Solution Manual For Electricity Principles and Applications By Richard Fowler and Mitchel Schultz

Solution Manual For Electricity Principles and Applications By Richard Fowler and Mitchel Schultz

 Parts and Equipment List  Learning Outcomes  Answers to Student Textbook Questions o Chapter Review Questions and Problems o Critical Thinking Questions o Homework Problems  Answers to Experiments Manual

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1

Section

1 Parts and Equipment List Materials such as lamps, resistors, insulating tape, plastics, wood, and switches are required as specified for various projects. These materials are consumable; quantities depend on the number and type of projects assigned.

Miscellaneous materials, such as hook-up wire, nails, magnets, ferrite rods, and paper, are required throughout the Experiments Manual.

Such materials are nonconsumable and can be used repeatedly.

INSTRUMENTS Qty.

Qty.

1 DMM and/or VOM 1 multimeter, l0-MΩ input resistance* 1 ammeter, ac, 5 mA, ±3%* 1 ammeter, ac, 300 or 500 mA* 1 ammeter, ac, 10 A* 1 voltmeter, dc panel, 5 V, 1000 Ω/V

2 wattmeters, 1000 W, 300 V, 7 A 1 oscilloscope with external trigger 1 signal generator, 20-Hz to 150-kHz range, 5-V sinusoidal output Multisim

RESISTORS Qty.

Qty.

1 2 Ω, 10 W, ±10% 1 2.2 Ω, ½ W, ±5% 1 4.7 Ω, ½ W, ±5% 1 6 Ω, 50 W % (Dale RH-50) 1 10 Ω, ½ W, ±5% 1 22 Ω, ½ W, ±5% 1 220 Ω, ½ W, ±10% 1 470 Ω, ½ W, ±5% 1 470 Ω, ½ W, ±10% 2 1000 Ω, ½ W, ±5% 1 1000 Ω, ½ W, ±10% 1 1000 Ω, 1 W, ±2% 1 1000 Ω, 1 W, ±5% 1 2200 Ω, ½ W, ±10% 1 4700 Ω, ½ W, ±5% 1 4700 Ω, ½ W, ±10% 2 10 kΩ, ½ W, ±5% 1 10 kΩ, ½ W, ±2%

1 22 Ω, 1 W, ±10% 1 25 Ω, 12 W, ±10% 1 33 Ω, ½ W, ±5% 1 47 Ω, ½ W, ±5% 1 51 Ω, ½ W, ±5% 2 100 Ω, ½ W, ±5% 1 220 Ω, ½ W, ±5% 1 20 kΩ, ½ W, ±1% 1 20 kΩ, ½ W, ±5% 2 200 kΩ, ½ W, ±5% 1 470 kΩ, ½ W, ± 10% 2 10 MΩ, ½ W, ±5% 1 potentiometer, 0 to 10 kΩ, 2 W, ±5%, linear 1 potentiometer, 0 to 10 Ω, ½ W, ±10%, linear

CAPACITORS

POWER SUPPLIES

Qty.

Qty.

Qty.

1 1000 pF, 100 WV dc, ±10% 1 1500 pF, 100 WV dc, ±10% 1 0.005 µF, 100 WV dc, ±10% 1 0.01 µF, 100 WV dc, ±10%

0.47 µF, 100 WV dc, ±10% 1 1 µF, 100 WV dc, ±10% 2 1000 µF, 50 WV dc, electrolytic

1 power supply, variable, 0 to 20 V dc 1 power supply, 6 to 8 V dc, 2 A (rectified output of a 6-V transformer will do)

assorted dry cells, 1.5 V and 6 V

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2

*These meters are not needed if the DMM and/or VOM has the appropriate ac ranges.

MISCELLANEOUS Qty.

Qty.

2 switches, SPDT, 1 A, 125 V 1 switch, SPST, 1 A, 125 V 3 lamps, 150 W, 120 V, incandescent, with lamp holders 4 lamps, miniature no. 47 (6.3 V, 0.15 A), with holders 3 lamp, neon, NE-2 1 LED, T 1 3/4 3 transformers, dual 115-V primary; dual 6-V, 2-A secondary (Magnetek/Triad F-106Z, Stancor P-6376, or equivalent) 1 transformer, variable, 0 to 120 V, 7A 1 soft grease pencil 1 heater or hair dryer, 500 W, 120 V

3 fuses, 0.5 A, with holders 1 loudspeaker, 8 Ω 2 chokes, 2.5 mH, >100 mA, <40 Ω (J.W. Miller 6302 or similar) 1 choke, 0.5 H, 300 mA dc, 30-Ω dc resistance (Triad C36X or equivalent) 1 zener diode, 8.2 V, 500 mW 1 transistor, 2N4126 1 zener diode, 14 V, 1 W, ±5% 1 motor, capacitorstart, ⅓- or ½-hp, 1725-rpm, 115-V, reversible 1 motor, split-phase, ⅓- or ½-hp, 1725-rpm, 115-V, reversible

Parts and equipment list continued on the next page.

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3

Section

2 Learning Outcomes Upon completion of this course, the student will be able to: 1. Measure current, voltage, resistance, and power in ac and dc circuits.

2. Calculate energy, power, voltage, current, resistance, and cost of energy in ac and dc circuits.

3. Construct electric circuits which fulfill the specifications given in a schematic diagram.

4. Troubleshoot circuits to locate shorts, opens, and overloads.

5. Measure inductance, capacitance, impedance, and phase shift.

6. Analyze ac circuits to determine current, voltage, reactance, impedance, phase angle, and power.

7. Identify components used in ac circuits as to type, rating, and applications.

8. Connect and change configurations of threephase ac transformers and circuits.

9. Analyze complex circuits by reducing them to equivalent circuits.

More specific performance objectives for each chapter are listed below. The objectives are stated in terms of what the student will be able to do after completing each chapter.

Learning Outcomes Introduction to Powers of 10 Express any number in scientific or engineering notation List the metric prefixes and their corresponding powers of 10 Change a power of 10 in engineering notation to

Convert between metric prefixes Add and subtract numbers expressed in powers of 10 notation Multiply and divide numbers expressed in powers of 10 notation Determine the reciprocal of a power of 10 Find the square of a number expressed in powers of 10 notation Find the square root of a number expressed in powers of 10 notation Enter numbers written in scientific and engineering notation into your calculator

1

BASIC CONCEPTS

1. List four kinds of energy into which electric energy can be converted.

2. Graphically illustrate an atom and label its parts.

3. List and explain two commercial uses of static electricity.

4. Identify the base unit of energy.

2 ELECTRICAL QUANTITIES AND UNITS 1. Explain current flow in solids, liquids, and gases.

2. List and explain the factors that determine the resistance of an object.

3. List four sources of voltage.

4. List and explain the base units of charge, current, voltage, resistance, and power.

5. Convert base units of electric quantities to subunits and multiple units.

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4

its corresponding metric prefix

3 BASIC CIRCUITS, LAWS, AND MEASUREMENTS

6 COMPLEX-CIRCUIT ANALYSIS

1. List the essential parts of a circuit and explain the purpose of each part.

2. Draw schematic diagrams from physical circuits.

3. Construct simple circuits by following schematic diagrams.

4. Draw symbols for fuses, circuit breakers, switches, resistors, batteries, and generators.

5. Calculate the value of current, voltage, resistance, power, and energy in a circuit.

6. Determine the cost of operating various electric devices.

7. Measure voltage, current, and resistance.

1. Identify complex circuits.

2. Select appropriate techniques for analyzing a given complex circuit.

3. Explain how a current source differs from a voltage source.

4. Reduce a complex circuit to a simple equivalent circuit.

5. Convert from a voltage source to a current source and vice versa.

6. Use simultaneous-equations techniques to solve sets of loop equations.

4

CIRCUIT COMPONENTS

1. Measure the specific gravity of a cell to determine its state of charge.

2. List the major difference among commonly used dry cells.

3. Explain the operation of a flasher lamp.

4. Specify component values for simple circuits.

5. Determine resistance values by using the standard color code.

6. Interpret switch symbols and specifications.

7. Interpret fuse and circuit-breaker specifications.

8. Measure wire sizes.

5

MULTIPLE-LOAD CIRCUITS

1. Explain and measure voltage in series circuits.

2. Explain and measure current in parallel circuits.

3. Simplify and solve problems involving complicated series-parallel circuits.

4. Determine maximum power transfer.

5. Explain the relationship between power transfer and efficiency.

7 MAGNETISM AND ELECTROMAGNETISM 1.

2.

3.

4.

5.

Demonstrate the existence of magnetic poles.

Make temporary and permanent magnets.

Control the strength of an electromagnet.

Demagnetize tools and materials.

Explain the operation of simple magnetic devices.

8 ALTERNATING CURRENT AND VOLTAGE 1. Graphically illustrate a sine wave, a square wave, and a sawtooth wave.

2. List and explain three ways of specifying the amount of alternating current.

3. List the advantages of alternating current over direct current.

4. Explain the relationship between period, frequency, and cycle.

5. List the advantages of three-phase alternating current over single-phase alternating current.

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5

9

POWER IN AC CIRCUITS

1.

2.

3.

4.

Use phasors to show phase relationships.

Calculate ac power in resistive circuits.

Calculate power in out-of-phase circuits.

Use the properties of a right triangle to determine the power factor of a circuit.

5. Explain the difference between apparent power and true power.

6. Explain the importance of power factor.

10

CAPACITANCE

1. Determine the reactance of a capacitor at any specified frequency.

2. Describe the behavior of a capacitor in ac and dc circuits.

3. Write specifications for ordering capacitors from electrical supply distributors.

4. List several types of capacitors and describe their major uses.

5. Calculate current and voltage distributions in capacitive circuits.

6. Explain phase shift in a capacitor.

7. Determine when a capacitor is open or shorted.

8. Calculate RC time constants.

11 INDUCTANCE 1.

2.

3.

4.

Identify various types of inductors.

Write complete specifications for inductors.

List uses of inductors.

Determine reactance of inductors at various frequencies.

5. Calculate total inductance for series and parallel combinations of inductances.

12 TRANSFORMERS 1. Determine the efficiency of a transformer.

2. Connect three-phase transformers in delta and wye connections.

3. Use a transformer for impedance matching.

4. Connect secondary winding in series-aiding or series-opposing configurations to obtain different voltages.

5. List and explain three common uses of transformers.

6. Explain why transformers have voltampere ratings.

13 R, C, AND L CIRCUITS 1. Determine resonant frequency.

2. Calculate impedance for a variety of combinations of R, C, and L circuits.

3. Graphically and mathematically add phasors representing electrical quantities.

4. Determine phase angles for circuits containing different combinations of R, C, and L. 5. List and explain the characteristics of four basic types of filters.

14 ELECTRIC MOTORS 1. Identify various types of motors.

2. Explain how a single-phase motor produces torque.

3. Explain and compare the characteristics of single-phase induction motors.

4. Calculate synchronous speed, horsepower, and torque for various motors.

5. Explain the operating characteristics of series, shunt, and compound dc motors.

6. Explain the operation of a universal motor.

7. Interpret and use the information given on

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6

the nameplate of a motor.

15 INSTRUMENTS AND MEASUREMENTS

16 RESIDENTIAL WIRING CONCEPTS

1. Explain the operation of several types of meter movements.

2. Explain how a digital meter can measure voltage, current, resistance, inductance, capacitance, and frequency.

3. Determine values of shunt and multiplier resistors.

4. Extend the range of voltmeters and ammeters.

5. Determine when significant voltmeter or ammeter loading will occur.

6. Measure three-phase power in balanced and unbalanced loads.

7. Measure frequency.

8. Measure impedance and reactance.

9. Measure inductance and capacitance.

1.

2.

3.

4.

5.

6.

7.

8.

Explain how electric energy is distributed.

Explain the purpose of a GFCI. Explain the purpose of an AFCI. Determine when a conductor with white insulation is not a neutral conductor.

Determine which conductor to use in the return leg of a switching circuit.

Explain the operation of low-voltage controlled circuits.

Explain how a feeder circuit is wired.

Explain how a lighted-handle switch operates.

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7

Section

3 Answers to Student Textbook Questions A CHAPTER REVIEW QUESTIONS AND PROBLEMS Chapter Review Questions – Solutions 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

d a b c c a a d b c c b a d c b c a b d 10 is the base and 7 is the exponent a. A form of powers of 10 notation in which a number is expressed as a number between 1 and 10 times a power of 10. b. A form of powers of 10 notation in which a number is expressed as a number between 1 and 1000 times a power of 10 that is a multiple of 3. c. Numbers that are written in standard form without using any form of

23. Because most electrical quantities (voltage, current, resistance, power, etc.) are specified in terms of the metric prefixes. Since the metric prefixes are direct substitutes for the powers associated with engineering notation, it makes sense that it is the preferred form of powers of 10 notation in electronics.

24. a. milli (m) b. kilo (k) c. micro (µ) d. mega (M) e. nano (n) f. giga (G) g. pico (p) h. tera (T) 25. a. Hertz (Hz) is the unit and f is the symbol b. Volt (V) is the unit and V is the symbol c. Watt (W) is the unit and P is the symbol d. Ohm (Ω) is the unit and R is the symbol e. Farad (F) is the unit and C is the symbol f. Henry (H) is the unit and L is the symbol g. Ampere (A) is the unit and I is the symbol

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8

powers of 10 notation.

CHAPTER 1 QUESTIONS 1-1.

1-2.

1-3.

1-4.

1-5.

1-6.

1-7.

1-8.

1-9.

1-10.

1-11.

1-12.

1-13.

1-14.

F T T F T F F B C D D B B C

2-6.

2-7.

2-8.

2-9.

2-10.

2-11.

2-12.

2-13.

2-14.

2-15.

2-16.

2-17.

2-18.

2-19.

2-20.

PROBLEMS 1-1.

1-2.

1-3.

1-4.

1-5.

1-6.

64% 240 J 240 J 30, 350 J 840 J 275 J

CHAPTER 2 QUESTIONS 2-1.

2-2.

2-3.

2-4.

2-5.

F T F F T

F T T T F F C B D A D A B Resistivity of the material, length of the object, cross-sectional area of the object, and temperature of the object a. V b. A c. R d. Ω e. W f. Q g. mA h. hp

PROBLEMS 2-1.

2-2.

2-3.

2-4.

2-5.

2-6.

2-7.

2-8.

2-9.

2-10.

2-11.

3.9 kWh or 3900 Wh or 14,040,000 J 62.2% 36 W 133.3 W 63 J 200 W 90 s 4.2 A 2 kC 49.7% 0.035 Ω or 35 mΩ

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9

Introduction to Powers of 10 1. 3.5 × 106

16. 1.82 × 104

2.

2. 6.78 × 102

17. 6.4 × 105

3.

3. 1.6 × 108

18. 1.1 × 10−2

4.

4. 5.5 × 10−4

19. 1.75 × 10−9

5.

5. 1.5 × 10−1

20. 3.2 × 1012

6.

6. 9.42 × 10−12

21. 0.000165

7. 2.27 × 10

3

22. 560,000

8.

8. 4.21 × 10

4

23. 863

9.

9. 3.3 × 10

−2

24. 0.00315

10. 6.0 × 10

−6

25. 0.0000000017

7.

11. 7.77 × 10

7

26. 4,650,000

12. 1.0 × 102

27. 1,660

13. 8.7 × 101

28. 0.025

14. 1.8 × 10−3

29. 0.0000000000033

15. 9.5 × 10−8

30. 92,100

Section I-2 Engineering Notation and Metric Prefixes 31. 5.5 × 103

46. 100 × 10−6

32. 5.5 × 10−3

47. 2.57 × 1012

33. 6.2 × 106

48. 20 × 103

34. 150 × 103

49. 70 × 10−6

35. 99 × 103

50. 2.5 × 103

36. 10 × 10−3

51. 1 kW

37. 750 × 10−6

52. 10 kΩ

38. 550 × 10−3

53. 35 mV

39. 10 × 106

54. 50 µA

40. 3.2 × 10−9

55. 1 µF

41. 68 × 10

−6

56. 1.57 MHz

42. 92 × 10

9

57. 2.2 MΩ 3

58. 162 kV

−12

59. 1.25 GHz

43. 270 × 10 44. 18 × 10

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10

45. 450 × 10−9

60. 330 pF

61. 250 µA

66. 240 V

62. 61 pF

67. 4.7 Ω

63. 500 mW

68. 1 mH

64. 2.2 kΩ

69. 50 µW

65. 180 kΩ

70. 100 pA

Section I-3 Converting Between Metric Prefixes 71. 55 mA

81. 7.5 mA

72. 10,000 pF

82. 1,000 µA

73. 0.0068 µF

83. 100,000 W

74. 1,490 kHz

84. 50,000 kW

75. 22 µF

85. 4.7 nF

76. 1.5 mH

86. 0.56 µF

77. 1,500 kΩ

87. 1.296 GHz

78. 2,200 MHz

88. 50,000 µH

79. 39 kΩ

89. 7,500,000 pF

80. 5.6 MΩ

90. 220 GΩ

Section I-4 Addition and Subtraction Involving Powers of 10 Notation 91. 7.5 × 104

97. 5.0 × 107

92. 9.5 × 106

98. 1.5 × 10−2

93. 5.9 × 10−10

99. 1.45 × 10−2

94. 2.5 × 10−2

100. 5.0 × 102

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11

95. 2.15 × 10−3

101. 2.6 × 104

96. 2.0 × 102

102. 4.6 × 10−3

Section I-5 Multiplication and Division Involving Powers of 10 Notation 103. 1.8 × 106

109. 2.5 × 104

104. 1.0 × 10−2

110. 3.0 × 10−3

105. 3.0 × 109

111. 1.25 × 102

106. 1.43 × 10−1

112. 2.5 × 10−1

107. 1.0 × 10−5

113. 5.0 × 107

108. 5.0 × 10−6

114. 1.1 × 10−2

Section I-6 Reciprocals With Powers of 10 115. 10−4 116. 104 117. 10−1 118. 108 119. 107 120. 1013 121. 10−15 122. 10−18

Section I-7 Squaring Numbers Expressed in Powers of 10 Notation 123. 2.5 × 107 124. 6.25 × 10−14 125. 8.1 × 1011 126. 4.9 × 1011 127. 1.44 × 10−16 128. 6.4 × 10−19

Section I-8 Square Roots of Numbers Expressed in Powers of 10 Notation 129. 2.0 × 10−2 130. 7.07 × 102 Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

12

131. 6.0 × 10−6 132. 2.21 × 10−1 133. 3.87 × 10−2 134. 5.92 × 10−3

Section I-9 The Scientific Calculator 135. Enter the problem using the following keying sequence: 1

5

EXP

EXP

3

=

±

3

×

1

•

2

.

The calculator will display the answer as 18.000 × 1000.

136. Enter the problem using the following keying sequence: 6

0

÷

1

•

5

EXP

3

= .

The calculator will display the answer as 40.000 × 10−03.

137. Enter the problem using the following keying sequence: 1

2

÷

1

0

EXP

3

= .

The calculator will display the answer as 1.200 × 10−03.

138. Enter the problem using the following keying sequence: 5

EXP

EXP

3

±

3

×

1

2

0

= .

The calculator will display the answer as 600.000 × 1000.

139. Enter the problem using the following keying sequence: 6

•

5

EXP

4

+

2

5

EXP

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13

3

= .

The calculator will display the answer as 90.000 × 1003.

140. Enter the problem using the following keying sequence: 2 EXP

•

5 ±

EXP 6

±

4

-

5

0

= .

The calculator will display the answer as 200.000 × 10−06.

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14

CHAPTER 3 QUESTIONS 3-1.

3-2.

3-3.

3-4.

3-5.

3-6.

3-7.

3-8.

3-9.

3-10.

3-11.

3-12.

3-13.

T T F F F F T T F F T F Conductors Insulator Energy source Load

3-14.

3-15.

3-16.

3-17.

3-18.

3-19.

Current Resistance 3½ digit meter VOM The VOM range indicates how much to multiply the scale reading by. The DMM just lists the maximum value the range can indicate.

3-20. The DMM shifts the decimal point on the readout when the range is changed. For example, changing from the 100-ohm range to the 10-ohm range shifts the decimal point on the readout one place to the left.

PROBLEMS 3-1.

3-2.

52.5 Ω 4.17 A

3-7.

3-8.

3-9.

3-10.

11 c /kWh 96 Ω 3.8 V 4.34 V

CHAPTER 4 QUESTIONS 4-1.

4-2.

4-3.

4-4.

4-5.

4-6.

4-7.

4-8.

4-9.

4-10.

4-11.

4-12.

4-13.

4-14.

4-15.

4-16.

4-17.

4-18.

4-19.

4-20.

4-21.

4-22.

4-23.

4-24.

4-25.

4-26.

4-27.

4-28.

4-29.

4-30.

4-31.

4-32.

4-33.

4-34.

4-35.

4-36.

T T F F F T F T F F F T F F F F F F T F F A A C D C 4.7 Ω 226 Ω Thermal and magnetic Current rating, voltage rating, and fusing characteristic A circuit with abnormally low resistance Slow-blow or longtime-lag Resettable fuse a. decrease b. decrease Rheostat LED

PROBLEMS

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15

3-3.

3-4.

3-5.

3-6.

4-3.

4-4.

4-5.

4-6.

4-7.

2640 W 1200 Wh 14.4 c 40 V 2W 9.72 W, 11.88 W 29.9 mA 945 Wh W P P  2 A × 9 V  18 J/s 3.6´ 104 J t= = 2´ 103 s = 33 minutes 18 J/s & 20 seconds P = IV and t =

CHAPTER 5 QUESTIONS

5-1.

5-2.

5-3.

5-4.

5-5.

5-6.

5-7.

5-8.

5-9.

5-10.

5-11.

5-12.

5-13.

5-14.

5-15.

5-16.

5-17.

5-18.

5-19.

5-20.

5-21.

5-22.

5-23.

5-24.

5-25.

5-26.

Parallel Parallel Most Decreases Series T F F F F T F T T F T F T F F D A Decrease Increase Decrease No. IR1 can never be less than IR2 or IR3 and R3 is larger than either R2 or R1. Since P = I2R, PR1 will be the largest of the three powers.

5-27. (a) Zener divider and regulator.

4-1.

4-2.

5-4.

5-5.

5-6.

5-7.

5-8.

5-9.

5-10.

5-11.

5-12.

4230 Ω 0.2 Ω 1250 Ω, 28 W 20 Ω, 60 Ω, 120 Ω 125 Ω, 625 Ω, 16 W 2.17 mA 7.54 mA 37.55 mW 50.55 mW ID1 = 8.75 mA, IT = 15 mA, PD1 = 43.75 mW Efficiency = 10.4%

CHAPTER 6 QUESTIONS 6-1.

6-2.

6-3.

6-4.

6-5.

6-6.

6-7.

6-8.

6-9.

6-10.

6-11.

6-12.

6-13.

Network Kirchhoff’s voltage Incorrect Two F T F T F Three Three Superposition theorem a. Yes b. Yes c. Yes

PROBLEMS 6-1.

6-2.

6-3.

6-4.

6-5.

6-6.

6-7.

6-8.

2A + 7B − 3C = 7 A = 2, B = 3 a. 36 V b. 24 V 36 V IN = 8 mA RN = 2 kΩ A = 34.15, B = –21.85, and C = –9.69 VR3 = 4.5 V, IR3 = 0.3 A VR3 = 2.5 V, VR4 = 4.5 V, IR1 = 0.3 A, IR2 = 0.2 A

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16

6-9.

PROBLEMS 5-1.

5-2.

5-3.

6-10.

6-11.

20 V, 60 Ω, 30 W, 0.05 S 100 mA, 1000 Ω, 150 mA, 3 mS 20 V, 0.1 A, 40 V IR2 = 0.9 A, VR2 = 18 V 24 = 10I1 + 20I1 + 20I2 = 30I1 + 20I2 22.5 = 20I2 + 20I1 + 15I2 = 20I1 + 35I2

CHAPTER 7 QUESTIONS 7-1.

7-2.

7-3.

7-4.

7-5.

7-6.

7-7.

7-8.

7-9.

7-10.

7-11.

7-12.

7-13.

7-14.

7-15.

7-16.

7-17.

7-18.

7-19.

7-20.

7-21.

7-22.

7-23.

7-24.

7-25.

T F F F T F F T F T Domains Temporary magnetic Armature Tesla Weber Ampere-turn Ampere-turn/meter Generator Transformer Iron Flux A A C D

PROBLEMS 7-1.

7-2.

7-3.

7-4.

7-5.

7-6.

7-7.

900 A · t 166.7 tesla 450 A · t/m 0.002 Wb/(A · t · m) 200 turns 1.29 A 0.32 weber

IR1 = 0.05 A, IR2 = 0.03 A, IR3 = 0.02 A, VR4 = 30 V

CHAPTER 8 QUESTIONS 8-1.

8-2.

8-3.

8-4.

8-5.

8-6.

8-7.

8-8.

8-9.

8-10.

8-11.

8-12.

8-13.

8-14.

8-15.

8-16.

8-17.

8-18.

8-19.

8-20.

8-21.

8-22.

8-23.

8-24.

8-25.

8-26.

8-27.

T F T T F T T T F F T T T T F T Sine wave Hertz, Hz Period, T Balanced, neutral Sine (or symmetical) Cycle per second Neutral The number of lines of flux cut per unit of time Flux density of field poles, number of conductors in the armature, speed at which the armature rotates Smoother rectified dc, more efficient use of copper, less complex motors with more constant torque, more constant load on the generator Fluctuating

PROBLEMS 8-1.

8-2.

8-3.

8-4.

8-5.

8-6.

100 Hz, 0.01 s 0.005 s 99 V 108 Vav 60 Hz Ip–p = 12 A

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7-8.

7-9.

7-10.

7-11.

3A 25 A · t/m 60 T 1.008 × 10–3 Wb/(A · t · m)

CHAPTER 9 QUESTIONS 9-1.

9-2.

9-3.

9-4.

9-5.

9-6.

9-7.

9-8.

9-9.

9-10.

9-11.

9-12.

9-13.

9-14.

9-15.

9-16.

9-17.

9-18.

9-19.

9-20.

9-21.

9-22.

9-23.

9-24.

9-25.

9-26.

T F T F F F F T F Moving or rotating The phase angle Lag Lead, 90 Reactance Voltampere 90 Theta True, apparent Resistive, reactive T Apparent, greater 20 0.9 because, with equal voltage, less current is required to provide equal amounts of power No, because 1.414 × 1.414 does not equal 1 × 106 cos θ P = IV cos θ

PROBLEMS 9-1.

9-2.

9-3.

9-4.

9-5.

9-6.

0 W, 480 VA, 90° 5.4 A 1287 W, 1680 VA, 0.766 or 76.6% 48.2° 192.5 V, 577.5 VA 14.5 mW

8-7.

8-8.

8-9.

8-10.

8-11.

8-12.

70.7 V 60 Hz 52 V 9.8 Ap-p 30 V 3A

CHAPTER 10 QUESTIONS 10-1. T 10-2. T 10-3. F 10-4. T 10-5. T 10-6. F 10-7. T 10-8. F 10-9. F 10-10. F 10-11. T 10-12. T 10-13. F 10-14. F 10-15. F 10-16. T 10-17. Farad 10-18. F 10-19. 0.0056 10-20. C 10-21. Lead, 90 10-22. Inversely, capacitance, frequency 10-23. Five 10-24. Distance between plates—decrease distance; area of plates—increase area; type of dielectric material—use material with a higher dielectric constant; temperature—increase temperature if using material with positive temperature coefficient 10-25. In the dielectric; electrical 10-26. Distance between plates and voltage across plates 10-27. Two capacitors are connected in series with like polarities connected together 10-28. Capacitance, tolerance, voltage rating, type, lead configuration 10-29. It blocks the dc component of a waveform and passes the ac (fluctuating part) of the waveform

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9-7.

9-8.

9-9.

9-10.

9-11.

