INSTRUCTOR’S SOLUTIONS 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) (b)
= 0.4 × 103 – 2 = 0.4 × 101 = 4.0 = 0.5 × 10−6 – (−8) = 0.5 × 102 = 50 S1-1
= 2.1 × 108 – (−5) = 2.1 × 1013
(c)
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
21. (a) 3.0 × 10−6 F = 3.0 μF (b) 3.3 × 106 Ω = 3.3 MΩ (c) 350 × 10−9 A = 350 nA 22. (a) 5.0 μA = 5.0 × 10−6 A (b) 43 mV= 43 × 10−3 V (c) 275 kΩ = 275 × 103 Ω (d) 10 MW = 10 × 106 W
Section 1-3:
Metric Unit Conversions
23. (a) (5.0 mA) (1.0 × 103 μA/mA) = 5.0 × 103 μA = 5000 μA (b) (3200 μW)(1 × 10−3 W/μW) = 3.2 mW (c) (5000 kV)(1 × 10−3) MV/kV = 5.0 MV (d) (10 MW)(1 × 103 kW/MW) = 10 × 103 kW = 10,000 kW
24. (a)
=
= 1.0 × 103 = 1000
(b)
=
= 0.05 × 106 = 50,000
(c)
=
= 0.02 × 10−3 = 2.0 × 10−5
(d)
=
= 155 × 10−6 = 1.55 × 10−4
25. (a) 50 mA + 680 μA = 50 mA + 0.68 mA = 50.68 mA (b) 120 kΩ + 2.2 MΩ = 0.12 MΩ + 2.2 MΩ = 2.32 MΩ (c) 0.02 μF + 3300 pF = 0.02 μF + 0.0033 μF = 0.0233 μF
26. (a)
=
= 0.820
(b)
=
= 5000
(c)
=
= 500
Section 1-4:
Measured Numbers
27. (a) 1.00 × 103 has 3 significant digits (b) 0.0057 has 2 significant digits (c) 1502.0 has 5 significant digits (d) 0.000036 has 2 significant digits (e) 0.105 has 3 significant digits (f) 2.6 × 105 has 2 significant digits 28. (a) 50,505 ≈ 50,500 = 50.5 × 103 (b) 220.45 ≈ 220 S1-3
(c) 4646 ≈ 4650 = 4.65 × 103 (d) 10.99 ≈ 11.0 (e) 1.005 ≈ 1.00 (using round to even)
S1-4
Chapter 2:
VOLTAGE, CURRENT, AND RESISTANCE
BASIC PROBLEMS Section 2-2:
Electrical Charge
1.
Q = (charge per electron)(number of electrons) = (1.6 × 10 −19 C/e)(50 × 1031e) = 80 × 1012 C
2.
(6.25 × 1018 e/C)(80 × 10−6 C) = 5 × 1014 e
3.
The magnitude of the charge on a proton (p) is equal to the magnitude of the charge on the electron (e). Therefore, (1.6 × 10−19 C/p)(29 p) = 4.64 × 10−18 C
4.
(1.6 × 10−19 C/p)(17 p) = 2.72 × 10−18 C
Section 2-3: 5.
Voltage
(a) V =
=
= 10 V
(b) V =
=
= 2.5 V
(c) V =
=
= 4.0 V
6.
V=
=
= 5.0 V
7.
V=
=
= 20 V
8.
W = VQ = (12 V)(2.5 C) = 30 J
9.
V=
=
= 12.5 V
Section 2-4: 10. I =
=
Current
= 20 mA
11. (a) I =
=
= 75 A
(b) I =
=
= 20 A
(c) I =
=
= 2.5 A
S1-5
12. I =
=
= 0.2 A
13. I =
t=
=
= 2.0 s
14. Q = I × t = (1.5 A)(0.1 s) = 0.15 C
Section 2-5:
Resistance
15. A: Blue, gray, red, silver: 6800 Ω ± 10% B: Orange, orange, black, silver: 33 Ω ± 10% C: Yellow, violet, orange, gold: 47,000 Ω ± 5% 16. A: Rmin = 6800 Ω − 0.1(6800 Ω) = 6800 Ω − 680 Ω = 6120 Ω Rmax = 6800 Ω + 680 Ω = 7480 Ω B: Rmin = 33 Ω− 0.1(33 Ω) = 33 Ω− 3.3 Ω = 29.7 Ω Rmax = 33 Ω + 3.3 Ω = 36.3 Ω C: Rmin = 47,000 Ω− (0.05)(47,000 Ω) = 47,000 Ω− 2350 Ω = 44,650 Ω Rmax = 47,000 Ω + 2350 Ω = 49,350 Ω 17. (a) 1st band = red, 2nd band = violet, 3rd band = brown, 4th band = gold (b) 330 Ω: orange, orange, brown, (B) 2.2 kΩ: red, red, red (D) 39 kΩ: orange, white, orange (A) 56 kΩ: green, blue, orange (L) 100 kΩ: brown, black, yellow (F) 18. (a) 36.5 Ω ± 2% (b) 2.74 kΩ ± 0.25% (c) 82.5 kΩ ± 1% 19. (a) Brown, black, black, gold: 10 Ω ± 5% (b) Green, brown, green, silver: 5,100,000 Ω ± 10% = 5.1 MΩ ± 10% (c) Blue, gray, black, gold: 68 Ω ± 5% 20. (a) 0.47 Ω ± 5%: Yellow, violet, silver, gold (b) 270 kΩ ± 5%: Red, violet, yellow, gold (c) 5.1 MΩ ± 5%: Green, brown, green, gold 21. (a) Red, gray, violet, red, brown: (b) Blue, black, yellow, gold, brown: (c) White, orange, brown, brown, brown:
28,700 Ω ± 1% = 28.7 kΩ ± 1% 60.4 Ω ± 1% 9310 ± 1% = 9.31 kΩ ± 1%
22. (a) 14.7 kΩ ± 1%: brown, yellow, violet, red, brown (b) 39.2 Ω ± 1%: orange, white, red, gold, brown (c) 9.76 kΩ ± 1%: white, violet, blue, brown, brown S1-6
23. (a) 220 = 22 Ω (b) 472 = 4.7 kΩ (c) 823 = 82 kΩ (d) 3K3 = 3.3 kΩ (b) 560 = 56 Ω (f) 10M = 10 MΩ 24. 500 Ω, equal resistance on each side of the contact.
