≅ = ∆t ∆t dt T dt T which yields y (n ) − y (n − 1) x (n ) − x (n − 1) y (n ) + τ −τ =0 T T Solving for y(n), we obtain y (n ) = a1 y (n − 1) + b0x (n ) + b1x (n − 1) in which a1 = b0 = −b1 = E6.25 τ /T 1 + τ /T (a) Solving Equation 6.58 for d and substituting values, we obtain fs 104 = = 10 d = 2fnotch 2 × 500 (b) Repeating for fnotch = 300 Hz, we have 226 d = fs 10 4 = = 16.67 2fnotch 2 × 300 However, d is required to be an integer value so we cannot obtain a notch filter for 300 Hz exactly for this sampling frequency. (Possibly other more complex filters could provide the desired performance.) Problems P6.1 The fundamental concept of Fourier theory is that all signals are sums of sinewaves of various amplitudes, frequencies, and phases. P6.2 A MATLAB program to create the plot is: t = 0:2e-6:2e-3; vt = ones(size(t)); for n=1:2:19 vt = vt +(8/(n*pi)^2)*cos(2000*n*pi*t); end plot(t,vt) The resulting plot is: t (ms) P6.3 A MATLAB program to create the plot is: t = 0:2e-6:2e-3; vfw = (2/pi)*ones(size(t)); for n=2:2:60 vfw = vfw + (4/(pi*(n-1)*(n+1)))*(-1)^(1+ n/2)*cos(2000*n*pi*t); end 227 plot(t,vfw) The resulting plot is: P6.4 A MATLAB program to create the plots is: t = -0.5:1e-3:1.5; vhw = (1/pi)*ones(size(t)); vhw = vhw + 0.5*cos(2*pi*t); for n=2:2:4 vhw = vhw + (2/((n-1)*(n+1)*pi))*(-1)^(n/2+1)*cos(2*n*pi*t); end subplot(2,2,1) plot(t,vhw) axis([-.5 1.5 -0.5 1.5]) for n=6:2:50 vhw = vhw + (2/((n-1)*(n+1)*pi))*(-1)^(n/2+1)*cos(2*n*pi*t); end subplot(2,2,2) plot(t,vhw) axis([-.5 1.5 -0.5 1.5]) The resulting plots are 228 P6.5 A MATLAB program to create the plots is: t = 0:2e-6:2e-3; vst = ones(size(t)); for n=1:3 vst = vst - (2/(n*pi))*sin(2000*n*pi*t); end subplot(2,2,1) plot(t,vst) axis([0 2e-3 -0.5 2.5]) for n=4:50 vst = vst - (2/(n*pi))*sin(2000*n*pi*t); end subplot(2,2,2) plot(t,vst) axis([0 2e-3 -0.5 2.5]) The resulting plots are: 229 P6.6 The transfer function shows how a filter affects the amplitude and phase of input components as a function of frequency. It is defined as the output phasor divided by the input phasor as a function of frequency. To determine the complex value of the transfer function for a given frequency, we apply a sinusoidal input of that frequency, wait for the output to achieve steady state, and then measure the amplitude and phase of both the input and the output using voltmeters, oscilloscopes or other instruments. Then, the value of the transfer function is computed as the ratio of the output phasor divided by the input phasor. We change the frequency and repeat to determine the transfer function for other frequencies. P6.7 Filters (and other systems described by linear time-invariant differential equations) separate the input into sinusoidal components with various frequencies, modify the amplitudes and phases of the components, and then add the modified components to create the output. P6.8* The given input signal is v in (t ) = 5 + 2 cos(2π2500t + 30 o ) + 2 cos(2π7500t ) This signal has a component v in 1 (t ) = 5 with f = 0, a second componentv in 2 (t ) = 2 cos(2π2500t + 30 o ) with f = 2500 , and a third component v in 3 (t ) = 2 cos(2π7500t ) with f = 7500 . From Figure P6.8, we find the transfer function values for these frequencies: H (0) = 2, H (2500 ) = 1.75∠ − 45o , and H (7500 ) = 1.25∠ − 135o The dc output is v out 1 = H (0)v in 1 = 10 The phasors for the sinusoidal input components are Vin 2 = 2∠30 o and Vin 3 = 2∠0 o Multiplying the input phasors by the transfer function values , results in: Vout 2 = Vin 2 × H (2500 ) Vout 3 = Vin 3 × H (7500 ) = 3.5∠ − 15o = 2.5∠ − 135o The corresponding output components are: v out 2 (t ) = 3.5 cos(2π 2500t − 15o ) v out 3 (t ) = 2.5 cos(2π 7500t − 135o ) Thus, the output signal is v out (t ) = 10 + 3.5 cos(2π 2500t − 15o ) + 2.5 cos(2π 7500t − 135o ) 230 P6.9 The solution is similar to that for Problem 6.8. The answer is: v out (t ) = 8 + 7.5 cos(2π 5000t − 120 o ) V P6.10 The solution is similar to that for Problem 6.8. The answer is: v out (t ) = 12 + 3.4 cos(6000πt − 54 o ) + 5.6 cos(12000πt − 108o ) V P6.11* The phasors for the input and output are: Vin = 2∠ − 25 o and Vout = 1∠20 o The transfer function for f = 5000 Hz is H (5000 ) = Vout Vin = 0.5∠45 o P6.12* The input has a peak value of 5 V and reaches a positive peak at 1 ms. Since the period is 4 ms, the frequency is 250 Hz and 1 ms corresponds to a phase angle of θin = −(td /T ) × 360° = −(1 / 4) × 360° = −90° . Thus, the phasor for the input is Vin = 5∠ − 90° . Similarly, the phasor for the output is Vout = 15∠ − 135° . Then, the transfer function for a frequency of 250 Hz is H (250) = P6.13* Vout = 3∠ − 45° Vin The triangular waveform is given in Problem P6.2 as 8 8 cos(6000πt ) + L v t (t ) = 1 + 2 cos(2000πt ) + π (3π ) 2 8 + cos(2000nπt ) + L (nπ ) 2 in which n assumes only odd integer values. This waveform has components with frequencies of 0 (dc), 1000 Hz, 3000 Hz, etc. The transfer function shown in Figure P6.13 is zero for all components except the dc term for which we have H (0) = 2. Thus, the dc term is multiplied by 2 and all of the other terms are rejected. Thus, v o (t ) = 2 . P6.14* Given v in (t ) = Vmax cos(2πft ) t v out (t ) = ∫Vmax cos(2πft )dt = 0 Vmax sin(2πft ) 2 πf The phasors are 231 Vin = Vmax ∠0 o Vout = Vmax ∠ − 90 o 2πf The transfer function is −j V 1 H (f ) = out = ∠ − 90 o = Vin 2πf 2πf Plots of the magnitude and phase of this transfer function are: P6.15 The input signal given in Problem P6.5 is: v st (t ) = 1 − 2 π sin(2000πt ) − −L− 2 nπ 2 2 sin(4000πt ) − sin(6000πt ) 2π 3π sin(2000nπt ) − L The frequencies of the various components are f = 0, 1000, 2000, 3000, and so forth. The transfer function H (f ) is zero for all components except for the one with a frequency of 3000 Hz. For the 3000-Hz component, we have: Vout = Vin × H (3000 ) = [(2 3π )∠90 o ] × 5∠0 o = (10 3π )∠90 o Thus, the output is: v out (t ) = (10 3π ) cos(6000πt + 90 o ) = −(10 3π ) sin(6000πt ) 232 P6.16 From Figure P6.16, we see that the period of the signals is 10 ms. Therefore, the frequency is 100 Hz. Because the input reaches a positive peak at t = 2 ms, it has a phase angle of θ in = −(td /T ) × 360° = −(2 / 10) × 360° = −72° The output reaches its peak at 4 ms, and its phase angle is θ out = −(td /T ) × 360° = −(4 / 10) × 360° = −144° The transfer function is V 2∠ − 144° H (100) = out = = 2∠ − 72° Vin 1∠ − 72° P6.17 The input signal is v in (t ) = 2 + 3 cos(1000πt ) + 3 sin(2000πt ) + cos(3000πt ) which has components with frequencies of 0, 500, 1000, and 1500 Hz. We can determine the transfer function at these frequencies by dividing the corresponding output by the input. Thus, we have 2∠30 o H (0) = 3 / 2 = 1.5 H (500) = = 0.6667 ∠30 o o 3∠0 0 3∠0 o H (1000) = = 0 H ( 1500 ) = =3 1∠0 o 3∠ − 90 o P6.18 v in (t ) =Vmax cos(2πft ) v out (t ) = Vmax cos(2πft ) +Vmax cos[2πf (t − 10 −3 )] v out (t ) = Vmax cos(2πft ) +Vmax cos(2πft − 2πf 10 −3 ) The phasors are Vin = Vmax ∠0 o Vout =Vmax [1 + 1∠ − 2πf / 1000] The transfer function is V H (f ) = out = 1 + 1∠ − 2πf / 1000 = 1 + exp( − j 2πf / 1000) Vin in which the angle is expressed in radians. A MATLAB program to plot the