30.6 mW 2.5 A 72 Ω Rphase = 23 Ω, Pt = 2494 W Papp = 4200 W, PF = 0.917 or 91.7%, θ = 23.6°, % eff. = 87.2%

10-3.

10-4.

482 Ω a. 5 µF b. 318.3 Ω c. 157 mA d. 62.8 mA 10-5. a. 1.2 µF b. 1326.3 Ω c. 20 V 10-6. 173 V 10-7. 5.94 ms 10-8. 1592 10-9. 20 µF 10-10. 0.48 A 10-11. 6 V 10-12. 18 V

CHAPTER 11 QUESTIONS 11-1. F 11-2. T 11-3. F 11-4. F 11-5. F 11-6. F 11-7. T 11-8. T 11-9. T 11-10. T 11-11. F 11-12. T 11-13. Litz 11-14. Henry 11-15. H 11-16. Ohm 11-17. XL 11-18. Opposes 11-19. bemf 11-20. Directly

10-30. When the conductors are parallel to each other and the circuit frequency is high

PROBLEMS 10-1.

10-2.

14 J 112.5 mC

11-31. Axis orientation, shielding, and physical separation 11-32. Core diameter—increase diameter; number of turns—use more turns; spacing of turns—decrease the spacing; type of core—use higher-permeability material 11-33. Inductance, tolerance, current, voltage, dc resistance, and quality 11-34. Q decreases

PROBLEMS 11-1. 0.05 s 11-2. 14, 137 Ω; 1.4 mA 11-3. 123 11-4. 10 H, 3.2 mA, 32 V 11-5. 2.92 mH 11-6. 1.59 H 11-7. 14 V 11-8. 0.06 H 11-9. 1.67 H 11-10. 4.2 kΩ, 5.7 mA 11-11. 90.4 V

CHAPTER 12 QUESTIONS 12-1. F 12-2. T 12-3. F 12-4. T 12-5. F 12-6. T 12-7. F 12-8. F 12-9. F 12-10. T 12-11. T 12-12. F

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11-21. Mutual inductance 11-22. Leads, 90 11-23. Coils, reactors, chokes 11-24. Twice 11-25. Lenz’s 11-26. Inductance 11-27. Electric, magnetic 11-28. dc, ohmic 11-29. Inductive kick 11-30. Impedance 12-24. Energizing 12-25. Secondary 12-26. Isolation 12-27. Autotransformer 12-28. Maximum power transfer 12-29. Resistive 12-30. Current, voltage 12-31. Four 12-32. Delta, wye 12-33. Eddy current 12-34. Hysteresis 12-35. Smaller 12-36. Isolation, irnpedance matching, and changing voltage levels 12-37. They are the same 12-38. When the transfer of power is more important than efficiency

PROBLEMS 12-1.

12-2.

12-3.

12-4.

21 V, 2 A 200 turns 75% No, because the secondary is designed to carry no more than 6.52 A. Or, the apparent power of the load exceeds the true power rating of the transformer.

12-5. 4 kΩ 12-6. 4 A, 8 V 12-7. 35 W 12-8. 140 turns 12-9. 1200 VA 12-10. 10 A 12-11. 20 V at 2 A 12-12. 208 V

12-13. T 12-14. Mutual inductance 12-15. Henry 12-16. Primary, secondary 12-17. Coefficient of coupling 12-18. Leakage 12-19. Up 12-20. Core, coils 12-21. Copper 12-22. Hysteresis, eddy current 12-23. Resistive 13-12. T 13-13. T 13-14. Resistance, reactance 13-15. Ohm 13-16. Quality, frequency 13-17. Z 13-18. Leads 13-19. Lags 13-20. Decrease, increase, decrease, increase 13-21. Decrease, increase, increase, remain the same 13-22. Voltage 13-23. Impedance 13-24. Inductor 13-25. Unity or one 13-26. Zero 13-27. Inductive 13-28. Band-pass, band-reject, low-pass, and high-pass 13-29. Band-pass and band-reject

PROBLEMS 13-1.

13-2.

13-3.

13-4.

13-5.

13-6.

CHAPTER 13

5.375 kHz a. 2147 Ω b. 33.54 V c. 625 mW d. 0.838 a. 981 Ω b. 40.8 mA c. 889 mW d. 57° 1806 Ω 59.32 kHz, 1.06 kHz a. 1262 Ω b. 23.8 mA c. 63 V d. 65 kHz

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QUESTIONS 13-1. F 13-2. F 13-3. T 13-4. T 13-5. F 13-6. F 13-7. T 13-8. T 13-9. T 13-10. F 13-11. F

CHAPTER 14 QUESTIONS 14-1. F 14-2. F 14-3. F 14-4. T 14-5. F 14-6. T 14-7. F 14-8. F 14-9. F 14-10. T 14-11. F 14-12. F 14-13. F 14-14. F 14-15. B 14-16. D 14-17. C 14-18. D 14-19. B 14-20. ac, dc, and universal 14-21. sfhp, fhp, and ihp 14-22. National Electrical Manufacturers Association 14-23. Starting or locked-rotor, stall or breakdown or pull-out, and running or full-load 14-24. 240 V, because the higher voltage is within the tolerance of the motor and it provides better efficiency, requires less line current, produces greater starting torque, and keeps the motor temperature lower 14-25. Totally enclosed fan-cooled and totally

13-7.

a. 1196 Ω b. 25.1 mA c. 13.3 mA d. 4.47° 13-8. Z = 2913 Ω, PF = 0.412 13-9. Q = 50 13-10. Z = 1810 Ω 13-11. Vout = 31.4 mV when fin = 200 Hz Vout = 864.6 mV when fin = 10 kHz 13-12. Vout = 62.8 mV when fin = 200 Hz Vout = 953 mV when fin = 10 kHz

PROBLEMS 14-1. 220 V, 73%, yes 14-2. 2.28 lb-ft 14-3. 750 rpm 14-4. 5.6% 14-5. 115°C 14-6. 1140 r/min 14-7. 127 V 14-8. 0.208 Ω 14-9. 0.238 Ω 14-10. 29.7 A 14-11. 1.875 hp 14-12. two pole, ½ hp

CHAPTER 15 QUESTIONS 15-1. d’Arsonval 15-2. Iron-vane 15-3. Rheostat 15-4. Multiplier 15-5. Shunt 15-6. Shorting-type 15-7. VOM 15-8. Frequency 15-9. Clamp-on ammeter 15-10. Voltage 15-11. Frequency 15-12. Time 15-13. Voltage 15-14. cemf 15-15. Time 15-16. Rectifier 15-17. F

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enclosed nonventilated 14-26. Generator action 14-27. Use an auxiliary winding, a shaded-pole, or a differential reluctance pole 14-28. Equal amplitude and 90º out of phase 14-29. Variable-speed means the speed is continuously variable over a wide range of speeds, while dual-speed means two discrete full-load speeds 14-30. Because it has a constant-amplitude rotating field in the stator 14-31. Three-phase motor 14-32. A brushless dc motor 14-33. No

15-18. T 15-19. F 15-20. T 15-21. T 15-22. F 15-23. F 15-24. F 15-25. A 15-26. A 15-27. B 15-28. A 15-29. B 15-30. Current-voltage method and equivalentresistance method

PROBLEMS

16-16. Neutral 16-17. Service drop 16-18. Distribution panel, service head, meter box 16-19. Grounding conductor 16-20. Red, black 16-21. White 16-22. Two, one 16-23. Three 16-24. Feeder 16-25. High-voltage three-phase 16-26. 14-3/G 16-27. Earth ground 16-28. It is 120 V referenced to ground or neutral and 240 V referenced to the other hot conductor.

16-29. White, black, red 16-30. Two 16-31. By the dark-colored screw 16-32. When the black conductor is needed for the return leg of the switch circuit 16-33. Excessive ground-fault current; excessive imbalance between neutral and hot currents 16-34. Arcing anywhere in the protected part of the circuit 16-35. Mechanical latching 16-36. To prevent disruptions of the neutral conductors if the duplex receptacle is removed.

16-37. To the neutral connector on the GFCI breaker 16-38. To the neutral bus

15-1. 1.2 MΩ 15-2. Yes 15-3. 4 kΩ 15-4. 34 and 38 mA 15-5. 52.63 Ω 15-6. 99 kΩ 15-7. 10 kΩ V 15-8. 250 kΩ 15-9. 0.2 µF 15-10. 4.7 kΩ 15-11. 4.70 kΩ or 4700 Ω 15-12. 400 mA, or 0.4 A; two places, right 15-13. 1200 W, 600 W

CHAPTER 16 QUESTIONS 16-1. F 16-2. T 16-3. T 16-4. F 16-5. F 16-6. F 16-7. F 16-8. T 16-9. T 16-10. F 16-11. F 16-12. T 16-13. F 16-14. F

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16-15. T

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B CRITICAL THINKING QUESTIONS Introduction to Powers of 10 – Solutions Critical Thinking I-1 a. 1 × 101 b. 1 × 104 c. 1 × 10−6

CHAPTER 1 1-1. Some topics that could be discussed are: a. Home entertainment—no TV, compact discs, radios, tapes, computer games, etc., unless battery operated units are available. b. Heating and air-conditioning—many heating systems need electricity for control circuits and/or forced air. c. Inoperative elevators in high-rise buildings. d. Inadequate and expensive alternative lighting sources may restrict length and/or type of evening activities. e. Electric water heater inoperative—no hot showers. f. No electrical appliances or tools to aid in household chores or maintenance. g. No refrigeration—need to go to market every day for fresh food.

1-2. Production of electric energy has environmental consequences. Hydroelectric plants restrict water flow and may reduce fish population. Fossil-fuel plants add pollutants to the atmosphere. And, atomic plants produce radioactive waste.

1-3. The added cost in making the small motor more efficient may more than offset the savings in operating cost resulting from the increased efficiency. The additional energy needed to make the small motor more efficient may exceed the energy saved by higher efficiency during the useful life of the motor.

1-4. The summer months. During the colder weather, the heat generated by the

1-5. The energy and cost involved in making a device more efficient may exceed the cost and energy savings during the expected life of the more efficient device.

CHAPTER 2 2-1. The efficiency formula based on power assumes that the system draws power from the source only when the system is producing power. Thus, the time factor cancels out:

Pout Wout / t Wout = = Pin Win / t Win 2-2. For a solid conductor to carry current, free electrons must return to valence electrons.

Part of the energy of the free electron is converted to heat energy when the free electron is captured by an ion and returns to the lower-energy level of a valence electron.

2-3. By definition, resistance is the electrical property that converts electric energy to heat energy. Also, free electrons produce heat when they are captured by ions. More free electrons must be returned to valence electrons in a long conductor than in a short conductor for the current to travel through the conductor.

2-4. By substituting equivalent units, it can be shown that the unit is the watt.

N-m´ A JA JA J = = = = W C C As s

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inefficient lights reduces the amount of coulomb Note: Ampere = ; therefore, heat needed from other sources. During the second hotter weather, heat from lights makes the coulomb = ampere-second house even hotter.

2-5. By substituting equivalent units and 3-6. A = C/s and A2 = C2/s2 canceling out seconds, we have:

Joule watt-second watt = = Coulomb ampere-second ampere coulomb Note: Ampere = ; therefore, second coulomb = ampere-second 2-6. Yes. Some people still have a practical knowledge of the power a horse can produce in pulling a load, so this gives them a point of reference. No. Few people have much knowledge of the power of a horse. If electric motors and combustion engines were rated in watts, people would soon develop some reference point such as a lkW motor.

2-7. 450 W Wout = 27 Wh × 3600 s/h = 97,200 Ws t = 18 min × 60 s/min = 1080 s Pout = 97,200 Ws/1080 s = 90 W Pin = Pout/eff. = 90 W/0.2 = 450 W

CHAPTER 3

ΩW=

V J/C Js = = A C/s C2

J=

C2 Js s ´ ´ =J s2 C2 1

3-7. Since an open switch physically interrupts a circuit, the ammeter could be connected to the terminals of the open switch. The ammeter then completes the circuit, and the current will flow though the lamp and the meter.

3-8. Calculated resistance is 240 Ω. The current through the lamp when measuring resistance is extremely small compared to the current when 120 V is applied to it.

Thus, the measured resistance is the “cold” resistance, and the calculated resistance is the “hot” resistance. The temperature coefficient of the filament material causes the resistance to increase as the material gets hotter. Since the difference between the temperatures of the “hot” and “cold” conditions is extreme, so is the difference between the resistances.

3-9. The power would quadruple because P = V 2IR and doubling the value of V will quadruple the value of V 2.

3-1. Watt = ampere × volt, J/s = C/s × J/C, J/s = J/s 3-2. No. All conductive materials have some resistance. In these circuits, it must have been assumed that the resistance of the CHAPTER 4 wires, fuse, and switch was so small, compared to the resistance of the load, that 4-1. R = V/I = 5.49 V/3.4 A = 1.6 Ω it could be ignored.

P = IV = 3.4 A × 5.49 V = 18.7 W 3-3. The internal resistance must be very low.

4-2. I would expect the internal resistance to Otherwise, inserting the meter into the increase for several reasons: circuit would significantly reduce the circuit a. Carbon has a negative temperature current. coefficient, and cell current flows in the 3-4. I would buy a DMM for the following carbon rod. reasons: b. Chemicals tend to be less active at a. No “ohms adjustment” is required. lower temperatures, so the b. No selection of an appropriate scale is depolarizing agent would be less required. effective. c. No figuring of major and minor division Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

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3-5.

4-4.

4-5.

4-6.

value is required. d. No cell and/or battery needs to be changed.

W = cost ÷ rate = 120 /c ÷ 8 /c /kWh = 15 kWh P = W ÷ t = 15 kWh ÷ 6 h = 2.5 kW R = V 2 ÷ P = 240 V 2 ÷ 2500 W = 23 Ω One might consider these environmental issues: a. Environmental damage or change caused by obtaining the raw materials used in manufacturing the cells. b. Environmental consequences of the manufacturing process itself. c. Problems involved in safely disposing of used cells.

With alternating current, the voltage periodically reduces to zero as the current periodically reverses. Since some minimum critical values of current and voltage are needed to maintain an arc (ionized air), it takes less direct current (constant voltage and current) than alternating current to maintain an arc between the switch contacts.

The glow in the lamp is caused by the positive neon ion giving off light energy when it returns to a neon atom/molecule.

This occurs only at the negative electrode (i.e., the electrode connected to the negative terminal of the power source) of the lamp where the positive ion picks up an electron.

Rmin = 100 Ω – 5 Ω = 95 Ω Iload = 50 V/(95 Ω + 5 Ω) = 0.5 A Rmax = 0.5 A2 × 95 Ω = 23.75 W Rmin = 50 Ω − 5 Ω = 45 Ω Iload = 60 V/(45 Ω + 3 Ω) = 1.25 A Vload = 1.25 A × 45 Ω = 56.25 V

4-3. Since the terminal voltage was considered to be independent of load current, it must have been assumed that the internal resistance of the power source was insignificant compared to the other resistance(s) in the circuit.

CHAPTER 5 5-1. Several additional applications are: a. Refrigerator—the interior lamp and the compressor motor are in parallel and operate independently of one another. b. Kitchen range—each burner and oven is in parallel with the other, and each can be independently operated. c. Numerous industrial machines have motors, lights, and heaters operated in parallel.

5-2. RT,min = 150 − (150 × 0.05) = 142.5 Ω VLT =

142.5´ 90 = 84.54 V 151.7

PT,max = 84.542/142.5 = 50.15 W 5-3. Yes. Even though the transfer efficiency is only 50 percent, it is desirable to transfer maximum power because of the cost and complexity of producing the relatively small amount of audio power available from the amplifier.

4-7.

5-4. No. The 50 percent efficiency would be very costly because of the high power involved, and the generator would rapidly burn out 4-8. as it tries to dissipate 50 kW. 5-5. Substituting G = 1/R into GT = G1 + G2 yields l/RT = l/R1 + l/R2. Now, taking the 4-9. reciprocal of both sides of the formula 1.728´ 107 J yields W= = 4.8 kWh = 4800 Wh 3.6´ 106 J/kWh 1 RT = 1/ R1 + 1/ R2 W = VIt 5-6. Series. In a parallel circuit, connecting the voltmeter across any component can Specify any arbitrary value of time in hours. increase IT and PT only a slight amount. In a t = 1.6 h series circuit, connecting the voltmeter to a Solve for I: component converts the circuit to a seriesCopyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

26

I=

4800 Wh W = = 125 A Vt 24 V ´ 1.6 h

Cell capacity = It = 125 A × 1.6 h = 200 Ah (If t is specified as 1 h, the calculated current equals the ampere-hour capacity.)

4-10. Vchange = 9 V − 8.8 V = 0.2 V Ichange = 500 mA − 250 mA = 250 mA = 0.25 A Rinternal = 0.2 V ÷ 0.25 A = 0.8 Ω

5-8. Calculate IT when all resistors are on the low end of their tolerances. Repeat for the high end of their tolerances. Open the switch and measure IT by connecting an ammeter to the switch terminals. Using the tolerance of the ammeter, determine the range of the true current for the measured values. Is the range of the true current outside the range of the calculated current?

5-9. Turn on the circuit and measure the voltage across each resistor to be sure that the fuse, switch, battery, and connecting wires are not faulty. Turn off the circuit and remove the fuse. While being careful not to burn a finger, quickly and gently touch each resistor with a damp finger to determine whether it has been dissipating power.

5-10. VRint = 0.04 Ω × 150 A = 6 V Vload = 24 V – 6 V = 18 V PT = 24 V × 150 A = 3600 W Pload = 18 V × 150 A = 2700 W eff. = 2700 W ÷ 3600 W = 0.75 % eff. = 75% 5-11. R3 = 3.33 kΩ R2 = 8.0 kΩ

CHAPTER 6 6-1. No. Both sources cannot be the dominant source; nor can both sources be charging.

6-2. Yes. In the bridge circuit (Fig. 6-1 in the text), IR5 = 0 when R1/R4 = R2/R3. Another example would be the circuit in Fig. 6-5 in the text. In this circuit, IR3 = 0 when B1 = B2 and Rl = R2. Also, IR3 = 0 when B1/Rl = B2/R2.

6-3. No. The higher-capacity battery would have

parallel circuit because of the meter’s internal resistance. If circuit resistances are high, the voltmeter can significantly change the circuit.

5-7. The power dissipated by R1 and R3 would be 44 W and 57 W until one of the resistors opened. Resistor R3 would probably open because it would have the largest overload.

6-5. VR1 = 2 A × 10 Ω = 20 V VR2 = 1.2 A × 20 Ω = 24 V 20 V + 24 V = 44 V 6-6. VR2 = 2 A × 10 Ω = 20 V VR3 = 0.8 A × 15 Ω = 12 V VR5 = 0.2 A × 30 Ω = 6 V 44 V = 20 V + 12 V + (–6 V) + 18 V = 44 V

CHAPTER 7 7-1. The force of magnetic attraction between the aligned domains must not be as great as the forces of the random motions of the atoms and molecules of which the material is composed.

7-2. From the discussion of “Transformer Action” and “Flux—The Weber,” one can write (and solve) the formula: induced voltage number of turns 15 V = = 0.5 Wb/s 30 t

Rate of flux change =

mmf ; so, mmf = H1 = 50 A ×t/m´ 7-3. l 0.45 m = 22.5 A ×t mmf 22.5 A ×t Next, mmf = IN ; so, I = = N 85 t = 265 mA 7-4. m Relativepermeability(mr ) = ;therefore, m = mair H=

mr mair = 5000´ 1.26´ 10- 6 = 0.0063 Wb/(A ×t ×m)

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a larger effective plate (electrode) area, which would reduce its internal resistance; thus, for a given load, its terminal voltage would be greater.

6-4. An equivalent-circuit voltage source has its greatest efficiency for the smallest load current (largest load resistance), while an equivalent-circuit current source has its best efficiency for the largest load current.

As the load resistance approaches zero, the voltage-source efficiency approaches zero, while the current-source efficiency approaches 1.

7-6. Some limitations that might be discussed: a. The weight it can lift varies with the shape of the object. The smaller the air gap between the object and the electromagnet, the more the magnet can lift. b. Nonmagnetic materials cannot be moved with the electromagnet. c. Permanent-magnetic material will release from the electromagnet when it is turned off only if the weight of the material overcomes its attraction for the temporary-magnet material used in the electromagnet.

CHAPTER 8 8-1.

8-2.

8-3. Double the speed of rotation and halve the

B ; therefore, B = mH = 0.0063 Wb/ H (A ×t ×m) ´ 50 A ×t/m = 0.315 T Next, m =

7-5. Direct current. With alternating current, the strength of the permanent magnet would depend on the strength of the mmf at the instant the switch was turned off. Thus the strength of the group of magnets would vary randomly from zero to maximum. Also, the switch would have to be turned on and off for each magnet. With direct current, each magnet would have the same strength and the switch could be left on all the time.

8-6. Any single-phase motor connected to the line that connects to the open coil would be inoperative. All other single-phase motors, whether they are connected line-to-line or line-to-neutral, would operate. Again, the three-phase motors would not start.

8-7. The load phase current will be 1.73 times as large as the source phase current.

8-8. The source phase current will be 1.73 times as large as the load phase current.

8-9. The general formula for the frequency of a generator can be developed from the explanation of the operation of a generator.

The formula is: f = speed of rotation × pairs of poles per revolution × cycles per pair of poles When speed is in revolutions per second, the unit for f becomes:

r pairsof poles cycles cycles ´ ´ = = Hz s r pairsof poles s Since one pair of poles always produces one cycle, the numerical value of cycles per pair of poles will always equal 1. Hence, it is dropped from the formula. Because all “pairs of poles” are passed each revolution, the numerical value of pairs of poles per revolution will always equal the number of pairs of poles. Thus, the “revolution” is dropped from pairs of poles per revolution.

Finally, the “divide by 60” in the working formula converts revolutions per minute to revolutions per second.

Therefore, the unit for f in

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flux density of the field.

8-4. It will have no effect on the 208-V loads.

The effect on the 120-V loads will depend on the number and size of loads. If only one load is connected, it will be inoperative. If two or three equal loads are connected to different lines, they will operate normally. If two or three unequal loads are connected, the lightest load will receive too much voltage and will probably burn out.

8-5. There should be no effect on the singlephase motors. The motor(s) connected across the open coil will receive the correct voltage from the other two coils. There is the possibility that the other coils may be overloaded. The three-phase motors would not start. b.

Fig. 8-2 Output voltage plotted against time.

r/min ´ pairsof poles 60 will be hertz. f=

8-10. a.

Fig. 8-1 Output voltage plotted against degrees of rotation.

9-7. Voltage. In a series circuit there is only one current and the total voltage splits into Vresistance and Vreactive.

9-8. Current. In a parallel circuit, all voltages are the same and the total current splits into the branch currents.

9-9. No. Only the magnitudes of current and voltage and the phase angle are used in calculating power.

CHAPTER 10

10-1. C1,2 = (2 × 3)/(2 + 3) = 1.2 µ CT = 1.2 + 1.5 = 2.7 µF XCT = 1/(6.28 × 100 × 2.7 × 10−6) = 590 Ω CHAPTER 9 10-2. 6 µF; the formula for series capacitors was 9-1. Yes. One coil (electromagnet) could be solved for C2 as follows: connected across the source so that its mmf CT = C1C2/(C1 + C2)—multiply by (C1 + C2) would follow the voltage. The other coil C1CT + C2CT = C1C2—subtract C2CT and could be connected in series with the load factor out C2 so that its mmf would follow the current.

C1CT = C2(C1 – CT)—divide by (C1 – CT) and Then the direction and magnitude of the solve force between the two electromagnets C2 = C1CT/(C1 – CT) = (2 × 1.5)/(2 – 1.5) = would be a function of both the current and 6 µF the voltage.

10-3. 40.5 percent. The following procedure was 9-2. The motors should be loaded to rated used: value. A fully loaded motor is more resistive Known: Q = Q1 = QT for a series circuit, Q = and less reactive than a lightly loaded VC, and CT = C1C2/(C1 + C2) motor. The higher the PF, the more efficient Solution: Q1 = QT; so by substitution V1C1 = the system.

VT CT = VT × C1C2/(C1 + C3) Dividing by C1 yields V1 = VT × C2/(C1 + C2) 9-3.

Assign 100 V as VT so that V1 will equal the percentage of VT on C1 V1 = 100 × 0.68/(1 + 0.68) = 40.5 V Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

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10-4. 26.7 percent 10-5. Yes; the current in the leads of the capacitor periodically reverses direction as the capacitor charges and discharges.

10-6. Known: F = C/V and V = J/C We can drop the 0.5 since we are interested only in units.

9-4.

C J J C J J ´ ´ = ´ ´ V C C J/C C C C2 J J = ´ ´ =J J C C

W = CV 2 =

9-5. Pout = 5/8 hp = 0.625 hp × 746 W/hp = 466.25 W Pin = 4 A × 240 V × 0.72 = 691.2 W eff = 466.25 W/691.2 W = 0.675 9-6. No. Either way, the voltage is the reference and the current phasor is in the same place.

The cosine is the same for 290° and 70°.

10-8. Since voltage and capacitance are inversely proportional, the 8-μF capacitor will drop 3/11 of the source voltage and the 3-μF capacitor will drop 8/11 of the source voltage.

Proof: For this series circuit we know that Q1 = Q2 = QT, VT = VC1, + VC2, Q1 = C1VC1, and Q2 = C2VC 2. Therefore, C1VC1 = C2VC 2 and VC1 = (C2VC 2)/C1. Now, substituting into the VT formula and solving for VC 2 yields: VT = [(C2VC 2)/C1] + VC 2 VTC1 = C2VC 2 + VC 2C1 VTVC1 = VC 2(C2 + C1) VT C1 VC2 = C2 + C1 Solving for the data given:

VC2 =

40 V ´ 3 mF 3 = ´ 40 V = 10.91 V 8 mF + 3 mF 11

10-9. XC = VC/IT, because VC would be part VR and part VX. VT = VC1 + VC 2 because VC1 may not be in phase with VC 2.

10-10. Known:

10-7. C = Q/V = 0.3 C/500 V = 600 μF W = 0.5 CV 2 = 0.5 × 600 × 10−6 × 5002 = 75 J P = W/t = 75 J/O.0005 s = 150 kW

CHAPTER 11 11-1. The ferrite slug inductor would have the larger range because the difference between the µ of air and ferrite is much greater than between air and brass.

11-2. An inductor shield needs to shield the inductor from stray magnetic fields.

This can be done only with high µ (low reluctance) materials.

11-3. The swinging choke would have the larger inductance because the air gap in the core of the smoothing choke would increase the reluctance of its core. This higher reluctance would reduce the flux change per unit of current charge.

11-4. Since the same current must flow through both inductors, the two series inductors would be equivalent to one inductor with a larger core and more turns than either of the series inductors.

11-5. Adding a parallel inductor does not change the voltage across or the current through the original circuit inductor. The new inductor increases the total current, and thus it must reduce the total opposition (reactance). This is the same way parallel resistors affect total current and total opposition.

11-6. Every time the field reverses, the magnetic domains must turn 180º to realign with the field. Overcoming the opposition of the atomic/molecular structure to this realignment converts some electric energy

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C=

Q ,VT = VC1 = VC2 , and QT = Q1 + Q2 V

Therefore, Q Q + Q2 CV + C2VT CT = T = 1 = 1 T VT VT VT  C1 + C2 10-11. Known: Q C = , VT = VC1 + VC2 , and QT = Q1 = Q2 V Therefore,

CT = =

QT QT = VT VC1 + VC2

to heat energy.

20 V V = 154ΩW 11-7. XL = L = I L 0.13 A XL = 2p fL Rearranged:

f=

11-8. We can use either Xl, or L because they are directly proportional (XL = 2πfL).

46 mA ´ 0.2 H I L 11-9. I L2 = T 1 = L 1 + L 2 0.2 H + 0.15 H = 26.3 mA

QT 1 = QT / C1 + QT / C2 1/ C1 + 1/ C2

10-12. Yes. Both C1 and C2 would be charging through R1. Thus, T would increase and it would take longer for C1 to charge from 70 V to 80 V. This would increase the period of the waveform.