Section 2-6:
The Electric Circuit
25. There is current through Lamp 2. 26. See Figure 2-1.
Section 2-7:
Basic Circuit Measurements
27. See Figure 2-2(a).
28. See Figure 2-2(b).
S1-7
29. Position 1: V1 = 0 V, V2 = VS Position 2: V1 = VS, V2 = 0 V 30. See Figure 2-3. 31. On the 600 V DC scale: 250 V 32. R = (10)(10 Ω) = 100 Ω 33. (a) 2(100 Ω) = 200 Ω (b) 15(10 MΩ) = 150 MΩ (c) 45(100 Ω) = 4500 Ω 34. See Figure 2-4.
Section 2-8:
Green Technology
35. The RoHS directive addresses the environmental responsibility of eliminating or reducing the use of specific hazardous substances in proudcts and manufacturing processes. 36. The WEEE supplements the RoHS directive in addressing the disposal of electric and electronic assemblies containing hazardous substances that could create environmental or health issues.
ADVANCED PROBLEMS 37. I = Q = I × t = (2.0 A)(15 s) = 30 C V=
=
= 33.3 V S1-8
38. I = Q= −2
Q= I=
= 9.184 × 10 C =
= = 0.367 A
39. Total wire length = 100 ft Resistance per 1000 ft = (1000 ft)(6 Ω/100 ft) = 60 Ω Smallest wire size is AWG 27, which has 51.47 Ω/1000 ft 40. (a) 4R7J = 4.7 Ω ± 5% (b) 560KF = 560 kΩ ± 1% (c) 1M5G = 1.5 MΩ ± 2% 41. The circuit in (b) can have both lamps on at the same time. 42. There is always current through R5. 43. See Figure 2-5. 44. See Figure 2-5.
45. See Figure 2-6.
S1-9
OHM’S LAW, ENERGY, AND POWER
Chapter 3:
BASIC PROBLEMS Ohm’s Law
Section 3-1: 1.
I is directly proportional to V and will change the same percentage as V. (a) I = 3(1.0 A) = 3.0 A (b) I = 1.0 A − (0.8)(1.0 A) = 1.0 A − 0.8 A = 0.2 A (c) I = 1.0 A + (0.5)(1.0 A) = 1.0 A + 0.5 A = 1.5 A
2.
(a) When the resistance doubles, the current is halved from 100 mA to I = V/(2R) = 0.5(V/R) = 50 mA. (b) When the resistance is reduced by 30%, the current increases from 100 mA to I = V/(0.7R) = 1.429(V/R) = (1.429)(100 mA) ≈ 143 mA (c) When the resistance is quadrupled, the current decreases from 100 mA to I = V/(4R) = 0.25(V/R) = 25 mA.
3.
Tripling the voltage triples the current from 10 mA to 30 mA, but doubling the resistance halves the current to 15 mA.
Ohm’s Law
Section 3-2: 4.
5.
(a) I =
=
= 5.0 A
(b) I =
=
= 1.5 A
(c) I =
=
= 0.5 A
(d) I =
=
= 2.0 µA
(e) I =
=
= 1.5 A
(a) I =
=
= 3.33 mA
(b) I =
=
= 550 μA
(c) I =
=
= 588 μA
(d) I =
=
= 0.50 A = 500 mA
(e) I =
=
= 6.60 mA
S1-10
6.
I=
7.
(a) I =
=
= 2.50 mA
(b) I =
=
= 2.27 µA
(c) I =
=
= 8.33 mA
8.
9.
=
= 1.2 A
Orange, violet, yellow, gold, brown = 37.4 Ω ± 1% I=
=
= 0.321 A = 321 mA
I=
=
= 642 mA = 0.642 A
0.642 A is greater than 0.5 A, so the fuse will blow. 10. (a) V = IR = (2.0 A)(18 Ω) = 36 V (b) V = IR = (5.0 A)(47 Ω) = 235 V (c) V = IR = (2.5 A)(620 Ω) = 1550 V (d) V = IR = (0.6 A)(47 Ω) = 28.2 V (e) V = IR = (0.1 A)(470 Ω) = 47 V 11. (a) V = IR = (1.0 mA)(10 Ω) = 10 mV (b) V = IR = (50 mA)(33 Ω) = 1.65 V (c) V = IR = (3.0 A)(4.7 kΩ) = 14.1 kV (d) V = IR = (1.6 mA)(2.2 kΩ) = 3.52 V (e) V = IR = (250 μA)(1.0 kΩ) = 250 mV (f) V = IR = (500 mA)(1.5 MΩ) = 750 kV (g) V = IR = (850 μA)(10 MΩ) = 8.5 kV (h) V = IR = (75 μA)(47 Ω) = 3.53 mV 12. VS = IR = (3.0 A)(27 Ω) = 81 V 13. (a) V = IR = (3.0 mA)(27 kΩ) = 81 V (b) V = IR = (5.0 μA)(100 MΩ) = 500 V (c) V = IR = (2.5 A)(47 Ω) = 117.5 V
14. (a) R =
=
= 5.0 Ω
(b) R =
=
= 200 Ω
(c) R =
=
= 10 Ω
(d) R =
=
= 2.75 Ω
(e) R =
=
= 300 Ω
15. (a) R =
=
= 2.0 kΩ S1-11
(b) R =
=
= 3.5 kΩ
(c) R =
=
= 2.0 kΩ
(d) R =
=
= 100 kΩ
(e) R =
=
= 1.0 MΩ
16. R =
=
= 3.0 kΩ
= 4.0 Ω
17. (a) R =
=
(b) R =
=
= 3.0 kΩ
(c) R =
=
= 0.2 MΩ = 200 kΩ
18. I =
=
= 0.82 A = 820 mA
Section 3-3: 19. P =
=
Energy and Power = 2.6 W
20. Since 1.0 J/s = 1.0 W, then 350 J/s = 350 W
21. P =
=
(
)(
)=
= 0.417 J/s = 417 mW
22. (a) 1000 W = 1.0 × 103 W = 1.0 kW (b) 3750 W = 3.75 × 103 W = 3.75 kW (c) 160 W = 0.16 × 103 W = 0.16 kW (d) 50,000 W = 50 × 103 W = 50 kW 23. (a) 1,000,000 W = 1.0 × 106 = 1.0 MW (b) 3.0 × 106 W = 3.0 MW (c) 15 × 107 W = 150 × 106 W = 150 MW (d) 8700 kW = 8.7 106 W = 8.7 MW 24. (a) 1.0 W = 1000 × 10−3 W = 1000 mW (b) 0.4 W = 400 × 10−3 W = 400 mW (c) 0.002 W = 2.0 × 10−3 W = 2.0 mW (d) 0.0125 W = 12.5 × 10−3 W = 12.5 mW 25. (a) 2.0 W = 2,000,000 µW (b) 0.0005 W = 500 µW S1-12
(c) 0.25 mW = 250 µW (d) 0.00667 mW = 6.67 µW 26. (a) 1.5 kW = 1.5 × 103 W = 1500 W (b) 0.5 MW = 0.5 × 106 = 500,000 W (c) 350 mW = 350 × 10−3 W = 0.35 W (d) 9000 µW = 9000 × 10−6 W = 0.009 W
27. P =
in watts
V=
in volts
I=
in amperes
P = VI = (
)( ) =
Therefore, (1.0 V)(1.0 A) = 1.0 W.