magnitude of this transfer function is f = 0:1:2000; Hmag = abs(1 + exp((-i*2*pi/1000)*f)); plot(f,Hmag) A plot of the magnitude of the transfer function is: 233 P6.19 v in (t ) =Vmax cos(2πft ) t v out (t ) = 1000 ∫ V t −1e −3 max cos(2πft )dt 1000Vmax [sin(2πft ) − sin(2πft − 2πf 10 −3 )] 2πf The phasors are v out (t ) = Vin = Vmax ∠0 o Vout = 1000Vmax [1∠ − 90 o − 1∠( −90 o − 2πf / 1000)] 2πf The transfer function is H (f ) = − j 1000[1 − 1∠( −2πf / 1000)] 1000[1 − exp( − j 2πf / 1000)] Vout = = Vin 2πf j 2πf in which the angle is expressed in radians. A MATLAB program to plot the magnitude of this transfer function is f = 0:1:2000; Hmag = abs((1000)*((1 - exp(-j*2*pi*f./1000))./(j*2*pi*f))); plot(f,Hmag) The resulting plot of the magnitude of the transfer function is: 234 P6.20 Given v in (t ) = Vmax cos(2πft ) v out (t ) = d [v in (t )] = −2πfVmax sin(2πft ) dt The phasors are Vin =Vmax ∠0 o Vout = 2πfVmax ∠90 o The transfer function is H (f ) = Vout = 2πf∠90 o = j 2πf Vin Plots of the magnitude and phase of this transfer function are: 235 P6.21 The circuit diagram is: The half-power frequency is: 1 fB = 2πRC Plots of the magnitude and phase of the transfer function are shown in Figure 6.8 in the text. P6.22 The circuit diagram of a first-order RL lowpass filter is: The half-power frequency is: fB = R 2πL Plots of the magnitude and phase of the transfer function are identical to those shown in Figure 6.8 in the text. P6.23* The phase of the transfer function is given by Equation 6.11 in the text: ∠H (f ) = − arctan(f fB ) Thus, f = −fB tan[∠H (f )]. 236 For ∠H (f ) = −1 o , we have f = 0.01746fB . For ∠H (f ) = −10 o , we have f = 0.1763fB . For ∠H (f ) = −89 o , we have f = 57.29fB . P6.24 The time constant is given by τ = RC and the half-power frequency is 1 1 . Thus, we have fB = fB = . 2πRC 2πτ P6.25* The half-power frequency of the filter is 1 fB = = 500 Hz 2πRC The transfer function is given by Equation 6.9 in the text: 1 H (f ) = 1 + j (f fB ) The given input signal is v in (t ) = 5 cos(500 πt ) + 5 cos(1000πt ) + 5 cos(2000 πt ) which has components with frequencies of 250, 500, and 1000 Hz. Evaluating the transfer function for these frequencies yields: 1 H (250 ) = = 0.8944∠ − 26.57 o 1 + j (250 500 ) H (500 ) = 0.7071∠ − 45 o H (1000 ) = 0.4472∠ − 63.43o Applying the appropriate value of the transfer function to each component of the input signal yields the output: v out (t ) = 4.472 cos(500 πt − 26.57 o ) + 3.535 cos(1000 πt − 45 o ) + 2.236 cos(2000 πt − 63.43o ) P6.26 The transfer function is given by Equation 6.9 in the text: 1 H (f ) = 1 + j (f fB ) The given input signal is v in (t ) = 3 + 2 sin(800πt + 30 o ) + 5 cos(20 × 103 πt ) which has components with frequencies of 0, 400, and 10,000 Hz. Evaluating the transfer function for these frequencies yields: 237 H (0 ) = 1 =1 1 + j (0 200 ) H (400 ) = 0.4472∠ − 63.43o H (20 × 103 ) = 0.0100∠ − 89.43o Applying the appropriate value of the transfer function to each component of the input signal yields the output: v out (t ) = 3 + 0.8944 cos(1000πt − 123.43o ) + 0.0500 cos(20 × 103 πt − 89.42o ) P6.27 Rearranging Equation 6.8 in the text yields: C = P6.28 1 2πfB R = 1 2π (10 )5000 3 = 31.83 nF To achieve a reduction of the 50-kHz component by a factor of 200, we must have 1 1 = H (50 × 103 ) = 2 200 1 + ((50 × 103 ) fB ) Solving, we find fB = 250.0 Hz At 2 kHz, the resulting value of the transfer function is H (f ) = 1 1 + j (f fB ) = 1 = 0.124∠ − 82.88o 1 + j (2000 250 ) Thus, the 2-kHz component is changed in amplitude by a factor of 0.124. P6.29 The period of a 5-kHz sinusoid is 200 µs. The phase shift corresponding to a delay of 30 µs is θ = −54 o . (We give the phase shift as negative because the output lags the input.) Equation 6.11 gives the phase shift of the first-order filter. Thus, we can write:  5000   = −54 o ∠H (f ) = − arctan  fB  fB = 3633 Hz 238 P6.30* The circuit seen by the capacitance is: The open-circuit or Thévenin voltage is Vt = V1 − V2 2000 1000 − Vin 2000 + 1000 1000 + 2000 Thus, we obtain: 1 Vt = Vin 3 Zeroing the source, we have = Vin (1) The Thévenin resistance is 1 1 + Rt = 1 1000 + 1 2000 1 2000 + 1 1000 = 1333 Ω Thus, the equivalent circuit is: As in the text, this circuit has the transfer function: Vout 1 = Vt 1 + j (f fB ) 1 1 where fB = = = 11.94 Hz 2πRt C 2π1333 × 10 −5 239 (2) Using Equation (1) to substitute for Vt in Equation (2) and rearranging, we have 13 Vout = Vin 1 + j (f fB ) A sketch of the transfer-function magnitude is: P6.31 The input and output voltages given have a frequency of 10 kHz. Dividing the output phasor by the input phasor, we obtain the transfer function: V 0.2∠ − θ H (10 4 ) = out = = 0.040∠ − θ Vin 5 Using Equation 6.10 from the book, we have 1 = 0.040 H (10 4 ) = 1 + (10 4 / fB ) 2 Solving, we obtain fB = 400.3 Hz. Then using Equation 6.11, we have  10 4   = −θ ∠H (f ) = − arctan f B   o θ = 87.70 P6.32 (a) First, we find the Thévenin equivalent for the source and resistances. The open-circuit voltage is given by vt (t ) = v oc (t ) = v s (t ) RL Rs + R + RL 240 In terms of phasors, this becomes: RL Rs + R + RL Vt = Vs (1) Zeroing the source, we find the Thévenin resistance: Rt = 1 1 RL + 1 (R + Rs ) Thus, the original circuit has the equivalent: The transfer function for this circuit is: Vout 1 = Vt 1 + j (f fB ) 1 where, fB = 2πRt C (2) Using Equation (1) to substitute for Vt in Equation (2) and rearranging, we have: H (f ) = (b) RL Vout 1 = × Vs Rs + R + RL 1 + j (f fB ) Evaluating for the circuit components given, we have: Rt = 980 Ω fB = 812.0 Hz H (f ) = 0.02 1 + j (f fB ) A MATLAB program to plot the transfer-function magnitude is: foverfb=0:0.01:3; Hmag=abs(0.02./(1 + i*foverfb)); plot(foverfb,Hmag) axis([0 3 0 0.02]) 241 (3) The resulting plot is: P6.33 (a) Applying the voltage-division principle, we have: H (f ) = R2 (R1 + R2 ) R (R + R2 ) R2 Vout = = = 2 1 Vin R1 + R2 + j 2πfL 1 + j 2πfL (R1 + R2 ) 1 + j (f fB ) where fB = (R1 + R2 ) (2πL ) (b) Evaluating for the component values given, we have: fB = 1.061 MHz H (f ) = 0. 5 1 + j (f fB ) A sketch of the transfer function magnitude is: 242 P6.34 For the 20-kHz signal, the transfer function magnitude is H (f ) = Vout Vin = 0. 5 × 2 = 0.1 = 5× 2 1 1 + (f / fb ) 2 = 1 1 + (20 × 10 3 / fb )2 Solving, we find fb = 2.01 kHz. Then for the 150-kHz signal, we have V × 2 V 1 = H (f ) = out = orms 3 Vin 5× 2 1 + [150 × 10 /(2.01 × 103 ]2 which yields Vorms = 67.00 mV. P6.35 With an assumed velocity v = Vm cos(2πft ) , the force is f =m dv + kv = −m 2πfVm sin(2πft ) + kVm cos(2πft ) dt The phasors for the velocity and the force are V = Vm ∠0° and F = jm 2 π fV m + kV m Then the transfer function is H (f ) = where fB = Vm V 1/k 1/k = = = F jm 2πfVm + kVm 1 + jm 2πf / k 1 + jf / fB k is the half-power frequency. 