11-10. No. Impedance is composed of resistance and reactance, and reactance always causes current and voltage to be out of phase.

11-11. Change in I = 60 mA − 20 mA = 40 mA = 0.04 A Time for change = 2 ms = 0.002 s cemf produced = 0.5 V L (in H) = 0.5 V/(0.04 A/0.002 s) = 0.5 V/(20 A/s) = 0.025 H = 25 mH

12-5.

12-6.

12-7.

CHAPTER 12 12-1. With 100 percent coupling the secondary voltage would be 40/5 = 8 V. Since 6 V is 75 percent of 8 V, the percentage of coupling is also 75 percent. Thus, the coefficient of coupling is 0.75.

12-2. If the window size was increased while the cross-sectional area of the core was held constant, the length of the flux path would be increased. This would increase the reluctance and reduce the flux for a given mmf. If the cross-sectional area of

154 W XL = = 613 Hz 2p L 6.28´ 0.04 H

12-8.

12-9.

If the two windings are series-opposing, then: Effective turns on “new primary” = 600 – 30 = 570 Turns per volt = 570/100 = 5.7 Voltage produced by “old 8-V” secondary = 40/5.7 = 7.02 V The new voltage would be 7.02 V Any imbalance in the voltage of the three windings; this imbalance could be caused by one winding having one too many or one too few turns or by one leg of the core having a larger air gap (more leakage flux) than the other legs.

Because there is no path for current flow in a wye-connected secondary until a load is connected Secondary power = 120 V × 6 A × 0.75 = 540 W Primary power = 540 W/0.9 = 600 W Primary apparent power = 230 V × 3 A = 690 VA PF = 600/690 = 0.87 The secondary voltage would be reduced because of the change in the turns ratio.

Because of the increased primary inductance, the energizing current would be reduced. Angle theta would increase because adding turns increases reactance proportionately more than it increases resistance.

The 10-Ω resistive load would produce the larger change because the resistance uses

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31

the core was reduced to accommodate the larger windows, the same reduction of flux would result.

The flux per unit of mmf could be held constant by increasing the cross-sectional area enough to compensate for the increased length of the flux path. Of course, this larger core would be heavier and more expensive.

12-3. The power rating is based on the currentcarrying capacity of the secondary coil wire. If the load is partly reactive, the PF will be less than 1 and the current will exceed rated value before the rated power is reached.

12-4. Primary turns = 5 × 120 = 600 6-V secondary turns = 5 × 6 = 30 8-V secondary turns = 5 × 8 = 40 If the two windings are series-aiding, then: Effective turns on “new primary” = 600 + 30 = 630 Turns per volt = 630/100 = 6.3 Voltage produced by “old 8-V” secondary = 40/6.3 = 6.35 V The new voltage would be 6.35 V would draw 2.5 A from the 20-V source; so the source power would be 50 W and the secondary load power would be 25 W. For case b: The results, without matching, are Pload = 5 W and % eff. = 33 percent.

With matching, the results are Pload = 5.62 W and % eff. = 50 percent.

Calculations used to determine results with matching were: Zratio = 10 Ω/5 Ω = 2:1 Tratio = 2 = 1.414:1 Vpri = Vsource/2 = 15 V/2 = 7.5 V Vsec = 7.5 V/1.414 = 5.3 V Pload = V2/R = 5.32/5 = 5.62 W 12-11.

power and the reactance does not. The secondary power comes from the primary so theta decreases and Ipri increases to furnish the power required by the secondary.

12-10. For case a: Without matching, the 20 W of source power provides 16 W of load power for an efficiency of 80 percent (16 W/20 W = 0.8).

In Chapter 5 of the text it was shown that matching impedance produced maximum power transfer but an efficiency of 50 percent. So, with matching, the efficiency would be reduced from 80 percent to 50 percent. The matching transformer would need a 1 : 2 turns ratio to reflect 4 Ω to the primary. The 4 Ω in series with the 4 Ω of internal resistance I T 2 = I R2 + I L2,so

I L = I T 2 - I R2 = 29.582 - 12.52 = 26.81mA V 15 XL = L = = 559.5 W I L 26.81mA XL = 2p fL, so X 559.5 L= L = = 178 mH 2p f 6.28´ 500 13-4. XL = 2πfL = 6.28 × 500 × 0.1 = 314 Ω PF = cos θ = 0.82, so θ = 34.92, so tan θ = 0.698 opposite XC - XL tan q = = , so adjacent R XC = (R × tan θ + XL = (1000 × 0.698) + 314 = 1012 Ω 1 XC = , so 2p fC 1 1 C= = 2p fXC 6.28´ 500´ 1012 = 0.314 mF The required capacitance is 0.314 µF. 13-5. P = IR2R, so I R = P/ R = 0.0954/1000 = 9.77 mA V I R = I T and Z = T , so IT

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32

Z=

25 V = 2559 W 9.77 mA

CHAPTER 13

Z=

R2 + XL2, so

13-1. The factors are the resistance, the capacitance, and the frequency. To minimize the phase shift, use large values of resistance and capacitance and don’t operate at low frequencies.

13-2. tan 25 = 0.4663 = tan θ 1 1 XC = = 2p fC 6.28´ 200´ 0.68´ 10- 6 = 1170.9ΩW

XL = Z 2 - R2 = = 2356 W

opposite XC = , so adjacent R 1170.9 WΩ XC R= = = 2511ΩW tan q 0.4663 The required resistance is 2511 Ω.

13-3. cos 65° = 0.4226 = cos θ 15 V V IR = R = = 12.5 mA R 1200 WΩ I cos q = R , so IT 12.5 mA I IT = R = = 29.58 mA cos q 0.4226 1 = 13-7. fr = 6.28 LC tan q =

1 - 3

6.28 0.1´ 10 ´ 0.001´ 10- 6

XL = 2p fL, so X 2356 L= L = = 1.5 H 2p f 6.28´ 250 The required inductance is 1.5 H. 1 , so 13-6. fr = 6.28 LC 2 2 æ ö æ 1 ö 1 ÷ ÷ ç ç ÷ ÷ C= ç /L = ç /0.01 ÷ ÷ ÷ çè6.28 f ø çè6.28´ 450´ 103 ÷ ø r

 12.52 pF The required capacitance is 12.52 pF.

14-2.

= 503.5 kHz

XL = 6.28 fL = 6.28´ 503,500´ 0.0001 = 316.2 W Q=

13-8.

XL 316.2 = = 15.8 R 20

13-9.

YX = 144- 81 = 7.94 Y1 = Y3 + YX = 18 – 7.94 = 10.06 Y1 = 10.06 m Using data from question 13-8 yields

The required features would include: a. A tachometer to monitor the motor rpm so that the motor can be loaded to rated speed b. A variable braking mechanism so that the load on the motor can be varied c. A torque arm of known length connected to the braking yoke d.

14-3.

The quality of the series LC circuit is 15.8 122 = 92 + YX2

YX2 =122 – 92

25592 - 10002

14-4.

A device to measure the force produced at the end of the torque arm

Interchanging the capacitors would have no effect on the starting characteristics.

However, the running torque would be severely reduced, and more important, the electrolytic capacitor would rapidly overheat and might explode.

Shaded-pole motors are inefficient because current continuously flows in the shading coils. This current produces a continuous PR loss while the motor is running, but it provides very little, if any,

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33

Y1 = Y3 + YX = 18 + 7.94 = 25.94 Y1 = 25.94 m 13-10. We cannot use this formula because R and X must be added as phasors as well as reciprocally.

13-11. RC filters are used more often because the values of L needed at audio frequencies are generally large, expensive, and of fairly low quality.

14-5.

14-6.

CHAPTER 14 14-1.

14-8.

Pmax = (0.75 hp × 746 W/hp × 1.1)/0.82 = 750.5 W Imax = 750.5 W/(0.77 × 115 V) = 8.5 A cmmin = 300 cm/A × 8.5 A = 2550 cm From the wire table, 16-gage wire will handle the current if the voltage drop is not excessive From the wire table, R = 4.016 Ω/1000 ft Wire length = 95 ft × 2 = 190 ft Rwire = 4.016 Ω/1000 ft × 190 ft = 0.763 Ω Vdrop = 8.5 A × 0.763 Ω = 6.5 V Vterm = 114 V – 6.5 V = 107.5 V Vmin term = 120 V – (0.1 × 120 V) = 108 V Since Vterm is less than Vmin term, the 16gage wire has too much voltage drop.

Therefore, use 15-gage wire.

14-7.

The differential compound motor should have less torque because the series portion of the field would cancel more of 15-2. the parallel portion of the field flux as the armature current increased to provide more torque.

The differential compound motor 15-3. should have better speed regulation because increasing the load would cause more armature current; this in turn would decrease the field flux and cause the armature to maintain or increase speed to produce the required cemf.

15-4.

0.7376 TS P 14-9. = so 746 5252 746´ 0.7376 TS 550.26 TS P= = = 0.1048 TS 5252 5252

running torque.

Pout = 2.8 hp × 746 W/hp = 2088.8 W Pin = 2088.8/0.87 = 2400.9 W I = P/V cos θ = 2400.9/(220 × 0.82) = 13.31 A Vdrop = 225 V – 220 V = 5 V Rwire = 5 V/11.58 A = 0.432 Ω 0.432 Ω/200 ft = 2.159 Ω/1000 ft From the wire table, 13 gage has 2.003 Ω per 1000 ft; so 13 gage will suffice.

The low frequency would not produce the required reactance in the windings so the motor would draw excessively large current. If the motor did not overheat with no load, it certainly would with a full load. Of course, the motor would not operate at rated speed.

This capacitor must carry many amperes of start-winding current when the motor is first turned on. To do this, it has to have a low reactance of 60 Hz; therefore, it must have a large capacitance. To obtain a large capacitance with reasonable cost and size, the capacitor must be an ac electrolytic. An electrolytic capacitor of this capacitance has fairly low Q, and the I 2R loss would cause it to overheat if operated continuously.

Assuming no error in the voltmeter, the measured voltage could be as small as 3.38 V or as large as 4.14 V It is nonlinear because the force between magnetic poles follows the inverse square law. Thus, the further the vanes are separated, the smaller the force created by a given mmf.

Low-temperature-coefficient shunt and multiplier resistors are necessary if a meter is to be accurate when used in environments in which the ambient temperature may range from –70° to +130°F. The ac ranges have lower sensitivity because the alternating current must be rectified before the d’Arsonval meter movement will respond. Ten volts of rectified alternating current does not yield the equivalent of 10 V of direct

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P = 0.1049 TS where P is in watts T is in Newton-meters S is in revolutions per minute 14-10. Since the beam is in balance the torque on each end must be equal. The torque on the right side is: T = 135 lb × 2 ft = 270 lb-ft The weight on the left side must be: 270 lb-ft torque Weight = = = 77.1lb arm length 3.5 ft

current in terms of deflecting a meter movement. Thus, the multiplier for an ac voltage range will be smaller than for an equivalent dc voltage range.

9V 15-5. For zero ohms, RT = = 45 kWΩ 0.2 mA 9V For 15 percent deflection, RT = 0.2 mA ´ 0.15

 300 kΩ

15-6.

CHAPTER 15 15-1.

Input resistance (RI) = 3 kΩ/V × 5 V = 15 kΩ R1(min) = 10 kΩ − (0.1 × 10 kΩ) = 9 kΩ R2(max) = 10 kΩ + (0.1 × 10 kΩ) = 11 kΩ

9 kW´Ω 1 5 k WΩ = 5.625 kWΩ 9 kW+ Ω 1 5 k WΩ 11kW´Ω 1 5 k WΩ R2,I = = 6.346 kWΩ 11kW+ Ω 1 5 k WΩ 10 V ´ 5.625 kWΩ VR1(measured) = = 3.38 V 5.625 kW+ Ω 1 1 k WΩ 10 V ´ 6.346 kWΩ VR2 (measured) = = 4.14 V 6.346 kW+ Ω 9 k WΩ 43.27 V 4.7 mWΩ = 46.73 V R1,m 46.73 V´ 4.7 MWΩ R1,m = = 5.076 MWΩ 43.27 V R1,I =

15-7.

15-8.

VR2 = VT - VR1 , m = 90 V - 46.73 V = 43.27 V 16-2.

What value of Rm in parallel with R1 equals 5.076 MΩ?

RR 1 m where R1 = 6.8 MWΩand RT = R1 + Rm 5.076 MW Therefore: The meter’s internal resistance is 20 MΩ.

RT =

15-9.

1 = 2 ms = time for 500 kHz one count, and one count should represent 1 pF.

Time for one cycle =

R15% = 300 kΩ – 45 kΩ = 255 kΩ 15 percent deflection represents 255 kΩ tI CV 3.3´ 10- 6 ´ 5 C = ; therefore t = = V I 2´ 10- 3  8.25 ms The time required is 8.25 ms.

The voltage at the terminals of a cell or battery is a function of the load current and the remaining charge in the cell.

Different ranges of the ohmmeter require different load currents, so the cell voltage is different for each range.

For calculation, R1 = 6.8 MΩ, R2 = 4.7 MΩ, Rm = meter resistance, and Rl,m = R1 in parallel with Rm. We know that for series circuits V and R are directly proportional and VT = V1 + V2.

16-3.

Cost for lighted handles is: Cost = W × rate = 1.9272 kWh × l0 /c /kWh = 19.3 /c Cost for 200-W lamp is: t = 2 h/day × 365 days = 730 h W = Pt = 200 W × 730 h = 146 kWh Cost = W × rate = 146 kWh × lO /c /kWh = 1460 /c Total cost is: 19.3 /c + 1460 /c = 1479.3 /c = $14.79 % of cost attributable to handles is: % of cost = (l9.3 /c ÷ 1479.3 /c ) × 100 = 1.3% In the simple circuit, you know the off position of the switch even when the light bulb is burned out. In the three-way circuit, you can’t determine the off

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35

It V Therefore: A 5-µA current source is required.

RT R1 + RR 1 m = RR 1 m RT R1 = RR R 1 m T Rm = Rm( R1 - RT ) R R 5.076 MW´Ω 6.8 M WΩ Rm = T 1 = = 20 MWΩ n C=

¨

16-4.

CV 1pF´ 10 V I= = = 5 mA t 2 ms 15-10. Worst case is R1 = 100 kΩ – (0.1 × 100 kΩ) = 90 kΩ, R2 = 100 kΩ + (0.1 × 100 kΩ) = 110 kΩ, and Vmeter reading = VR1 – (0.02 × 50 V) = ?

Vmin meter reading = 45 V – (0.02 × 50 V) = 44 V

VR1 =

100 V ´ 90 kWΩ VT R1 = = 45 V R1 + R2 90 kW+ Ω 1 1 0 k WΩ

CHAPTER 16 16-1.

16-6.

16-7.

Time handles are lit = 24 h/day – 2 h/day = 22 h/day Total time = 22 h/day × 365 days = 8030 h I for two switches = 2 × 1 mA = 2 mA W for two switches is: P = IV = 2 mA × 120 V = 240 mW W = Pt = 240 mW × 8030 h = 1927.2 Wh is a good reason to always trip the circuit breaker on a branch circuit before repairing any part of it!

All conductors have some resistance.

Thus, they create heat when current flows. The 40 percent rule ensures that conductors will not get too hot when conducting their rated current. Also, when a conduit run has its maximum allowable number of 900 bends, pulling conductors with more than a 40 percent fill would be difficult.

Two possible solutions are shown below.

The first solution requires operating two valves at each location. Both solutions are

16-5.

position when the bulb is burned out. So, when the three-way circuit is on and the fixture is improperly wired, the screw shell of the fixture and the screw of the bulb will be hot (120 V with respect to ground). These parts might be accidentally contacted when a bulb is changed in a dark area.

If the neutral conductor ever lost continuity, the equipment enclosure, if physically isolated from earth ground, would be hot with respect to ground. The internal resistance of the equipment is so small compared to one’s body resistance that nearly all of the source voltage would appear across the body when the grounded body made contact with the equipment enclosure. For the above situation to develop in a three-wire circuit, the equipment would have to be shorted and the equipment-grounding wire would have to be interrupted.

Some reasons not to switch the neutral are: a. When the neutral is opened by a switch, the electrical circuitry in all the equipment connected to the rest of the branch circuit is left ungrounded. b. Some people will repair or replace a light fixture after turning the light off with the light switch. If the neutral is switched, both leads of the fixture will still be hot with respect to ground. A severe or fatal shock could result. This shown in the off position and can be changed to the on position from either location.

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37

C Answers to Homework Problems Introduction to Powers of 10 – Homework Solutions 1. a. 3.3 × 104 b. 1.8 × 10−5 c. 2.257 × 106 d. 6.8 × 10−2

9. a. 2.4 × 104 b. 5.0 × 10−7 c. 2.0 × 10−4 d. 3.35 × 10−3

2. a. 275 b. 0.0000934 c. 510,000,000 d. 0.000000466

10. a. 105 b. 10−9 c. 10−12 d. 1015

3. a. 22 × 106 b. 1.5 × 10−3 c. 75.2 × 103 d. 25 × 10−9

11. a. 2.5 × 109 b. 1.6 × 10−3

4. a. 25 µV b. 470 kΩ c. 5 mA d. 18 kW

12. a. 4.472 × 104 b. 4.0 × 10−2 13. a. Enter the problem using the following keying sequence:

5. a. 25,000 µA b. 1,200 kΩ c. 1.5 nF d. 0.033 µF 3

6. a. 6.2 × 10 b. 7.0 × 10−4 3

7. a. 1.35 × 10 b. 5.85 × 10−6 8. a. 7.2 × 10−3 b. 1.5 × 10−5 c. 3.2 × 106 d. 2.64 × 10−2

4.0 EXP 5 × 25 EXP ± 2 = The calculator will display the answer as 100.000 × 103 in engineering notation. b. Enter the problem using the following keying sequence: 550 EXP 4 + 1.5 EXP ± 7 = The calculator will display the answer as 5.500 × 106 in engineering notation (Because 550 × 104 is so much larger than 1.5 × 10−7, the answer is basically 5.5 × 106)

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38

Chapter 1 Homework Problems – Solutions 1.

2.

3.

4.

5.

6.

7.

8.

9.

Work can be defined as a force moving through a distance whereas energy can be defined as the ability or capacity to do work. Energy is independent of work because it exists whether or not any work is done. Work, however, requires energy.

The symbol for both work and energy is W. The base unit of work and energy is the joule (J). (a) The electron possesses a negative charge whereas (b) the proton possesses a positive charge.

Work (W) = force × distance W = 50 newtons × 100 meters = 5000 newton-meters Since 1 J = 1 newton-meter, 5000 newtonmeters = 5,000 joules (J) Work (W) = force × distance W = 200 newtons × 250 meters = 50,000 newton-meters Since 1 J = 1 newton-meter, 50,000 newton-meters = 50,000 joules (J) In a light bulb, electrical energy is converted into both heat and light energy.

Another example of energy conversion is an electric stove or oven. In an electric stove or oven most of the electrical energy is converted into heat energy while only a small portion is converted into light energy.

Another example of energy conversion is a motor. In a motor, electrical energy is converted into rotating mechanical energy.

Useful energy out % efficiency = × 100 Total energy in Useful energy out % efficiency = × 100 Total energy in 5250 J = × 100 7460 J = 70.38%

10.

11.

12.

13.

14.

15.

20% =

300 J

× 100 Total energy in 300 J Total energy in = × 100 20 = 1500 J A valence electron is an electron orbiting in the outermost shell of an atom whereas a free electron is a valence electron that has been temporarily separated from an atom.

The two basic particles that exist in the nucleus of an atom are the proton and the neutron.

When an atom loses one or more valence electrons the atom has more protons in its nucleus than it does orbiting electrons.

When this occurs, the atom exhibits a positive charge and is called a positive ion.

Conversely, when an atom gains one or more valence electrons the atom has more orbiting electrons than it does protons in its nucleus. When this occurs, the atom exhibits a negative charge and is called a negative ion.

When a neutral object picks up or gains additional electrons it has an overall net charge that is negative. Conversely, when a neutral object gives up or loses some of its electrons it has an overall net charge that is positive.

Static charge, either negative or positive, is electric charge that is stationary or motionless.

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39

Chapter 2 Homework Problems – Solutions 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

1 C = 6.25 × 1018 electrons.

Electric current is the unified movement of electric charge, either electrons or ions.

The electron.

Both positive and negative ions.

I The ampere (A).

1C 1A= 1s Q 5C I= Therefore, I = = 2.5 A T 2s Voltage, also known as potential difference, is an electrical pressure or force that causes current to flow.

Potential energy is energy at rest. Kinetic energy is energy in motion or energy doing work or being converted into another form.

The volt (V).

1 Joule 1 Volt = 1 Coulomb Three common sources of voltage are electric generators, electric chemical cells and solar cells.

Resistance is the opposition a material offers to the flow of electric current.

The symbol for resistance is R. (a) The resistance of a conductor is low and (b) the resistance of an insulator is high.

Covalent bonding refers to the sharing of valence electrons.

The resistance of a semiconductor material lies between that of conductors and insulators.

The temperature coefficient of a material refers to the changes in its resistance for a given change in temperature.

Resistors are electronic components that are manufactured to have a specific value of resistance. They are available in a wide variety of shapes and sizes and are available in a wide range of resistance values. They are used to control the current in a circuit.

21.

22.

23.

24.

25.

26.

27.

Power can be defined as the rate of using energy or doing work.

The symbol for power is the uppercase letter P. The base unit of power is the Watt (W).

1J 1W= 1s W 250 J P= so P = = 500 W t 0.5s Begin by determining the total number of seconds in a 5-hour period.

3,600 s t=5h× = 18,000 s h Since 1.5 kW = 1,500 W Then, W=P×t = 1,500 W × 18,000 s = 27 × 106 Watt-seconds or 27 × 106 J P % eff = out × 100 Pin =

28.

29.

30.

100 W

× 100 165 W = 60.61 % P % eff = out × 100 Rearranging to solve Pin for Pout gives us: P  %eff Pout = in 100 250 W ×72 % Pout = = 180 W 100 (a) The resistance of a conductor increases in direct proportion to its length. (b) The resistance of a conductor decreases as its diameter and cross-sectional area increase.

Power companies charge their customers according to the amount of energy they use.

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40

Chapter 3 Homework Problems – Solutions 1.

2.

3.

The four essential parts of an electric circuit are; a. An energy source to provide the voltage needed to force current through the circuit. b. Conductors through which current can flow. c. Insulators to confine the current to the desired paths. d. A load to control the amount of current and convert the electric energy taken from the energy source.

A schematic diagram is a drawing that uses only symbols to show how electronic components are connected together.

Electron flow refers to the direction electrons move or flow in a circuit. With electron flow, electrons flow from the negative side of the voltage source, through the external circuit and return to the positive side of the voltage source.

Conventional current flow assumes that positive charges leave the positive side of the voltage source, flow through the external circuit and return to the negative side of the voltage source.

7.

8.

9.

10.

11.

12.

13.

14.

15.

4.

5.

6.

See figure 3-1. a. I = 12 mA b. R = 2 kΩ c. V = 72 V a. I = 300 μA b. R = 900 kΩ c. V = 50 V

a. P = 1.8 kW b. I = 15 A c. V = 16 V a. P = 480 mW b. I = 150 mA c. V = 2.5 V a. P = 800 W b. P = 25 W c. P = 19.8 W a. P = 66.67 mW b. P = 225 mW c. P = 2.813 μW W = Pt Where P = V × I = 12 V × 20 A = 240 W W = 240 W × 5 h = 1,200 Wh or 1.2 kWh Cost = rate × energy = 12¢/kWh × 250 kWh = $0.12 × 250 kWh The cost is $30.00 Begin by calculating the power, P. P=V×I = 120 V × 15 A = 1.8 kW Next calculate the energy in kWh. #kWh = 1.8 kW × 48 h = 86.4 kWh And finally, calculate the cost.

Cost = rate × energy = 12¢/kWh × 86.4 kWh = $10.368 or $10.37 Measuring current is much more difficult (and more inconvenient) as compared to measuring either voltage or resistance. The reason is that to measure current, the circuit must be opened and the meter must be connected in series with the moving electric charges.

When measuring resistance, power must be removed from the circuit.

Chapter 4 Homework Problems – Solutions 1. A secondary cell can be recharged whereas a

13. Fig. 4-47

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41

primary cell cannot.

2. W = VIt = 12.6 V × 200 Ah = 2520 Wh or 2.52 kWh V - VLT 9V - 8.7V 3. Ri  NLT  2Ω IL 150 mA 4. a. 2.1 V b. For a fully charged lead-acid cell, the specific gravity ranges from 1.21 to 1.28.

The typical automotive battery is fully charged at a specific gravity of about 1.26. c. Specific gravity is measured with a hydrometer.

5. Four notable characteristics of a nickelcadmium cell are: a. Good low-temperature operating characteristics. b. High initial cost but low operating cost. c. Very low internal resistance which results in a minimal drop in output voltage at high load currents. d. Nearly constant output voltage as the cell is discharged by a moderate load.

6. Longer shelf life and higher energy-to-weight ratio is the main advantage of lithium cells over other types of cells.

P 4W 7. I    666.67 mA 6V V 8. P = VI = 120 V × 625 mA = 75 W 9. The four broad categories into which resistors can be categorized are: Fixed, variable, adjustable and tapped.

10. The power rating of a resistor is determined primarily by its physical size and the type of material used to make the resistor.

V 2 15.81 V 2 11. P    250 mW. Therefore, the R 1 kW power rating should ideally be twice that, which is 500 mW or ½ W. 12. Examples of four different types of resistors are; Carbon-composition, wire-wound, deposited-film and surface mount although several other types of resistors exist.

a. R = 100 Ω, ±10% b. R = 47 kΩ, ±5% c. R = 56 Ω, ±5% d. R = 820 kΩ, ±10% e. R = 3.9 Ω, ±10% f. R = 1 MΩ, ±5% 14. Reading from left to right, the colored bands for each resistor are: 2.2 kΩ; Red, red, red, silver 33 kΩ; Orange, orange, orange, gold 10 kΩ; Brown, black, orange, silver 5.6 Ω; Green, blue, gold, silver 2.2 MΩ; Red, red, green, gold 910 Ω; White, brown, brown, gold 15. Varistor.

16. See Fig. 4-1.

17. The two most important switch ratings are its current rating and its voltage rating.

18. Circular mils (cmil or CM are abbreviations for the circular mil area of round wire.)

19. There are three important fuse ratings or characteristics; it’s current rating, it’s voltage rating and it’s blowing characteristic (fastblow, medium-blow or slow-blow).

20. The main advantage of a circuit breaker versus a fuse is that a circuit breaker can be reset whereas a fuse must be replaced.

Chapter 5 Homework Problems – Solutions Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

42

1. The current, IT, is the same everywhere in a series circuit because there is only one path for current flow.

2. a. RT = R1 + R2 + R3 = 1 kΩ + 2 kΩ + 3 kΩ = 6 kΩ

VT RT 120 V = 6 kW

b. IT =

= 20 mA Note: IT = IR1 = IR2 = IR3 c. VR1 = IT × R1 = 20 mA × 1 kΩ = 20 V VR2 = IT × R2 = 20 mA × 2 kΩ = 40 V VR3 = IT × R3 = 20 mA × 3 kΩ = 60 V d. PR1 = VR1 × IT = 20 V × 20 mA = 400 mW PR2 = VR2 × IT = 40 V × 20 mA = 800 mW PR3 = VR3 × IT = 60 V × 20 mA = 1.2 W e. PT = PR1 + PR2 + PR3 = 400 mW + 800 mW + 1.2 W = 2.4 W 3. a. If R2 becomes open, there is no longer a complete path for current flow. Therefore, the series current, IT drops to zero. b. The voltage drops across resistors R1 and R3 is zero with R2 open. Mathematically this is proven as follows: VR1 = 0 A × 1 kΩ = 0 V Also, VR3 = 0 A × 3 kΩ = 0 V

c. The voltage drop across R2 equals 120 V. This can be proven with Kirchhoff’s voltage law. Kirchhoff’s voltage law states that in a series circuit the total voltage, VT must equal the sum of the individual voltage drops.