28. P =
=
= 1.0 W
1.0 W × 1.0 s = 1.0 W∙s = 1.0 J 1.0 Wh = 1.0 W ∙ 1.0 h = 1.0 W × 3600 s = 3600 W∙s = 3600 J 1.0 kWh = 1000 × 3600 J = 3,600,000 J = 3.6 × 106 J
Section 3-4:
Power in an Electric Circuit
29. P = VI = (5.5 V)(3.0 mA) = 16.5 mW 30. P = VI = (115 V)(3.0 A) = 345 W 31. P = I2R = (500 mA)2(4.7 kΩ) = 1.18 kW Test 32. P = I2R = (100 × 10−6 A)2(10 × 103 Ω) = 1.0−4 × 10 W = 100 μW
33. P =
=
34. P =
=
(
)
(
)
= 5.81 W
= 0.0402 W = 40.2 mW
35. P = I2R R = R=
(
)
= 25 Ω
36. 5.0 × 106 W for 1 minute = (5.0 × 103)(1.0 kW) for 1 minute = 5.0 × 10 3 kW∙min (5.0 × 103 kW∙min)(
37. 6700 Ws = (
) = 83.3 kWh )(
) = 0.00186 kWh = 1.86 × 10−3 kWh S1-13
38. 50 W = (50 W)(
) = 0.05 kW
(50 W)(12 h) = (0.05 kW)(12 h) = 0.60 kWh
39. I =
40. P =
(
)
= 0.156 W = 156 mW
W = Pt
W = (156 mW)(90 h)(
Section 3-5:
) = 50,625 J = 50.6 kJ
The Power Rating of Resistors
41. P = I2R = (10 mA)2(6.8 kΩ) = (10 × 10−3 A)2(6.8 × 103 Ω) = 680 × 10−3 W = 680 mW Use the next higher standard power rating that is greater than 680 mW, which is 1.0 W. 42. If the 8.0 W resistor is used, it will be operating in a marginal condition. To allow for a safety margin of 20% greater than 8.0 W, use the 12 W resistor.
Section 3-6:
Energy Conversion and Voltage Drop in a Resistance
43. (a) + at the top and – at the bottom of the resistor (b) + at the bottom and – at the top of the resistor (c) + on the right and – at the left of the resistor
Section 3-7: 44. VOUT = √
Power Supplies and Batteries √(
)(
) = 7.07 V
45. Ampere-hour rating is (1.5 A)(24 h) = 36 Ah
46. I =
= 8.0 A
47. I =
= 13.5 mA
48. PLOST = PIN – POUT = 500 mW – 400 mW = 100 mW % efficiency = (
)×100% = (
)×100% = 80%
49. POUT = (efficiency)(PIN) = (0.85)(5.0 W) = 4.25 W
S1-14
Section 3-8:
Introduction to Troubleshooting
50. The 4th bulb from the left is open. 51. It should take five (maximum) resistance measurements.
ADVANCED PROBLEMS 52. Assume that the total consumption of the power supply is the input power plus the power lost. POUT = 2.0 W % efficiency = ( PIN = (
)×100% )×100% = (
)×100% = 3.33 W
Energy used in 24 hours = PIN × t = (3.33 W)(24 h) = 80 Wh 53. Rf =
= 150 Ω
54. Measure the current with an ammeter connected as shown in Figure 3-1. Then calculate the unknown resistance with from the formula R =
.
55. Calculate I for each value of V:
I1 =
=0A
I7 =
= 600 mA
I2 =
= 100 mA
I8 =
= 700 mA
I3 =
= 200 mA
I9 =
= 800 mA
I4 =
= 300 mA
I10 =
= 900 mA
I5 =
= 400 mA
I11 =
= 1.0A
I6 =
= 500 mA
S1-15
The graph of V vs. I is a straight line, as shown in Figure 3-2. This indicates a linear relationship between I and V.
= 200 Ω
56. R = (a) I =
= 7.4 mA
(b) I =
= 10 mA
(c) I =
= 15 mA
(d) I =
= 20 mA
(e) I =
= 50 mA
57. R1 =
= 0.5 Ω
R2 =
= 1.0 Ω
R3 =
= 2.0 Ω
58. V2 =
(
)( (
) )
= 6.0 V for the new value
The voltage decreased by 4.0 V, from 10 V to 6.0 V. 59. The current increase is 50%, so the increase in voltage also must be 50%. Therefore, the increase in voltage is (0.5)(20 V) = 10 V The new value of voltage V2 = 20 V + 10 V = 30 V. (
)(
60. The wire resistance RW = (a) I =
)
= 0.154 Ω.
= 59.91 mA
(b) VR = (59.91 mA)(100 Ω) = 5.991 V (c)
= 6.0 V − 5.991 V = 0.009 V
For one length of wire, V =
= 0.0045 V = 4.5 mV. S1-16
61. 300 W = 0.3 kW 30 days = (30 days)(24 h/day) = 720 h Energy = (0.3 kW)(720 h) = 216 kWh
62.
= 48.39 kWh/day P= (
)(
) = 2.02 kW
63. The minimum power rating you should use is 12 W so that the 10 W power dissipation does not exceed the rating.
64. (a) P =
(
(b) W = Pt = (
)
= 14.4 W )(
)(
) = 0.48 Wh
(c) If the resistor remains connected for an additional minute the power dissipation neither increases nor decreases, because power dissipation is not time-dependent. 65. A fuse with a rating less than 2.5 A must be used. A 2.0 A fuse is recommended.