2πm P6.36 A logarithmic frequency scale is one for which equal distances correspond to multiplying frequency by the same factor. A linear frequency scale is one for which equal distances correspond to adding the same amount to the starting frequency. P6.37 A notch filter rejects components with frequencies in a narrow band while passing components with frequencies higher or lower than those in the rejection band. One application is to eliminate 60-Hz power line interference from audio signals. P6.38 The primary advantage of converting transfer-function magnitudes to decibels before plotting is that both very large and very small values can be read from the plot with relatively small percentage error. 243 P6.39 The passband of a filter is the range of frequencies of the components that are passed to the output with relatively little change in amplitude. P6.40* (a) We have: 20 log H (f ) = −10 log H (f ) = −0.5 H (f ) = 10 −0.5 = 0.3162 (b) Similarly, 20 log H (f ) = 10 log H (f ) = 0.5 H (f ) = 10 0.5 = 3.162 P6.41* (a) (b) P6.42 log(100 ) + log(3000 ) 2 or equivalently: fa = 100 × 3000 = 547.7 Hz 100 + 3000 fb = = 1550 Hz 2 log(fa ) = To convert to decibels, we take 20 times the common logarithm of the transfer function. Thus, we have: 20log(0.5) = -6.021 dB 20log(2) = +6.021 dB 20 log(1 / 2 ) = −3.010 dB 20 log( 2 ) = 3.010 dB P6.43 (a) (b) (c) (d) P6.44 If the output terminals of one filter are connected to the input terminals of a second filter, we say that the filters are cascaded. P6.45 On a linear scale, the frequencies are 2, 14, 26, 38, 50 Hz. (We add (50 − 2)/4 = 12 Hz to each value to obtain the next value.) On a logarithmic scale, the frequencies are 2, 4.472, 10, 22.361, 50 Hz. (We multiply each value by 4 50 / 2 = 2.236 to obtain the next value.) 1600 Hz is one octave higher than 800 Hz. 200 Hz is two octaves lower than 800 Hz. 8 Hz is two decades lower than 800 Hz. 8 kHz is one decade higher than 800 Hz. 244 P6.46* (a) (b) The overall transfer function is the product of the transfer functions of the filters in cascade: 1 H (f ) = H 1 (f ) × H 2 (f ) = [1 + j (f fB )]2 H (f ) = H (f3dB ) = (f3dB 1 1 + (f fB ) 1 1 = 2 2 1 + (f3dB fB ) fB ) = 2 − 1 2 f3dB = fB P6.47 (a) 2 2 − 1 = 0.6436fB We have 10 Nd × 20 = 4500 . Taking the common logarithm of both sides, we have: Nd + 1.301 = 3.653 Nd = 2.352 decades (b) Similarly, in octaves, 2Noct × 20 = 4500 . Taking the common logarithm of both sides, we have: Noct log(2) + 1.301 = 3.653 Noct = 7.813 octaves P6.48 We have H (f ) = H 1 (f )H 2 (f ) and H (f ) dB = H 1 (f ) dB + H 2 (f ) dB . For these formulas to be valid, H 1 (f ) must be the transfer function of the first filter with the second attached. P6.49 For filters in cascade, the transfer functions in decibels are added. H (f1 ) dB = H 1 (f1 ) dB + H 2 (f1 ) dB = −20 dB P6.50 A Bode plot is a plot of the magnitude of a transfer function in decibels versus frequency using a logarithmic scale for frequency. P6.51 The slope of the high-frequency asymptote is -20 dB/decade. The slope of the low-frequency asymptote is zero. The asymptotes meet at the half-power frequency fB . 245 P6.52* H (f ) = 100 100 = 2 1 + j (f 1000 ) 1 + (f 1000 ) H (f ) dB = 20 log(100 ) − 20 log 1 + (f 1000 ) 2 = 40 − 20 log 1 + (f 1000 ) 2 This is similar to the transfer function treated in Section 6.4 in the text except for the additional 40 dB constant. The half-power frequency is fB = 1000 . The asymptotic Bode plots are: P6.53 Because the transfer functions in decibels add for cascaded systems, the slope becomes 60 dB/decade. P6.54 This is a first-order lowpass RC filter. The break frequency is: 1 fB = = 5.895 MHz 2πRC The Bode plots are like Figures 6.15 and 6.16 in the text. P6.55 H (f ) = H (f ) = 10 1 − j (f 500 ) 10 1 + (f 500 ) 2 H (f ) dB = 20 log(10 ) − 20 log 1 + (f 500 ) 2 = 20 − 20 log 1 + (f 500 ) 2 246 This is similar to the transfer function magnitude treated in Section 6.4 in the text except for the additional 20 dB constant. The half-power frequency is fB = 500 . The phase is + arctan(f fB ) rather than − arctan(f fB ) . The asymptotic Bode Plots are: P6.56 (a) Solving for the input voltage, we have t v in (t ) = v out (t ) + 200π∫v out (t )dt 0 t = A cos(2πft ) + 200π∫ A cos(2πft )dt 0 t 200π A sin(2πft ) 0 2πf 100 = A cos(2πft ) + A sin(2πft ) = A cos(2πft ) + f (b) Then the transfer function for the system is H (f ) = Vout 1 A = = Vin A − jA(100 / f ) 1 − j (100 / f ) (c) For f >> 100, the asymptote is constant at 0 dB. For f 0), the equivalent circuit is a short circuit. Thus in the equivalent circuit, the voltage source is zero and the resistance is zero. In the reverse bias region (0 > vD > -10 V) the equivalent circuit is an open circuit. Thus in the equivalent circuit, the voltage is indeterminate and the resistance is infinite. In the reverse breakdown region (0 > iD), the equivalent circuit consists of a 10-V voltage source in series with zero resistance. 10.46* The equivalent circuits for each segment are shown below: 384 For the circuit of Figure P10.46a, we can determine by trial and err (or by a load-line analysis) that the device operates on the middle line segment. Thus, the equivalent circuit is: 15 − 0.756 = 4.68 mA 3000 + 44.4 v = 0.756 + 44.4i = 0.964 V i = For the circuit of Figure P10.46b, we can determine by trail and err that the device operates on the upper right-hand line segment. Thus, the equivalent circuit is: 2 + 6.8 = 10.4 mA 50 + 800 v = 800i − 6.8 = 1.48 V i = P10.47* For small values of iL, the Zener diode is operating on line segment C of Figure 10.19, and the equivalent circuit is 385 Writing a KCL equation at node A, we obtain: v L − 13 v L − 6 + + iL = 0 100 12 Solving we obtain v L = 6.75 − 10.71iL This equation is valid for v L ≥ 6 V. When 0 ≤ v L ≤ 6 V, the Zener diode operates on line segment B, for which the Zener is modeled as an open circuit and we have v L = 13 − 100iL Plotting these equations results in P10.48 (a) Assuming that the diode is an open-circuit, we can compute the node voltages using the voltage-division principle. 100 200 v1 = 4 = 2V v2 = 4 = 1V 200 + 200 300 + 100 Then, the voltage across the diode is v D = v 1 − v 2 = 1 V. Because vD is greater than Vf = 0.7 V, the diode is not operating as an open circuit. (b) Assuming that the diode operates as a voltage source, we can use KVL to write: v 1 − v 2 = 0.7 Placing a closed surface around the diode to form a super node and writing a KCL equation gives v1 − 4 v1 v − 4 v2 + + 2 + =0 200 200 300 100 Solving these equations, we find v 1 = 1.829 V and v 2 = 1.129 V. Then, writing a KCL equation at node 1 gives the diode current. 