VT = VR1 + VR2 + VR3 Solving for VR2 gives us: VR2 = VT – VR1 – VR3 Inserting values gives us: VR2 = 120 V – 0 V – 0V = 120 V d. Since the current, IT is zero, the power dissipated by all three resistors is 0 W. PR1 = PR2 = PR3 = 0 W e. The total power, PT is also 0 W. 4. a. If resistor R2 becomes short-circuited, its resistance decreases to 0 Ω. This results in a decrease in the circuits total resistance, RT. RT = 1 kΩ + 0 Ω + 3 kΩ = 4 kΩ The reduction in total resistance causes the total current, IT to increase.

VT RT 120 V = 4 kW

IT =

= 30 mA b. The voltage drops across resistors R1 and R3 increase since more current through the same resistance results in a higher voltage drop.

VR1 = IT × R1 = 30 mA × 1 kΩ = 30 V VR3 = IT × R3 = 30 mA × 3 kΩ = 90 V

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c. The voltage drop across resistor R2 is 0 V since its resistance is 0 Ω.

VR2 = IT × R2 = 30 mA × 0 Ω =0V d. PR1 = VR1 × IT = 30 V × 30 mA = 900 mW PR2 = VR2 × IT = 0 V × 30 mA =0W PR3 = VR3 × IT = 90 V × 30 mA = 2.7 W e. PT = PR1 + PR2 + PR3 = 900 mW + 0 W + 2.7 W = 3.6 W 5. Kirchhoff’s voltage law states that the total voltage, VT equals the sum of the individual resistor voltage drops. Expressed mathematically; VT = VR1 + VR2 + VR3.

In problem 2, VT = 20 V + 40 V + 60 V = 120 V 6. a. VR1 = 12 V – 6.3 V = 5.7 V. The required current is 150 mA. Therefore, R1 is calculated as follows;

5.7V VR1   38 Ω IT 150 mA b. A 39 Ω resistor would be suitable in this application. c. First, calculate the power dissipated by R1 using the original value of 38 Ω calculated in part a.

R1 

PR1 

5.7 V 2 VR12   855 mW R1 38 W

To provide a safety factor of 2, choose a 39 Ω, 2-W resistor in this application.

7. a. The voltage across the rheostat, R in Fig. 5-43 is calculated as follows; VR = 17 V – 12 V = 5 V

Since R is adjusted to provide a current of 125 mA, then the value of R is calculated as follows; R=

5V VR   40 Ω I T 125 mA

b. PR 

5V 2 VR2   625 mW R1 40 W

8. RT = 300 Ω IT = 80 mA VR1 = 9.6 V VR2 = 14.4 V PR1 = 768 mW PR2 = 1.152 W PT = 1.92 W 9. RT = 9 kΩ IT = 2 mA VR1 = 3 V VR2 = 4.8 V VR3 = 10.2 V PR1 = 6 mW PR2 = 9.6 mW PR3 = 20.4 mW PT = 36 mW

R1 × VT RT 1.5 kW = × 18 V 9 kW

10. VR1 =

=3V 11. R3 = 6.8 kΩ VR1 = 54 V VR2 = 30 V VT = 220 V RT = 11 kΩ

VRL RL 7.5 V = 50 W

12. a. IL =

= 150 mA 9V - 7.5V b. RB = = 10 Ω 150 mA c. PRB = IL2 × RB (Note: IL = IT) = 150 mA2 × 10 Ω

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= 225 mW PRL = IL2 × RL = 150 mA2 × 50 Ω 1.125 W PT = PRB + PRL = 225 mW + 1.125 W = 1.35 W d. When RB = RL, maximum power is transferred from the source to the load.

Since RB = 10 Ω, RL must equal 10 Ω. e. PL(max) = 2.025 W 13. The voltage is the same across all branches in a parallel circuit because each branch is connected directly across the terminals of the voltage source.

VT R1 36V = 120 W

14. IR1 =

= 300 mA

VT R2 36V = 180 W

IR2 =

= 200 mA IT = IR1 + IR2 = 300 mA + 200 mA = 500 mA

VT IT 36V = 500 mA

RT =

= 72 Ω PR1 = 36 V × 300 mA = 10.8 W PR2 = 36 V × 200 mA = 7.2 W PT = PR1 + PR2 = 10.8 W + 7.2 W = 18 W 15. IR1 = 1 A IR2 = 2 A IR3 = 3 A IT = 6 A RT = 20 Ω

16. Kirchhoff’s current law states that the total current, IT equals the sum of the individual branch currents. Expressed mathematically; IT = IR1 + IR2 + IR3.

In Fig. 5-49; IT = 1 A + 2 A + 3 A =6A 17. a. 5 A b. Assume IX is used to designate the current in the wire connecting resistors R1 and R2.

Then; IX = I2 + I3 =2A+3A =5A 18. IR1 = 12 mA IR2 = 6 mA IR3 = 24 mA R1 = 2 kΩ R2 = 4 kΩ RT = 571.43 Ω VT = 24 V 1 1 1 19. Use the reciprocal formula RT = + + .

1 1 1 R3 R1 R2

1 1 1 + + 1 1 1 3.3 kW 150 W 330 W = 100 Ω R1´ R2 20. Use the formula RT = .

R1+ R2 RT =

100 W ´ 150 W 100 W + 150 W = 60 Ω 1 1 21. G1 = = = 8.33 mS R1 120 W RT =

1 1 = = 16.67 mS R2 60 W 1 1 G3 = = = 25 mS R3 40 W G2 =

GT = G1 + G2 + G3

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PR1 = 120 W PR2 = 240 W PR3 = 360 W PT = 720 W

= 8.33 mS + 16.67 mS + 25 mS = 50 mS

G3 × IT VR4 = 18 V GT IR2 = 60 mA 25 mS = ×6A IR3 = 40 mA 50 mS PT = 3.6 W =3A 25. RT = 300 Ω 23. a. No, R1 and R2 are not in series. The reason IT = 120 mA is that the current in R1 equals the sum of VR1 = 12 V the currents in resistors R2 and R3.

VR2 = 24 V b. No, R3 and R4 are not in series. The reason VR3 = 14.4 V is VR4 = 9.6 V that the current in R4 equals the sum of the IR2 = 24 mA currents in resistors R2 and R3.

IR3 = IR4 = 96 mA c. Yes, R1 and R4 are in series because both PT = 4.32 W resistors have the same current.

26. a. RT = 2.5 kΩ I1 = I4 = I2 + I3.

IT = 6 mA d. Yes, R2 is in parallel with R3 because both VR1 = 6 V resistors are connected across the same VR2 = 9 V two points. R2 and R3 will therefore have IR1 = 6 mA the same voltage.

IR2 = 6 mA e. No, R2 is not in parallel with the total b. RT = 2 kΩ voltage, VT. The reason is that R2 is not IT = 7.5 mA connected directly across the terminals of VR1 = 7.5 V the voltage source. Resistors R1 and R4 will VR2 = 7.5 V drop a portion of the total voltage, VT IR1 = 7.5 mA across them leaving the remaining voltage IR2 = 5 mA across resistors R2 and R3, which are in IRLoad = 2.5 mA parallel. In other words, V2 ≠ VT so they 27. IR1 = 6 mA cannot be in parallel.

IRLoad = 3 mA 24. RT = 360 Ω ID1 = 3 mA IT = 100 mA VR1 = 12 V VR2 = 6 V VR3 = 6 V 22. IR3 =

Chapter 6 Homework Problems – Solutions 1. a. 5x + 4y = 21 b. 4x = 16 c. 8x + y = 15 2. a. x = 5 and y = −1 b. x = 4 and y = 2 c. x = 1.857 and y = 0.1428 3. a. x = 1, y = 3, z = 5 b. x = −2, y = 4, z = 1

IT = 1.949 A VR1 = 5.665 V VR2 = 3.959 V VR3 = 6.335 V VR4 = 8.041 V VR5 = 1.706 V RT = 6.156 Ω 5. IR1 = 6.923 mA

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46

4. IR1 = 629.444 mA IR2 = 1.32 A IR3 = 1.056 A IR4 = 893.444 mA IR5 = 426.396 mA

6. The voltage at the junction of resistors R1, R2 and R3 (the node voltage) is 2.455 V. IR1 = 3.818 A IR2 = 6.273 A IR3 = 2.455 A 7. IR1 = 5.455 mA IR2 = 49.09 mA IR3 = 43.636 mA VR1 = 1.091 V VR2 = 4.909 V VR3 = 13.091 V 8. IR1 = 300 mA IR2 = 250 mA IR3 = 250 mA IR4 = 300 mA VR1 = 15 V VR2 = 10 V VR3 = 5 V VR4 = 3 V 9. A constant (ideal) voltage source has no internal resistance.

10. The terminal voltage with a 6 Ω load is approximately 11.813 V. 11. a. See Fig. 6-1.

b. IL = 2 mA and VRL = 3 V 12. a. See Fig. 6-2.

IR2 = 10.769 mA IR3 = 3.846 mA VR1 = 13.846 V VR2 = 16.154 V VR3 = 3.846 V

13. A constant (ideal) current source has infinite internal resistance.

14. a. ISC = 1 mA and RS = 10 kΩ b. See Fig. 6-3.

c. IL = 666.67 µA and VRL = 3.33 V 15. a. See Fig. 6-4.

b. IL = 666.67 mA and VRL = 8 V 16. a. See Fig. 6-5.

b. IL = 1.818 mA and VRL = 1.091 V c. IR1 = 9.091 mA IR2 = 7.273 mA VR1 = 10.909 V VR2 = 13.091 V 17. IN = 16 mA RN = 1.5 kΩ 18. VTH = 5 V RTH = 100 kΩ

b. IR3 = 1.25 A, VR3 = 6.25 V Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

47

c. IR1 = 375 mA IR2 = 875 mA VR1 = 3.75 V VR2 = 8.75 V

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Chapter 7 Homework Problems – Solutions 1. a. The north geographic pole has south b. Reluctance is the opposition to magnetic magnetic Polarity. flux. b. The south geographic pole has north 14. The number of magnetic flux lines that cut magnetic Polarity. across the coil turns as well as the rate at 2. Magnetic flux or just flux. which the magnetic flux lines cut across the 3. Unlike magnetic poles attract each other and coil turns. like magnetic poles repel each other.

15. a. Ampere-turn (A•t) 4. North, south. b. Weber 5. Perpendicular or at right angles. c. A•t/m 6. If you wrap your fingers around the coil in the d. Tesla same direction as the current is flowing, your e. There are no units for relative permeability, thumb will point in the direction of the north µr. It is unitless. pole.

16. a. mmf = 20 A•t 7. Attract, repel. b. mmf = 50 A•t 8. In magnetic materials, more electrons spin in c. mmf = 300 A•t one direction than in the other. Thus, not all d. mmf = 160 A•t the magnetic fields are cancelled, and a 17. 500 A•t/m molecule of the material possesses a weak 18. φ = 0.05 Wb magnetic field. These molecules then group 19. µ = 0.003 Wb/(A•t/m) together to form small magnetic domains that 20. a. The type of core material. have magnetic fields with definite poles. b. The size and shape of the core material.

9. A permanent magnet retains a high c. The number of turns on the coil. percentage of its magnetism after it is d. The amount of current in the coil. removed from a strong magnetizing field 21. Rotational force is called torque. whereas a temporary magnet only retains a 22. The commutator and brushes. small amount or percentage of its 23. A solenoid consists of a moveable plunger, a magnetism after it is removed. coil (electromagnet) and an iron core.

10. To increase the magnetizing force (mmf) 24. No, a DC relay cannot be operated from an of an electromagnet, either increase the alternating current. The reason is that the number of coil turns or increase the amount relay will chatter with alternating current of current flow. because the flux in the core builds up and 11. A magnetic material is said to be saturated if decays to zero several times a second. an increase in the magnetizing force (mmf) 25. A remotely located relay can be controlled no longer increases the amount of magnetic using a low-voltage, low-current power flux in the material. supply. The low-voltage, low-current supply is 12. An alternating current produces a connected to just the relay coil. The low continuously varying mmf whose magnetic current produces minimal losses in the polarity periodically reverses. As the strength conductors carrying current to the relay coil. of the magnetic field (mmf) decreases, so too Then the relay contacts, which are located does the strength of the magnetic field in the right at the load, can control a much higher material being demagnetized. voltage and/or current load without having 13. a. The term residual magnetism refers to the excessive losses in the conductors supplying amount of flux that remains in a given power to the load. material after it is removed from a magnetic field.

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49

Chapter 8 Homework Problems – Solutions 1. Pulsating direct current periodically reduces to zero whereas fluctuating direct current does not.

2. The amplitude variations of a sine wave vary continuously whereas the amplitude of a square wave switches abruptly between two distinct values.

3. a. One cycle includes the variations between two successive points having the same value and varying in the same direction. b. An alternation is one-half cycle. The positive half-cycle is referred to as the positive alternation whereas the negative half-cycle is referred to as the negative alternation. c. The period of an AC waveform is the length of time required to complete one cycle. d. The frequency of an AC waveform is the rate at which cycles are produced.

4. The base unit of frequency is the Hertz, abbreviated Hz. One hertz equals one cycle per second.

5. a. T = 8.33 mS b. T = 200 µS c. T = 2.5 µS d. T = 40 nS 6. a. f = 1 kHz b. f = 5 kHz c. f = 20 Hz d. f = 1 MHz 7. a. Vav = 19.11 V b. Vpp = 60 V c. Vrms = 21.21 V 8. a. VP = 33.94 V b. VPP = 67.88 V c. Vav = 21.62 V 9. a. VP = 169.56 V b. VPP = 339.12 V c. Vrms = 119.88 V 10. The rms value of a sine wave produces the same heating effect as DC. 11. a. The amount of induced voltage and/or current in a conductor is a function of the amount of flux cut per unit of time. b. The direction of an induced current and/or polarity of an induced voltage in a conductor depends on the direction the

12. a. 90° b. 180° c. 270° d. 360° 13. The three main parts of an AC generator are: a. conductor loop (usually several coil windings) b. armature c. slip rings and brushes 14. f = 120 Hz 15. f = 180 Hz 16. 600 r/min 17. a. It is easier to change the voltage level of AC as compared to DC. b. AC motors are less complex than DC motors because they do not require brushes and commutators.

18. Three-phase AC power systems have the following advantages over single-phase AC power systems. a. Three-phase AC power systems provide more efficient use of copper wire. b. Three-phase AC motors provide a more constant load on the generator. c. Three-phase motors are less complex and they provide a more constant torque. d. Three-phase AC power produces a smoother DC voltage and current when rectified.

19. 120° 20. 0 V 21. Three-phase AC generators are configured in either the wye (Y) or delta (∆) configuration.

22. Zero.

23. a. 240 V b. 240 V c. 30 A d. 30 A e. 51.962 A 24. a. 277 V b. 277 V c. 20 A d. 20 A e. 479.78 V (Approximately 480 V) 25. 30° 26. When the load is unbalanced.

27. Vphase = 230.94 V

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50

conductor is moving and the polarity 28. Iphase = 31.75 A of the magnetic field.

29. A three-phase full-wave rectifier produces a smoother DC output because its output never 30. Because the load on a three-phase AC drops below 0.866 of its peak value. generator never drops to zero. The more uniform load provides for smoother mechanical operation of the generator.

Chapter 9 Homework Problems – Solutions 1. Two waveforms are said to be in phase if both c. See Fig. 9-1(c) waveforms reach their positive and negative peaks and zero values at the same time.

V 2 120 V 2 2. P = = = 480 W R 30 W 3. P = I2R = (50 mA)2 × 100 Ω = 250 mW 4. P = 100 mW (Convert 28.28 VPP to rms) 5. a. 90° (I leads V) b. 90° (V leads I) c. 0° (V and I are in phase) 6. 0 W 7. Vectors show a direction in space whereas a phasor varies in time.

8. Negative power means that the load is 10. a. The resistive voltage is 86.6 V and the returning power to the source during a reactive voltage is 50 V. portion of the AC cycle. b. The resistive voltage is 25 V and the 9. a. See Fig. 9-1(a) reactive voltage is –43.3 V. c. The resistive current is 20 A and the reactive current is −15 A. 11. VR = 0.6 VT and VX = 0.8 VT 12. The power input to the motor is 4.157 kW. 13. The power used by the motor is 1.56 kW. 14. Apparent power is always greater the true power when both resistance and reactance are present.

15. A wattmeter measures true power.

16. The base unit of apparent power is the b. See Fig. 9-1(b) voltampere (VA).

17. a. Papp = 4.8 kVA b. Papp = 3.12 kVA 18. θ = 51.32° 19. P = 176.78 W 20. a. PF = 1 in a purely resistive circuit. b. PF = 0 in a purely reactive circuit.

21. It is important to keep the power factor (PF) as high as possible so the current in the power lines is kept to its lowest possible value.

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51

Keeping the current as low as possible significantly reduces the I2R power losses in the power line conductors.

22. a. I = 282.05 A b. I = 235.04 A c. I = 2.895 A 25. a. Papp = 10.4 kVA b. PF = 0.779 approximately. c. θ = 38.845°

23. a. PT = 4.5 kW b. Vphase = 120 V 24. a. PLoad = 33.33 kW b. VLine = 277 V

Chapter 10 Homework Problems – Solutions 9. Q  CV 1. Electrons are repelled from the negative  10 µF × 50 V terminal of the battery and flow through the  500 µC wire connected to one of the capacitor plates.

Q Electrons accumulate on this plate of the 10. C = V capacitor. At the same instant, electrons on 2.5 mC the opposite plate of the capacitor are = 10 V attracted to the positive terminal of the battery. These electrons move through the  250 nF lead connected to the positive battery Q 11. V = terminal. The net result is that one plate of C the capacitor has an excessive number of 1 mC = electrons and the other plate has a deficiency.

0.01 mF The oppositely charged plates result in a  100 V potential difference or voltage across the 12. a.

C  0.04 µF capacitor. When the capacitor voltage equals b. C  0.027 µF the battery voltage, the capacitor stops c. C  82,000 pF charging and there is no current flow. d. C  0.1 nF 2. No, electrons cannot flow through the 13. The three factors controlling the capacitance, dielectric of a capacitor because it is an C of a capacitor are; insulator. a. The area of the plates.

3. When a capacitor is fully charged, the b. The distance between the plates. capacitor voltage and battery voltage are c. The type of dielectric material used. equal. Because these voltages are opposing, a 14. Electrolytic capacitors must be connected state of equilibrium exists and current cannot with the correct polarity otherwise they may flow in either direction. explode.

4. The electric field between the plates of a 15. Common types of capacitors include; capacitor distorts the orbital path of electrons a. Disk ceramic within the dielectric material. The modified b. Mica Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

52

orbital path raises the energy level of these electrons which in turn is how a capacitor stores energy.

5. Energy is stored in the electric field of the capacitor.

6. For a capacitor, the DCWV (direct current working voltage) rating is the maximum voltage a capacitor can be charged to without breaking down. A capacitor is said to breakdown when its dielectric is punctured.

7. The base unit of capacitance is the farad (F).

8. One farad is the amount of capacitance that stores one coulomb of electric charge when the capacitor voltage is one volt.

c. Film and/or paper d. ceramic 16. W  0.5 CV2  0.5 × 1000 μF × 502 V  1.25 J 17. 63.2% 18. It takes five RC time constants for a capacitor to fully charge or discharge.

resistance losses and dielectric losses. The dielectric resistance appears as a parallel resistance whereas the plate resistances and dielectric losses appear as a series resistance.

27. CT = 40 nF 28. CT = 2 nF or 0.002 μF 19. Sometimes capacitors are connected in series 29. a. XC1 = 5 kΩ, XC2 = 2.5 kΩ and XCT = 7.5 kΩ b. IT = 2 mA because the voltage rating of a single c. VC1 = 10 V and VC2 = 5 V capacitor is less than the applied voltage.

30.

Because capacitance and capacitive reactance With two or more capacitors in series, the are inversely proportional. applied voltage is divided up between the 31. CT = 0.8 µF capacitors. Then each capacitor receives a 32.

CT = 10 nF voltage which is less than its own individual 33. a. XC1 = 1 kΩ, XC2 = 500 Ω and XCT = 333.33 Ω voltage rating. (It should b. IC1 = 20 mA, IC2 = 40 mA and ICT = 60 mA be noted that the distribution of voltage is 34. a. For medium and high values of capacitance, dependent on the individual capacitance the meter pointer will deflect towards zero values.) and then return to a high reading.

20. When a capacitor is used in a sine wave AC b.

If the capacitor is open, the meter pointer circuit, its current, I leads the voltage, V by a will not deflect. phase angle of 90°. c.

If the capacitor is leaky the meter pointer 21. Capacitive reactance can be defined as a will read a high resistance such as a few capacitors opposition to alternating current. thousand ohms.

22. a. The symbol for capacitive reactance is XC. 35.

Stray capacitance exists between; b. The unit of capacitive reactance is the a.

An insulated conductor and a metal chassis. ohm (Ω). b. Two adjacent conductors in a cable.

23. XC = 338.63 Ω. c. The turns in a coil of wire.

24. I = 1 mA d. The input and output leads of a transistor.

25. An ideal capacitor consumes no power. The reason is that a capacitor stores energy for one‐quarter cycle and then returns it during the next quarter‐cycle. In other words, it’s positive and negative power are equal and therefore the resultant power is zero.

26. The equivalent series resistance (ESR) of a capacitor represents all of the losses in a capacitor. The losses are classified as

Chapter 11 Homework Problems – Solutions 1. Inductance can be defined as an electrical

coil turns as compared to when the wire was

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53

property that opposes any change in the magnitude of the current in a circuit.

2. a. When a changing magnetic field induces a voltage in its own coil turns, it is referred to as self‐inductance. The inductance of an inductor is called self‐inductance. b. When a changing magnetic field in one conductor induces a voltage in another nearby conductor it is referred to as mutual inductance. A transformer operates on the principle of mutual‐inductance.

3. When a straight conductor is wrapped in the form of a coil, the magnetic flux surrounding one coil turn induces a voltage in multiple turns of the coil. The overall effect is that there are more magnetic field lines cutting across the

straight.

4. Lenz’s law states that a counter electromotive force (cemf) always has a polarity that opposes the force that created it.

5. Energy is stored in the magnetic field of an inductance.

6. The base unit of inductance is the henry (H).

7. One henry is the amount of inductance that produces one volt of induced voltage in itself when the current changes at the rate of 1 A/s.

8. L = 50 mH 9. The physical factors which affect the inductance, L of an inductor include; a. The type of core material. b. The number of turns of wire. c. The spacing between the turns of wire. d. The diameter of the coil (or core).

23. When an inductor is used in an AC circuit, the inductor voltage leads the inductor current by 10. The most common way to adjust the a phase angle of 90°. inductance of a variable inductor is to adjust 24. a. An ideal inductor does not consume or use the position of the core material. any power because the inductor stores 11. Noninductive wire‐wound resistors are energy for one‐quarter cycle and then wound so that the cemf of one‐half of the returns it during the next quarter‐cycle. In turns of other words, it’s positive and negative wire cancels the cemf of the other half. This power are equal and therefore the technique significantly minimizes the self‐ resultant power is zero. inductance of wire‐wound resistors. b. A practical or nonideal inductor consumes 12. When the magnetic field strength increases or uses some power because all inductors beyond a certain point, the core becomes have some internal resistance. The internal saturated and its permeability decreases. resistance means there will be some I2R Since magnetic field strength is dependent on power losses in the inductor. There are also current, higher current levels cause the core losses in the core of a typical inductor. to become saturated. This in turn reduces its 25.

At 5 kHz, XL equals approximately 1.037 kΩ. inductance at higher current levels.

X 13. Some common inductor ratings are its; Q= L a. DC or ohmic resistance rating.

R b. Current rating.

1.037 kW = = 41.5(Approximately) c. Voltage rating.

25 W d. The quality rating.

26. Because of the skin effect, the coil’s e. Tolerance. resistance (in the denominator) increases as 14. When the switch in a DC circuit is closed, the the frequency increases. The end result is that current increases gradually due to the the increase in the coils resistance offsets the inductors cemf opposing the build-up of increase in inductive reactance (in the current. numerator) at higher frequencies.

15. Inductive kick refers to the high amount of 27. LT = 20 µH cemf that is generated when an inductive 28. LT = 400 µH circuit is opened.

29. LT = 12.5 mH 16. a. The symbol for inductive reactance is XL. 30. LT = 102 mH b. The unit of measure for inductive reactance Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

54

is the ohm (Ω).

17. a. XL increases at higher frequencies because the current is changing faster. This in turn produces a higher cemf and more reactance. b. XL increases with more inductance because there is more cemf at a given frequency and in turn more reactance.

18. XL = 2 kΩ 19. a. XL doubles to 4 kΩ b. XL doubles to 4 kΩ c. XL cuts in half to 1 kΩ d. XL decreases by a factor of four to 500 Ω 20. L = 125 mH 21. f = 795.775 kHz 22. First determine the value of XL. 24 V V XL = L = = 4.8 kW I L 5 mA X Next, solve for L using the equation L  L 2p f XL  15.28 mH approximately

31. a. LT = 250 mH b. XL1 = 2 kΩ, XL2 = 3 kΩ, XLT = 5 kΩ c. IT = 4.8 mA d. VL1 = 9.6 V, VL2 = 14.4 V 32. a. LT = 60 mH b. XL1 = 2 kΩ, XL2 = 3 kΩ, XLT = 1.2 kΩ c. IL1 = 12 mA, IL2 = 8 mA, IT = 20 mA 50 mH L = 5 ms 33. T = = = R 10 kW 34. a. I = 632 µA b. I = 865 µA c. I = 1 mA (Approximately) 35. a. Axis orientation. b. Physical separation. c. Shielding.

1. A transformer cannot step‐up or step‐down a DC voltage because the magnetic field associated with a steady DC current is motionless or stationary. Without motion, the magnetic field lines cannot cut across the coil windings to induce a voltage in either the primary or secondary windings of the transformer.

2. For an iron‐core laminated transformer the coefficient coupling, k equals 1 or unity. This is due to the fact that all of the magnetic flux is concentrated in the high permeability core on which both coils are wound.

16 mWb = 0.64 3. k = 25 mWb

10. Eddy-current losses can be minimized by laminating the iron core.

11. Ploss = 90.32 W 12. a. 90° b. 0° 13. The magnetizing force produced by the secondary current opposes the magnetizing force produced by the primary current. This results in a weakening of the magnetic force produced by the primary current. In order for the magnetizing force to maintain the value required to produce the induced voltages, the primary current, Ipri must increase.

14. % Efficiency = 89.4% (Approximately) 15. To minimize flux leakage from the core, the I and E laminations are rotated 180° every few layers. This construction technique breaks up the joint between the I and E laminations so that there is no continuous air gap.

16. Transformers used at high frequencies typically do not use a core. These are called air-core transformers. Some transformers used at higher frequencies, however, do use a magnetic core. The type or make-up of the core is usually chosen so that core losses are

Chapter 12 Homework Problems – Solutions

4. L M =

L1´ L2 = 0.25 10 mH ´ 15 mH

= 3.062 mH

5. Nsec = 5 turns/V × 24 V = 120 turns V 6. I pri = sec ´ I sec Vpri

13 kV ´ 520 A 161 kV  42 A (Approximately) 7. When transmitting power over long-distance =

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transmission lines, it is desired to keep the minimized at higher frequencies. current as low as possible. This can be done 17. Isolation transformers have equal primary and by increasing the voltage to a very high level. secondary voltages. Therefore, if an isolation At higher voltages, the current does not have transformer has a primary voltage of 120 V, to its secondary voltage also equals 120 V. be nearly as high to transmit the desired 18. In an isolation transformer, there is no power. Because the current is much less at electrical connection between the primary higher voltages, the I2R losses in the and secondary windings. (Technically transmission line conductors is significantly speaking, reduced. This allows the use of smaller any transformer with separate primary and conductor size which in turn results in secondary windings is an isolation considerable cost savings when one realizes transformer.) With no electrical connection, that transmission lines often span the the earth ground connection in the primary is distance of several hundred miles. effectively isolated from the secondary.

