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Chapter 01
Chapter 01
Chapter 01
Chapter 01
Chapter 01
Chapter 01
Chapter 01
Chapter 01
Chapter 01
Chapter 01
Chapter 01
Chapter 01
Chapter 01
Chapter 01
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Chapter 01

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  • 1. Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. CHAPTER 1 ExercisesE1.1 Charge = Current × Time = (2 A) × (10 s) = 20 C dq (t ) dE1.2 i (t ) = = (0.01sin(200t) = 0.01 × 200cos(200t ) = 2cos(200t ) A dt dtE1.3 Because i2 has a positive value, positive charge moves in the same direction as the reference. Thus positive charge moves downward in element C. 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 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.E1.6 (a) pa (t ) = v a (t )ia (t ) = 20t 2 10 10 10 20t 3 20t 3 w a = ∫ pa (t )dt = ∫ 20t dt = 2 = = 6667 J 0 0 3 0 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 10 w b = ∫ pb (t )dt = ∫ (20t − 200)dt = 10t 2 − 200t 0 = −1000 J 0 0 1
  • 2. Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.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 AE1.8 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 VE1.10 Elements E and F are in parallel; elements A and B are in series. ρLE1.11 The resistance of a wire is given by R = . Using A = πd 2 / 4 and A substituting values, we have: 1.12 × 10 −6 × L 9. 6 = ⇒ L = 17.2 m π (1.6 × 10 − 3 )2 / 4E1.12 P =V 2 R ⇒ R =V 2 / P = 144 Ω ⇒ I = V / R = 120 / 144 = 0.833 AE1.13 P =V 2 R ⇒ V = PR = 0.25 × 1000 = 15.8 V I = V / R = 15.8 / 1000 = 15.8 mAE1.14 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
  • 3. Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. ProblemsP1.1 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 Eight subdivisions of EE are: 1. Communication systems. 2. Computer systems. 3. Control systems. 4. Electromagnetics. 5. Electronics. 6. Photonics. 7. Power systems. 8. Signal Processing.P1.3 (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. (d) The voltage across a closed switch is zero. The current through the switch can be any value depending of the circuit. 3
  • 4. Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. (e) Direct current is constant in magnitude and direction with respect to time. (f) Alternating current varies either in magnitude or direction with time.P1.4 (a) A conductor is anagolous to a frictionless pipe. (b) An open switch is anagolous to a closed valve. (c) A resistance is anagolous to a constriction in a pipe or to a pipe with friction. (d) A battery is analogous to a pump. 1 coulomb/sP1.5 Electrons per second = = 6.25 × 1018 1.60 × 10 coulomb/electron −19P1.6* 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 = (5 A) × (3 s) = 15 C. dq (t ) dP1.7* i (t ) = = (2 + 3t ) = 3 A dt dtP1.8 (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: 4
  • 5. Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 0.005 0.005 0.005 (b) Q = ∫ i (t )dt = ∫ 10 sin(200πt )dt = (10 / 200π ) cos(200πt ) 0 0 0 = 0.0318 C 0.01 0.01 0.01 (c) Q = ∫ i (t )dt = ∫ 10 sin(200πt )dt = (10 / 200π ) cos(200πt ) 0 0 0 =0 C ∞ ∞P1.9* Q = ∫ i (t )dt = ∫ 2e −t dt = −2e −t | 0 = 2 coulombs ∞ 0 0 dq (t ) dP1.10 i (t ) = dt = dt (3 − 3e −2t ) = 6e −2t AP1.11 The number of electrons passing through a cross section of the wire per second is 15 N = = 9.375 × 1019 electrons/second 1.6 × 10 − 19 The volume of copper containing this number of electrons is 9.375 × 1019 volume = = 9.375 × 10 − 10 m3 10 29 The cross sectional area of the wire is πd 2 A= = 3.301 × 10 − 6 m2 4 Finally, the average velocity of the electrons is volume u = = 0.2840 mm/s AP1.12 The electron gains 1.6 × 10 −19 × 9 = 14.4 × 10 −19 joulesP1.13* Q = current × time = (5 amperes) × (36,000 seconds) = 1.8 × 10 5 coulombs Energy = QV = (1.8 × 10 5 ) × (12) = 2.16 × 10 6 joules 5