4 − v1 v iD = − 1 = 1.714 mA 200 200 386 Because the diode current is positive, the diode operation is consistent with the model. P10.49 Half-wave rectifier with a capacitance to smooth the output voltage: Full-wave circuits: P10.50 The peak value of the ac source is Vm = 20 2 = 28.28 V. Thus the PIV is 28.28 V and the peak current is 282.8 mA. P10.51 The diode is on for VB ≤ Vm sin(ωt ) . Substituting values and solving, we find that during the first cycle after t = 0 the diode is on for arcsin(12 / 24) = π / 6 ≤ ωt ≤ π − arcsin(12 / 24) = 5π / 6 The current is given by 24 sin( ωt ) − 12 = 12 sin( ωt ) − 6 A i (t ) = 2 The charge passing through the circuit during the first cycle is 5 π / 6ω 5 π / 6ω 12 3 − 4π  12  = Q1 = ∫ [12 sin(ωt ) − 6]dt = − cos(ωt ) − 6t  ω  ω  π / 6ω π / 6ω The average current is the charge passing through the circuit in 1 second (or 60 cycles). Also, we have ω = 120 π Thus 12 3 − 4π 12 3 − 4π 6 3 I avg = 60 = 60 = − 2 = 1.308 A 120π ω π Then the time required to fully charge the battery is 100 T = = 76.45 hours 1.308 387 P10.52 The dc output voltage is equal to the peak value of the ac source, which is v L = 20 2 = 28.28 V. The load current is i L = v L / RL = 282.8 mA. The charge that passes through the load must also pass through the diode. The charge is Q = i LT = 0.2828 / 60 = 4.714 mC. The peak inverse voltage is 56.57 V. P10.53 (a) The integral of Vm sin( ωt ) over one cycle is zero, so the dc voltmeter reads zero. (b) Vavg t =T / 2 T /2 T  1  Vm V 1   = m =  ∫Vm sin(ωt )dt + ∫ 0dt  = − cos(ωt ) T  0 π t =0 T /2  T  ω (c) Vavg = T /2 T  2V 1  ω + −Vm sin(ωt )dt  = m V sin( t ) dt ∫ m ∫ π T 0 T /2  P10.54* The output voltage waveform is: The peak voltage is approximately 10 V. Assuming an ideal diode, the ac source must have a peak voltage of 10 V. The circuit is: The capacitance required is given by Equation 10.10 in the text. I T 0.1 (1 60 ) = 833 µF C = L = Vr 2 P10.55 As in Problem P10.54, the peak voltage must be 10 V. For a full-wave rectifier, the capacitance is given by Equation 10.12 in the text: 388 C = I LT 0.1 (1 60 ) = = 417 µF 2Vr 2 (2 ) The circuit diagram is: P10.56 As in Problem P10.54, the peak voltage must be 10 V. For a full-wave rectifier, the capacitance is given by Equation 10.12 in the text: C = I LT 0.1 (1 60 ) = = 417 µF 2Vr 2 (2 ) The circuit diagram is: P10.57 If we allow for a forward diode drop of 0.8 V, the peak ac voltage must be 10.8 V. Otherwise, the circuit is the same as in the solution to Problem P10.54. P10.58* For a half-wave rectifier, the capacitance required is given by Equation 10.10 in the text. I T 0.25 (1 60 ) = 20833 µF C = L = Vr 0.2 389 For a full-wave rectifier, the capacitance is given by Equation 10.12 in the text: I T 0.25 (1 60 ) = 10416 µF C = L = 2Vr 2 ( 0.2 ) P10.59 (a) The current pulse starts and ends at the times for which v s (t ) = VB 20 sin(200πt ) = 12 Solving we find that sin −1 (0.6) T tstart = = 1.024 ms and tend = − tstart = 3.976 ms 2 200π Between these two times the current is 20 sin(200πt ) − 12 i (t ) = 80 A sketch of the current to scale versus time is (b) The charge flowing through the battery in one period is Q= tend ∫ t i (t )dt = start tend ∫ t start 20 sin(200πt ) − 12 dt 80 t 1 12t  end  cos(200πt ) − = − 80 tstart  800π Q = 194 µ C Finally, the average current is the charge divided by the period. Q 194 × 10 −6 = = 19.4 mA I avg = T 10 × 10 −3 P10.60 (a) With ideal diodes and a large smoothing capacitance, the load voltage equals the peak source voltage which is Vm = 12 V. Then the PIV is 2Vm = 24 V. 390 (b) Here again with ideal diodes and a large smoothing capacitance, the load voltage equals the peak source voltage which is Vm = 12 V. However the PIV is only Vm = 12 V. P10.61 (a) The circuit operates as three full-wave rectifiers with a common load and shared diodes. Thus, the load voltage at any instant is equal to the largest source-voltage magnitude. The plot of the load voltage is T is the period of the sinusoidal sources. (b) The minimum voltage occurs at t =T / 12 and is given by Vmin = Vm cos ωT / 12 = Vm cos(π / 6) = 0.866Vm . Thus the peak-to-peak ripple is 0.134Vm. The average load voltage is given by 1 T Vavg = ∫ v L (t )dt T 0 However, since v L (t ) has 12 intervals with the same area, we can write: T / 12 1 12 T / 12 ( ) Vavg = v t dt = Vm cos(ωt )dt L T / 12 ∫0 T ∫0 6 sin(π / 6) Vavg =Vm ≅ 0.955Vm π (c) To produce an average charging current of 30 A, we require Vavg = 12 + 0.1 × 30 = 15 V (d) In practice, we would need to allow for forward drops of the diodes, drops across the slip rings, and resistances of the stator windings and wiring. 391 P10.62 A clipper circuit removes or clips part of the input waveform. An example circuit with waveforms is: We have assumed a forward drop of 0.6V for the diode. P10.63 Refer to Figure P10.63 in the book. When the source voltage is negative, diode D3 is on and the output vo(t) is zero. For source voltages between 0 and 10 V, none of the diodes conducts and vo(t) = vs(t). Finally when the source voltage exceeds 10 V, D1 is on and D2 is in the breakdown region so the output voltage is 10 V. The waveform is: 392 P10.64 P10.65 P10.66 393 P10.67 P10.68 A clamp circuit adds or subtracts a dc component to the input waveform such that either the positive peak or the negative peak is forced to assume a predetermined value. An example circuit that clamps the positive peak to +5 V is shown below: We have allowed a forward drop of 0.6 V for the diode. P10.69 This is a clamp circuit that clamps the positive peaks to zero. 394 P10.70* Refer to the circuit shown in Figure P10.70 in the book. If the output voltage attempts to become less than -5 V, the Zener diode breaks down and current flows, charging the capacitance. Thus the negative peak is clamped to -5 V. The input and output waveforms are: P10.71 The capacitor C1 and diode D1 act as a clamp circuit that clamps the negative peak of v A (t ) to zero. Thus, the waveform at point A is: 395 Diode D2 and capacitor C2 act as a half-wave peak rectifier. Thus, the voltage across RL is the peak value of v A (t ) . Thus, v L (t ) ≅ 2Vm . This is called a voltage-doubler circuit because the load voltage is twice the peak value of the ac input. The peak inverse voltage is 2Vm for both diodes. P10.72* A suitable circuit is: P10.73 A suitable circuit is: 396 P10.74 (a) A suitable circuit is: We choose the resistors R1 and R2 to achieve the desired slope. R2 1 Slope = = 3 R1 + R2 Thus, choose R1 = 2R2 . For example, R1 = 2 kΩ and R2 = 1 kΩ. (b) A suitable circuit is: Other resistor values will work, but we must make sure that D2 remains forward biased for all values of vin , including vin = −10 V . To achieve the desired slope (i.e., the slope is 0.5) for the transfer characteristic, we must have R1 = R2 . 397 P10.75* A suitable circuit is: We must choose the time constant RC >> T, where T is the period of the input waveform. P10.76 A suitable circuit is: We must choose R1 to ensure that the 5.6-V Zener is in the breakdown region at all times and choose the time constant R2C >>T , where T is the period of the input waveform. P10.77 IDQ represents the dc component of the diode current with no signal applied to the circuit, and id(t) represents the changes from the Q-point current when the signal is applied. Furthermore iD(t) is the total diode current. Thus we have iD (t ) = IDQ + id (t ) = 4 + 0.5 cos(200πt ) 398 P10.78 The small signal equivalent circuit of a diode is a resistance known as the dynamic resistance. The dynamic resistance is the reciprocal of the slope of the iD versus vD characteristic at the operating point. P10.79 Dc sources voltage sources are replaced by short circuits in a small-signal equivalent circuit. By definition the voltage across a dc voltage source is constant. Thus, even if there is ac current flowing through the dc source the ac voltage across it is zero as is the case for a short circuit. P10.80 We should replace dc current sources by