19. % Efficiency = 50%.

P 8. %Efficiency = sec ´ 100 2 æNpri ÷ ö Ppri ç ÷ 20. Z pri = çç ÷ ´ Z sec 3.6 kW è Nsec ÷ ø = ´ 100 2 4 kW æ2ö = ççç ÷ ÷ ÷ ´ 50 W   90% è1ø 9. The three main power losses in a transformer  200 Ω are; N pri Z pri 400 W 7.07 a. Hysteresis losses = = = 21.

Nsec Z sec 8W 1 b. Eddy current losses 2 c. Copper I R losses 27. If the transformer leads having opposite instantaneous polarity are connected in N pri Z pri 3 kW 7.75 = = = 22.

(Approximately) parallel, excessive current will flow and the Nsec Z sec 50 W 1 transformer will burn out. If the transformer 23. Never apply more than the rated voltage to leads are connected in series with incorrect the primary and never draw more than the polarity the secondary voltage will be 0 V. rated current from the secondary.

28. Transformers with off-center tapped windings 24. The measured secondary voltage from an are used to adjust impedance matching ratios unloaded transformer is usually about 5-10% in audio transformers. In power transformers, higher than its rated value. Therefore, the tapped secondaries are useful for adjusting measured secondary voltage will be the charging rate of a battery charger or the approximately 25.2 V to 26.4 V. output of an arc welder.

25. a. Isec (max) = 416.67 A 29. To check for correct phasing in a delta b. Ipri (max) = 7.692 A connection, open the connection between 26. Connect one lead of each secondary winding two phase windings and measure the voltage. together and measure the voltage across the If the meter measures 0 V, the windings are two open leads. If the voltage measures 0 V, properly configured. If the meter measures a the leads connected together have the same high voltage the leads are not configured instantaneous polarity. The two open leads correctly. Reverse the lead connections on also have the same instantaneous polarity. one phase at a time until the meter measures When connecting the secondaries in parallel, 0 V. the leads with the same instantaneous 30. A wye‐connected secondary provides equal polarity must be connected together. When line voltages when properly configured. If the connecting the secondaries in series, the line voltages are unequal, reverse one leads with opposite polarity must be winding at a time until all line voltages are the Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

56

connected together. In this case, the two open leads serve as the secondary voltage.

same (balanced).

Chapter 13 Homework Problems – Solutions 1. Impedance can be defined as the total opposition to the flow of sine wave alternating current.

2. a. The symbol for impedance is the letter Z. b. The base unit of impedance is the ohm (Ω).

3. a. In Fig. 13‐35a, IT = 25 A b. In Fig. 13‐35b, IT = 10.817 A c. In Fig. 13‐35c, IT = 5 A 4. a. XC = 5.305 kΩ b. Z = 6.248 kΩ c. IT = 3.84 mA d. VC = 20.37 V e. VR = 12.67 V f. θ = −58.11° g. True power, PT = 48.69 mW

7. a. XC = 7.234 kΩ b. IC = 1.382 mA c. IR = 1.786 mA d. IT = 2.26 mA e. Z = 4.425 kΩ f. θ = 37.73° g. True power, PT = 17.87 mW 8. See Fig. 13‐2

9. a. XC = 3.183 kΩ b. IC = 31.42 mA c. IR = 21.28 mA d. IT = 37.95 mA e. Z = 2.635 kΩ f. θ = 55.89° g. True power, PT = 2.13 W 10. a. Z decreases b. IT increases c. θ decreases (Becomes less negative)

5. See Fig. 13-1.

6. a. XC = 338.63 Ω b. Z = 759.65 Ω c. IT = 19.746 mA d. VC = 6.69 V e. VR = 13.43 V f. θ = −26.47° g. True power, PT = 265.14 mW

e. Z = 1.28 kΩ f. True power, PT = 1.136 W g. PF = 0.581 15. See Fig.13-4.

16. a. Z increases b. IT decreases c. θ increases 17. a. Z increases b. IT decreases c. θ decreases (Becomes less negative) 18. Yes, VL and VC can be much larger than VT in a series RCL circuit. However, the net reactive voltage phasor (VX) and the resistor voltage phasor (VR) still combine to equal the total voltage, VT.

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11. a. Z decreases b. IT increases c. θ increases 12. a. XL = 282.74 Ω b. Z = 391 Ω c. IT = 92.07 mA d. VL = 26.03 V e. VR = 24.86 V f. True power, PT = 2.29 W g. PF = 0.691 13. See Fig. 13‐3.

14. a. XL = 1.57 kΩ b. IL = 31.85 mA c. IR = 22.73 mA d. IT = 39.13 mA

25. A parallel resonant circuit is often called a tank circuit.

26. a. Maximum impedance, minimum source current, high circulating inductive and capacitive currents and a phase angle of almost zero degrees. b. Minimum impedance, maximum source current, high inductive and capacitive voltages and a phase angle of almost zero degrees.

27. a. Inductive b. Capacitive 28. a. Capacitive b. Inductive 29. If the components in a parallel resonant circuit were ideal, the phase angle would be zero degrees and the power factor (PF) would equal one or unity.

30. The bandwidth of a resonant circuit refers to the range of frequencies in which the circuit provides 70,7% or more of its maximum

19. Yes, IL and IC can be much larger than IT in a parallel RCL circuit. However, the net reactive current phasor (IX) and the resistor current phasor (IR) still combine to equal the total current, IT. 20. a. XL = 628.32 Ω b. XC = 1.061 kΩ c. Z = 1.09 kΩ d. IT = 91.74 mA e. VR = 91.74 V f. VL = 57.64 V g. VC = 97.33 V 21. a. XL = 1.256 kΩ b. XC = 795.8 Ω c. IL =15.92 mA d. IC = 25.13 mA e. IR = 35.71 mA f. IT = 36.88 mA g. Z = 542.3 Ω 22. An LC circuit is resonant when XL = XC. 23. fr = 2 MHz 24. a. The resonant frequency, fr decreases. b. The resonant frequency, fr increases. c. The resonant frequency, fr stays the same.

c. A band-pass filter only passes a narrow band of frequencies. d. A band-reject filter blocks a narrow band of frequencies.

35. a. See Fig.13-5a. b. See Fig.13-5b. c. See Fig.13-5c. d. See Fig.13-5d.

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response.

31. The higher the Q, the narrower the bandwidth and the lower the Q the wider the bandwidth.

32. a. BW = 200 kHz b. BW = 80 kHz c. BW = 40 kHZ 33. a. fr = 2.5 MHz b. Z = 10 Ω c. Q = 39.27 d. BW = 63.66 kHz 34. a. A low-pass filter passes the lower frequencies but eliminates the higher frequencies. b. A high-pass filter passes the higher frequencies but eliminates the lower frequencies.

Chapter 14 Homework Problems – Solutions 1. DC motors, AC motors and universal motors 3. Torque motors. are the three major categories of motors.

4. Synchronous motors.

2. a. ihp refers to motors rated above 1 hp.

5. Voltage rating. b. fhp refers to motors rated from 1/20 hp 6. Vmin = 105.3 V and Vmax = 128.7 V to 1 hp.

7. No, because the minimum voltage, Vmin for c. sfhp refers to motors rated at less 110 V, ± 5% is 104.5 V. than 1/20 hp.

8. The minimum and maximum voltages over which this motor is designed to operate are; Inserting values, Vmin = 108 V and Vmax = 132 V 373 W % eff. = × 100 The minimum and maximum voltages over 686.4 W which the line voltage varies are; = 54.34% Vmin = 114 V and Vmax = 126 V 16. As the load on a motor increases, its power Under full load (25 A), 7.95 V is dropped factor increases. An unloaded or lightly across the 200-ft length of copper wire. The loaded motor is highly inductive with a phase calculations are: Vdrop = 0.318 Ω × 25 A = 7.95 V. angle close to 90°. As the load on the motor Considering the case in which the line voltage increases, it becomes less inductive and the Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

59

is at its minimum value; phase angle decreases significantly. As the Vmotor = Vmin(supply) – Vdrop phase angle decreases the power factor = 114 V – 7.95 V increases. With a higher power factor, a given = 106.05 V amount of current delivers more power The minimum voltage this motor is without increasing the copper losses. This, in guaranteed to operate at is 108 V. Therefore, turn increases the motor’s efficiency. it is not guaranteed that this motor will have 17. The frame number on a motor indicates its the minimum voltage required to operate size, i.e., height, length and shaft diameter normally under these conditions. etc.

9. If a motor is operated at a voltage higher than 18. The larger the frame number, the larger the what it is rated for, it tends to produce slightly horsepower rating. more noise.

19. Open motors have enclosures with ventilating 10. The current rating on the motor nameplate openings which allow surrounding air to be designates its running current. The running forced through the enclosure to cool the current is the current the motor draws when motor windings. Totally enclosed motors do the motor is loaded to its rated horsepower not allow free exchange between surrounding and is operated at its rated voltage, frequency air and air within the enclosure. and temperature.

20. The two main parts of a squirrel-cage motor 11. The maximum allowable temperature include the stator and rotor. rise is 120°.

21. The rotor bars are skewed in a squirrel-cage 12. The service factor (SF) rating is a number that motor to allow the bars to enter the magnetic can be multiplied by the horsepower rating to flux of one tooth of the stator before leaving determine the maximum continuous the flux of the adjacent tooth. horsepower the motor can provide.

22. The names of the two windings in the stator 13. Pmax = Prated × SF in a split-phase motor are; = 0.5 hp × 1.15 a. Run winding = 0.575 hp b. Start winding Therefore, a ½ hp motor with SF = 1.15 will 23. a. Run winding not be able to meet demands of a 6 hp load, b. Run winding even though it is occasional. c. Run winding 24. To reverse the direction of rotation of a split5252P 5252 ´ 0.5 hp 14. T = = = 0.741 lb-ft phase motor, reverse the lead connections of S 3540r / min either the run or start windings.

P 15. % eff. = out × 100 120 f 25. Synchronous speed = Pin no.of poles where Pout and Pin calculated as follows; 120´ 60 Hz 746W = Pout = ½ hp × = 373 W 4 hp = 1800 r/min.

Pin = 208 V × 3.3 A × 0.71 = 686.4 W 26. Percent slip = synchronousspeed- ratedspeed × 100 synchronousspeed 31. Shaded-pole motors are only used for subfractional and fractional horsepower 1200 r /min - 1120 r / min = × 100 applications because they are very inefficient, 1200 r / min have a low power factor and a very weak = 6.67% starting torque.

27. In a split-phase motor, the starting current is 32.

In a brush-type DC motor, the stator winding higher than the running current because once Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

60

the motor is running, the rotor flux interacts with the stator windings. This interaction increases the effective inductance of the run winding and helps to reduce the current required to run the motor at its normal operating speed. This, along with the removal of the start-winding current, is the reason why the run current is only a fraction of the start current in a split-phase motor.

28. The advantages of a capacitor-start motor versus a split-phase motor are; a. Greater starting torque. b. Less starting current per unit of starting torque.

29. The permanent-split capacitor motor uses a smaller capacitor.

30. The advantages of a permanent-split capacitor motor versus a capacitor-start motor include; a. Higher power factor (PF). b. Requires less current to operate. c. More efficient. d. Quieter. e. Capable of variable-speed operation.

is called the field winding and the rotating winding is called the armature winding.

33. When the load is increased, the armature current must increase so that a stronger armature flux will provide more torque as it reacts with the field flux.

34. By controlling the field current with a series rheostat.

35. The speed of a brushless DC motor can be controlled by controlling both the voltage and current the phase windings receive.

36. Five advantages of a brushless motor are; a. Much quieter operation. b. Can operate at higher rpms. c. No electromagnetic interference produced by sparking brushes. d. Can operate in environments that may produce explosive fumes or dust. e. More efficient operation.

37. The advantages of three-phase motors versus single-phase motors include; a. Higher efficiency. b. They are available in a much wider range of power ratings. c. They produce less noise. d. They provide a constant shaft torque.

Chapter 15 Homework Problems – Solutions 1. When measuring voltage, a DMM is actually 6. A thermocouple meter is used to measure measuring the time it takes for a capacitor to radio frequency currents. charge to the value of the measured voltage.

7. Two important ratings of an analog meter An internal capacitor is charged by a constantmovement are its full-scale current rating (Im) current source, so the voltage across the and its internal resistance (Rm). capacitor rises linearly over time.

8. In parallel.

2. Constant current.

9. a. Rs = 50 Ω 3. R = 4.7 kΩ b. Rs = 12.5 Ω 4. The most commonly used type of meter c. Rs = 5.55 Ω movement in an analog meter is the d. Rs = 0.505 Ω D’Arsonval movement.

10. In series.

5. The iron-vane meter can operate on either 1 1 11. Sensitivity = = = 50 kΩ/V. alternating or direct current.

I m 20 mA 12. a. Rm = 19 kΩ b. Rm = 99 kΩ c. Rm = 199 kΩ d. Rm = 499 kΩ e. Rm = 999 kΩ

20. The ohmmeter scale of an analog meter is nonlinear. This is because equal changes in meter pointer deflection do not produce equal changes in the value of resistance being displayed on the ohmmeter scale. When

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f. Rm = 1.999 MΩ 13. The input resistance of a voltmeter is W designated Rv. Rv = rating × Vrange.

V a. Rv = 100 kΩ b. Rv = 200 kΩ c. Rv = 500 kΩ d. Rv = 2 MΩ e. Rv = 5 MΩ f. Rv = 20 MΩ 14. Current or ammeter loading occurs if the ammeters internal resistance noticeably contributes to the overall resistance of the circuit when making current measurements.

In low resistance circuits, adding the ammeter’s resistance may reduce the current below the value that exists without the meter present. In this case, we say that current or ammeter loading has occurred. Since current meters generally have very low resistance, current or ammeter loading is not an issue when measuring currents in high resistance circuits.

15. Voltmeter loading occurs when the voltage division in a circuit is altered by the presence of the meter. The parallel equivalent resistance of the voltmeter and the circuit resistance across which the voltage is being measured, should be approximately the same with or without the meter present. If the voltmeter resistance is much higher (10 to 20 times higher) than the circuit resistance, the voltage division is not significantly altered.

The problem with measuring voltages in high resistance circuits is that the meters resistance may not be high enough. This, in turn, results in voltmeter loading.

16. a. V1 MΩ = 20 V b. V1.5 MΩ = 30 V 17. a. V1 MΩ = 12.5 V b. V1.5 MΩ = 18.75 V 18. a. V1 kΩ = 20 V b. V1.5 kΩ = 30 V 19. a. V1 kΩ = 20 V (Approximately) b. V1.5 kΩ = 30 V (Approximately)

measured. When both voltmeters read the same voltage, the unknown impedance and

examining the circuitry of an analog ohmmeter, notice it is a simple series circuit.

The current in the circuit is calculated as V I= . Since the voltage, V is constant, I is RT inversely proportional to R, which is a nonlinear relationship. In other words, equal changes in the resistance, RT do not produce equal changes in the current, I. Since the calibration of the ohmmeter scale is based on current, the scale is nonlinear.

21. An insulation tester, or Megger as it is often called, applies a high voltage across the insulator and in turn measures the amount of leakage current flowing through it. The insulator’s resistance is tested at a voltage higher than what the insulator is normally subjected to.

22. A Wheatstone bridge is said to be balanced when the voltage division in each leg or branch of the bridge is identical. When the bridge is balanced, the voltage measured between points A and B will be zero volts.

(See Fig. 15-21).

23. Four connections are required when connecting a wattmeter into a circuit.

24. a. Voltage b. Current 25. A wattmeter measures true power.

26. The ratings associated with a wattmeter include its voltage rating, its current rating and its power rating.

27. The two-wattmeter method.

28. One of the simplest ways to measure an unknown impedance is to measure its voltage and current and then apply Ohm’s law. Both the voltmeter and current meter must be able to accurately measure their respective quantities at the frequency where the impedance is to be determined. Another technique for measuring an unknown impedance is to use the circuit shown in Fig. 15-27. It is important to note that the frequency of the AC source must be set to the frequency at which the impedance is to be 29. Inductance and capacitance can be measured with an AC bridge.

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series resistance both have the same value. If the variable resistance has a calibrated dial, the value of the unknown impedance can be read directly the dial.

Chapter 16 Homework Problems – Solutions 1. The purpose of stepping up the voltage to a 9. The neutral/ground bus in the main very high level is to keep the current to its distribution panel is connected to ground lowest possible value on the transmission line using one or two different grounding conductors. At higher voltages, less current is conductors. One grounding conductor goes to required to transmit the desired power. With a grounding electrode (8-ft-long metal rod) 2 lower currents, the I R power losses in the that is driven into the ground outside the transmission line conductors are significantly building. If the pipe from the water main is reduced. Also, lower currents allow for the metal, a second grounding conductor use of smaller transmission line conductors connects to it. which results in considerable cost savings.

10. The ground system, in conjunction with all the 2. There are two hot conductors, which are equipment grounding conductors in the insulated, and one neutral conductor, which is branch circuits, is to help protect the wiring uninsulated or bare. system from lightning strikes and other types 3. The neutral conductor is grounded of high voltage surges.

4. Under normal circumstances the grounding 11. The cable used in residential house wiring is conductor carries no current. nonmetallic (NM)-sheathed cable.

5. a. 120 V 12. a. 14 AWG. b. 240 V b. 12 AWG. c. 0 V c. 10 AWG. 6. The three main elements in the service 13. a. 14 AWG. entrance to a home or residence are; b. 12 AWG. a. The service head. c. 10 AWG. b. The meter box. d. 8 AWG or larger. c. The main distribution panel.

14. a. Hot conductor.

7. The three main elements in the main b. Hot conductor distribution panel of a service entrance are; c. Neutral conductor. a. The main circuit breaker, which also serves d. Grounding conductor. as the disconnect from power company 15. Black tape must be put around the white service. conductor in this situation to indicate that the b. A parallel hot bus system that connects to conductor is a hot. the output of the main circuit breaker.

16. The hot conductors can be any color except c. The neutral/ground bus which provides a white, natural gray or green. common connection point for all neutral 17. If an appliance requires both 120-V and 240-V conductors, grounding conductors and the to operate, the NM sheathed cable must have bonding jumper conductors. four conductors. The insulation on three of 8. In a typical residential service entrance, the the conductors in the sheathed cable will be ampere rating of the main circuit breaker is black, red and white. Another bare (or green typically 100 to 200 A. insulated) wire will also be in the NM cable.

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18. a. Brass-colored screws connect to the hot side of the 120-V AC power line voltage.

33. It is creating a ground fault between the hot b. Light-colored screws connect to the neutral conductor and the grounding conductor. The side of the 120-V AC power line voltage. ground fault current is typically 5 mA. 19. The wide slot is the neutral slot and the 34. A GFCI should be tested monthly. narrow slot is the hot slot.

35. No, that is the job or role of the circuit 20. The white neutral wire and the bare (or breaker protecting the branch circuit. A GFCI green) grounding conductor must have a protects all other outlets that it feeds but it continuous path between the main does not protect anything in the line that distribution panel and the connecting loads. feeds it.

This is an absolute requirement in residential 36. Any time there is an imbalance of 5 mA or house wiring systems. more between the hot and neutral 21. An ordinary on-off wall switch is sometimes conductors. referred to as a snap switch.

37. The advantage of using a GFCI in the main 22. So that if the hot part of the switch comes distribution panel is that it protects into contact with the metal frame of the everything in the branch circuit including the switch outlet box, the circuit breaker will trip cable that feeds the first outlet box. The and remove power from the switch. disadvantage is that it takes up extra space in 23. The white neutral conductor in the NM cable the panel. They also cost more than ordinary should connect to the light-colored screw or conventional circuit breakers. And finally, it whereas the black hot conductor should is much more convenient to reset a GFCI if it is connect to the dark-colored screw. right in front of you, rather than go all the 24. In a 120-V branch circuit the black hot and way to the main distribution panel, which is white neutral conductors carry exactly the usually located in some remote location like a same amount of current under normal basement or garage. circumstances. The bare or green grounding 38. A GFCI can detect a short between the neutral conductor will not carry any current until a conductor and the grounding conductor ground-fault occurs. whereas an ALCI cannot. Otherwise they 25. When the jumper bar on the hot-side of the function very similarly. duplex receptacle is removed, the two 39. LCDIs are typically found in a special cord set receptacles can be controlled independently and are factory installed in the appliance in or separately. which it is protecting. The cord set contains 26. Switched receptacles are common when it is both the plug and the cord. desired to control a table lamp from a wall 40. An AFCI detects abnormalities in the shape of switch while still providing continuous power the 60 Hz sinusoidal waveform associated to the other receptacle in the duplex with 120-V AC power line voltage. receptacle.

41. The test button on an AFCI is often a blue 27. A three-way switch is actually a single-pole color. double throw (SPDT) switch.

42. In bedrooms and sleeping areas.

28. A dark or blackish-colored screw identifies the 43. A feeder circuit transfers power from a main pole connection on a three-way switch. panel to a subpanel. The subpanel may be 29. Four. located in a detached garage as an example.

30. A four-way switch is required.

The feeder cable could be buried between the 31. a. Off. house and garage in this example. b. On. 44. No, in a subpanel, there is a separate bus for 32. The incoming wires that bring power to the the neutral conductors and the equipment GFCI are connected to the line screw grounding conductors. This is because the terminals whereas the wires that feed the neutral wires can only be grounded in the other outlets connect to the load screw main distribution panel. In the subpanel, the Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

64

terminals.

neutral bus must be insulated from the metal box of the subpanel.

45. A mechanical latching relay is used in low-voltage control circuits so that only a pulse of current is required to open or close 47. Electronic control circuits do not use cables the relay. like traditional low-voltage control circuits.

46. Typically, No. 22 to No. 18 gage wire is used in Instead, they use coded high-frequency radio low voltage circuit wiring. signals for electronic control of the circuits.

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65

Section

4 Answers to Experiments Manual GENERAL NOTES TO THE INSTRUCTOR The tests for the various chapters may be too long for one class period. They can be broken into two or three tests by having the students do every other question or every third question. This would still provide broad coverage of the material because of the way in which the tests are arranged. By splitting the test in this fashion, part of it can be used as a pretest before the other activities are undertaken. The rest of the test can then be a posttest given after the other activities are completed. If you want to give partial credit when students use the correct procedure in solving a problem, you may choose to have the students attach their worksheets to the tests.

CHAPTER 1 TEST 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

F F T T T F F T F C B A D A D B B Given: Distance = 2.1 m, force = 1.8 N Find: Work Known: W = force × distance Solution: W = 1.8 N × 2.1 m = 3.78 J

Answer: W = 3.78 J 19. Given: Win = 900 J, Wout = 190 J Find: Efficiency W Known: %eff. = out ´ 100 Win 190 J Solution: %eff. = ´ 100 = 21.1 900 J Answer: Efficiency = 21.1% 20. Given: Win = 10,867 J, % eff. = 37 Find: Wout W Known: %eff. = out ´ 100 Win 10,867 J´ 37 Solution:Wout = = 4021J 100 Answer: Output energy = 4021 J 21. Heat 22. Static 23. Negative 24. Negative

1-1 LAB EXPERIMENT: STATIC CHARGES

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66

1. Experiments with static charges are hard to control because of atmospheric conditions and variations in materials. Therefore, you should try this experiment with the materials to be used by the students.

Leather and a glass rod have been a dependable combination for producing strong static charges.

2. a. The charge on the rod induced an opposite charge on the surface of the paper ball. The opposite charges then attract each other. b. When the paper touched the rod, part of the charge on the rod was transferred to the paper ball.

The ball and the rod then possess the same kind of charge and repel each other.

4. a. Yes. They both received a charge from the same rod. b. The principle of attraction can be used to remove dust from air or to uniformly distribute paint or abrasive material. The principle of repulsion can be used to arrange abrasive particles on backing paper.

CHAPTER 2 NOTES TO THE INSTRUCTOR 1. The advanced problems of experiment 2-2 are probably too difficult for many students. However, the mathematically inclined student will enjoy them.

2. Chapter 2 introduces many new terms and symbols. Of necessity, it is long and involved. The classroom activity can be broken up by introducing the laboratory activities as soon as possible.

3. Experiment 2-1 introduces students to a variety of conductors and insulators.

4. Experiment 2-3 is intended to introduce the student to the variety of technical information included in parts catalogs as well as those located on multiple websites. Being exposed to this mass of information may help the student to see the importance of learning technical terminology in class.

TEST 1. Negative 2. Liquids, gases 3. Alternating 4. Kinetic 5. 6.25 × 1018 6. Compound 7. Specific resistance 8. Horsepower 9. Temperature coefficient 10. Ions 11. C 12. D 13. B 14. C 15. B 16. D 17. A 18. D 19. B

20. B 21. C 22. C 23. D 24. C 25. C 26. D 27. B 28. B 29. C 30. A 31. B 32. D 33. A 34. C 35. C 36. A 37. A

38. Given: Voltage = 12 V, charge = 2000 C Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

67

Find: Energy Known: W = VQ Solution: W = 12 V × 2000 C = 24,000 J Answer: Energy = 24,000 J 39.

Given: Charge = 4 C, time = 2 s Find: Current

Known: I =

Q t

4C = 2 C/s = 2 A 2s Answer: Current = 2 A Solution: I =

40.

Given: Resistivity = 0.004 Ω·cm at 20C Length = 250 cm, CSA = 0.02 cm2 Find: Resistance at 20C

KL A Solution: R = 0.004 Ω·cm × 250 cm/0.02 cm2 = 50 Ω Answer: Resistance at 20C is 50 Ω Given: Power = 500 W, time = 3 h Find: Energy Known: W = Pt Solution: W = 500 W × 3 h = 1500 Wh Answer: Energy is 1500 Wh Given: Power out = 602 W, W = 51,600 J, t = 60 s Find: Efficiency Known: R=

41.

42.

Known: Pin =

Pout %eff.

1 hp = 746 W

Solution: Pout = 34 hp ´ 746 W/hp = 559.5 W Pin =

43.

44.

45.

46.

559.5 W = 999.1W 0.56

Answer: Efficiency = 70% Given: Pout = ¾ hp, % eff. = 56 Find: Pin Answer: Input power = 999.1 W 0.21 2 × 106 0.0000053

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68

47.

48.

49.

68 3000 2.2

2-1 LAB EXPERIMENT: CONDUCTORS, INSULATORS, AND RESISTANCE 2.

3.

5.

6.

Yes In Table 2-1, aluminum and copper are conductors. The other materials should be in the Poor Conductor or Insulator column. a. No b. No c. No P Known: %eff. = out ´ 100 Pin W Pin = t 51,600 J Solution: Pin = = 860 W 60 s 602 %eff. = ´ 100 = 70 860 a. Yes (dimly) b. Yes c. Yes d. Yes e. Yes, as the wire went deeper into the solution, more surface area was available to interact with the ions. f. Yes, as the wires approached each other, the ions had less distance to travel. g. Decrease h. Increase

2-2 ADVANCED PROBLEMS 2-1. Given: Pout = 100 W, efficiency = 65% Find: Pin P Known: %eff. = out ´ 100 Pin P Therefore Pin = out ´ 100 %eff.