Multisim Troubleshooting Problems 66. R is open. 67. No fault. 68. R1 is shorted. 69. Lamp 4 is shorted. 70. Lamp 6 is open.
S1-17
Chapter 4:
SERIES CIRCUITS
BASIC PROBLEMS Section 4-1:
Resistors in Series
1.
See Figure 4-1 for possible connections.
2.
The groups of series resistors are R1, R2, R3, R9 and R4 R13, R7, R14 and R16 R6, R8 and R12 R10, R11, R15 and R5 See Figure 4-2.
3.
R1-8 = R13 + R7 + R14 +R16
4.
= 68 kΩ + 33 kΩ + 47 kΩ + 22 kΩ = 170 kΩ R2-3 = R12 + R8+ R6 = 10 Ω + 18 Ω + 22 Ω = 50 Ω
Section 4-2: 5.
Total Series Resistance
RT = 82 Ω + 56 Ω = 138 Ω
S1-19
6.
(a) RT = 560 Ω + 1.0 kΩ = 1560 Ω = 1.56 kΩ (b) RT = 47 Ω + 33 Ω = 80 Ω (c) RT = 1.5 kΩ + 2.2 kΩ + 10 kΩ = 13.7 kΩ (d) RT = 1.0 kΩ + 1.8 kΩ + 100 kΩ + 1.0 MΩ = 1,102,800 Ω (round to 1.10 MΩ)
7.
(a) RT = 1.0 kΩ + 4.7 kΩ + 2.2 kΩ = 7.9 kΩ (b) RT = 10 Ω + 10 Ω + 12 Ω + 1.0 Ω = 33 Ω (c) RT = 1.0 MΩ + 560 kΩ + 1.0 MΩ + 680 kΩ + 10 MΩ = 13.24 MΩ See Figure 4-3.
8.
RT = 12(5.6 kΩ) = 67.2 kΩ
9.
RT = 6(47 Ω) + 8(100 Ω) + 2(22 Ω) = 282 Ω + 800 Ω + 44 Ω = 1126 Ω
10. RT = R1 + R2 + R3 + R4 + R5 R5 = RT − (R1 + R2 + R3 + R4) = 20 kΩ− (4.7 kΩ + 1.0 kΩ + 2.2 kΩ + 3.9 kΩ) = 20 kΩ − 11.8 kΩ = 8.2 kΩ 11. (a) R1−8 = R13 + R7 + R14 + R16 = 68 kΩ + 33 kΩ + 47 kΩ + 22 kΩ = 170 kΩ (b) R2−3 = R12 + R8 + R6 = 10 Ω + 18 Ω + 22 Ω = 50 Ω (c) R4−7 = R10 + R11 + R15 + R5 = 2.2 kΩ + 8.2 kΩ + 1.0 kΩ + 1.0 kΩ = 12.4 kΩ (d) R5−6 = R1 + R2 + R3 + R9 + R4 = 220 Ω + 330 Ω + 390 Ω + 470 Ω + 560 Ω = 1.97 kΩ 12. RT = R1−8 + R2−3 + R4−7 + R5−6 = 170 kΩ + 50 Ω + 12.4 kΩ + 1.97 kΩ = 184.42 kΩ
Section 4-3: 13. I =
=
Current in a Series Circuit = 0.1 A
14. I = 5.0 mA at all points in the circuit. S1-20
Application of Ohm’s Law
Section 4-4:
15. (a) RT = 2.2 kΩ + 5.6 kΩ + 1.0 kΩ = 8.8 kΩ I=
=
= 625 µA
(b) RT = 1.0 MΩ + 2.2 MΩ + 560 kΩ = 3.76 MΩ I=
= 4.26 µA
The ammeters are connected in series. See Figure 4-4.
16. (a) V1 = (
)VS = (
)5.5 V = 1.375 V
V2 = (
)VS = (
)5.5 V = 3.500 V
V3 = (
)VS = (
)5.5 V = 0.625 V = 625 mV
(b) V1 = (
)VS = (
)16 V = 4.26 V
V2 = (
)VS = (
)16 V = 9.36 V
V3 = (
)VS = (
)16 V = 2.38 V
17. (a) RT = 3(470 Ω) = 1410 Ω I=
34.0 mA
(b) VR = IR = (34.0 mA)(470 Ω) = 16 V (c) Pmin = I2R = (340 mA)2(470 Ω) = 0.545 W = 545 mW. A 1.0 W resistor is recommended. 18. RT = Reach =
=
= 5.0 kΩ = 1.25 kΩ
S1-21
Section 4-5:
Voltage Sources in Series
19. See Figure 4-5.
20. The total voltage is 6.0 V + 6.0 V + 6.0 V – 6.0 V = 12 V
Kirchhoff’s Voltage Law
Section 4-6:
21. VS = 5.5 V + 8.2 V + 12.3 V = 26 V 22. VS = V1 + V2 + V3 + V4 + V5 V5 = VS – (V1 + V2 + V3 + V4) = 20 V − (1.5 V + 5.5 V + 3.0 V + 6.0 V) = 20 V – 16 V = 4.0 V 23. (a) By Kirchhoff’s Voltage Law: 15 V = 2.0 V + V2 + 3.2 V + 1.0 V + 1.5 V + 0.5 V V2 = 15 V – (2.0 V + 3.2 V + 1.0 V + 1.5 V + 0.5 V) = 15 V – 8.2 V = 6.8 V (b) VR = 8.0 V, V2R = 16 V, V3R = 24 V, and V4R = 32 V See Figure 4-6(b).