  • 6. Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.P1.14* (a) P = -vaia = -20 W Energy is being supplied by the element. (b) P = vbib = 50 W Energy is being absorbed by the element. (c) P = -vcic = 40 W Energy is being absorbed by the element.P1.15 The amount of energy is W = QV = (3 C) × (10 V) = 30 J. Because the reference polarity 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.P1.16 Q = w V = (600 J) (12 V) = 50 C . To increase the chemical energy stored in the battery, positive charge should move from the positive terminal to the negative terminal, in other words from a to b. Electrons move from b to a.P1.17 p (t ) = v (t )i (t ) = 20e −t W ∞ Energy = ∫ p (t )dt = −20e −t | 0 = 20 joules ∞ 0 The element absorbs the energy.P1.18 ( a) p (t ) = v ab iab = 50 sin(200πt ) W 0.005 0.005 0.005 (b) w = ∫ p (t )dt = ∫ 50 sin(200πt )dt = (50 / 200π ) cos(200πt ) 0 0 0 = 0.1592 J 0.01 0.01 0.01 (c) w = ∫ p (t )dt = ∫ 50 sin(200πt )dt = (50 / 200π ) cos(200πt ) 0 0 0 =0 J 6
  • 7. Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Cost $40P1.19* Energy = = = 400 kWh Rate 0.1 $/kWh Energy 400 kWh P 555.5 P = = = 555.5 W I = = = 4.630 A Time 30 × 24 h V 120 40 Reduction = × 100% = 7.20% 555.5P1.20 (a) P = 60 W delivered to element A. (b) P = 60 W taken from element A. (c) P = 60 W delivered to element A.P1.21* (a) P = 60 W taken from element A. (b) P = 60 W delivered to element A. (c) P = 60 W taken from element A.P1.22 The power that can be delivered by the cell is p = vi = 0.12 W. In 75 hours, the energy delivered is W = pT = 9 Whr = 0.009 kWhr. Thus the unit cost of the energy is Cost = (0.50) /( 0.009) = 55.56 $/kWhr which is 556 times the typical cost of energy from electric utilities.P1.23 The current supplied to the electronics is i = p /v = 50 / 12.6 = 3.968 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 = 100 / i = 25.2 hours. The energy delivered by the battery is W = pT = 50(25.2) = 1260 wh = 1.26 kWh. Neglecting the cost of recharging, the cost of energy for 300 discharge cycles is Cost = 75 /(300 × 1.26) = 0.1984 $/kWh.P1.24 The currents in series-connected elements are equal.P1.25 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. 7
  • 8. Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.P1.26* Elements A and B are in series. Also elements E and F are in series.P1.27 (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 ib = ia + ic = 3 + 1 = 4 A . Also we found earlier that id = −ic = −1 A.P1.28* 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, 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.P1.29* 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 AP1.30 We are given ia = −1 A, ic = 3 A, i g = 5 A, and ih = 1 A. Applying KCL, we find ib = ic + ia = 2 A ie = ic + ih = 4 A id = if − ia = 7 A if = i g + ih = 6 AP1.31 (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 c and substituting values, we find v c = 7 V. Also we have v b = −v a = −2 V. 8
  • 9. Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.P1.32* 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.P1.33 We are given v a = 5 V, v b = 7 V, vf = −10 V, and v h = 6 V. Applying KVL, we find v d = v a + v b = 12 V v c = −v a − vf − v h = −1 V v e = −v a − v c + v d = 8 V v g = ve − v h = 2 V v b = vc + ve = 7 VP1.34* Applying KCL and KVL, we have ic = ia − id = 1 A ib = −ia = −2 A v b = v d − v a = −6 V vc = vd = 4 V The power for each element is PA = −v a ia = −20 W PB = v b ib = 12 W PC = v c ic = 4 W PD = v d id = 4 W Thus, PA + PB + PC + PD = 0P1.35 (a) In Figure P1.28, elements C, D, and E are in parallel. (b) In Figure P1.33, no element is in parallel with another element. (c) In Figure P1.34, elements C and D are in parallel.P1.36 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 5 − 15 − v ac = 0 v ac = −10 V 9
