Instructor’s Manual
For
Electronics Fundamentals: Circuits, Devices,
and Applications
9th Edition
Thomas L. Floyd
David M. Buchla
Gary D. Snyder
Part 1. Solutions to End-of-Chapter Problems
Part 2. Solutions to Application Assignments
Part 3. Solutions to Circuit Simulation Problems
Part 4. Laboratory Solutions for Experiments in Electronics Fundamentals:
Circuits, Devices, and Applications by David M. Buchla
, PART ONE
Solutions to End-of-Chapter Problems
,Chapter 1: QUANTITIES AND UNITS
Section 1-1: Scientific and Engineering Notation
1. (a) 3000 = 3.0 × 103
(b) 75,000 = 7.5 × 104
(c) 2,000,000 = 2.0 × 106
2. (a) = 0.002 = 2.0 × 10−3
(b) = 0.0005 = 5.0 × 10−4
(c) = 0.0000002 = 2.0 × 10−7
3. (a) 8400 = 8.4 × 103
(b) 99,000 = 9.9 × 104
(c) 0.2 × 106 = 2.0 × 105
4. (a) 0.0002 = 2.0 × 10−4
(b) 0.6 = 6.0 × 10−1
(c) 7.8 × 10−2 (already in scientific notation)
5. (a) 2.5 × 10−6 = 0.0000025
(b) 5.0 × 102 = 500
(c) 3.9 × 10−1 = 0.39
6. (a) 4.5 × 10−6 = 0.0000045
(b) 8.0 × 10−9 = 0.000000008
(c) 4.0 × 10−12 = 0.0000000000040
7. (a) (9.2 × 106) + (3.4 × 107) = (9.2 × 106) + (34 × 106) = 4.32 × 107
(b) (5.0 × 103) + (8.5 × 10−1) = (5.0 × 103) + (0.00085 × 103) = 5.00085 × 103
(c) (5.6 × 10−8) + (4.6 × 10−9) = (56 × 10−9) + (4.6 × 10−9) = 6.06 × 10−8
8. (a) (3.2 × 1012) – (1.1 × 1012) = 2.1 × 1012
(b) (2.6 × 108) – (1.3 × 107) = (26 × 107) – (1.3 × 107) = 24.7 × 107
(c) (1.5 × 10−12) – (8 × 10−13) = (15 × 10−13) – (8 × 10−13) = 7 × 10−13
9. (a) (5.0 × 103)(4.0 × 105) = 5.0 × 4.0 × 103 + 5 = 20 × 108 = 2.0 × 109
(b) (1.2 × 1012)(3 × 102) = 1.2 × 3 × 1012 + 2 = 3.6 × 1014
(c) (2.2 × 10−9)(7 × 10−6) = 2.2 × 7 × 10− 9 − 6 = 15.4 × 10−15 = 1.54 × 10−14
10. (a) = 0.4 × 103 – 2 = 0.4 × 101 = 4.0
(b) = 0.5 × 10−6 – (−8) = 0.5 × 102 = 50
S1-1
, (c) = 2.1 × 108 – (−5) = 2.1 × 1013
11. (a) 89,000 = 89 × 103
(b) 450,000 = 450 × 103
(c) 12,040,000,000,000 = 12.04 × 1012
12. (a) 2.35 × 105 = 235 × 103
(b) 7.32 × 107 = 73.2 × 106
(c) 1.333 × 109 (already in engineering notation)
13. (a) 0.000345 = 345 × 10−6
(b) 0.025 = 25 × 10−3
(c) 0.00000000129 = 1.29 × 10−9
14. (a) 9.81 × 10−3 = 9.81 × 10−3
(b) 4.82 × 10−4 = 482 × 10−6
(c) 4.38 × 10−7 = 438 × 10−9
15. (a) 2.5 × 10−3 + 4.6 × 10−3 = (2.5 + 4.6) × 10−3 = 7.1 × 10−3
(b) 68 × 106 + 33 × 106 = (68 + 33) × 106 = 101 × 106
(c) 1.25 × 106 + 250 × 103 = 1.25 × 106 + 0.25 × 106 = (1.25 + 0.25) × 106 = 1.50 × 106
16. (a) (32 × 10−3)(56 × 103) = 1792 × 10(−3 + 3) = 1792 × 100 = 1.792 × 103
(b) (1.2 × 10−6)(1.2 × 10−6) = 1.44 × 10(−6 – 6) = 1.44 × 1012
= 1.50 × 106
−3 −3
(c) (100)(55 × 10 ) = 5500 × 10 = 5.5
17. (a) = 22.7 × 10−3
9
(b) = 0.2 × 10(3 – (−6)) = 0.2 × 10 = 200 × 106
(c) = 0.848 × 10(3 – 3) = 0.848 × 100 = 848 × 10−3
Section 1-2: Units and Metric Prefixes
18. (a) 89,000 Ω = 89 × 103 = 89 kΩ
(b) 450,000 Ω = 450 × 103 = 450 kΩ
(c) 12,040,000,000,000 Ω = 12.04 × 10 12 = 12.04 TΩ
19. (a) 0.000345 A = 345 × 10−6 A = 345 μA
(b) 0.025 A = 25 × 10−3 A = 25 mA
(c) 0.00000000129 A = 1.29 × 10 −9 A = 1.29 nA
20. (a) 31 × 10−3 A = 31 mA
(b) 5.5 × 103 V = 5.5 kV
(c) 20 × 10−12 F = 20 pF