Section 4-7: 24. (
Voltage Dividers
)100% = 4.4%
25. (a) VAB = ( (b) VAB = (
)12 V = 3.84 V )8.0 V = (
)8.0 V = 6.77 V
S1-22
26. VA = VS = 15 V VB = (
)VS = (
)15 V = 10.6 V
VC = (
)VS = (
)15 V = 2.62 V
27. Vmin = (
)VS = (
)12 V = 3.80 V
Vmax = (
)VS = (
)12 V = 9.38 V
28. RT = R + 2R + 3R + 4R + 5R = 15R VR = (
)9.0 V = ( )9.0 V = 0.6 V
V2R = (
)9.0 V = ( )9.0 V = 1.2 V
V3R = (
)9.0 V = ( )9.0 V = 1.8 V
V4R = (
)9.0 V = ( )9.0 V = 2.4 V
V5R = (
)9.0 V = ( )9.0 V = 3.0 V
29. V5.6k = 10 V (by measurement) I= (
) = 1.79 mA
V1.0k = (1.79 mA)(1.0 kΩ) = 1.79 V V560 = (1.79 mA)(560 Ω) = 1.00 V V10k = (1.79 mA)(10 kΩ) = 17.9 V
Section 4-8:
Power in Series Circuits
30. PT = 5(5.0 mW) = 250 mW 31. RT = 5.6 kΩ + 1.0 kΩ +560 Ω + 10 kΩ = 17.16 kΩ P = I2RT = (1.79 mA)2(17.16 kΩ) = 0.0547 W = 54.7 mW
Section 4-9:
Voltage Measurements
32. Voltage from point A to ground (G): VAG = 10 V Resistance between A and G: RAG = 5.6 kΩ + 5.6 kΩ + 1.0 kΩ + 1.0 kΩ = 13.2 kΩ Resistance between B and G: RBG = 5.6 kΩ + 1.0 kΩ + 1.0 kΩ = 7.6 kΩ Resistance between C and G: RCG = 1.0 kΩ + 1.0 kΩ = 2.0 kΩ VBG = (
)10 V = (
)10 V = 5.76 V
VCG = (
)10 V = (
)10 V = 1.52 V
VDG = (
)10 V = (
)10 V = 0.758 V = 758 mV S1-23
33. Measure the voltage at point A with respect to ground and the voltage at point B with respect to ground. The difference of the two (VR2) is: VR2 = VA − VB 34. RT = R1 + R2 + R3 + R4 + R5 = 560 kΩ + 560 kΩ + 100 kΩ + 1.0 MΩ + 100 kΩ = 2.32 MΩ VT = 15 V VA = (
)VT = (
)15 V = 11.38 V = 11.4 V
VB = (
)VT = (
)15 V = 7.76 V
VC = (
)VT = (
)15 V = 7.11 V
VD = (
)VT = (
)15 V = 674 mV
35. VAC = VA – VC = 11.38 V – 7.11 V = 4.27 V 36. VCA = VC – VA = 7.11 V – 11.38 V = −4.27 V
Section 4-10:
Troubleshooting
37. (a) Zero current indicates an open. R4 is open since all the voltage is dropped across it. (b)
=
= 33.3 mA
R4 and R5 have no effect on the current. There is a short from A to B. 38. RT = 10 kΩ + 8.2 kΩ + 12 kΩ + 2.2 kΩ + 5.6 kΩ = 38 kΩ The meter reads about 28 kΩ. It should read 10 kΩ more, or 38 kΩ. The 10 kΩ resistor is shorted.
ADVANCED PROBLEMS 39. RT =
= 3.0 kΩ
R3 = 3.0 kΩ – (680 Ω + 1.0 kΩ + 270 Ω + 270 Ω) = 3.0 kΩ – 2.22 kΩ = 780 Ω 40. RT = 3(5.6 kΩ) + 1.0 kΩ + 2(100 Ω) = 18 kΩ. One possibility is to have three 5.6 kΩ resistors, one 1.0 kΩ resistor, and two 100 Ω resistors. 41. VA = 10 V and RT = 22 kΩ + 10 kΩ + 47 kΩ + 12 kΩ + 5.6 kΩ = 96.6 kΩ VB = VA − V27k = 10 V − (
)10 V = 10 V – 2.28 V = 7.72 V
VC = VB – V10k = 7.72 V − (
)10 V = 7.72 V – 1.04 V = 6.69 V
VD = VC – V47k = 6.69 V − (
)10 V = 6.69 V – 4.87 V = 1.82 V
VE = VD – V12k = 1.82 V − (
)10 V = 1.82 V – 1.24 V = 0.580 V = 580 mV
VF = 0 V 42. V2 = IR2 = (20 mA)(100 Ω) = 2.0 V R5 =
=
= 330 Ω S1-24
R6 =
=
(
)
= 280 Ω
V6 = IR6 = (20 mA)(280 Ω) = 5.6 V V1 = VS – (20 V + V6) = 30 V – (20 V + 5.6 V) = 4.4 V R1 =
= 220 Ω
=
V3 + V4 = 20V – V2 – V5 = 20 V – 2.0 V – 6.6 V = 11.4 V V3 = V4 =
= 5.7 V
R3 = R4 =
= 285 Ω
=
43. VS = IT = (250 mA)(1.5 kΩ) = 375 V Inew = 250 mA – 0.25(250 mA) = 250 mA – 62.5 mA = 188 mA Rnew =
= 2000 Ω
=
2000 Ω – 1500 Ω = 500 Ω must be added to the existing circuit reduce I by 25%.
44. P = I2R I = √ =√
Imax = √
= 0.646 A = 64.6 mA
Since all resistors in a series circuit have the same current, use the largest value of R to determine the maximum allowable current because the largest resistance dissipates the most power. Thus, the 120 Ω resistor burns out first. 45. (a) PT =
W+
W+
I= √
=√
W = 0.125 W + 0.25 V + 0.5 W = 0.875 W. = 19.1 mA
(b) VS = ITRT = (19.1 mA)(2400 Ω) = 45.8 V (c) R1/8 = R1/4 = R1/2 =
=
(
(
) )
=
(
= 343 Ω
= 686 Ω )
= 1371 Ω = 1.37 kΩ
46. See Figure 4-7.
S1-25
47. See Figure 4-8. The total resistance is RT = 20 kΩ + 20 kΩ+ 180 kΩ + 20 kΩ = 240 kΩ The current for all potentiometer settings is IT =
= 5.0 mA
Using the voltage divider formula, when the wiper is at its minimum setting (the bottom of the potentiomenter) VOUT = (
)120 V = 10 V.
When the wiper is at its maximum setting (the top of the potentiometer) VOUT = (
)120 V = (
)120 V= 100 V
(Changed figure)
48. See Figure 4-9. R1 = R2 = R3 = R4 =
(
)
= 5.4 kΩ
=
(
)
(
)
=
= 9.9 kΩ
=
= 6.52 kΩ
= 8.18 kΩ
A series of standard resistor values must be used to approximately achieve each resistance. One possibility using 2% or 5% resistors is as follows: R1 = 5100 Ω + 300 Ω = 5.4 kΩ R2 = 7500 Ω + 2400 Ω = 9.9 kΩ R3 = 3600 Ω + 2700 Ω + 220 Ω= 6.52 kΩ R4 = 7500 Ω + 680 Ω = 8.18 kΩ The highest power dissipation is in the 7500 Ω resistors. P = (1.0 mA)2(7500 Ω) = 7.5 mW.