100 Solution: Pin = ´ 100 = 153.85 W 65 Answer: Pin = 153.85 W 2-2. Given: Pout = 1.0 hp, W = 6.1 × 106 J, t = 1 h Find: Efficiency P Known: %eff. = out ´ 100 Pin W P= ,1hp = 746 W t Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

69

6.1´ 106 J = 1694.4 W 3600 s Pout = 746 W/hp´ 1.0 hp = 746 W 746 W %eff. = ´ 100 = 44 1694.4 W Answer: Efficiency = 44% 2-3. Given: Pout = 120 W, efficiency = 60%, t = 45 min Find: W P Known: %eff. = out ´ 100, W = Pt Pin 120 W P Solution: P = out ´ 100 = ´ 100 %eff.

60 = 200 W 60 s t = 45 min´ = 2700 s min W = 200 W × 2700 s = 5.4 × 105 Ws Answer: Energy = 5.4 × 105 J 2-4. Given: V = 20 V, Q = 2000 C, P = 3.7 W Find: t to use 50 percent of W Known:W = QV ° W t= P Solution:W = 2000 C´ 20 V = 40,000 J 50 percent of W = 0.5´ 40,000 J = 20,000 J 20,000 J t= = 5405.4s = 1.5 h 3.7 W Answer: It will take 1.5 h to use half the battery's energy.

2-5. 60 C Solution: Pin =

2-3 LAB EXERCISE: TEMPERATURE COEFFICIENTS OF RESISTORS 2.

The answers here will vary greatly, depending on the source. The low temperature coefficient (tc) will probably be less than 10 ppm/C° for precision resistors. The high tc will probably be more than 500 ppm/C° for carbon resistors. a. The answer will again vary with the source. Most students will record “no.” b. In many electric devices the small change in resistance due to a change in temperature has no significant effect on the operation of the device.

CHAPTER 3 NOTES TO THE INSTRUCTOR 1.

2.

If possible, use 5 or 1% resistors for the experiments in this chapter.

Students have not been introduced to the color code yet. If the resistors used are color-coded, students will have to be told which resistor has which value.

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70

3.

Lab experiment 3-5 requires the use of square roots. However, the problems are designed so that the answers are whole numbers. Thus, the square roots are easy to do. Some of the problems require the use of the definitions of base units developed in Chap. 2.

Lab experiment 3-4 can be done with either physical components or with Multisim. If you are using physical components, the fuse (which is open) and the lamp can be encased in heatshrinkable tubing so that faults are not visible. Of course, some holes will be needed in the tubing covering the lamp so that the students can tell when it is working. These components can be labeled (Ll and Fj) for use in future troubleshooting experiments.

TEST 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

T T F F F F T F F T T T F F C A C B

19.

20.

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

31.

32.

33.

34.

35.

36.

A D 65 V 6.5 mA 0.68 V 3.4 mA 2500 Ω 0.6 A 4V 9Ω 17.64 W 105.84 Wh 24.3 cents 1.43 A 375 W 12.5 V 12 /c /kWh 360 Ω

3-1 LAB EXERCISE: WORKING WITH SCHEMATIC DIAGRAMS 1.

a and b

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71

2.

and

3.

4.

5.

6.

No. The meter still reads the source voltage.

Yes, a load (lamp) has a voltage across it only when current flows.

Current from the source might damage the ohmmeter.

3-2 LAB EXPERIMENT: MEASURING VOLTAGE, CURRENT, AND RESISTANCE 1.

2.

3.

a. Approximately 1.5 V and approximately 6 V b. Approximately 1.5 V c. Yes, because the DMM has more digits to the right of the decimal point on the lower range, or each minor division of the scale on the VOM represents a smaller voltage on the lower range. a. Approximately 1000 Ω b. Approximately 1000 Ω c. The × 100 range, because a minor division around the 10 mark on the scale represents less resistance than a minor division around the 100 mark. Or, the 2-k range, because the DMM has more digits to the right of the decimal point on the 2-k range than on the 20-k range. d. Yes, because the meter is measuring the resistance of both the resistor and the body. a. Approximately 6 mA b. 6 mA c. Yes. Yes unless there is a faulty component, instrument, or source

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72

3-3 LAB EXPERIMENT: OHM'S LAW 1. a. 0.006 A = 6 mA b. 6 mA ± the tolerances of the meter, resistor, and battery c. Yes 2. a. The current should decrease to one-tenth of its former value. b. 0.6 mA c. Yes 3. a. The current in Fig. 3-7(a) should be four times greater than the current in Fig. 3-7(d). b. 1.5 mA c. Yes

3-4 LAB EXPERIMENT: TROUBLESHOOTING 1.

2.

The fuse is open. With the switch closed, I measured 6 V across the fuse and the voltage should be zero unless the fuse is open.

Yes. The lamp could be shorted. The switch is not defective because the voltage across the fuse was zero when the switch was opened.

3-5 ADVANCED PROBLEMS 3-1. Solution: P = I 2R P R P and I = = R

Therefore, I 2 =

360 W 40 WΩ

= 9= 3A Answer: Current = 3 A 3-2. Solution: W = Pt W 0.25 kWh Therefore, P = = = 0.05 kW t 5h = 50 W

V2 R Therefore, V 2 = PR V = PR = 50 W´ 98 WΩ P=

V = 4900 = 70 V Answer: Voltage = 70 V Q 30 C = 1.5 A 3-3. Solution: I = = t 20 s P = I2R = 1.52 × 5 = 11.25 W Answer: Power = 11.25 W W 30 J = = 10 V 3-4. Solution:V = Q 3C P = IV = 2 A × 10 V = 20 W Answer: Power = 20 W

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73

NOTES TO THE INSTRUCTOR 1.

2.

3.

Experiment 4-1 can be completed with Multisim or with physical components.

The values given in Table 4-1 will vary widely depending on the type, size, and condition of the dry cell. However, the general trend of the changes in internal resistance should be present.

The 4.7-Ω resistor in experiment 4-3 must be at least a ½-W resistor.

TEST 1. Primary 2. Two 3. Polarization 4. Lithium 5. Lead-acid 6. Internal resistance 7. Hydrometer 8. Decreases 9. Instrument 10. T 11. F 12. T 13. T 14. T 15. T 16. F 17. F 18. F 19. F 20. T 21. T 22. F 23. T 24. F 25. F 26. F 27. T 28. T

29. F 30. F 31. F 32. T 33. F 34. T 35. T 36. F 37. F 38. D 39. C 40. A 41. A 42. B 43. 0.51 Ω ±5% 44. 8.2 MΩ ±10% 45. 100 Ω ±1% 46. 20.5 kΩ ±2% 47. 19.1 V 48. 33.9 W 49. 2 W 50. 2.5 A 51. 5.04 Ω 52. 4.8 kWh 53. 1 Ω 54. 1 W 55. 131 mW 56. 240 mW

4-1 LAB EXPERIMENT: LAMP SWITCHING CIRCUIT 1.

2.

3.

4.

The center, or middle, terminal a. Yes b. Yes c. Yes Yes, if the fuse is open, the source voltage appears across the fuse.

Yes

4-2 LAB EXPERIMENT: RESISTANCE OF A LAMP 1.

Approximately 6 Ω. Value will vary with the type of ohmmeter.

2. a. 6 V b. 150 mA Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

74

c. d. a. b. a. b.

3.

4.

400 Much less 42 Ω Step 2 0.945 W 0.9 W

4-3 LAB EXPERIMENT: INTERNAL RESISTANCE 1.

See Table 4-1.

TABLE 4-1

Measured and Calculated Values for Fig. 4-3

Load, Ω

V measured, V

N.L.

1.584

100

8.

9.

10.

11.

I calculated,

V Int.R, mV

Rint, Ω

1.535

15.35

49

3.19

47

1.530

32.55

54

1.66

22

1.516

68.91

68

0.99

10

1.507

150.7

77

0.51

4.7

1.479

314.7

105

0.33

mA

No The resistor would try to dissipate about 0.7 W, so it would be ruined.

Answers depend on values recorded in Table 4-1.

No. The cell voltage is less for the 4.7- Ω than for the 47-Ω resistor.

4-4 LAB EXPERIMENT: CHARACTERISTICS OF AN LED CIRCUIT 4.

5.

6.

7.

8.

9.

10.

8.0 V, 2.0 V 17 mA ±1 mA Between 16 and 18 mA 470 Ω ±50 Ω Yes 1 mA, 1.8 V The values of both VD1 and ID1 are reduced. However, VD1 only reduced from 2.0 V to 1.8 V while IR1 (which is also ID1) reduced from 17 mA to 1 mA.

4-5 LAB EXERCISE: RESISTOR COLOR CODE, TOLERANCE AND STANDARD VALUES 1.

3.

4.

5% 10% 430 470 470 560 510 560 a. No, because the −5 percent value of 750 and the +5 percent value of 680 overlap b. No c. Yes Answers will vary depending on which resistors are chosen.

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75

4-6 LAB EXERCISE: USING WIRE CHARTS AND TABLES 1.

2.

3.

a. 16 gage b. Because the longer cord would have too much resistance with the smaller conductor Approximately 0.56 Ω Approximately 274 cmil/A

CHAPTER 5 NOTES TO THE INSTRUCTOR 1.

2.

3.

4.

5.

6.

If a VOM is used in these experiments, it should have a sensitivity of at least 20 kΩ/V. The answer to the “measured values” in these activities is given as a range. This range assumes the students are using 5% resistors and 2% meters. Under these conditions, the probability of a reading falling outside the stated range is very small.

Experiments 5-5, 5-6, 5-7, 5-8, 5-10, 5-11, and 5-12 can be done with physical components or with Multisim There are two columns under “Measured Value” in Tables 5-1 to 5-4 in the Experiments Manual.

Results obtained with Multisim should be listed under the “Multisim” column, and those measured with physical components go under the “Hardware” column.

These same tables (5-1 to 5-4) in the Instructor's Manual contain only one column for “Measured Value” because the values measured using Multisim will be the same as the calculated values.

Experiment 5-10 (troubleshooting) uses two resistors that are not their stated values. In Fig. 514(a), R6 is actually a 4.7-kΩ resistor, and in Fig. 5-14(b), R3 is really a l-kΩ resistor. Each resistor in this experiment should be covered with heat-shrinkable tubing and labeled with its Rn number.

Experiment 5-11 (troubleshooting) uses four resistors that are not their stated values. In Fig. 515(a), R5 is really 4.7 kΩ and R15 is 2.2 Ω (a zero-ohm resistor would be even better for R15 if you have one). In Fig. 5-15(b), R6 is actually 2.2 kΩ and R16 is 10 MΩ. An open resistor would be even better for R16 and can be made by filing into a wire-wound or spiral-film resistor. However, the l0MΩ resistor will drop all but 1.5 mV of the 15.2 V that drops across R4 + R5. As in experiment 5-10, all resistors should have their coded values hidden.

TEST 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

Parallel Series Parallel Decrease Less Low Decreases Highest T T F F T F F

17.

18.

19.

20.

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

31.

F T F F C D B A A A C B C C B

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76

16.

F

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77

32.

33.

34.

35.

36.

37.

38.

39.

VR1 = 32 V, R2 = 5 kΩ, R3 = 9.5 kΩ, PT = 0.36 W, RT = 22.5 kΩ IR1 = 0.4 A, IR2 = 0.3 A, R2 = 40 Ω, PT = 18 W, PR1 = 4.8 W RT = 12 Ω, IT = 0.5 A, PT = 3 W, GT = 83.3 mS VR1 = 6 V, VR3 = 2 V, IR1 = 0.8 A, R1 = 7.5 Ω, RT = 10 Ω RT = 13 kΩ, IT = 1.9 mA, PT = 48 mW R1 and R2; R3 and R4 None; but the two parallel groups are.

VR1 = 30 V, VR3 = 36 V IR4 = 4 mA, IT = 10 mA and RT = 6.6 kΩ

5-1 LAB EXPERIMENT: SERIES CIRCUITS—GENERAL PROPERTIES 1. a.

Yes, because they have the same current and they are identical lamps (same resistance) b.

No, because each lamp receives only one-half its rated current and voltage c.

Yes d.

VL1 = 3 V, VL1 = 3 V, VB1 = 6 V e.

Yes f.

105 mA g.

Yes h.

Yes 2. a.

More brilliant, because the full source voltage would be applied to it b.

Yes c.

It would become more brilliant 3. a.

It reduced to 0 because the series circuit has an open. b.

0 V, because no current is flowing through L2 c.

0V d.

Approximately 6 V, because VT = VL1 + VL2 and VL2 is 0 V e.

6V f.

Yes 4. a.

It is more brilliant. b.

L1 because it is lit quite brightly. It is brighter than it was in Fig. 5-5(a) when it dropped 3 V. c.

L1 d.

Dimmer e.

R1 f.

Greater than g.

Yes

5-2 LAB EXPERIMENT: PARALLEL CIRCUITS—GENERAL PROPERTIES 1.

2.

a. b. c. d. e. f. a. b. c. d.

Yes No, because the terminals on the L1 socket are still connected to the battery.

No No No Yes Decrease Decrease L2 Yes

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78

5-3 LAB EXPERIMENT: SERIES-PARALLEL CIRCUITS—GENERAL PROPERTIES 1.

2.

a. b. c. d. e. a. b. c. d. e. f. g. h. i. j.

L1 L3 and L4 Because the current for both L2 and L3 has to flow through L1 Because L2 has twice as much voltage across it as either L3 or L4 Because they are in series Brighter Brighter Shorting L1 applies the full 6 V to L2 Brighter Dimmer Shorting L2 also shorts out L3 and L4. Of course, L1 receives 6 V and lights at full brilliancy.

Dimmer Dimmer Because RT is decreased, which causes more IT, and IT flows through L1 Because it receives the same voltage as L2

5-4 LAB EXERCISE: POLARITIES AND MEASUREMENTS 1.

2.

3.

b. c. a. b. c. a. b. c. d.

Positive 70 V Positive 30 V IR6 or IR3 + IR4 Negative Right to left IT Negative lead IR1 Positive lead

5-5 LAB EXPERIMENT: SERIES CIRCUITS—CURRENT, VOLTAGE, AND RESISTANCE 1.

2.

a. b. c. d. e. f. g. h. i. j. a.

20 kΩ 18 to 22 kΩ 0.8 to 1.2 mA 1 mA Yes VR1, = 9 to 11 V and VR2 = 9 to 11 V Yes VR1 = 10 V and VR2 = 10 V 20 m W Yes See Table 5-1.

TABLE 5-1

Electrical Quantities for Fig. 5-10(b)

Quantity

Measured Value

Calculated Value

RT

27 to 33 kΩ

30 kΩ

IT

0.6 to 0.74

0.67 mA

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79

mA

3.

VR1

6 to 7.4 V

6.7 V

VR2

12 to 14.6 V

13.3 V

b. c. d. e. f. a. b. c. d. e. f.

Yes Yes 40 V PR1 = 0.0044 W and PR2 = 0.0089 W Yes Yes, because RT has increased only from 30 to 31 kΩ Yes Yes About 0.65 V Less than Increase

5-6 LAB EXPERIMENT: PARALLEL CIRCUITS—CURRENT, VOLTAGE, AND RESISTANCE 1.

a. b. c. d.

Yes No No See Table 5-2.

TABLE 5-2

2.

e. f. g. a.

Electrical Quantities for Fig. 5-11(a)

Quantity

Measured Value

Calculated Value

RT

4.4 to 5.6 kΩ

5 kΩ

IR 1

1.8 to 2.2 mA

2 mA

IR 2

1.8 to 2.2 mA

2 mA

IT

3.7 to 4.3 mA

4 mA

VR1

20 V

20 V

VR2

20 V

20 V

Yes, within tolerances PR1 = 36 to 44 mW, PR2 = 36 to 44 mW, PR3 = 74 to 86 mW Yes, within tolerances See Table 5-3.

TABLE 5-3

Electrical Quantities for Fig. 5-11(b)

Quantity

Measured Value

Calculated Value

RT

830 to 90 Ω

909 Ω

IR 1

1.8 to 2.2 mA

2 mA

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80

IR 2

18 to 22 mA

20 mA

IT

20 to 24 mA

22 mA

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81

3.

4.

b. c. d. a. b. c. d. a. b. c. d.

No Yes R2 RT = 800 to 940 Ω, IT = 21 to 25 mA, IR3 = 0.8 to 1.2 mA Yes Yes 19 to 23 mA (1) circuit c, (2) circuit c, (3) circuit c Less than Less than (1) R3, (2) R2, (3) R2

5-7 LAB EXPERIMENT: SERIES-PARALLEL CIRCUIT—CURRENT, VOLTAGE, AND RESISTANCE 1.

Less 2.

Less 3.

See Table 5-4.

TABLE 5-4

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

Electrical Quantities for Fig. 5-12

Quantity

Measured Value

Calculated Value

RT

3.9 to 4.8 kΩ

4.34 V

VR1

20 V

20 V

VR3

2.3 to 2.9 V

2.6 V

VR4

15.8 to 19 V

17.4 V

IR 1

1.8 to 2.2 mA

2 mA

IR 2

2.3 to 2.9 mA

2.61 mA

IR 3

1.6 to 1.9 mA

1.74 mA

IR 4

0.8 to 1.0 mA

0.87 mA

IT

4.1 to 5.1 mA

4.61 mA

82 to 102 mW 92 mW Probably not Because of resistor tolerance, meter tolerance, and error in reading the meter Approximately 17.4 V Negative Yes No No Yes It would increase.

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82

5-8 LAB EXPERIMENT: ZENER VOLTAGE DIVIDER AND REGULATOR 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

8.06 V a. 3.94 V b. 3.92 V c. Yes, within tolerances of equipment.

4.72 mA 4.79 mA Yes a. It should decrease to about 4 mA. b. Very little, unless the zener voltage changes dramatically. a. 7.90 V b. 0.18 mA c. 8.73 mA 8.2% 8.11 V Increase a. 2.7% b. 0.62%

5-9 ADVANCED PROBLEMS 5-1.

5-2.

5-3.

5-4.

5-5.

Five 113.3 V 2Ω 20 A, 6 Ω IR2 = 0.15 A, R2 = 66.7 Ω, R3 = 15 Ω, VR3 = 6 V, R4 = 66.7 Ω, PT = 6.4 W

5-10 SELF-DESIGNED CIRCUITS 1.

a. Any two resistors with a 10 to 1 ratio as long as the higher resistor is greater than 700 Ω. (This derates the power rating of the highest resistor by about 50 percent.) b. A parallel circuit with a 100-Ω resistor or a 220-Ω resistor and a source voltage no greater than 5 V. c. One possibility would be a series combination of a 100-Ω resistor and a 1000-Ω resistor paralleled by a 2200-Ω resistor. The source voltage should not exceed 17.4 V if the 1000-Ω resistor is to dissipate less than 0.25 W.

5-11 LAB EXPERIMENT: TROUBLESHOOTING—ONE FAULTY COMPONENT 1.

a. R6 b. The resistance is about one-half its nominal value. c.

Nominal Value (Calculated)

Measured Value

IT

1.8 mA

1.72 V

VR4

2.60 V

3.79 V

VR s

5.56 V

8.10 V

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83

VR6

11.8 V

8.10 V

RT

16.9 kΩ

11.63 kΩ

RT determined with measured values of IT and VT is low. The measured value of VRs is equal to the measured value of VR6. Therefore, actual value of R6 must be about 4.7 kΩ. Calculating with measured value indicates that R6 = 8.10 V ÷ 1.72 mA = 4.7 kΩ.

2.

a. R3 b. Resistance is about one-fifth of its nominal value. c.

Nominal Value (Calculated)

Measured Value

IT

3.75 mA

4.19 mA

VR4

8.25 V

9.21 V

VR5

5.87 V

8.89 V

VR3

5.87 V

1.89 V

VR6

11.80 V

10.80 V

RT

5.3 kΩ

4.8 kΩ

Measured VR5 and VR3 show that either R5 increased in value or R3 decreased. Since RT measured is less than RT nominal, R3 had to decrease rather than R5 increasing. Calculations using verified values show that I R5 = 8.89 V ÷ 4.7 kΩ = 1.89 mA = I R3 and R3 = 1.89 V ÷ 1.89 mA = 1 kΩ.

5-12 LAB EXPERIMENT: TROUBLESHOOTING—TWO FAULTY COMPONENTS 1.

a. R5 and R15 b. R5 decreases to about 5 kΩ, and R15 is shorted. c.

Nominal Value (Calculated)

Measured Value

IT

1.66 mA

2.04 m

VR6

7.80 V

9.56 V

VR15

6.10 V

1.15 mV

VR10

6.10 V

10.44 V

VR5

9.22 V

7.11 V

VR4

2.98 V

3.33 V

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RT

12.0 kΩ

9.8 kΩ

The ratio of the measured values of VR5 and VR4 shows that either R4 or R5 is faulty. The measured value of VR10 equals VR4 + VR5 and VR15 = 0 V, so either R15 is shorted or R10 is open. If R10 is open, then I T = I R6 = I R5 = I R4 and either VR4 = 2.04 mA × 2.2 kΩ = 4.49 V or VR5 = 2.04 mA × 6.8 kΩ = 13.87 V. Since neither is close to the measured value, R15 must be shorted. Because R15 is shorted, I R10 = 10.4 V ÷ 20 kΩ = 0.52 mA and I R5 = I R6 - I R10 = 2.04 mA – 0.52 mA = 1.52 mA. Now, R4 = 3.33 V ÷ 1.52 mA =

2.

2.19 kΩ, so R4 is not defective. Finally, R5 = 7.11 V ÷ 1.52 mA = 4.68 kΩ. Therefore, R5 is defective because its nominal value is 6.8 kΩ. a. R16 and R6 b. R16 is open and R6 is about one-half its stated value. c.

Nominal Value (Calculated)

Measured Value

IT

2.96 mA

2.53 mA

VR16

4.18 V

14.40 V

VR3

1.9 V

0.00 V

VR2

1.07 V

2.53 V

VR5

5.01 V

11.90 V

VR6

13.9 V

5.57 V

RT

6.75 kΩ

7.91 kΩ

Voltages across R2 and R5 have the correct ratio, so R2 and R3 are either both faulty or both all right.

Since measurements show that either R16 is open or R3 is shorted, both R2 and R5 must be good resistors. Measured values show that R6 = 5.57 V ÷ 2.53 mA = 2.2 kΩ, so R6 is defective. Knowing R5 is not defective yields IR5 = 11.90 V ÷ 4.7 kΩ = 2.53 mA. This shows that IR5 = IR6 and IR3 = 0; therefore, R16 has to be open rather than R3 being shorted.

CHAPTER 6 NOTES TO THE INSTRUCTOR 1.

2.

When measuring the R5 of the constant-current source of experiment 6-2, use an ohmmeter range that has an internal voltage supply of less than 14 V. (Some VOMs use a 30-V battery on the × 100kΩ range.)

The “unknown voltage source” for experiment 6-5 should be a 1.5-V D cell in series with a 100-Ω resistor. The cell and resistor should be mounted in a box so that only the two output terminals are visible to the student.

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3.

4.

If a VOM is used in experiment 6-5, it should have at least a 10-kΩ/V rating.

The approach given in the answers to experiment 6-6 is somewhat unconventional, but it is fairly simple. It should be a challenge for many students.

Unless you have precision (l percent) resistors available, I would not recommend doing this experiment with real components. However, if you elect to use real components, R4 should be a 2.2-Ω resistor and all resistor values should be camouflaged.

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86

TEST 1. Multiple-source 2. Constant-current 3. Norton’s 4. Thevenin’s, Norton’s 5. Three 6. T 7. T 8. T 9. F 10. F 11. F 12. F 25. IR3 = 0.8 A, VR3 = 24 V, IR1 = 1.8 A 26. IR2 = 1.38 A, IR3 = 0.06 A

13. F 14. T 15. T 16. F 17. F 18. F 19. F 20. F 21. T 22. 2X + 6Y + 4Z = 7 23. A = 3, B = 2, C = 6 24. RTH = 7.5 Ω, VTH = 47.5 V 27. IN = 0.3 A, RN = 8 Ω 28. 41.4 mA

6-1 LAB EXPERIMENT: ANALYZING A BRIDGE CIRCUIT 1.

2.

3.

Negative VR5 = 1.33 V, IR5 = 0.28 mA a. 1.3 V ±10% b. Yes, yes c. 0.28 mA ±10% d. No e. Yes 4. VR2 = 2.07 V, IR2 = 2.07 mA 5. VR2 = 2.1 V ±10%, IR2 = 2.1 mA ±10% 6. IR3 = 1.79 mA 7. IR3 = 1.8 mA, yes 8. VR4 = 16.6 V 9. VR4 = 16.6 V ±10% 10. Increases 11. IR5 = 0.21 mA when R5 is 6.8 kΩ 12. Yes

6-2 LAB EXPERIMENT: THE CONSTANT-CURRENT SOURCE 2.

3.

4.

5.

6.

7.

8.

>> 1 MΩ, yes 1 mA ±10%, 2.2 V ±10% 1 mA ±10%, 0.1 V ±10%, no, yes 1 mA ±10%, 10 V ±10% Yes 0.72 mA ±20% (approximately 70 percent of the current recorded in the previous steps) a. Within limits, the current from the terminals is independent of the load resistance. b. Within limits, the voltage at the terminals increases as the load resistance increases. c. The internal resistance of a constant current source is very high. d. When the limits of a constant current source are exceeded, the current drops.

6-3 LAB EXPERIMENT: EQUIVALENT CIRCUITS—NORTON’S AND THEVENIN’S Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

87

1.

See Table 6-1.

TABLE 6-1 Measured Currents and Voltages for the Circuits of Fig. 6-6* Value of RL 1000 Ω

2200 Ω

4700 Ω

Type of circuit

VRL

IR L

VRL

IR L

VRL

IR L

Complex [Fig. 6-6(a)]

0.87 V

0.87 mA

1.65 V

0.75 mA

2.76 V

0.59 mA

Norton’s [Fig. 6-6(b) and (c)]

0.87 V

0.87 mA

1.65 V

0.75 mA

2.76 V

0.59 mA

Thevenin’s [Fig. 6-6(d) and (e)]

0.87 V

0.87 mA

1.65 V

0.75 mA

2.76 V

0.59 mA

*Recorded values may vary ±15% depending on component tolerances.

2. RN = 6.7 kΩ, IN = 1 mA 3. 1 mA ±10%, yes 5. Yes 7. Yes 8. RTH = 6.7 kΩ, VTH = 6.7 V 9. V = 6.7 V ±10%, yes 10. V = 6.7 V, yes 11. Yes 12. Yes

6-4 LAB EXPERIMENT: THE SUPERPOSITION THEOREM 1. 1 mA, right to left 2. 1 mA ±10%, yes 3. 1.2 mA ±10%, left to right 4. 2.2 mA ±10%, right to left 5. No, no 6. Yes 7. Yes 8. No 9. Increase 10. 3.5 mA ±10%

6-5 LAB EXPERIMENT: ANALYSIS OF A VOLTAGE SOURCE 1. 1.55 V ±10% 2. 1.1 V ±10% 3. 100 Ω ±10% 4.

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88

6. 1.3 V ±10%, yes 7. 15.5 mA ±10% 8. 1.4 mA ±10% 9. 1.4 mA ±10%, yes

6-6 LAB EXPERIMENT: TROUBLESHOOTING A BRIDGE 2.

3.

R4 Shorting out any resistor in Fig. 6-9 reduces the circuit to a series-parallel circuit. This makes it easy to calculate the total current. When a given resistor is momentarily shorted out with a jumper wire, and the measured total current does not agree with the calculated total current, that resistor is not the faulty resistor. When the two currents (measured and calculated) agree, the faulty resistor has been found.

Rather than calculating all five possible currents before making any measurement, I did one calculation and then measured the current for the same shorted resistor. Luckily, I found the faulty resistor on the second round.

Calculated and measured values follow: With R6 shorted: 10´ 4.7 R4,5 = = 3.20 kW 14.7 R3,4,5 = 1+ 3.20 = 4.20 kW

R2,3,4,5 = RT = IT =

1´ 4.20 = 0.808 kW 5.20

15 = 18.56 mA 0.808

Measured IT = 21.0 mA; so R6 is not faulty.