open circuits in a small-signal equivalent circuit. The current through a dc current source is constant. Thus, the ac current must be zero even if we apply an ac voltage. Zero current for a non-zero applied voltage implies that we have an open circuit. P10.81* A plot of the device characteristic is: Clearly this device is not a diode because it conducts current in both directions. The dynamic resistance is given by: −1  di  8 rD =  D  = 2 3vD  dvD  A plot of the dynamic resistance versus vD is: 399 P10.82 We are given iD = −10 −6 (1 + v D 5 ) 3 for − 5 V < vD < 0 A plot of this is: The dynamic resistance is: −1 4  di  v rD =  D  = 1.67 × 106 ×  1 + D  5   dvD  To find the dynamic resistance at a given Q-point, we evaluate this expression for vD = VDQ . For IDQ = −1.0 mA, we have VDQ = −4.5 V and rD = 167 Ω. For IDQ = −10.0 mA, we have VDQ = −4.77 V and rD = 7.48 Ω. 400 P10.83 We are given vD(t) = 5 + 0 .01cos(ωt) V and iD(t) = 3 + 0.2cos(ωt) mA. The dynamic resistance is the ratio of the ac voltage amplitude to the ac current amplitude. v 0.01 rD = d = = 50 Ω 0.2 × 10 −3 id The Q-point results if we set the ac signals to zero. Thus, we have VDQ = 5 V and I DQ = 3 mA P10.84 Dynamic resistance is given by −1  di  dv rD =  D  = D diD  dvD  Because voltage is constant for changes in current, the dynamic resistance is zero for an ideal Zener diode in the breakdown region. P10.85* To find the Q-point, we ignore the ac ripple voltage and the circuit becomes: Thus, we have: 8−5 = 150 mA I sQ = 20 I LQ = 5 100 = 50 mA IDQ = I sQ − I LQ = 100 mA The small-signal or ac equivalent circuit is: where rD is the dynamic resistance of the Zener diode. Using the voltage-division principle, the ripple voltage across the load is v Lac = Vripple × Rp R + Rp 401 where Rp = 1 is the parallel combination of the load resistance 1 RL + 1 rD and the dynamic resistance of the diode. Substituting values, we find VLac = 10 × 10 −3 = 1 × Rp 20 + Rp Solving, we find Rp = 0.202 Ω . Then, we have: Rp = 0.202 = 1 which yields rD = 0.202 Ω . 1 100 + 1 rD Practice Test T10.1 (a) First, we redraw the circuit, grouping the linear elements to the left of the diode. Then, we determine the Thévenin equivalent for the circuit looking back from the diode terminals. Next, we write the KVL equation for the network, which yields VT = RT iD + v D . Substituting the values for the Thévenin voltage and resistance, we have the load-line equation, 3 = 200iD + v D . For iD = 0 , we 402 have v D = 3 V which are the coordinates for Point A on the load line, as shown below. For v D = 0 , the load-line equation gives iD = 15 mA which are the coordinates for Point B on the load line. Using these two points to plot the load line on Figure 10.8, we have The intersection of the load line and the diode characteristic gives the current at the operating point as iD ≅ 9.6 mA. (b) First, we write the KCL equation at the top node of the network, which yields iD + v D / 25 = 40 mA. For iD = 0 , we have v D = 1 V which are the coordinates for Point C on the load line shown above. For v D = 0 , the load-line equation gives iD = 40 mA which plots off the vertical scale. Therefore, we substitute iD = 20 mA , and the KCL equation then yields v D = 0.5 V . These values are shown as point D. Using Points C and D we plot the load line on Figure 10.8 as shown above. The intersection of the load line and the diode characteristic gives the current at the operating point as iD ≅ 4.2 mA. T10.2 If we assume that the diode is off (i.e., an open circuit), the circuit becomes 403 Writing a KCL equation with resistances in kΩ, currents in mA, and v − 12 v x − ( −16) + = 0 . Solving, we find that voltages in V, we have x 1 2 v x = 2.667 V. However, the voltage across the diode is v D = v x , which must be negative for the diode to be off. Therefore, the diode must be on. With the diode assumed to be on (i.e. a short circuit) the circuit becomes Writing a KCL equation with resistances in kΩ, currents in mA and v − 12 v x − ( −16) v x + + = 0 . Solving, we find that voltages in V, we have x 1 2 4 404 v x = 2.286 V. Then, the current through the diode is v iD = i x = x = 0.571 mA. Of course, a positive value for iD is consistent 4 with the assumption that the diode is on. T10.3 We know that the line passes through the points (5 V, 2 mA) and (10 V, 7 mA). The slope of the line is -1/R = -∆i/∆v = (-5 mA)/(5 V), and we have R = 1 kΩ. Furthermore, the intercept on the voltage axis is at v = 3 V. Thus, the equivalent circuit for the device is T10.4 The circuit diagram is: Your diagram may be correct even if it is laid out differently. Check to see that you have four diodes and that current flows from the source through a diode in the forward direction then through the load and finally through a second diode in the forward direction back to the opposite end of the source. On the opposite half cycle, the path should be through the other two diodes and through the load in the same direction as before. Notice in the diagram that current flows downward through the load on both half cycles. 405 T10.5 An acceptable circuit diagram is: Your diagram may be somewhat different in appearance. For example, the 4-V source and diode B can be interchanged as long as the source polarity and direction of the diode don't change; similarly for the 5-V source and diode A. The parallel branches can be interchanged in position. The problem does not give enough information to properly select the value of the resistance, however, any value from about 1 kΩ to 1 MΩ is acceptable. T10.6 An acceptable circuit diagram is: The time constant RC should be much longer than the period of the source voltage. Thus, we should select component values so that RC >> 0.1 s. T10.7 We have kT 1.38 × 10 −23 × 300 VT = = = 25.88 mV q 1.60 × 10 −19 nVT 2 × 25.88 × 10 −3 = rd = = 10.35 Ω I DQ 5 × 10 −3 The small-signal equivalent circuit for the diode is a 10.35 Ω resistance. 406 CHAPTER 11 Exercises E11.1 (a) A noninverting amplifier has positive gain. Thus v o (t ) = Avv i (t ) = 50v i (t ) = 5.0 sin(2000πt ) (b) An inverting amplifier has negative gain. Thus v o (t ) = Avv i (t ) = −50v i (t ) = −5.0 sin(2000πt ) E11.2 RL Vo 75 = Avoc = 500 = 375 Vi Ro + RL 25 + 75 RL Ri V 2000 75 500 Avs = o = Avoc = = 300 25 + 75 Vs Rs + Ri Ro + RL 500 + 2000 R I 2000 = 10 4 Ai = o = Av i = 375 × Ii 75 RL Av = G = Av Ai = 3.75 × 10 6 E11.3 Recall that to maximize the power delivered to a load from a source with fixed internal resistance, we make the load resistance equal to the internal (or Thévenin) resistance. Thus we make RL = Ro = 25 Ω. Repeating the calculations of Exercise 11.2 with the new value of RL, we have RL Vo 25 = Avoc = 500 = 250 Vi 25 + 25 Ro + RL R I 2000 = 2 × 10 4 Ai = o = Av i = 250 × Ii 25 RL Av = G = Av Ai = 5 × 10 6 407 E11.4 By inspection, Ri = Ri 1 = 1000 Ω and Ro = Ro 3 = 30 Ω. Vo 3 Ri 3 Ri 2 = Avoc1 Avoc2 A Vi 1 Ro1 + Ri 2 Ro2 + Ri 3 voc3 V 2000 3000 Avoc = o3 = 10 20 30 = 5357 100+ 2000 200+ 3000 Vi 1 Avoc = E11.5 Switching the order of the amplifiers of Exercise 11.4 to 3-2-1, we have Ri = Ri 3 = 3000 Ω and Ro = Ro 1 = 100 Ω Vo1 Ri 2 Ri 1 Avoc2 A = Avoc3 Vi 3 Ro3 + Ri 2 Ro2 + Ri 1 voc1 V 2000 1000 Avoc = o1 = 30 20 10 = 4348 300+ 2000 200+ 1000 Vi 3 Avoc = E11.6 E11.7 Ps = (15 V) × (1.5 A) = 22.5 W Pd = Ps + Pi − Po = 22.5 + 0.5 − 2.5 = 20.5 W P η = o × 100% = 11.11% Ps The input resistance and output resistance are the same for all of the amplifier models. Only the circuit configuration and the gain parameter are different. Thus we have Ri = 1 kΩ and Ro = 20 Ω and we need to find the open-circuit voltage gain. The current amplifier with an open-circuit load is: 408 Avoc = E11.8 v ooc Aiscii Ro AiscRo 200 × 20 = = = =4 1000 vi Ri ii Ri For a transconductance-amplifier model, we need to find the shortcircuit transconductance gain. The current-amplifier model with a shortcircuit load is: Gmsc = iosc Aiscii Aisc 100 = = = = 0. 