S1-26
All resistors must be at least 1/10 W.
49. The groups are: R1, R7, R8, and R10 R2, R4, R6, and R11 R3, R5, R9, and R12 See Figure 4-10. R1-4 = R10 + R1 + R7 + R8 = 1.0 kΩ + 2.2 kΩ + 560 Ω + 470 Ω = 4.23 kΩ R2-3 = R6 + R2 + R4 + R11 = 3.3 kΩ + 4.7 kΩ + 5.6 kΩ + 10 kΩ = 23.6 kΩ R5-6 = R5 + R3 + R9 + R12 = 3.9 kΩ + 1.0 kΩ + 8.2 kΩ + 6.8 kΩ = 19.9 kΩ
50. Position 1: RT = R1 + R3 + R5 = 510 Ω + 820 Ω + 680 Ω = 2.01 kΩ Position 2: RT = R1 + R2 + R3 + R4 + R5 = 510 Ω + 910 Ω + 820 Ω + 750 Ω + 680 Ω = 3.67 kΩ 51. Position A: RT = R1 + R2 + R3 + R4 = 220 Ω + 470 Ω + 510 Ω + 1.0 kΩ = 2.2 kΩ I=
=
= 5.46 mA S1-27
Position B: RT = R2 + R3 + R4 = 470 Ω + 510 Ω + 1.0 kΩ = 1.98 kΩ I=
=
= 6.06 mA
Position C: RT = R3 + R4 = 510 Ω + 1.0 kΩ = 1.51 kΩ I=
=
= 7.95 mA
Position D: RT = R4 = 1.0 kΩ I=
=
= 12.0 mA
52. Position A: RT = R1 = 1.0 kΩ I=
=
= 9.00 mA
Position B: RT = R1 + R2 + R5 = 1.0 kΩ + 33 kΩ + 22 kΩ = 56 kΩ I=
=
= 161 µA
Position C: RT = R1 + R2 + R3 + R4 + R5 = 1.0 kΩ + 33 kΩ + 68 kΩ + 27 kΩ + 22 kΩ = 151 kΩ I=
=
= 59.6 µA
53. First, find the value of R5 with the switch in Position D. 6.0 mA = R5 =
– 1.8 kΩ = 1.2 kΩ
Position A: RT = 5.38 kΩ I=
= 3.35 mA
V1 = (3.35 mA)(1.8 kΩ) = 6.02 V V2 = (3.35 mA)(1.0 kΩ) = 3.35 V V3 = (3.35 mA)(820 Ω) = 2.74 V V4 = (3.35 mA)(560 Ω) = 1.87 V V5 = 4.00 V Position B: RT = 5.38 kΩ − 560 Ω = 4.82 kΩ I=
= 3.73 mA
V1 = (3.73 mA)(1.8 kΩ) = 6.72 V V2 = (3.73 mA)(1.0 kΩ) = 3.73 V V3 = (3.73 mA)(820 Ω) = 3.06 V V5 = 4.50 V S1-28
Position C: RT = 4.0 kΩ I=
= 4.50 mA
V1 = (4.50 mA)(1.8 kΩ) = 8.10 V V2 = (4.50 mA)(1.0 kΩ) = 4.50 V V5 = 5.40 V Position D: RT = 3.0 kΩ I=
= 6.00 mA
V1 = (6.00 mA)(1.8 kΩ) = 10.8 V V5 = 7.20 V Note: The voltage approach also can be used. 54. See Figure 4-10. The results in the table are correct. 55. Yes, because RT should be 19.9 kΩ. R3 and R5 each are shorted. Refer to Figure 4-10. 56. Yes, a short exists between the points indicated in Figure 4-10. 57. (a) R11 burned open due to excessive power dissipation because it had the largest resistance value. (b) Replace R11 (10 kΩ). (c) RT = 47.7 kΩ Imax = √
=√
= 7.07 mA
VFAIL = ImaxRT = (7.07 mA)(47.7 kΩ) = 338 V
Multisim Troubleshooting Problems 58. R1 is open. 59. R6 is shorted. 60. R2 is open. 61. Lamp 4 is open. 62. No fault. 63. The 82 Ω resistor is shorted.
S1-29
Chapter 5:
PARALLEL CIRCUITS
BASIC PROBLEMS Section 5-1:
Resistors in Parallel
1.
See Figure 5-1 for possible representation of connections.
2.
R1, R2, and R5 are not in parallel. See Figure 5-2 for possible representation of connections.
Section 5-2:
Total Parallel Resistance
3.
From Problem 2: RT = R6 || (R5 + R1 + R2) || R3 || R4 || R7 || R8 = 15 kΩ || 108 kΩ || 10 kΩ || 12 kΩ || 68 kΩ || 56 kΩ = 3.43 kΩ
4.
RT =
5.
(a) R = 47 Ω || 56 Ω = 25.6 Ω (b) R = 560 Ω || 1.0 kΩ = 359 Ω (c) R = 1.5 kΩ || 2.2 kΩ || 10 kΩ = 819 Ω (d) R = 1.0 kΩ || 2.2 MΩ || 1.0 MΩ || 470 kΩ = 996 Ω
6.
(a) RT =
=
(b) RT =
=
(c) RT =
=
= 557 kΩ
7.
RT =
= 2.0 Ω
8.
RT1 =
= 3.0 Ω
( ( (
)( )( )(
) )
= 1.50 kΩ
= 18.2 Ω )
= 892 Ω
S1-31
RT2 =
= 10 Ω
RT3 =
= 5.0 Ω = 1.58 Ω
RT =
Section 5-3: 9.
Voltage in a Parallel Circuit
V1 = V2 = V3 = V4 = 12 V IT =
=
= 20 mA
The total current divides equally among the four equal resistors. I 1 = I 2= I 3 = I 4 = I 5 =
= 5.0 mA
10. The resistors are all in parallel across the source. Each voltmeter is measuring the voltage across a resistor, so each meter indicates 100 V.