With R4 shorted: 4.7´ 4.7 R5,6 = = 2.35 kW 9.4 R2,5,6 = 1+ 2.35 = 3.35kW

RT = R3,2,5,6 =

1´ 3.35 = 0.77 kW 4.35

15 = 19.48 mA 0.77 Measured IT = 19.50 mA, so R4 is faulty.

IT =

6-7 LAB EXPERIMENT: TROUBLESHOOTING COMPLEX CIRCUITS

1.

a. IR1 = 0.692 A I2 = 0.846 A IR3 = 1.538 A VR2 = 16.923 V VR3 = 23.077 V b. R2 is faulty. c. It has considerably less resistance (12 Ω) than its rated value of 20 Ω.

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89

d. The measured source voltages were correct, but VR1 and VR2 were too small and VR3 was too large.

Next, I disconnected one end of R3 and measured VR1 at 4.545 V. 4.545 V Then I calculated I R1 = = 0.455 A. 10 W Then I calculated IR1 (with R3 disconnected) using the values specified in Fig. 6-10. This value was: 40 V - 30 V 10 V I R1 = = = 0.333 A 10 W+ 20 W 30 W This shows that either R1 or R2 or both have less resistance than specified.

Next, I reconnected R3, disconnected R2, and measured VR1 at 12.0 V. Now the calculated value of IR1 is (using measured VR1) 12 V I R1 = = 1.2 A 10 W Recalculating IR1 using only values specified in Fig. 6-10 yields: 30 V I R1 = = 1.2 A 25 W This shows the stated value of R1 is correct and, therefore, R2 has less than its stated resistance.

The actual resistance of R2 can be determined by using the value of IR1 determined with the measured voltage of R1 when R3 was disconnected. This value was 0.455 A. The needed calculations are: VR2 = V2 - V1 - VR1 = 40 V - 30 V - 4.545 V = 5.455 V R2 =

2.

VR2 I R2

=

5.455 V = 11.99 W 0.455 A

a. I R1 = 1.167 A

I R2 = 0.083 A I R3 = 1.250 A

I R2 = 2.5 V I R3 = 12.5 V b. R3 is faulty. c. The resistance of R3 has decreased to 5 Ω. d. The measured values of both I R1 and I R2 are greater than the calculated values obtained with the voltages and resistances given in Fig. 6-11. Using measured currents for R1 and R2 and Kirchhoff ʼs I law, I determined that I R3 = I R1 + I R2 = 1.444 A + 0.222 A = 1.66 A This current is also larger than expected. The voltages across the resistors are calculated to be: VR3 = 1.666 A × 10 Ω = 16.66 V

VR2 = 0.222 A × 30 Ω = 6.66 V VR1 = 1.444 A × 15 Ω = 21.66 V Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

90

Using the specified source voltage, the calculated resistor voltages and Kirchhoff ʼs V law, I discovered that the loop of V1, VR1 , and VR3 violates Kirchhoff’s law as does the loop of V2, VR2 , and VR3 . However, the outer loop (V1, V2, VR1 , and VR2 ) satisfies Kirchhoffʼs V laws. Therefore, I can conclude that the resistance of R3 is less than specified.

The true value of R3 was determined by these calculations: VR3  30 V − 21.66 V  8.34 V, or

VR3  15 V − 6.66 V  8.34 V and R3 =

8.34 V = 5.02 W 1.666 A

CHAPTER 7 NOTES TO THE INSTRUCTOR 1.

2.

3.

4.

The battery specified for experiment 7-1 provides enough current for observable results. If you plan to use a substitute source, check its effects on the experiment. If the substitute power source provides too much current, use a series resistance.

The no. 36 wire for experiment 7-1 can be obtained by removing a strand from a piece of stranded test lead wire.

The powdered iron slug from a discarded adjustable coil can be used for the ferrite rod in experiment 7-2.

The 2-Ω, 10-W resistor in experiment 7-3 is not needed if you have power supplies that provide about 2 A in the 2- to 3-V range. A center-tapped 6.3-V transformer works for the ac supply. The same transformer can be full-wave rectified to provide the dc supply. No filtering is required.

TEST 1. F 2. F 3. T 7. T 8. T 9. T 10. T 11. F 12. F 13. F 14. T 15. F 16. T 17. F 18. F 19. T 20. T 21. T 22. T

4. F 5. T 6. F 30. Parallel 31. Decreases 32. Temporary 33. Magnetomotive force (mmf) 34. Flux 35. Flux density 36. Commutator 37. B 38. C 39. A 40. D 41. C 42. A 43. C 44. C 45. D

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23. F 24. T 25. Lodestone 26. Magnetic flux 27. South 28. Attract 29. Decrease

46. A 47. A 48. 400 tesla 49. 800 A·t 50. 300 A·t/m 51. 0.0033 Wb/(A·t·m) 52. 1.3A µr = 450

7-1 LAB EXPERIMENT: MAGNETIC FIELDS AROUND CONDUCTORS 2.

3.

4.

a. Opposite b. Repel c. Repel a. Yes b. No c. Attract d. The magnetic lines of force (flux) which exist around the conductors when they carry current Repel, because the currents are flowing in opposite directions

7-2 LAB EXPERIMENT: MAGNETIC MATERIALS, POLES, AND FIELDS 1.

2.

3.

4.

5.

a. Iron and ferrite b. No c. Temporary a. Yes b. Permanent a. Yes a. Opposite b. Yes c. No a. Yes b. Larger c. The air it replaces

7-3 LAB EXPERIMENT: ELECTROMAGNETISM 1.

2.

3.

a. Yes b. Yes c. Demagnetized d. No a. Yes, because the flux from the electromagnet travels through the relatively low reluctance of the nails b. Most (or all) small nails were released from the magnet, because the large nail made a temporary magnet. c. Yes d. Residual magnetism a. Yes

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92

4.

5.

b. No, because the first circuit (with the lamp) magnetized the nail well above its residual magnetism level c. No d. No a. Opposite of 4b b. Opposite of 4a c. Opposite of 4b (same as 4a) a. No b. No, because wood and air have about the same permeability c. Yes, because iron has a much higher permeability than either air or wood. The reluctance of the magnetic circuit has been reduced; therefore, more flux is created. d. Yes, because the magnetomotive force has increased by a factor of 10. e. The nail is rapidly pulled all the way into the coil. f. Because the reluctance of the nail is very low compared with that of air. By drawing the nail into the coil, the flux has the lowest reluctance path and the shortest path. g. The nail fell out of the coil. h. Because the magnetomotive force was reduced to zero. i. Yes j. Change the dc supply to an ac supply.

CHAPTER 8 NOTES TO THE INSTRUCTOR Experiment 8-4 is to be done using Multisim.

TEST 1. T 2. F 3. T 4. F 5. T 6. T 7. T 8. T 9. T 10. T 11. F 23. f 24. Hertz 25. Hz 26. Frequency 27. Vertical 28. Electrical degrees, horizontal 29. Period 30. 360

12. F 13. T 14. T 15. T 16. T 17. T 18. F 19. F 20. T 21. T 22. T 34. C 35. B 36. 0.002 s or 2 ms 37. 2500 Hz 38. 283 V 39. 200 V 40. 17 A 41. 50 Hz

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93

31. 1.73, wye 32. C 33. B

42. 450 V 43. 3 A 44. 5.2 A

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94

8-1 LAB EXERCISE: AC WAVEFORMS AND TERMINOLOGY 1.

a.

See Fig. 8-1.

Fig. 8-1

2.

b. c. d. e. f. a. b. c. d. e. f.

A sine wave 0.036 s 27.8 Hz 49.49 V 44.59 V Sawtooth 4V 3V 2.0 ms 500 Hz 500 cycles

8-2 LAB EXERCISE: THREE-PHASE CONNECTIONS 1.

a.

See Fig. 8-3(a).

Fig. 8-3a

b. c. d.

17.3 V 17.3 V 10 V

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95

2.

a.

See Fig. 8-3(b).

Fig. 8-3b

b.

10 V

8-3 LAB EXPERIMENT: VIEWING AND MEASURING WAVEFORMS 1.

2.

3.

a. b. c. d. e. f. g. a. b. c. d. a. b. c. d.

2V 2.5 ms ±10% Decrease Yes Decrease Yes Decrease 3.5 V ±10% Yes Yes No, the reason depends on the type of DMM used by the student.

6.25 ms 160 Hz 25 V 17.7 V

8-4 LAB EXPERIMENT: FREQUENCY AND VOLTAGE 1.

a. b. c. d. e. f. g. h. i. j. k. l.

0.2 ms 5 kHz Yes l1.2 Vp–p or 5.6 Vp 3.96 mA Yes, Yes Vp Vp–p V Yes No Yes

8-5 LAB EXPERIMENT: OSCILLOSCOPE MEASUREMENTS 1.

a. b. c.

1 ms ±5% 1 kHz ±5% Yes

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96

d. e. f. g.

h.

30 Vp–p ±5% 10 Vp–p ±5% 20 Vp–p ±5% (1) 3.5 mA ±5% (2) 2 kΩ ±5% (3) 37.5 mW ±5% (4) 12.5 mW ±5% 3.18 mW ±5%

8-6 LAB EXPERIMENT: HALF-WAVE RECTIFICATION 1.

2.

3.

4.

18 V Sinusoidal pulses separated by a distance equal to half a cycle of the input ac.

56.56 V a. 707 V b. 60 Hz c. Yes d. The oscilloscope indicates 1000 Vpeak Vav = Vpeak × 0.637 = 1000 V × 0.637 = 637 V Vdc =

e.

Vav 637 V = = 318.5 V 2 2

Yes

8-7 LAB EXPERIMENT: THREE-PHASE CIRCUIT 3.

4.

5.

6.

7.

8.

a. 277 V b. 277 V c. 277 V d. 479.78 V e. 479.78 V f. 479.78 V g. 5.54 A h. 5.54 A i. 5.54 A j. 277 V k. 277 V l. 277 V Yes 50 Ω No V1ine = 1.732 Vphase a. 277 V b. 277 V c. 277 V d. 479.78 V e. 479.78 V f. 479.78 V g. 6.648 A h. 5.837 A i. 5.837 A

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97

9.

j. 249.3 V k. 291.84 V l. 291.84 V All of the line currents increased. VR1 decreased while VR2 and VR3 increased. Other voltages

remained unchanged.

10. a. 277 V b. 277 V c. 277 V d. 479.78 V e. 479.78 V f. 479.78 V g. 7.387 A h. 5.54 A i. 5.54 A j. 277 V k. 277 V l. 277 V 11. 1.847 A 12. Yes 13. The only change in measured values was that the line 1 current increased.

CHAPTER 9 NOTES TO THE INSTRUCTOR 1.

2.

3.

4.

The motor used in experiment 9-1 must be mounted on an adequate base so that it does not bounce around.

The three meters specified for experiment 9-1 can all be replaced by a multipurpose meter such as the Simpson Model 390 appliance tester.

Experiment 9-1 can be done without the variable transformer; however, the motor will put a severe overload on the ammeter and wattmeter until it comes up to rated speed. This overload can be eliminated by connecting a switch across the ammeter and the current coil of the wattmeter. The switch is left closed until the motor comes up to speed; then it is opened to read the current or power.

Also, without the variable transformer, the starting torque of the motor will cause the motor to jump when it is plugged into the outlet. Use insulated connectors and terminals for this experiment.

If possible, power this experiment from a GFCI outlet.

Experiments 9-3 and 9-4 are to be done using Multisim.

TEST 1.

2.

3.

4.

5.

6.

7.

8.

T F T T T T F T

15. T 16. F 17. T 18. T 19. C 20. B 21. B 22. A 23. 2160 W

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98

9. T 10. F 11. F 12. T 13. F 14. T 30. 9.2 A 31. 0.5, or 50%

24. (a) 25. 2.144 26. 159 W 27. 1440 W 28. 0.866, or 86.6% 29. 30° out of phase 32. 50°

9-1 LAB EXPERIMENT: TRUE POWER AND APPARENT POWER 1.

2.

a. b. c. d. e. a. b. c. d. e. f.

4.17 A ±10% 500 VA ±15% 500 W ±15% Yes, because the current and voltage are in phase 1, or 100% 4 to 8 A 420 to 900 VA 100 to 450 W No, because the current and voltage are not in phase for an inductive-type load 0.3 to 0.54 73° to 57°

9-2 ADVANCED PROBLEMS 9-1. a. b. c. d. e.

9-2.

V1 Yes Lagging No Inductive

9-3. a. b. c. d. e.

9-4. a. b. c. d.

1680 VA 1176 W 45.6° 4.9 A 5A 6A 0.797 37.1° 4.78 A

9-3 LAB EXPERIMENT: PHASE SHIFT 1.

a. b. c.

About 50° Decrease About 26°

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99

d.

About 90°

9-4 LAB EXPERIMENT: THETA AND POWER FACTOR 1.

a. b. c. d. e.

14.57 V 19.07 V More than 24 V, 14.57 mA 212.3 mW, 350 mVA, 52.6°, 60.7%

9-5 LAB EXPERIMENT: THREE-PHASE POWER 1.

2.

3.

4.

5.

Wye Wye One a. 54 V b. 129 W c. 500 Ω a.

b.

Vline 1.732 Pphase = Iline Vphase

c.

Rphase =

Vphase =

6.

7.

8.

9.

Vphase I line

a. 254 V b. 129 W c. 500 Ω Yes 129 W, one Yes

CHAPTER 10 NOTES TO THE INSTRUCTOR 1.

2.

Experiments 10-3 and 10-4 can be completed with either physical components or with Multisim.

Experiments 10-6, 10-7, and 10-8 are troubleshooting experiments designed specifically for Multisim. The hidden fault in Fig. 10-15 is 1.2 MΩ of leakage in C1. For Fig. 10-16, C1 is open. In Fig. 10-17, C2 is open and C3 is shorted.

TEST 1.

2.

3.

4.

5.

6.

7.

8.

9.

T F F F F F T T T

20. T 21. F 22. T 23. F 24. F 25. C 26. C 27. B 28. D

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100

10. F 11. F 12. F 13. T 14. F 15. F 16. F 17. T 18. F 19. T 39. XC 40. Farad 41. C 42. Capacitors 43. Coulomb per volt 44. Tantalum 45. 44.8 J 46. 1500 pF 47. 1809 Ω 48. 320 pF 49. 68.75 pF

29. C 30. A 31. C 32. B 33. Electrolytic 34. Capacitive reactance 35. Five 36. Inversely 37. Electrolytic 38. Metal oxide 50. 155 ms 51. a. 60 V b. 3316 Ω 52. a. 6.3 mA b. 18.8 mA 53. 33 nF 54. a. 4.19 mA b. 16.76 mA c. 1.33 µF d. 1193 Ω

10-1 LAB EXPERIMENT: CAPACITORS IN AC AND DC CIRCUITS 1.

2.

3.

4.

a. b. c. d. e. f. a. b. c. d. e. a. b. c. d. e. a. b. c. d. e.

Yes No, because the capacitor is fully charged Zero Yes 10 V A spark occurred as the capacitor dissipated its energy.

Yes Yes Approximately the source voltage Less than 1 V The lamp 3.3 V 6.7 V 3.3 V ±10% Yes Yes Yes No No Yes Inversely proportional to the size of the capacitor

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101

10-2 LAB EXPERIMENT: RESISTOR-CAPACITOR TIME CONSTANTS 1.

a. b. c. d. e. f.

11 s 6.32 V 3.68 V Yes 1.35 V. See Fig. 10-7(a).

Yes. See Fig. 10-7(b).

(volts) 3.7 ± 10%

22

1.4 ± 10%

33

0.5 ± 10%

44

0.2 ± 10%

55

0.07 ± 10% (a )

Fig. 10-7

2.

a. b. d. e. f. g.

VR

Time (seconds) 11

VC

(VT − VR) (volts) 6.30 ± 10% 8.60 ± 10% 9.50 ± 10% 9.80 ± 10% 9.93 ± 10%

11 s 3.68 V See Fig. 10-9(a).

See Fig. 10-9(b).

No Yes

Time (seconds) 11 22 33 44 55

Vo

(volts) 3.7 ± 10% 1.4 ± 10% 0.5 ± 10% 0.2 ± 10% 0.07 ± 10% (a )

Fig. 10-9

3.

a. b. c. d.

Decrease 5 to 6 s No Yes

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102

10-3 LAB EXPERIMENT: RELATIONSHIPS BETWEEN FREQUENCY, CAPACITANCE, AND REACTANCE 1.

2.

3.

4.

a. 2.0 mA ± 10% b. 1500 Ω ± 10% c. 1592 Ω d. Yes e. Yes a. It will double. b. It will decrease by 50%. c. 1 mA ±10% d. Yes e. 4 mA ±10% f. 4 mA ±10% g. Yes h. 8 mA ±10% a. Half as much (2 mA ±10%), because half as much capacitance provides twice as much reactance b. 1 mA ±10% c. Capacitive reactance is inversely proportional to capacitance. a. XC is one-fourth as great. b. Does not change. c. XC is four times as great.

10-4 LAB EXPERIMENT: SERIES AND PARALLEL CAPACITORS 1.

2.

3.

a. b. c. d. e. f. g. h. i. a. b. c. d. e. f. g. h. i. j. a. b. c.

0.333 µF 4782 Ω 1 mA Yes 1.7 V 3.3 V 1.7 V ±10%, 3.3 V ±10% Smallest Greater than 1.5 µF 1327 Ω 3.77 mA 3.8 mA ±10% Yes Yes C1 2.5 mA ±10%, 1.3 mA ±10% Yes Yes Parallel Series Parallel

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103

d.

Larger

10-5 ADVANCED PROBLEMS 10-1. a. 1.33 µF b. 2387.3 Ω c. 4.19 mA d. 6.67 V 10-2. a. 1250 Ω b. 1768.4 Ω c. 0.85 µF d. 34.1 mA 10-3. 1778 µF, 0.8 C 10-4. a. 12,500 µF b. 1000 J

10-6 LAB EXPERIMENT: TROUBLESHOOTING 1 1.

2.

b. 100 s c. 36.8 V d. 50 s With the voltmeter across R1, the time constant should be 50 s, and VR1 after IT should be 36.8 V. The measured value of VR1 was stabilized at 99.6 V after 50 s. The voltage across C1 must therefore be only 0.4 V. This indicates that C1 must be leaking (partially shorted). If R1 is not faulty, the value of the leakage resistance can be calculated using series circuit rules. The value of R1 was checked by putting the multimeter on the dc I function and connecting it across C1 while S1 was open. The 10µA reading indicated that R1 was not faulty. Thus, the leakage resistance was calculated as follows:

I T = I R1 = Rleakage =

VR1 R1

VRleakage IT

=

99.6 V = 19.92 mA 5 MW

=

0.4 V = 20 kW 19.92 mA

In summary, C1 has 20 kΩ of leakage resistance.

10-7 LAB EXPERIMENT: TROUBLESHOOTING 2 1.

2.

VT and IT, 120 V and 45.2 mA Calculations with data from the diagram give the following results: XC1 = 4423 Ω, XC2 = 6635 Ω, IC1 = 27.1 mA, IC2 = 18.1 mA, and IT = 45.2 mA. Measurements show that VT = 120 V as expected, but IT (measured with the ammeter across the open switch) was 18.3 mA. C1 must be open because IT (measured) = IT (calculated) − IC1. Capacitor C2 seems to be all right because the measured current equals its current.

It is possible, but not likely, that both capacitors are faulty and measured IT just happens to equal the calculated IC2.

10-8 LAB EXPERIMENT: TROUBLESHOOTING 3 1.

Calculations with given values resulted in the following: XCT = 54.1 kΩ, IT = 1.48 mA, VC1 = 9.4 V, VC2 = 47.1 V, and VC3 = 23.5 V.

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104

Measurements and conclusions: Measured VC1 = 0, so either C1 is shorted or the circuit is open (or both). Measured VC2 = 80 V, so C2 is open. Measured I = 12.73 mA with the ammeter across C2 and the switch closed.

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105

More calculations: Calculated I with C2 shorted by the ammeter is I=

80 V 80 V = = 3.59 mA XC1 + XC3 22.3 kW

So, either C1 or C3 must be shorted. Calculated I with C1 and C2 shorted is I=

80 V 80 V = = 5.03 mA XC3 15.9 kW

which is still small compared to measured I. Calculated I with C2 and C3 shorted is

I=

80 V 80 V = = 12.5 mA XC1 6.4 kW

which matches (within tolerances and round-off) the measured I. Final conclusions: C2 is open and C3 is shorted.

10-9 LAB EXPERIMENT: DETERMINING CAPACITANCE WHEN FREQUENCY IS KNOWN 2.

XC = 530 Ω IC = 56.6 mA VT = 30 V C = 1.5 µF

CHAPTER 11 NOTES TO THE INSTRUCTOR The values given in the data table in experiment 11-3 are average values obtained from many examples.

Individual values may be 15 percent above or below the tabulated values.

If you experience a problem with the circuits in experiments 11-3 and 11-4 overloading your signal generator, you can increase the resistance of each resistor by a factor of 5.

Experiments 11-6 to 11-9 are written for Multisim. Experiments 11-6 and 11-7 are repeats of experiments 11-3 and 11-4 except that resistors R1, R2, and R3 have been eliminated.

In experiment 11-8 the hidden fault is an open switch.

For experiment 11-9, the trouble is that L2 is open.

TEST 1. T 2. T 3. T 4. T 5. F 6. T 7. F 8. F 9. T 10. F

12. T 13. T 14. T 15. T 16. T 17. T 18. T 19. F 20. F 21. F

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106

11. T 23. F 24. Skin effect 25. cemf 26. Variable 27. Iron 28. Iron 29. Lags, 90 30. Skin effect 31. B 32. D 33. A 34. C 35. C 36. A 37. A 38. B 39. D 40. B 41. B

22.

42.

43.

44.

45.

46.

47.

48.

49.

50.

51.

52.

53.

F A 240 Ω 1571 Ω 78.5 2.4 H The 4-H inductor The 0.4-H inductor 0.7 H 17 mA 120.6 kHz 167.5 IT = 69.7 mA LT = 171.4 mH XLT = 430.6 Ω 54. LT = 0.54 H XLT = 1357 Ω IT = 22.12 mA I0.4H = 13.3 mA

11-1 LAB EXPERIMENT: INDUCTORS IN AC AND DC CIRCUITS 1.

2.

3.

4.

a. 30 Ω ±10% b. Yes c. Yes a. Yes b. Yes c. No, because the reactance of the inductor does not allow sufficient current to flow d. Yes e. Reactance f. Yes g. Yes h. Yes, because the reactance has been reduced i. 0.5 H The inductive circuit of Fig. 11-2(a). When the current rapidly decreases, the cemf (inductive kick) voltage greatly exceeds the source voltage. a. No b. 188 Ω c. 1 Ω d. Laminated iron

11-2 LAB EXPERIMENT: INDUCTIVE KICK 1.

a. b. c. d. e.

No No 1.5 V (less than 80 V) Yes About 80 V

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107

f. The rapidly collapsing current (and thus flux) in the inductor Yes

2.

3.

a. b. c.

About 240 V Yes About 160 times larger

11-3 LAB EXPERIMENT: RELATIONSHIP BETWEEN f AND XL 1.

2.

393 Ω a. It will double. b. It will decrease to half its former value. c. Yes 1536 Ω ±15%. See Fig. 11-7.

3.

Voltage across R1

Frequenc VR1 y (measured f )

Current through

Voltage across L1

R1 IR1

VL1

Inductive reactance

(measured )

XL1

Voltage across R1

Frequenc VR1 y (measured f )

Current through

Voltage across L1

R1 IR1

VL1

Inductive reactance

(measured )

XL1

384 Ω

25 kHz

1.25 V

0.0125 A

4.8 V

384 Ω

25 kHz

1.25 V

0.0125 A

4.8 V

50 kHz

0.65 V

0.0065 A

4.9 V

754 Ω

50 kHz

0.65 V

0.0065 A

4.9 V

754 Ω

100 kHz

0.30 V

0.003 A

5.0 V

1667 Ω

100 kHz

0.30 V

0.003 A

5.0 V

1667 Ω

Fig. 11-7

Fig. 11-7

11-4 LAB EXPERIMENT: SERIES AND PARALLEL INDUCTORS 1.

2.

3.

a. b. c. d. e. f. g. a. b. c. a. b. c. d. e. f. g. h.

393 Ω 0.62 V ±10% 12.2 mA ±10% 4.95 V ±10% 406 Ω (±15%) 1.25 mH 393 Ω 0.64 V ±10% and 0.64 V ±10% Yes No Yes, because they have equal reactances and equal currents 1570.8 Ω 2.5 V ±10% 2.5 V ±10% 0.32 V ±10% 3.2 mA ±10% 1563 Ω ±15% Yes

11-5 ADVANCED PROBLEMS 11-1. a. 27 mH b. 33.93 kΩ c. 18.85 kΩ d. 0.88 mA e. 0.53 mA Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

108

11-2. a. 5655 Ω b. 16,655 Ω c. 79.3 V d. 5.84 H 11-3. a. 1000 Ω b. 628 Ω c. 31.8 mA d. 415 Ω e. 1.65 mH

11-6 MULTISIM: L, f, AND XL RELATIONSHIPS 1.

2.

3.

4.

5.

a. 5 V, 12.57 mA b. 398 Ω c. 393 Ω The current will be half its former value (6.28 mA). The reactance will double (796 Ω). a. 5 V b. 6.28 mA c. 796 Ω 6.28 mA 6.28 mA

11-7 MULTISIM: L IN SERIES AND PARALLEL 1.

2.

3.

4.

a. b. c. d. e. f. g. h. a. b. c. d. e. f. a. b. a. b. c. d. e.

785 Ω 392.5 Ω 1.25 mH 392.5 Ω Yes 12.74 mA Yes No 5V 12.57 mA 398 Ω Yes 9.42 mA 9.43 mA XL1 = 785 Ω, XLT = 1570 Ω, IT = 3.18 mA, VL1 = 2.5 V Yes VT = 5 V; IT = 3.14 mA; VL1 = 2.5 V; VL2 = 2.5 V XL1 = 1592 Ω; XL1 = 796 Ω Yes VL1 = 1 V; IT = 1.27 mA Yes

11-8 LAB EXPERIMENT: TROUBLESHOOTING 1 1.

2.

XL1 = 565 Ω, XL2 = 942 Ω, IT = 9.95 mA, VL1 = 5.62 V The measured value of VL1 was zero, so either L1 is shorted or the circuit is open (or both). Next, VL2 was measured at zero volts, so the circuit is probably open. However, if F1 is open, L1 and/or L2

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109

could also be either shorted or open. Next, the voltage across F1 was measured, and it also indicated zero. This indicates that either S1 is open or there are multiple faults. Next, the voltage across S1 was measured at 15 V regardless of the position of the S1 toggle, so S1 is open. Both L1 and L2 cannot be shorted or F1 would have opened before S1. Therefore, it should be safe to put the ammeter across S1 and measure IT. This was done, and the meter indicated 9.82 mA. Thus, both inductors are all right, or both have changed values.

A final check was made with the ammeter across both S1 and L1. The current was 15.71 mA, which is correct for 15 V across the 942 Ω of reactance of L2. This final I measurement confirmed that the only fault was that S1 was open.

11-9 LAB EXPERIMENT: TROUBLESHOOTING 2 1.

2.

XL1 = 236 Ω, XL2 = 283 Ω, XL3 = 472 Ω, IT = 58.1 mA, VL1 = 13.7 V, IL2 = 36.4 mA VL1 measured 8 V, so some possibilities are: L1 increased in value (not likely), both L2 and L3 decreased in value, VT increased in value, or either L2 or L3 opened (most likely fault). Next, the measured current with the ammeter across S1 (and S1 open) was 33.52 mA. This showed that L1 was all right, because VL1 = IT × XL1 = 33.52 mA × 236 Ω = 7.91 V. Since XL3 is twice as large as XL1, L3 would drop 15.82 V if it also carried IT, which it would do if L2 were open. Because the values of VL1 and VL3 calculated with the measured current are so very close to the measured voltages across L1 and L3, I conclude that L2 is open.