2 S 500 vi Ri ii Ri The impedances are the same for all of the amplifier models, so we have Ri = 500 Ω and Ro = 50 Ω . E11.9 For a transresistance-amplifier model, we need to find the open-circuit transresistance gain. The transconductance-amplifier model with an open-circuit load is: Rmoc = v ooc Gmscv i Ro = = GmscRo Ri = 0.05 × 10 × 10 6 = 500 kΩ v i / Ri ii The impedances are the same for all of the amplifier models, so we have Ri = 1 MΩ and Ro = 10 Ω . E11.10 The amplifier has Ri = 1 kΩ and Ro = 1 kΩ . (a) We have Rs < 10 Ω which is much less than Ri , and we also have RL > 100 kΩ which is much larger than Ro . Therefore for this source and load, the amplifier is approximately an ideal voltage amplifier. 409 (b) We have Rs > 100 kΩ which is much greater than Ri , and we also have RL < 10 Ω which is much smaller than Ro . Therefore for this source and load, the amplifier is approximately an ideal current amplifier. (c) We have Rs < 10 Ω which is much less than Ri , and we also have RL < 10 Ω which is much smaller than Ro . Therefore for this source and load, the amplifier is approximately an ideal transconductance amplifier. (d) We have Rs > 100 kΩ which is much larger than Ri , and we also have RL > 100 kΩ which is much larger than Ro . Therefore for this source and load, the amplifier is approximately an ideal transresistance amplifier. (e) Because we have Rs ≅ Ri , the amplifier does not approximate any type of ideal amplifier. E11.11 We want the amplifier to respond to the short-circuit current of the source. Therefore, we need to have Ri > RL . Thus, we have an approximately ideal transconductance amplifier. A 100 Gmsc = vo = 6 = 10 −4 S Ro 10 P11.55* To sense the open-circuit voltage of a sensor, we need an amplifier with very high input resistance (compared to the Thévenin resistance of the sensor). To avoid loading effects by the variable load resistance, we need an amplifier with very low output resistance (compared to the smallest load resistance). Thus, we need a nearly ideal voltage amplifier with a gain of 1000. P11.56* The input resistance is that of the ideal transresistance amplifier which is zero. The output resistance of the cascade is the output resistance of the ideal transconductance amplifier which is infinite. An amplifier having zero input resistance and infinite output resistance is an ideal current amplifier. Also, we have Aisc = RmocGmsc . P11.57 To sense the short-circuit current of a sensor, we need an amplifier with very low input resistance (compared to the Thévenin resistance of the sensor). To avoid loading effects by the potenitally variable load resistance, we need an amplifier with very low output resistance (compared to the smallest load resistance). Thus, we need a nearly ideal transresistance amplifier. P11.58 To sense the short-circuit current of a sensor, we need an amplifier with very low input resistance (compared to the Thévenin resistance of the sensor). For the load current to be independent of the variable load resistance, we need an amplifier with very high output resistance (compared to the largest load resistance). Thus, we need a nearly ideal current amplifier. P11.59 The input resistance is that of the voltage amplifier which is infinite. The output resistance of the cascade is the output resistance of the 431 transconductance amplifier which is infinite. An amplifier having infinite input resistance and infinite output resistance is an ideal transconductance amplifier. Also, we have Gmsc - cascade = AvocGmsc . P11.60 The input resistance is that of the transconductance amplifier which is infinite. The output resistance of the cascade is the output resistance of the transresistance amplifier which is zero. An amplifier having infinite input resistance and zero output resistance is an ideal voltage amplifier. Also, we have Avoc-cascade = RmocGmsc . With the order reversed, the input resistance is that of the transresistance amplifier which is zero. The output resistance of the cascade is the output resistance of the transconductance amplifier which is infinite. An amplifier having zero input resistance and infinite output resistance is an ideal current amplifier. Also, we have Aisc-cascade = RmocGmsc . P11.61* To sense the source voltage with minimal loading effects, we need Ri >> Rs . To force a current through the load independent of its resistance, we need Ro >> RL . Thus, we need a nearly ideal transconductance amplifier. The equivalent circuit is: We have I L = Ri Ro Ri + Rs Ro + RL GmscVs . For the two given values of Rs , we require: 0.99 Ri = Ri Ri + 1000 Ri + 2000 Solving, we have Ri = 98 kΩ . For the two given values of RL , we require: 0.99 Ro = Ro Ro + 100 Ro + 300 Solving, we find: Ro = 19.7 kΩ 432 P11.62 We need Ri 0, then vo = +5. On the other hand, if vid < 0, then vo = -5. 541 The output waveform is P14.29 (a) This circuit has negative feedback. It is the voltage follower and has unity gain except that the output voltage cannot exceed 5 V. The output waveform is: (b) This circuit has positive feedback, and vo = +5 if the differential input voltage vid is positive. On the other hand, vo = -5 if vid is negative. In this circuit, we have v id = v o − v in Thus, the output waveform is: 542 P14.30 The inverting amplifier is shown in Figure 14.4 in the text and the voltage gain is Av = − R2 R1 . Thus to achieve a voltage gain magnitude of 2, we would select the nominal values such that R2nom = 2R1nom . However for 5%tolerance resistors, we have R1 min = 0.95R1nom R1 max = 1.05R1nom R2 min = 0.95R2nom R2 max = 1.05R2nom Thus we have Av min = − 0.95R2nom R2 min =− = −1.81 1.05R1nom R1 max Av max = − 1.05R2nom R2 max =− = −2.21 0.95R1nom R1 min Thus Av= 2 plus 10.5% minus 9.5%. P14.31 The noninverting amplifier is shown in Figure 14.11 in the text, and the voltage gain is Av = 1 + R2 R1 . Thus to achieve a voltage gain magnitude of 2, we would select the nominal values such that R2nom = R1nom . However for 5%-tolerance resistors, we have R1 min = 0.95R1nom R1 max = 1.05R1nom R2 min = 0.95R2nom R2 max = 1.05R2nom Thus we have Av min = 1 + 0.95R2nom R2 min =1+ = 1.905 1.05R1nom R1 max Av max = 1 + 1.05R2nom R2 max =1+ = 2.105 0.95R1nom R1 min Thus Av= 2 ± 5%. 