Application of Ohm’s Law
Section 5-4:
11. (a) RT = 33 kΩ || 33 kΩ || 33 kΩ = 11 kΩ IT =
= 909 µA
(b) RT = 1.0 kΩ || 3.9 kΩ || 560 Ω = 329 Ω IT =
= 76.1 mA
12. IT = I1 + I2 + I3 =
= 0.5 A + 0.5 A + 0.5 A = 1.5 A
13. (a) I1 =
= 179 µA
I2 =
= 455 µA
(b) I1 =
= 444 µA
I2 =
= 80 µA
14. RT =
=
= 2.0 kΩ
Reach = (4)(2.0 kΩ) = 8.0 kΩ
Section 5-5:
Kirchhoff’s Current Law
15. IT = 250 mA + 300 mA + 800 mA = 1350 mA = 1.35 A 16. IT = I1 + I2 + I3 + I4 + I5 I5 = IT − (I1 + I2 + I3 + I4) I5 = 500 mA − (50 mA + 150 mA + 25 mA + 100 mA) = 500 mA − 325 mA = 175 mA
S1-32
17. I2−3 = IT − (I1 + I4) = 50 mA − 35 mA = 15 mA I2 = I3 = 7.5 mA See Figure 5-3.
18. IT = I1 + I2 + I3 + I4 + I5 + I6 =4(0.5 A) + 2(1.2 A) = 4.4 A 19. (a) IT= I1 + I2 + I3 + I4 + I5 + I6 + I7 + I8 = 4(0.5 A) + 2(1.2 A) + 2(1.0 A) = 6.4 A (b) Iground = 6.4 A
Section 5-6:
Current Dividers
20. The 10 kΩ resistor has the highest current. 21. I1 = (
)IT = (
)3.0 A = 2.19 A
I2 = (
)IT = (
)3.0 A = 811 mA
22. (a) I1 = (
)IT = (
)10 µA = 6.875 µA = 6.88 µA (round to even)
I2 = (
)IT = (
)10 µA = 3.125 µA = 3.12 µA (round to even) = 516 Ω
(b) RT = I1 = (
)IT = (
)10 mA = 5.16 mA
I2 = (
)IT = (
)10 mA = 2.35 mA
I3 = (
)IT = (
)10 mA = 1.56 mA
I4 = (
)IT = (
)10 mA = 922 µA
Section 5-7:
Power in Parallel Circuits
23. PT = 5(40 mW) = 200 mW
S1-33
24. (a) RT = 1.0 MΩ || 2.2 MΩ = 688 kΩ PT = IT2RT = (10 µA)2(688 kΩ) = 68.8 µW (b) RT = 1.0 kΩ || 2.2 kΩ || 3.3 kΩ || 5.6 kΩ = 516 Ω PT = IT2RT = (10 mA)2(516 Ω) = 51.6 mW 25. P = VI Ieach =
=
= 0.625 A = 625 mA
IT = 6Ieach = 6(625 mA) = 3750 mA = 3.75 A
Section 5-8: 26. Ieach =
Troubleshooting
=
= 0.625 A = 625 mA
The current in each bulb is independent of the number of parallel bulbs. IT = 3.75 A – 0.625 A = 3.125 A 27. First, determine what the total current should be: RT = 220 Ω || 100 Ω || 1.0 kΩ || 560 Ω || 270 Ω = 47.5 Ω IT =
= 210 mA
The measured current is approximately 200 mA which is 10 mA less than it should be. Therefore, one of the resistors must be open. Ropen =
= 1.0 kΩ
=
The 1.0 kΩ resistor is open.
= 2.3 kΩ
28. RT = IT =
=
= 10.9 mA
The meter indicates 7.82 mA, which is approximately 3.1 mA less than the expected current.Therefore, a resistor must be open. = 8.06 kΩ. R3 = 8.2 kΩ ≈ 8.06 kΩ. Therefore, R3 must be open.
Ropen =
29. GT =
+
Gmeas =
+
= 8.52 mS
= 4.81 mS. This is less than expected, so a resistor is open.
Gopen = GT – Gmeas = 8.52 mS – 4.81 mS = 3.71 mS Ropen =
= 269 Ω ≈ 270 Ω = R2
=
Therefore, R2 is open. 30. GT = Gmeas =
+
+
+
= 29.1 mS
= 24.6 mS. This is less than expected, so a resistor is open.
Gopen = GT – Gmeas = 29.1 mS – 24.6 mS = 4.5 mS S1-34
Ropen =
= 221 Ω ≈ 220 Ω
=
Therefore, one of the 220 Ω resistors is open, but identifying which 220 Ω resistor requires more information.
ADVANCED PROBLEMS 31. VS = I1R1 = (1.0 mA)(50 Ω) = 50 mV R2 =
=
= 25 Ω
R3 =
=
= 100 Ω
I4 = IT – (I1 + I2 + I3) = 7.5 mA – 3.5 mA = 4.0 mA R4 =
= 12.5 Ω
=
32. VT = ITRT = (100 mA)(25 Ω) = 2500 mV = 2.5 V I220 =
=
= 11.4 mA
33. RT =
=
= 0.48R
I1 = (
)IT = (
)10 mA = 4.8 mA
I2 = (
)IT = (
)10 mA = 2.4 mA
I3 = (
)IT = (
)10 mA = 1.6 mA
I4 = (
)IT = (
)10 mA = 1.2 mA
34. (a) PT = IT2RT = (50 mA)2(1.0 kΩ) = 2.5 W Number of resistors = n =
=
= 10
All resistors are equal because each dissipates the same amount of power. (b) RT = R = nRT = 10(1.0 kΩ) = 10 kΩ (c) I =
=
= 5.0 mA
(d) VS = ITRT = (50 mA)(1.0 kΩ) = 50 V 35. (a) I2 = IT – I1 = 150 mA – 100 mA = 50 mA R1 =
=
= 100 Ω
R2 =
=
= 200 Ω
(b) P1 = PT – P2 = 2.0 W – 0.75 W = 1.25 W VS = R1 =
=
= 10 V =
=
= 80 Ω S1-35
R2 =
=
I1 =
= 133 Ω
=
=
=
= 125 mA
I2 = IT – I1 = 200 mA – 125 mA = 75 mA (c) I3 =
=
I2 = =
= =
= 10 mA =
= 14.7 mA
I1 = IT – (I2 + I3) = 50 mA – (14.7 mA + 10 mA) = 50 mA – 24.7 mA = 25.3 mA R1 =
=
= 395 Ω
=
36. (a) RT = R1 = 510 kΩ = 245 kΩ
(b) RT = R1 || R2 = (c) RT = R1 = 510 kΩ
= 193 kΩ
(d) RT = R1 || R2 || R3 =