S1-2
For
Electronics Fundamentals: Circuits, Devices,
and Applications
9th Edition
Thomas L. Floyd
David M. Buchla
Gary D. Snyder
Part 1. Solutions to End-of-Chapter Problems
Part 2. Solutions to Application Assignments
Part 3. Solutions to Circuit Simulation Problems
Part 4. Laboratory Solutions for Experiments in Electronics Fundamentals:
Circuits, Devices, and Applications by David M. Buchla
, PART ONE
Solutions to End-of-Chapter Problems
,Chapter 1: QUANTITIES AND UNITS
Section 1-1: Scientific and Engineering Notation
1. (a) 3000 = 3.0 × 103
(b) 75,000 = 7.5 × 104
(c) 2,000,000 = 2.0 × 106
2. (a) = 0.002 = 2.0 × 10−3
(b) = 0.0005 = 5.0 × 10−4
(c) = 0.0000002 = 2.0 × 10−7
3. (a) 8400 = 8.4 × 103
(b) 99,000 = 9.9 × 104
(c) 0.2 × 106 = 2.0 × 105
4. (a) 0.0002 = 2.0 × 10−4
(b) 0.6 = 6.0 × 10−1
(c) 7.8 × 10−2 (already in scientific notation)
5. (a) 2.5 × 10−6 = 0.0000025
(b) 5.0 × 102 = 500
(c) 3.9 × 10−1 = 0.39
6. (a) 4.5 × 10−6 = 0.0000045
(b) 8.0 × 10−9 = 0.000000008
(c) 4.0 × 10−12 = 0.0000000000040
7. (a) (9.2 × 106) + (3.4 × 107) = (9.2 × 106) + (34 × 106) = 4.32 × 107
(b) (5.0 × 103) + (8.5 × 10−1) = (5.0 × 103) + (0.00085 × 103) = 5.00085 × 103
(c) (5.6 × 10−8) + (4.6 × 10−9) = (56 × 10−9) + (4.6 × 10−9) = 6.06 × 10−8
8. (a) (3.2 × 1012) – (1.1 × 1012) = 2.1 × 1012
(b) (2.6 × 108) – (1.3 × 107) = (26 × 107) – (1.3 × 107) = 24.7 × 107
(c) (1.5 × 10−12) – (8 × 10−13) = (15 × 10−13) – (8 × 10−13) = 7 × 10−13
9. (a) (5.0 × 103)(4.0 × 105) = 5.0 × 4.0 × 103 + 5 = 20 × 108 = 2.0 × 109
(b) (1.2 × 1012)(3 × 102) = 1.2 × 3 × 1012 + 2 = 3.6 × 1014
(c) (2.2 × 10−9)(7 × 10−6) = 2.2 × 7 × 10− 9 − 6 = 15.4 × 10−15 = 1.54 × 10−14
10. (a) = 0.4 × 103 – 2 = 0.4 × 101 = 4.0
(b) = 0.5 × 10−6 – (−8) = 0.5 × 102 = 50
S1-1
, (c) = 2.1 × 108 – (−5) = 2.1 × 1013
11. (a) 89,000 = 89 × 103
(b) 450,000 = 450 × 103
(c) 12,040,000,000,000 = 12.04 × 1012
12. (a) 2.35 × 105 = 235 × 103
(b) 7.32 × 107 = 73.2 × 106
(c) 1.333 × 109 (already in engineering notation)
13. (a) 0.000345 = 345 × 10−6
(b) 0.025 = 25 × 10−3
(c) 0.00000000129 = 1.29 × 10−9
14. (a) 9.81 × 10−3 = 9.81 × 10−3
(b) 4.82 × 10−4 = 482 × 10−6
(c) 4.38 × 10−7 = 438 × 10−9
15. (a) 2.5 × 10−3 + 4.6 × 10−3 = (2.5 + 4.6) × 10−3 = 7.1 × 10−3
(b) 68 × 106 + 33 × 106 = (68 + 33) × 106 = 101 × 106
(c) 1.25 × 106 + 250 × 103 = 1.25 × 106 + 0.25 × 106 = (1.25 + 0.25) × 106 = 1.50 × 106
16. (a) (32 × 10−3)(56 × 103) = 1792 × 10(−3 + 3) = 1792 × 100 = 1.792 × 103
(b) (1.2 × 10−6)(1.2 × 10−6) = 1.44 × 10(−6 – 6) = 1.44 × 1012
= 1.50 × 106
−3 −3
(c) (100)(55 × 10 ) = 5500 × 10 = 5.5
17. (a) = 22.7 × 10−3
9
(b) = 0.2 × 10(3 – (−6)) = 0.2 × 10 = 200 × 106
(c) = 0.848 × 10(3 – 3) = 0.848 × 100 = 848 × 10−3
Section 1-2: Units and Metric Prefixes
18. (a) 89,000 Ω = 89 × 103 = 89 kΩ
(b) 450,000 Ω = 450 × 103 = 450 kΩ
(c) 12,040,000,000,000 Ω = 12.04 × 10 12 = 12.04 TΩ
19. (a) 0.000345 A = 345 × 10−6 A = 345 μA
(b) 0.025 A = 25 × 10−3 A = 25 mA
(c) 0.00000000129 A = 1.29 × 10 −9 A = 1.29 nA
20. (a) 31 × 10−3 A = 31 mA
(b) 5.5 × 103 V = 5.5 kV
(c) 20 × 10−12 F = 20 pF
S1-2