543 P14.32* The circuit diagram is:  R  io = − 1 + 1  iin R  2  Because of the summing-point constraint, we have vin = 0. Thus, Rin = 0. Because the output current is independent of RL, the output impedance is infinite. In other words looking back from the load terminals, the circuit behaves like an ideal current source. P14.33 By the voltage-division principle, we have vx = RT RT + (1 −T )R v in =Tv in Then, we can write v − v x v in (1 −T = ix = in R R v o = −Rix + v x = −v in (1 −T ) +Tv in = v in (2T − 1) ) 544 Thus, as T varies from 0 to unity, the circuit gain varies from -1 through to 0 to +1. P14.34 From the circuit we can write: vo 1 = v 3 i4 = i3 = vo 1 R Thus we have v 4 = v 3 = vo 1 v o 2 = v 3 + v 4 = 2v o 1 vo 1 vo 2 + =0 4R 4R v iin = in R v in v o 1 v o 2 + + =0 R 4R 4R v A1 = o 1 = − 4 3 v in v 2v A2 = o 2 = o 1 = 2A1 = − 8 3 v in v in iin + P14.35 Very small resistances lead to excessively large currents, possibly exceeding the capability of the op amp, creating excessive heat or overloading the power supply. 545 Very large resistances lead to instability due to leakage currents over the surface of the resistors and circuit board. Stray pickup of undesired signals is also a problem in high-impedance circuits. P14.36* To achieve high input impedance and an inverting amplifier, we cascade a noninverting stage with an inverting stage: The overall gain is: R + R2 R4 Av = − 1 × R1 R3 Many combinations of resistance values will achieve the given specifications. For example: R1 = ∞ and R2 = 0 . (Then the first stage becomes a voltage follower.) This is a particularly good choice because fewer resistors affect the overall gain, resulting in small overall gain variations. R4 = 100 kΩ, 5% tolerance. R3 = 10 kΩ, 5% tolerance. P14.37* A solution is: 546 P14.38 Use the inverting amplifier configuration: Pick R2nom = 10R1nom to achieve the desired gain magnitude. Pick R1nom > 10 kΩ to achieve input impedance greater than 10 kΩ. Pick R1nom and R2nom < 10 MΩ because higher values are impractical. Many combinations of values will meet the specifications. For example: (a) Use 5% tolerance resistors. R1 = 100 kΩ and R2 = 1 MΩ. (b) Use 1% tolerance resistors. R1 = 100 kΩ and R2 = 1 MΩ. (c) R2 =1 MΩ 1% tolerance. R1 = 95.3 kΩ 1% tolerance fixed resistor in series with a 10-kΩ adjustable resistor. After constructing the circuit, adjust to achieve the desired gain magnitude. P14.39 We use a noninverting amplifier and place a resistor in parallel with the input terminals to achieve the desired input impedance. R1 = 1 kΩ, 1% tolerance. Many combinations of values for R2 and R3 will meet the given specifications. For example: R2 = 1 kΩ, 1% tolerance. R3 = 9.09 kΩ, 1% tolerance. (These values result in a nominal gain of 10.09, which is within the specified range.) 547 P14.40 Here are two answers: Many other correct answers exist. P14.41* One possibility is to place unity-gain voltage follower circuits between the sources and the input terminals of the circuits designed for Problem P14.40. A better answer (because it requires fewer op amps) is: All resistors are ± 1% tolerance. P14.42 To avoid excessive gain variations because of changes in the source resistances, we need to have input resistances that are much greater than the source resistances. Many correct answers exist. Here is one possibility: 548 The fixed resistors should be specified to have a tolerance of ± 1% because they are more stable in value than 5% tolerance resistors. The adjustment procedure is: 1. Set v1 = 0 and v2 = + 1 V. Then, adjust the 2-kΩ potentiometer to obtain vo = 3 V. 2. Set v1 = 1 V and v2 = 0. Then, adjust the 1-kΩ potentiometer to obtain vo = -10. P14.43 To avoid excessive variations in Avs = v o v s because of changes in Rs, we need to have Rin >> Rs . Rin =100 kΩ is sufficiently large. Thus, a suitable circuit is R1 and R2 should be 1% tolerance resistors. P14.44 Imperfections of real op amps in their linear range of operation include: 1. Finite input impedance. 2. Nonzero output impedance. 3. Finite dc open-loop gain. 4. Finite open-loop bandwidth. 549 P14.45* Equation 14.34 states: ft = A0CLfBCL = A0OLfBOL Thus, for A0CL = 10, we have f 15 MHz fBCL = t = = 1.5 MHz 10 A0CL For A0CL = 100, we have fBCL = 150 kHz P14.46 Equation 14.23 gives the open-loop gain as a function of frequency: A0OL 200 × 103 AOL (f ) = = 1 + j (f fBOL ) 1 + j (f 5) For f = 100 Hz, we have 200 × 10 3 ( ) AOL 100 = 1 + j (100 5) AOL (100 ) = A0OL 1 + (f fBOL ) 2 = 9987 Similarly, we have AOL (1000 ) = 1000 AOL (10 6 ) = 1 P14.47 (a) From the circuit, we can write: v s = Rin is + Ro is + AOL (Rin is ) v o = Ro is + AOL (Rin is ) Dividing the respective sides of the previous equations yields: Ro + AOLRin v Avo = o = v s Rin + Ro + AOLRin Substituting values, we obtain: 25 + 10 5 × 10 6 Avo = 6 10 + 25 + 10 5 × 10 6 550 = 0.99999 (compared to unity for an ideal op amp) (b) Z in = vs = Rin + Ro + AOLRin is = 10 6 + 25 + 10 5 × 10 6 = 1011 Ω (compared to ∞ for an ideal op amp) (c) The circuit for determining the output impedance is: v i = −v x v − AOLv i v ix = x + x Rin Ro v 1 Zo = x = 1 + AOL 1 ix + Rin Ro Z o = 2.5 × 10 −4 Ω (versus Zo = 0 for an ideal op amp) P14.48 (a) From the circuit (shown in Figure P14.48 in the text), we can write: v s + vi vo + vi vi + + =0 R1 R2 Rin v o + v i v o − AOLv i + =0 R2 Ro Algebra results in: Avo = vo = vs − R2  1 R1 1 +  + 1 + 1  Ro R2 + R22    Rin  AOLR2 − Ro    R1 R2 Substituting values, we obtain: Avo = −9.9989 (compared to -10 for an ideal op amp) (b) From the circuit, we can write: 551 v s = Ri is − v i v i + (R2 + Ro )(v i Rin + is ) + AOLv i = 0 Algebra results in Z in = R2 + Ro vs = R1 + is 1 + AOL + (R2 + Ro ) Rin Substituting values, we obtain: Z in = 1.0001 kΩ (compared to 1 kΩ for an ideal op amp (c) To find the output impedance, we zero the input source and connect a test source to the output terminals. The circuit is: Rin′ 1 vx where Rin′ = R2 + Rin′ 1 R1 + 1 Rin v − AOLv i vx + x ix = R2 + Rin′ Ro v 1 Zo = x = AOLRin′ 1 1 ix + + R2 + Rin′ Ro Ro (R2 + Rin′ ) vi = − Substituting values, we obtain: Z o = 2.75 × 10 −3 Ω versus Z o = 0 for an ideal op amp P14.49 Equation 14.32 gives the closed-loop gain as a function of frequency: ACL (f ) = A0CL 1 + j (f fBCL ) However, the dc closed-loop gain is given as 10 so we have 10 ACL (f ) = 1 + j (f fBCL ) For f = 10 kHz, we have 552 ACL = ( 10 1 + 10 fBCL 4 ) 2 =9 Solving, we find fBCL = 20.65 kHz. Then the gain bandwidth product is ft = A0CLfBCL = 206.5 kHz = A0OLfBOL P14.50 Equation 14.32 gives the closed-loop gain as a function of frequency: ACL (f ) = A0CL 1 + j (f fBCL ) The phase shift is − arctan(f / fBCL ). Thus at 200 kHz, we have 10 o = arctan[(2 × 10 5 ) / fBCL ] which yields fBCL = 1.134 MHz. Then the gain bandwidth product is ft = A0CLfBCL = 11.34 MHz = A0OLfBOL . P14.51 Alternative 1: fBCL = ft 10 6 = = 10 kHz A0CL 100 ACL (f ) = 100 1 + jf 10 4 The closed-loop bandwidth is fBCL = 10 kHz . Alternative 2: 553 For each stage, we have fBCL = ft 10 6 = = 100 kHz and the 10 A0CL gain as a function of frequency is: 10 ACL (f ) = 1 + jf 10 5 The overall gain is 100 A (f ) = (1 + jf 105 )2 To find the overall 3-dB bandwidth, we have 100 100 = A (f3dB ) = 2 2 1 + (f3dB 10 5 ) Solving, we find that f3dB = 64.4 kHz Thus, the two-stage amplifier has wider bandwidth. P14.52* P14.53 The nonlinear limitations of real op amps include: 1. Limited output voltage magnitude. 