37. Imax = 0.5 A RT(min) =
= 30 Ω
=
GT(max) =
(
)
=
= 33.3 mS
G2 = GT(max) – G1 = 33.3 mS − R2 =
= 33.3 mS – 14.6 7S = 18.7 mS
= 53.7 Ω
=
38. Position A: RT = R1 || R2 || R3 = 560 kΩ || 220 kΩ || 270 kΩ = 99.7 kΩ IT =
=
= 241 µA
I1 =
=
= 42.9 µA
I2 =
=
= 109 µA
I3 =
=
= 89.0 µA
Position B: RT = R1 || R2 || R3 || R4 || R5 || R6 = 560 kΩ || 220 kΩ || 270 kΩ || 1.0 MΩ || 820kΩ || 2.2 MΩ = 78.7 kΩ IT =
=
= 305 µA
I1 = 42.9 µA I2 = 109 µA I3 89.0 µA I4 =
=
= 24.0 µA
I5 =
=
= 29.3 µA S1-36
I6 =
= 10.9 µA
=
Position C: RT = R4 || R5 || R6 = 1.0 MΩ || 820kΩ || 2.2 MΩ = 374 kΩ IT =
= 64.2 µA
=
I4 = 24.0 µA I5 = 29.3 µA I6 = 10.9 µA 39. From Kirchhoff’s current law, the total current into the room is equal to the current in the appliances: IHEATER + ILAMPS + IVACUUM = 8.0 A + 2(0.833 A) + 5.0 A = 14.7 A. The total current including the current for the vacuum cleaner is less than the capacity of the breaker, so the vacuum cleaner could be plugged in. 40. The circuit consists of three 120 Ω resistors in parallel, so that total resistance is given by RT =
= 40 Ω
=
For a total current IT = 200 mA, the source voltage VS is equal to VS = ITRT = (200 mA)(40 Ω) = 8.0 V 41. R1 || R2 || R5 || R9 || R10 || R12 = 100 kΩ || 220 kΩ || 560 kΩ || 390 kΩ || 1.2 MΩ || 100 kΩ = 33.6 kΩ R4 || R6 || R7 || R8 = 270 kΩ || 1.0 MΩ || 820 kΩ || 680 kΩ = 135 kΩ R5 || R11 = 330 kΩ || 1.8 MΩ = 279 kΩ 42. RT = RT(R1 + R2) = R1R2 RTR1 + RTR2 = R1R2 RTR1 = R1R2 – RTR2 RTR1 = R2(R1 – RT) R2 = R2 =
= (
)(
)
(
)
= 283 Ω
43. I1 = 1.5 mA – 1.2 mA = 0.3 mA VS = V1 = V2 = V3 = V4 = I1R1 = (0.3 mA)(1.0 kΩ) = 0.3 V I2 = 1.2 mA – 0.8 mA = 0.4 mA R2 =
=
= 0.75 kΩ = 750 Ω
I3 =
=
= 90.9 µA
I4 = 0.8 mA – I3 = 800 µA – 90.9 µA = 709 µA R4 =
= 423 Ω
=
44. VS = ITRT = (2.5 mA)(1.5 kΩ) = 3.75 V Inew = 2.5 mA + (0.25)(2.5 mA) = 2.5 mA + 6.25 mA = 3.13 mA RT(new) =
=
= 1.20 kΩ S1-37
RT(new) = RT || Rnew = = Rnew =
(
)
(
)
=
+
− −
= 833 µS − 667 µS = 167 µS
= 6.0 kΩ
45. RT = R1 || R2 || R3 = 4.7 kΩ || 10 kΩ || 10 kΩ = 2.62 kΩ IT(calc) =
= 10.3 mA
IT(meas) = 5.0 mA,which is 5.3 mA less than expected, so a resistor is open. The value of the open resistor that corresponds to the missing current is Ropen =
= 4.72 kΩ ≈ 4.7 kΩ
Therefore, the 4.7 kΩ resistor R1 is open. 46. Pins 1-2: RT = 1.0 kΩ || 3.3 kΩ = 767 Ω for a correct reading When one resistor is open the reading is either 1.0 kΩ or 3.3 kΩ Pins 3-4: RT = 270 Ω || 390 Ω = 160 Ω for a correct reading When one resistor is open the reading is either 270 Ω or 390 Ω Pins 5-6: RT = 1.0 MΩ || 1.8 MΩ || 680 kΩ || 510 kΩ = 201 kΩ for a correct reading R5 open: RT = 1.8 MΩ || 680 kΩ || 510 kΩ = 251 kΩ R6 open: RT = 1.0 MΩ || 680 kΩ || 510 kΩ = 226 kΩ R7 open: RT = 1.0 MΩ || 1.8 MΩ || 510 kΩ = 284 kΩ R8 open: RT = 1.0 MΩ || 1.8 MΩ || 680 kΩ = 330 kΩ 47. (a) One of the resistors has burned open because the power dissipation exceeded the 0.5 W resistor power rating. Because each resistor has the same voltage across it, the resistor with the smallest value will dissipate the most power first, per the power formula P = (b) P =
V= √
= √(
)(
.
) = 30 V.
(c) Replace the 1.8 kΩ resistor. For future operation, limit the circuit voltage to less than 30 V, or use a higher wattage rating for the 1.8 kΩ resistor. 48. (a) R1-2 = R1 || R2 || R3 || R4 || R11 || R12 = 10 kΩ || 2.2 kΩ || 2.2 kΩ || 3.3 kΩ || 18 kΩ || 1.0 kΩ = 422 Ω (b) R2-3 = R5 || R6 || R7 || R8 || R9 || R10 = 4.7 kΩ || 4.7 kΩ || 6.8 kΩ || 5.6 kΩ || 5.6 kΩ || 1.0 kΩ = 518 Ω (c) R3-4 = R5 || R6 || R7 || R8 || R9 || R10 = 518 Ω (d) R1-4 = R1 || R2 || R3 || R4 || R11 || R12 = 422 Ω 49. (a) R1-2 = (R1 || R2 || R3 || R4 || R11 || R12) + (R5 || R6 || R7 || R8 || R9 || R10) = 422 Ω + 518 Ω = 941 Ω (b) R2-3 = R5 || R6 || R7 || R8 || R9 || R10 = 518 Ω (c) R2-4 = R5 || R6 || R7 || R8 || R9 || R10 = 518 Ω (d) R1-4 = R1 || R2 || R3 || R4 || R11 || R12 = 422 Ω S1-38