  • 10. Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Similiarly, applying KVL to the loop abcda, substituting values and solving, we obtain: v ab − v cb + v cd + v da = 0 5 − 15 + v cd − 10 = 0 v cd = 20 VP1.37*P1.38P1.39 Four types of controlled sources are: 1. Voltage-controlled voltage sources. 2. Voltage-controlled current sources. 3. Current-controlled voltage sources. 4. Current-controlled current sources.P1.40 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 ρCu = (0.5) × (5.44 × 10 -8 ) (1.72 × 10 −8 ) = 1.58 Ω 10
  • 11. Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.P1.41P1.42P1.43* R= ( 1 )2 V = 1002 = 100 Ω P1 100 ( 2) 2 V 90 2 P2 = = = 81 W for a 19% reduction in power R 100P1.44 The power delivered to the resistor is p (t ) = v 2 (t ) / R = 2.5 exp( −4t ) and the energy delivered is ∞ ∞ ∞  2. 5  2. 5 w = ∫ p (t )dt = ∫ 2.5 exp(−4t )dt =  exp( −4t ) = = 0.625 J 0 0 − 4 0 4P1.45 The power delivered to the resistor is p (t ) = v 2 (t ) / R = 2.5 sin 2 (2πt ) = 1.25 − 1.25 cos( 4πt ) and the energy delivered is 10 10 10  1.25  w = ∫ p (t )dt = ∫ [1.25 − 1.25 cos( 4πt )]dt = 1.25t − sin( 4πt ) = 12.5 J 0 0  4π 0 11
  • 12. Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.P1.46 Equation 1.10 gives the resistance as ρL R = A (a) Thus, if the length of the wire is doubled, the resistance doubles to 1Ω. (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 0.125 Ω .P1.47 The power for each element is 20 W. The current source delivers power and the voltage source absorbs it.P1.48* 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.P1.49 This is a parallel circuit and the voltage across each element is 10 V positive at the top end. Thus, the current through the resistor is 10 V iR = = 2A 5Ω Applying KCL, we find that the current through the voltage source is zero. Computing power for each element, we find Pcurrent − source = −20 W 12
  • 13. Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Thus, the current source delivers power. PR = (iR ) 2 R = 20 W Pvoltage − source = 0P1.50* Applying Ohms 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, v2 v2 i3 = = 1 A , and i2 = = 0.5 A . Applying KCL, we have 5 10 i1 = i2 + i3 + 1 = 2.5 A . By Ohms law: v 1 = 5i1 = 12.5 V . Finally using KVL, we havev x = v 1 + v 2 = 17.5 V .P1.51 Ohm’s law for the 5-Ω resistor yields: i1 = 15 / 5 = 3 A. Then for the 10-Ω resistor, we have v 1 = 10i1 = 30 V. Using KVL, we have v 2 = v 1 + 15 = 45 V. Then applying Ohms law, we obtain i2 = v 2 / 10 = 4.5 A. Finally applying KCL, we have I x = i1 + i2 = 7.5 A. 13
  • 14. Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.P1.52* (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)P1.53 Applying KVL around the periphery of the circuit, we have − 18 + v x + 2v x = 0, which yields v x = 6 V. Then we have v 12 = 2v x = 12 V. Using Ohm’s law we obtain i12 = v 12 / 12 = 1 A and i x = v x / 2 = 3 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 = 2 A.P1.54 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. 14
  • 15. Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.P1.55 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.P1.56 (a) 10 = v 1 + v 2 (b) v 1 = 15i v 2 = 5i (c) 10 = 15i + 5i i = 0.5 A (d) P voltage −source = −10i = −5 W. (Power delivered by the source.) P15 = 15i 2 = 3.75 W (absorbed) P5 = 5i 2 = 1.25 W (absorbed)P1.57* v x = (4 Ω) × (1 A) = 4 V is = v x / 2 + 1 = 3 A Applying KVL around the outside of the circuit: v s = 3is + 4 + 2 = 15 V 15
  • 16. Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.P1.58 ix = −30 V/15 Ω = −2 A Applying KCL for the node at the top end of the controlled current source: is = ix / 2 − ix = −ix / 2 = 1 A The source labled is is an independent current source. The source labeled ix/2 is a current-controlled current source.P1.59 Applying Ohms law and KVL, we have 20 + 10ix = 5ix . Solving, we obtain ix = −4 A. The source labeled 20 V is an independent voltage source. The source labeled 5ix is a current-controlled voltage source. 16

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