2. Limited output current magnitude. 3. Limited rate of change of output voltage. (Slew rate.) 554 P14.54 The full-power bandwidth of an op amp is the range of frequencies for which the op amp can produce an undistorted sinusoidal output with peak amplitude equal to the guaranteed maximum output voltage. P14.55 If the ideal output, with a sinusoidal input signal, greatly exceeds the full-power bandwidth, the output becomes a triangular waveform. The slope of the triangle is equal to the maximum slew rate in magnitude. The triangle goes from the negative peak to the positive peak in half of the period. Thus, the peak-to-peak amplitude is Vp − p = SR ×T / 2 = 10 7 × 0.5 × 10 −6 = 5 V P14.56 The desired output voltage is v o (t ) = Vom sin(ωt ) and the rate of change of the output is dv o (t ) = ωVom cos(ωt ) dt The maximum rate of change of the output is dv o (t ) = ωVom dt max Thus, we require the slew rate to be at least as large as the maximum rate of change of the output voltage. SR = ωVom = 2π10 5 (5) = 3.14 V µs P14.57* (a) (b) fFP SR 10 7 = = = 159 kHz 2πVom 2π10 Vom = 10 V . (It is limited by the maximum output voltage capability of the op amp.) (c) In this case, the limit is due to the maximum current available from the op amp. Thus, the maximum output voltage is: Vom = 20 mA × 100 Ω = 2 V (d) In this case, the slew-rate is the limitation. v o (t ) = Vom sin(ωt ) dv o (t ) = ωVom cos(ωt ) dt 555 dv o (t ) = ωVom = SR dt max Vom = SR ω = (e) P14.58 10 7 = 1.59 V 2π10 6 To avoid slew-rate distortion, the op-amp slew-rate specification must exceed the maximum rate of change of the output-voltage magnitude. For the gain and input given in the problem, the output voltage is v o (t ) = 0 t ≤ 0 = 10t exp( −t ) t ≥ 0 The rate of change is dv o (t ) =0 t ≤0 dt = 10 exp( −t ) − 10t exp( −t ) t ≥ 0 The maximum value occurs at t = 0, and is 10 V/µs. Thus the required minimum slew-rate specification is 10 V/µs or 107 V/s. P14.59 To avoid slew-rate distortion, the op-amp slew-rate specification must exceed the maximum rate of change of the output-voltage magnitude. For a voltage follower, the gain is unity. For the input given in the problem, the output voltage is v o (t ) = 0 t ≤ 0 =t 2 =9 0 ≤t ≤ 3 3 ≤t The rate of change is 556 dv o (t ) / dt = 0 t ≤ 0 = 2t 0 ≤ t ≤ 3 = 0 3 [...]... 0.320 A (b) R1 and R2 are in series Furthermore, R3, and R4 are in series Finally, the two series combinations are in parallel Req 1 = R1 + R2 = 20 Ω v eq = 2 × Req = 20 V Req 2 = R3 + R4 = 20 Ω Req = i1 = v eq / Req 1 = 1 A 1 / Req 1 1 = 10 Ω + 1 / Req 2 i2 = v eq / Req 2 = 1 A (c) R3, and R4 are in series The combination of R3 and R4 is in parallel with R2 Finally the combination of R2, R3, and R4 is... + R4 Similarly, we find v 3 = 30 V and v 4 = 60 V 25 = 20 V (b) First combine R2 and R3 in parallel: Req = 1 (1 / R2 + 1 R3 ) = 2.917 Ω R1 = 6.05 V Similarly, we find R1 + Req + R4 Then we have v 1 = v s v2 = vs E2.4 Req = 5.88 V and v 4 = 8.07 V R1 + Req + R4 (a) First combine R1 and R2 in series: Req = R1 + R2 = 30 Ω Then we have Req R3 15 30 i1 = is = = 1 A and i3 = is = = 2 A R3 + Req 15 + 30... 2ix = 0 5 5 Then use ix = (10 − v 1 ) / 5 to substitute and solve We find v1 = 7.5 V 10 − v 1 = 0.5 A Then we have ix = 5 (b) Choose the reference node and node voltages shown: Then write KCL equations at nodes 1 and 2: v1 5 + v 1 − 2i y 2 +3= 0 v2 5 + v 2 − 2i y 10 30 =3 Finally use i y = v 2 / 5 to substitute and solve This yields v 2 = 11.54 V and i y = 2.31 A E2.16 >> clear >> [V1 V2 V3] = solve('V3/R4... 5i 1 + 10(i1 − i2 ) = 100 and 10(i2 − i1 ) + 7i2 + 3i2 = 0 Simplifying and solving these equations, we find that i1 = 10 A and i2 = 5 A The net current flowing downward through the 10-Ω resistance is i1 − i2 = 5 A To solve by node voltages, we select the reference node and node voltage shown (We do not need to assign a node voltage to the connection between the 7-Ω resistance and the 3-Ω resistance because... to the loop abca, substituting values and solving, we obtain: v ab − v cb − v ac = 0 15 + 7 − v ac = 0 v ac = 22 V Similiarly, applying KVL to the loop abcda, substituting values and solving, we obtain: v ab − v cb + v cd + v da = 0 15 + 7 + vcd + 10 = 0 vcd = −32 V 11 P1.47 (a) In Figure P1.36, elements C, D, and E are in parallel (b) In Figure P1.42, elements C and D are in parallel (c) In Figure P1.45,... we have vo = Po 8 = 8 V Then, we have i1 = vo / 32 = 0.25 A and io = vo / 8 = 1 A KCL gives 2000iin = i1 + io = 1.25 A Thus we have iin = 1.25 / 2000 = 625 µA Then, vin = 2iin = 1.25 mV, and finally we have I x = iin + vin / 5 = 875 µA 18 P1.72 (a) No elements are in series (b) Rx and the 8-Ω resistor are in parallel Also, the 3-Ω resistor and the 6-Ω resistor are in parallel Thus, the voltages across... resistors Thus we have: 1 Req = R1 + = 3Ω 1 / R2 + 1 / R3 + 1 / R4 (b) R3 and R4 are in parallel Furthermore, R2 is in series with the combination of R3, and R4 Finally R1 is in parallel with the combination of the other resistors Thus we have: 1 =5Ω Req = 1 / R1 + 1 /[R2 + 1 /(1 / R3 + 1 / R4 )] (c) R1 and R2 are in parallel Furthermore, R3, and R4 are in parallel Finally, the two parallel combinations are... leaving It is true because charge cannot collect at a node in an electrical circuit P1.32 A node is a point that joins two or more circuit elements All points joined by ideal conductors are electrically equivalent Thus, there are five nodes in the circuit at hand: P1.33 The currents in series-connected elements are equal P1.34* Elements E and F are in series P1.35 For a proper fluid analogy to electric... points along each pipe P1.36* At the node joining elements A and B, we have ia + ib = 0 Thus, ia = −2 A For the node at the top end of element C, we have ib + ic = 3 Thus, 9 ic = 1 A Finally, at the top right-hand corner node, we have 3 + ie = id Thus, id = 4 A Elements A and B are in series P1.37* We are given ia = 2 A, ib = 3 A, id = −5 A, and ih = 4 A Applying KCL, we find ic = ib − ia = 1 A ie... set consists of the KVL equation and any two of the KCL equations E2.14 (a) Select the reference node at the left-hand end of the voltage source as shown at right Then write a KCL equation at node 1 v 1 v 1 − 10 + +1 = 0 R1 R2 Substituting values for the resistances and solving, we find v1 = 3.33 V 10 − v 1 = 1.333 A Then we have ia = R2 (b) Select the reference node and assign node voltages as shown ... Fundamentals of Engineering Exam To be able to lead in the design of systems that contain electrical/ electronic elements To be able to operate and maintain systems that contain electrical/ electronic... effectively with electrical engineers P1.2 Broadly, the two objectives of electrical systems are: To gather, store, process, transport, and display information To distribute, store, and convert energy... series, and we have = which yields Rx = Ω / 20 + / Rx Rab = + = 10 Ω 40 P2.4* P2.5* The 20-Ω and 30-Ω resistances are in parallel and have an equivalent resistance of Req1 = 12 Ω Also the 40-Ω and