11-10 LAB EXPERIMENT: DETERMINING INDUCTANCE WHEN FREQUENCY IS KNOWN 2.

XL = 502 Ω IL = 79.6 mA VT = 40 V L = 0.4 H

CHAPTER 12 NOTES TO THE INSTRUCTOR 1.

2.

3.

4.

5.

A word of caution: The experiments in this chapter use 120- and 208-V sources. A shock received from either can be fatal. The equipment used for these experiments must be in good condition and have no exposed terminals, connections, and so on, in the 120- or 208-V part of the circuits.

The transformers used in this chapter should be mounted in an enclosure. The primary leads should terminate in banana jacks or other insulated connectors so that students do not accidentally make contact with them. Of course, the connectors or plugs must be numbered to correspond to the terminal numbers.

The primary fuse used in some of the activities should be mounted in the enclosure with the transformers. While the fuse is not essential in these circuits, it will prevent burning out transformers if students apply power to incorrectly phased transformers.

Answers for experiment 12-3 are based on using a 6.3-V transformer secondary as the power source. However, any low-voltage, high-current-capacity ac source will suffice. Of course, the quantitative answers given in this manual would have to be adjusted.

In experiment 12-4 the three single-phase transformers could be replaced by one three-phase transformer.

Experiment 12-3 can be done using either “hardware circuits” or “Multisim.”

Experiments 12-5 and 12-6 are for use only with Multisim. The circuit file for experiment 12-5 is Fig. 12-14. Resistor R3 in this circuit is partially shorted. For experiment 12-6, use the file for

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110

Fig. 12-15. The secondary of T3 is open in this circuit. Both files have the appropriate restrictions so that the students can make connections only between the DMM and the circuit test points.

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111

TEST 1.

2.

3.

4.

5.

6.

7.

8.

9.

19.

20.

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

31.

32.

33.

34.

35.

36.

37.

T T T T F T F F T F F T F T T F F T Magnetic flux Increased Step down Stepped up Step down Residual magnetism C D D A

10.

11.

12.

13.

14.

15.

16.

17.

18.

38.

39.

40.

41.

42.

43.

44.

45.

46.

47.

48.

49.

50.

51.

52.

53.

54.

55.

T T T T F T F F T A D D C A A D B C C 652.2 W 225 to 1 3150 Ω 1150 turns 12.5 V 5 turns per volt 84.7% 50%

12-1 LAB EXPERIMENT: TRANSFORMER EFFICIENCY AND POWER FACTOR 1.

2.

3.

4.

5.

a. b. a. b. c. a. b. c. d. a. b. a. b. c. d. e.

230 V No, because the incorrect phasing would effectively short out the primary 31 W ±5% 275 mA ±5% 6 V ±5% 33 W ±10% 0.939 20° Resistive 24 W ±5% 77% ±10% No No Decrease No Decrease

2-2 LAB EXPERIMENT: SERIES AND PARALLEL WINDINGS Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

112

1.

2.

3.

4.

5.

6.

a. b. c. d. a. b. c. a. b. a. b. c. a. b. c. d. a. b. c. d. e.

Series-aiding 12 V 13.8 V ±10% Yes Series-aiding 13.8 V ±10% Yes None Series-opposing 6.9 V ±10% Parallel No Series-aiding It will be half as great as before.

3.4 V ±5% No 3.6 V ±5% More Opposing Decrease 3.3 V ±5%

12-3 LAB EXPERIMENT: IMPEDANCE MATCHING

Load resistance (ohms) 220 470 1000 2200 4700

Load voltage measured (volts) 1.1 2.0 3.2 4.4 5.4 (a)

Load Power calculated (watts) P = V2/R 5.5  103 8.5  103 10.2  103 8.8  103 6.2  103

Load resistance (ohms)

Load voltage measured (volts)

Load Power calculated (watts) P = V2/R

2.2 4.7 10 22 47

0.06 0.15 0.25 0.35 0.45

1.6  103 4.8  103 6.3  103 5.6  103 4.3  103

Fig. 12-8

(b)

a. Yes b. 1000 Ω Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

113

2.

c. d. a. b. c. d. e. f. g. h.

Yes Yes 9.6 to 1 (approx. 10 to 1) 92 to 1 (approx. 100 to 1) The 10-Ω resistor Yes Greater by approximately 10 times Transformer losses in the matching transformer 2.5 Ω 2.5 Ω

12-4 LAB EXPERIMENT: THREE-PHASE TRANSFORMER CONNECTIONS 1.

2.

3.

a. b. c. d. e. f. g. h. a. b. c. d. e. f. g. h. i. j. k. a. b.

Yes Line current (meter A1) Yes Yes Decrease Stays the same Less than 0.2 V 12.4 V ±10% Yes Yes Yes Yes No Less than 1 mA No No Yes 140 mA ±5% Yes Yes No

12-5 LAB EXPERIMENT: TROUBLESHOOTING 1 1.

2.

3.

VR1 = 24 V, IR1 = 24 mA Vline 1 to line 2 = 41.52 V Vload star point to source star point = 0 V Yes The measured values of VR1, VR2, and VR3 were 36.54 V, 36.54 V, and 5.95 V respectively. This leads one to suspect either R3 or T3 as the faulty component. Next, the secondary voltage of T3 was measured at 24 V, so R1 must be below its rated value.

We can get an estimate of the true value of R1 by inserting an ammeter between load star point and source star point. When this was done, the ammeter indicated 218.2 mA. We know that (1) this current should be zero and (2) the current from R1 (24 mA) should exactly cancel the current from R2 and R3 when there is no circuit fault. Therefore, R1 is furnishing the 24 mA to cancel IR1 and IR2

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114

plus the 218.2 mA of neutral wire current. When the ammeter is connected, R1 is connected across the secondary of T3 so VR3 = 24 V. Thus, the true value of R1 will be:

R1 =

VR1 24 mA + 218.2 mA

=

24 V = 99.2 W 242.2 mA

12-6 LAB EXPERIMENT: TROUBLESHOOTING 2 1.

2.

3.

VR1 = 24 V, IR1 = 24 mA Vline I to line 2 = 41.52 V Vload star point to source star point = 0 V Yes 20.78 V, 20.78 V, and 12.04 V were measured across R1, R2, and R3 respectively. Most likely, the faulty component is either R3 or T3. The measured voltage across the secondary of T3 was only 12 V, so T3 must be faulty.

To help pinpoint the trouble with T3, the current between load and source star points was measured with the ammeter. (Remember that connecting the star points puts the secondary of each transformer across its load.) The current reading was 24 mA. Since this is the value of current that R3 should be furnishing to the star point to cancel IR1 + IR3 the secondary of T3 must be open.

12-7 LAB EXPERIMENT: THREE-PHASE CONNECTIONS 3.

There are many possible solutions to this design experiment. One possible solution is shown in Fig. 12-9 below.

Fig. 12-9

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115

CHAPTER 13 NOTES TO THE INSTRUCTOR 1.

2.

3.

Experiments 13-5 and 13-6 are designed specifically for Multisim.

The file for experiment 13-5 has no restriction on it, so the students can change the value of R. The file for experiment 13-6 is restricted so that the student can connect the DMM only to the test points provided in the circuit. The fault in the circuit is 1100 Ω of leakage in the capacitor.

TEST 1. T 2. T 3. T 4. T 5. F 6. F 7. T 8. F 9. T 10. T 11. T 12. F 13. T 14. T 15. F 16. T 17. T 18. F 19. F 20. T 21. F 22. T 23. F 24. T 25. F 26. F 27. T 28. D 29. B 30. C 31. D 32. B

33. D 34. B 35. A 36. 94.2 37. 133.3 38. 1.59 kHz 39. a. 3526 Ω b. 5.7 mA c. 57.9 mW d. 0.51 40. 673 kHz 41. a. 1548 Ω b. 6.6 mA c. 30.7° d. 12.9 mA 42. a. 103 Ω b. 0.63 c. 24 V d. Inductive 43. a. 43.8 mA b. 958.4 Ω c. Inductive d. 0.799 44. a. 2346 Ω b. 0.426 45. Z = 1230 Ω 46. Z = 59.5 Ω PF = 0.94 47. IT = 0.336 A P = 6.33 W

13-1 LAB EXPERIMENT: IMPEDANCE 1.

a. 995 Ω b. 1411 Ω c. 3.54 mA d. 3.54 V

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116

2.

3.

4.

5.

6.

e. 3.52 V f. 3.5 mA ±10% g. Yes h. VR = 3.5 V ±10%, VC = 3.5 V ±10% i. Yes j. 1430 Ω ±10% k. Yes a. Decrease b. Decrease c. Increase d. 2.7 mA ±10% e. 2.7 V ±10% f. 4.2 V ±10% g. Yes a. Decrease b. Decrease c. Increase d. 2.1 mA ±10% e. 2.1 V ±10% f. 4.5 V ±10% g. Yes a. 2117 Ω b. 1.89 mA c. 1.82 mA d. 2.62 mA e. 1527 Ω f. 2.60 mA ±10% g. 1538 Ω ±10% h. Yes i. 1.85 mA ±10% j. 1.70 mA ±10% k. Yes a. No change b Decreases c Decreases d. 1.70 mA ±10% e. 1.2 mA ±10% f. 2.2 mA ±10% g. Decreases a. 11.2 V ±10% b. 7.7 V ±10% c. 3.4 V ±10% d. Greater e. Greater f. Greater g. Less h. 7.2 mA ±10% i. 694 Ω ±10% j. 1069 Ω ±10%

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117

7.

a. b. c. d.

Decrease Decrease Decrease Decrease

13-2 LAB EXPERIMENT: RESONANCE, QUALITY, AND BANDWIDTH 1.

2.

3.

4.

a. b. c. d. e. a. b. c. d. a. b. c. d. e. f. g. a. b. c. d.

101 kHz 0.03 V ±20% 0.03 mA ±20% 167 kΩ ±20% Maximum 99.5 kHz ±20% 102.5 kHz ±20% 3 kHz ±20% 37 ±20% 101 kHz 22 V (depends on generator) 0.65 V (depends on generator) 34 ±20% Yes Yes Yes 1586 Ω 470 Within 10% Yes

13-3 LAB EXPERIMENT: FILTERS 1.

2.

a. b. c. d. a. b. c. d.

Low-pass 100 Hz Yes. See Fig. 13-8.

4.4 V ±10% High-pass 100 Hz Yes. See Fig. 13-10.

1500 Hz ±20% Frequency (hertz)

Output voltage (volts)

30

5.0

± 10%

60

5.0

± 10%

100

5.0

± 10%

300

4.9

± 10%

600

4.65

± 10%

1,000

4.15

± 10%

3,000

2.2

± 10%

6,000

1.25

± 10%

10,000

0.75

± 10%

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118

30,000

0.25

± 10%

60,000

0.08

± 10%

100,000

0.0

Fig. 13-8

Frequency (hertz)

Output voltage (volts)

30

0.05

± 10%

60

0.20

± 10%

100

0.35

± 10%

300

1.0

± 10%

600

1.8

± 10%

1,000

2.85

± 10%

3,000

4.5

± 10%

6,000

4.85

± 10%

10,000

4.95

± 10%

30,000

5.0

± 10%

60,000

5.0

± 10%

100,000

5.0

± 10%

Fig. 13-10

13-4 LAB EXPERIMENT: FILTERS 2 1.

A series RC low-pass filter 2. a. 318.5 Ω b. 31.8 Ω 3.

A 28-Vp–p, 100-kHz signal superimposed on a 28-Vp–p, l0-kHz signal 4.

Yes, because XC at 10 kHz is about ⅓ as large as R. 5. a. Yes b. Very little of either signal appears in the output. d. More of the 100-kHz signal will appear in the output.

13-5 LAB EXPERIMENT: MEASURING PHASE RELATIONSHIPS 1.

2.

a. b. c. d. e. f. g. h. a. b.

7V Lag 45° 7V Lead 90° Lead 45° Yes Yes

13-6 LAB EXPERIMENT: LC RESPONSE CURVES 1.

a. 158.7 kHz

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119

b. 159.2 kHz c. Yes d. Upper f = 175.7 kHz Lower f = 143.8 kHz e. 31.9 kHz f. g. h. i. j. k. l.

4.97 9.94, 15.9 kHz 9.86, 16.1 kHz Inversely 49.7, 3.19 kHz 48.1, 3.3 kHz The curve became sharper (narrower).

13-7 LAB EXPERIMENT: TROUBLESHOOTING 1.

2.

XC = 1224 Ω XL = 817 Ω Z = 454 Ω I = 33 mA VC = 40.5 V VL = 27.0 V VR = 6.6 V The measured voltages and current for the circuit were VC = 14.37 V, VL = 14.62 V, VR = 3.53 V, and I = 17.67 mA. The calculated oppositions using these measured values were R = 3.53 V/17.67 mA = 199.8 Ω, XC = 14.37/17.67 mA = 813.2 Ω, and XL = 14.62 V/17.67 mA = 827.4 Ω. Only XC was extremely different from the value found in step 1, so the capacitor must be faulty.

The small value of XC could be caused by a capacitance larger than the labeled value (not too likely). Or, the 813.2 Ω calculated with VC and IC is the impedance of a leaky capacitor rather than the reactance of a larger capacitor. We can check which is the case by calculating Z two ways: Z = VT / I T = 15 V/17.67 mA = 848.9 WΩ

Z = R2 + ( XL - XC )2 = 2002 + (827.4- 813.2)2 = 200.5ΩW Since the two calculated circuit impedances do not agree, we know that the capacitor is leaky and the 813.2 Ω calculated for the opposition of C is a combination of XC and R in parallel.

CHAPTER 14 NOTES TO THE INSTRUCTOR 1.

The locked-rotor torque of fhp motors can be measured by clamping a 1-ft or a 6-in torque arm to the motor shaft and measuring the force with a spring or digital scale. With a 1-ft arm, the scale reading will be in the 1- to 10-lb range. Be sure the motor is securely mounted and the outlet or test bench has a functioning circuit breaker. Allow ample time for the motor to cool before repeating the locked-rotor torque measurement. The torque measuring apparatus needs a restraining yoke over the torque arm to limit the travel of the arm if the motor is turning in the wrong direction.

Use a three-connector grounded cord and a grounded outlet for this activity. If possible, use a ground-fault protected outlet.

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120

2.

3.

Some split-phase motors have only three lead wires. One lead is internally connected to both a start and a run winding. These motors can not be reversed without modifying the wiring.

The two motors for these experiments should have the same current, speed, hp, and SF ratings.

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121

TEST 1. F 2. F 3. T 4. F 5. F 6. F 7. T 8. F 9. T 10. F 11. F 12. F 13. T 14. A 15. B 16. D 17. B 18. C 19. B

20. D 21. 0.625 hp 22. 0.863 hp 23. 1.52 lb-ft 24. 5% 25. 0.74 26. 1200 rpm 27. 66 ft 28. Shunt 29. Series 30. Shunt 31. Cumulative 32. Series 33. Armature and field 34. Cumulative compound 35. About 75% 36. 105°C 37. 76.7% 38. 90°C

14-1 LAB EXPERIMENT: REVERSING ROTOR ROTATION 1.

a. Open b. No, because the open start switch disconnects one side of the start winding from the power terminals c. 1.0 Ω (for a ½-hp motor) d. Run winding e. 3.5 Ω f. It is about 3.5 times as large. g. Yes

14-2 LAB EXPERIMENT: MEASURING MOTOR CHARACTERISTICS Current

Torque

Type of motor

Noload

Lockedrotor

Rated

Lockedrotor

Rated

Split-phase*

6A

50 A

7.8 A

1.9 lb-ft

1.52 lb-ft

Capacitor-start*

6A

35 A

7.8 A

4.6 lb-ft

1.52 lb-ft

*Data for ½-hp, 7, B-A, 1725-rpm motor.

Fig 14.1

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122

1.

2.

3.

4.

See Fig. 14-1. a. See Fig. 14-1. b. Because the effective inductance of the run winding is small and the start winding is in the circuit. c. The PF of the full-load current is much greater than that of the no-load current. Thus, the difference in power is much greater than the difference in current for the two load conditions. d. Yes, yes See Fig. 14-1. a. Split-phase, because the start winding has few turns and the run and start currents are only about 25° out of phase b. Capacitor-start, because the start and run currents are almost equal and about 90° out of phase c. Little or no difference, because the no-load currents, rated currents, and rated torques of the two motors are about equal and these parameters are independent of the start windings d. Because extended locked-rotor operation would overheat the windings. Also, in the capacitorstart motor, the capacitor would overheat and possibly vent or explode. e. Decrease, because, owing to the positive temperature coefficient of copper, the winding resistance would increase.

14-3 MULTISIM: SHUNT MOTOR CHARACTERISTICS 1.

a. 1800 rpm, 8.8 A, 115 V b. 1.1 Ω, 128 Ω c. Iarm is so small because the bearing friction is very small and it is the only load on this ideal motor. d. The speed decreases. e. No. It does not change because held is a function of the coil resistance. h. It also doubles. i. No. It does not double because the rpm decreases. j. See Fig. 14-2.

Load Torque in Newton-meters N.L.

1.0

2.0

3.0

4.0

4.917

6.0

7.0

Iarm, A

14.21 

1.79

3.58

5.369

7.159

8.800

10.74

12.53

Ifield, A

0.8984

0.8984

0.8984

0.8984

0.8984

0.8984

0.8984

0.8984

S, rpm

1965

1932

1898

1864

1831

1800

1764

1730

Pin, W

103.3

309.2

515.0

720.8

926.6

1115.3

1338.4

1544.3

2

I Rfield, W

103.3

103.3

103.3

103.3

103.3

103.3

103.3

103.3

2

0.0002 

3.52

14.1

31.7

56.4

85.2

126.9

172.7

I Rarm, W Pout, W

0

202.5

398.0

586.0

767.6

927.5

1109.2

1269.1

eff, %

0

65.5

77.2

81.3

82.8

83.2

82.8

82.8

Fig 14.2 Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

123

k. When operated at rated speed and Iarm. l. 1.24 hp m. Yes n. Yes. Because this ideal motor has no core or bearing friction losses.

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124

2.

a. 8.782 A, 183.1 mA, 8833 rpm b. Decrease. Because the field flux will increase. Thus, the armature can produce the needed cemf at a lower speed. c. 5.287 A, 305.1 mA, 5511 rpm d. Yes e. 1932 to 8833 rpm (over 1:4) f. No. Because the added R decreases the field flux so the armature current must increase to produce the required torque. Thus, Iarm will exceed rated value at higher speeds. g. 1865 to 2863 rpm h. 65.0 W, 79.7 W, 1044.2 W, 900.1 W i. Yes j. Zero. Because Iarm is already at rated value.

14-4 MULTISIM: SERIES MOTOR CHARACTERISTICS 1.

2.

a. 1765 rpm, 10 A, 1 Ω, 1 Ω b. 200 W c. 1200 W d. 1000 W e. 83.3% f. 5.41 Nm g. Yes a. 8.6 to 1 b. For the shunt motor, the ratio was 1.09 to 1 or about one-eighth that of the series motor. c. See Fig. 14-3.

0.1 I, A 1.361 S, rpm 15,210 Pin, W 163.3 Pout, W 159.4 2 I R, W 3.7 eEff, 97.6 %

Load Torque in Newton-meters 0.2 0.4 0.8 1.6 3.2 5.41 1.923 2.719 3.845 5.438 7.691 10.00 10,660 7436 5155 3542 2401 1765 230.8 326.3 461.4 652.6 922.9 1200 223.4 311.7 432.2 593.9 805.2 1001.1 7.4 14.8 29.6 59.1 118.3 200.0 96.8

95.5

93.7

91.0

87.2

83.4

7 11.38 1509 1365.6 1107.0 259.0 81.1

Fig 14.3

d. It decreases. Because the power loss in both coils is increasing and the speed is dropping rapidly. e. Yes, yes f. 1.34 hp

CHAPTER 15 NOTES TO THE INSTRUCTOR Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

125

1.

2.

3.

4.

5.

6.

The 5-V panel meter specified for experiment 15-1 can be replaced by nearly any low-voltage dc panel meter. Or, a low-voltage range of a VOM can be used. Of course, substitution may require a value of multiplier different from the 20 kΩ specified.

The answer for step 3b in experiment 15-2 is for a DMM with a l0-MΩ input resistance. For a 20-MΩ input resistance, the answer would be 6.7 to 7.7 V. The precision resistors and potentiometers specified for experiment 15-3 will be used again in experiment 15-4.

The answers given for experiment 15-5 are for cheap 3.2-Ω speakers. The answers on this experiment may vary considerably. They will be influenced by the type of speaker being used and by the enclosure, if any.

The 150-W lamps for experiment 15-6 must be properly mounted with appropriate, insulated terminals.

Experiments 15-3 and 15-4 can be done using Multisim. Experiment 15-7 is specifically designed for Multisim.

TEST 1. F 2. T 3. T 4. F 5. T 6. T 7. T 8. F 9. T 10. T 11. F 12. T 13. T 14. F 15. F 16. T 17. T 18. T 19. T 20. F 21. F 22. F 23. T 24. F 25. B

26. B 27. C 28. D 29. B 30. B 31. A 32. A 33. C 34. A 35. C 36. D 37. 5 kΩ 38. 3.3 kW 39. 2840 W 40. 0.0476 µF 41. 2 kΩ 42. 720 Ω 43. 263.2 Ω 44. 16.7 kΩ/V 45. 1 MΩ 46. 124.5 kΩ 47. a. voltage b. time c. cemf

15-1 LAB EXPERIMENT: EXTENDING THE RANGE OF A VOLTMETER 2.

3.

a. 5000 Ω b. 25 kΩ c. 20 kΩ Percentage of error entered by student in Table 15-1 on page 216 of the Experiments Manual should not exceed 5.

15-2 LAB EXPERIMENT: VOLTMETER LOADING Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

126

1.

2.

3.

a. 9 V b. 9 V c. No. No, because on the l0-V range both meters have input resistances greater than 20 kΩ d. VR1 = 8.5 to 9.5 V, VR2 = 8.5 to 9.5 V e. VR1 = 8.5 to 9.5 V, VR2 = 8.5 to 9.5 V f. Yes g. No a. No. Yes, because the VOM has only 200 kΩ of input resistance on the 10-V range b. VR1 = 5.5 to 6.5 V, VR2 = 5.5 to 6.5 V c. VR1 = 8.5 to 9.5 V, VR2 = 8.5 to 9.5 V d. The DMM e. No f. No a. VR5 = 0.3 to 0.4 V, VR6 = 0.3 to 0.4 V b. VR5 = 5.5 to 6.5 V, VR6 = 5.5 to 6.5 V c. Yes d. The DMM e. Yes, by calculating the voltage drops in the series-parallel circuit formed by the meter and the original series circuit

15-3 LAB EXPERIMENT: DC BRIDGE 1.

a. 2200 Ω b. 2200 ±10% c. 10 Ω, 980 Ω d. Yes e. 200, 19.6 kΩ f. Yes g. 200 kΩ h. Yes

15-4 LAB EXPERIMENT: AC BRIDGE 1.

2.

3.

a. 6667 Ω b. 6700 Ω ±15% c. Yes a. 0.01 µF b. Yes c. 3333 Ω d. 3400 Ω ±15% a. No b. Yes

15-5 LAB EXPERIMENT: MEASURING IMPEDANCE 1.

2.

a. 3.6 Ω b. 3.6 Ω (same as recorded in 1a) a. 4.2 Ω b. More c. Inductive d. 5.5 Ω e. 7.5 Ω; yes

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127

15-6 LAB EXPERIMENT: MEASURING THREE-PHASE POWER 1.

2.

a. 150 W b. 150 W ±10% c. Yes d. Yes e. Yes f. 450 W ±10% b. 225 W c. 225 W d. P1 = 225 W ±10%, P2 = 225 W ±10% e. Yes

15-7 MULTISIM: METER INTERNAL RESISTANCE 1.

I connected the two meters in series and applied 20 V across them. The readings were V1 = 6.667 V and V2 = 13.333 V. Therefore, 6.667 V 13.333 V 13.333 MWV = and Rinput = = 1MW Rinput 2 MW 13.333 V Note: There are many other ways to find Rinput. For example, V2 = 2Vi and Vi = V2 /2. Since all I are equal in a series circuit, Rint for Vi = Rint for V2 /2 = 2 MΩ /2 = 1 MΩ

I constructed a series circuit with the 20-V supply, the ammeter, and the 4.7-kΩ resistor. The current was 4.254 mA. Then I measured the voltage drop across the ammeter. It was 4.255 pV. Therefore,

Rint =

4.225 pV = 1nW 4.255 mA

15-8 MULTISIM: I AND R FUNCTIONS OF A DMM 1.

2.

a. 100 kΩ b. 99.999 kΩ c. 100 V d. 9.000 ± the tolerance of the resistor and the DMM. f. Yes g. 90.000 ± the tolerance of the resistor and the DMM. a. 6.1 mA b. 100 mA c. 6.111 d. Yes (within tolerances) e. No

CHAPTER 16 NOTES TO THE INSTRUCTOR 1. Experiments 16-1 through 16-3 could be done with physical components and, for safety, a lowvoltage supply. Many three-way switches can be disassembled by drilling out the rivets holding them together. After modification to create a fault, the switch can be reassembled using small Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

128

machine screws and nuts. Of course, the rivets should also be replaced on the switches without a fault.

2. The faults in the experiments are a short between the pole and throw of S1 in experiment 16-1, an open in the pole of S2 in experiment 16-2, and a short between the throws of S1 in experiment 16-3.

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129

TEST 1. F 2. F 3. T 4. T 5. T 6. F 7. F 8. F 9. F 10. T 11. F 12. F 13. T 14. T 15. F 16. F 17. T 18. T 19. T

20. F 21. Neutral 22. Two 23. One 24. Two 25. Feeder 26. Off 27. Green 28. GFCI 29. Neutral 30. Dark 31. 30 32. Service drop 33. Grounded 34. Parallel 35. Neon 36. Yes 37. Yes 38. The neutral pigtai

16-1 MULTISIM: TROUBLESHOOTING 1 Observation: The light can be turned off and on from either switch when the other switch is in the “right” position.

The light stays on when either switch is toggled when the other switch is in the “wrong” position.

Analysis: Analysis of the operation of the circuit reveals that: 1.

2.

3.

Any opens in the switch poles or throws would lead to some positions that would cause the light to stay off when the switch is toggled.

A short between the throws of either switch would keep the light on for all possible combinations of switch positions.

A short between a pole and a throw would allow the circuit to operate as observed above.

Measurement: Information gained by voltage measurements between the pole and throws of each switch does not identify which switch is shorted. A short in either switch can produce the same set of voltage measurements.

Conclusion: There is a short between the pole and throw of either S1 or S2.

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130

16-2 MULTISIM: TROUBLESHOOTING 2 Observation: The light will not turn on with any combination of switch positions.

Analysis: Some possibilities are an open fuse, an open in the hot or the neutral line, or an open in the pole of either switch.

Measurement: 120 V is measured between the pole of S2 and either throw of S2 when S1 is in the correct position. Zero voltage is measured between the pole of S1 and either throw of S1 when S2 is in either position.

Conclusion: The pole of S2 is open.

16-3 MULTISIM: TROUBLESHOOTING 3 Observation: No combination of switch positions will turn the light off.

Analysis: A short between the throws of one of the switches or a short between the lines connecting the throws of the two switches will cause the observed operation of the circuits.

Measurement: Voltage measurements can't reveal any more information than observation did.

Conclusion: A short exists between the throws of a switch or the throw-connecting lines.

16-4 MULTISIM: FOUR-WAY CIRCUIT 1.

b. Hopefully, yes c. Hopefully, none

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131

16-5 LAB EXPERIMENT: WIRING A FOUR-WAY CIRCUIT See Fig. 16-1 below.

Fig. 16-1 Wiring a four-way circuit.

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132