Introductory Circuit Analysis
Eleventh Edition
Columbus, Ohio
Robert L. Boylestad
Upper Saddle River, New Jersey
5.
15. a. 4.2 × 103 + 48.0 × 103 = 52.2 × 103 = 5.22 × 104
b. 90 × 103 + 360 × 103 = 450 × 103 = 4.50 × 105
c. 50 × 10 5 6 × 10 5 = 44 × 10 5 = 4.4 × 10 4
d. 1.2 × 103 + 0.05 × 103 0.6 × 103 = 0.65 × 103 = 6.5 × 102
16. a. (102)(103) = 105 = 100 × 103
b. (10 2)(103) = 101 = 10
c. (103)(106) = 1 × 109
d. (102)(10 5) = 1 × 10 3
e. (10 6)(10 × 106) = 10
f. (104)(10 8)(1028) = 1 × 1024
17. a. (50 × 103)(3 × 10 4) = 150 × 10 1 = 1.5 × 101
b. (2.2 × 103)(2 × 10 3) = 4.4 × 100 = 4.4
c. (82 × 106)(2.8 × 10 6) = 229.6 = 2.296 × 102
d. (30 × 10 4)(4 × 10 3)(7 × 108) = 840 × 101 = 8.40 × 103
18. a. 102/104 = 10 2 = 10 × 10 3
b. 10 2/103 = 10 5 = 10 × 10 6
c. 104/10 3 = 107 = 10 × 106
d. 10 7/102 = 1.0 × 10 9
e. 1038/10 4 = 1.0 × 1042
f. 2 10 / 100 = 101/10 2 = 1 × 103
19. a. (2 × 103)/(8 × 10 5) = 0.25 × 108 = 2.50 × 107
b. (4 × 10 3)/(60 × 104) = 4/60 × 10 7 = 0.667 × 10 7 = 6.67 × 10 8
c. (22 × 10 5)/(5 × 10 5) = 22/5 × 100 = 4.4
d. (78 × 1018)/(4 × 10 6) = 1.95 × 1025
20. a. (102)3 = 1.0 × 106 b. (10 4)1/2 = 10.0 × 10 3 c. (104)8 = 100.0 × 1030 d. (10 7)9 = 1.0 × 10 63
21. a. (4 × 102)2 = 16 × 104 = 1.6 × 105
b. (6 × 10 3)3 = 216 × 10 9 = 2.16 × 10 7
c. (4 × 10 3)(6 × 102)2 = (4 × 10 3)(36 × 104) = 144 × 101 = 1.44 × 103
d. ((2 × 10 3)(0.8 × 104)(0.003 × 105))3 = (4.8 × 103)3 = (4.8)3 × (103)3 = 110.6 × 109 = 1.11 × 1011
22. a. ( 10 3)2 = 1.0 × 10 6
b. 3 4 2 10 ) )(10 (10 = 10 2/103 = 1.0 × 10 5
c. 4 4 4 2 6 4 2 2 3 10 10 10 ) )(10 (10 10 ) (10 ) (10 = = = 1.0 × 10 8
d. 4 4 3 10 ) )(10 (10 = 107/10 4 = 1.0 × 1011
e. (1 ×
c.
(2
(16 ×
(310)1.6010(210)(810) (710)
× 9421/2 10 516 1010 (2710)(2.5610)(1610) 4910 (69.1210)(410)276.4810 49104910 ××× = × ××× == ×× = 5.64 × 104 = 56.4 × 103 +3
24. a. 6 × 103 = 0.006 × 10+6 3 3 b. 4 × 10 3 = 4000 × 10 6 +3 2 +3 +3
c. 50 × 105 = 5000 × 103 = 5 × 106 = 0.005 × 109 +2 3 3
d. 30 × 10 8 = 0.0003 × 10 3 = 0.3 × 10 6 = 300 × 10 9
25. a. 0.05 × 100 s = 50 × 10 3 s = 50 ms +3 +3
b. 2000 × 10 6 s = 2 × 10 3 s = 2 ms 3 3
c. 0.04 × 10 3 s = 40 × 10 6 s = 40 μs +3 +6
d. 8400 × 10 12 s ⇒ 0.0084 × 10 6 s = 0.0084 μs
e. 4 × 10 3 × 103 m = 4 × 100 m = 4000 × 10 3 m = 4000 mm +3 increase by 3 100
f. 260 × 103 × 10 3 m = 0.26 × 103 m = 0.26 km
26. a. 1.5 min ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ min 1 s 60 = 90 s b. 0.04 h
⎡ min 1 s 60 h 1 min 60 = 144 s
c. 0.05 s ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ s 10 s 1 6 μ = 0.05 × 106 μs = 50 × 103 μs
d. 0.16 m ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ m 10 mm 1 3 = 0.16 × 103 mm = 160 mm
t = 0.0568 mi 0.0167mi/s d υ = = 3.40 s
32. 30 mi5280 ft12 in.1 m1 h1 min h1 mi1 ft39.37 in.60 min60 s
= 13.41 m/s 33. ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢
38.
42. 6(4 + 8) = 72
43. (20 + 32)/4 = 13
44. 22 (812) + = 14.42
46.
47. ( ) 2 400/(610) + = 2.95
48. 205 × 10 6
49. 1.20 × 1012
6.667 × 106 + 0.5 × 106 = 7.17 × 106
50.
2.
3.
6. F =
a. F = 2 2 2 1 (10) 7.2 = r kQQ = 72 mN
b. Q1/Q2 = 1/2 ⇒ Q2 = 2Q1
7. V = 1.2 J 0.4 mC W Q = = 3 kV
8. W = VQ = (60 V)(8 mC) = 0.48 J
9. Q = V 16 J 96 = V W = 6 C
Q
12. I = 312 C (2)(60 s) Q t = = 2.60 A
13. Q = It = (40 mA)(0.8)(60 s) = 1.92 C
14. Q = It = (250 mA)(1.2)(60 s) = 18.0 C
15. t = mA 2 mC 6 = I Q = 3 s 16. 21.847 × 1018 electrons
17. Q = It = (4 mA)(90 s) = 360 mC 360 mC ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ × C 1 electrons 10 6.242 18 = 2.25 × 1018 electrons
18. I = s) (1.2)(60 C 86 = t Q = 1.194 A > 1 A (yes)
19. 0.84 × 1016 electrons ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ × electrons 10 6.242 C 1 18 = 1.346 mC I = ms 60 mC 346 1 = t Q = 22.43 mA
20. a. Q = It = (2 mA)(0.01 μs) = 2 × 10 11 C 2 × 10 11 C 6.2421018 electrons1 ¢ 1 Celectron
= 1.25 × 108 ¢ = $1.25 × 106 = 1.25 million
b. Q = It = (100 μA)(1.5 ns) = 1.5 × 10 13 C 1.5 × 10 13 C
(a) > (b)
21. Q = It = (200 × 10 3 A)(30 s) = 6 C V = C 6 J 40 = Q W = 6.67 V
22. Q = It = min) 5 (0 min C 420 ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ = 210 C V = C 210 J 742 = Q W = 3.53 V
23. Q = V 24 J 0.4 = V W = 0.0167 C I = 3 0.0167 C 510 s Q t = × = 3.34 A
24. I = h 40 Ah 200 (hours) rating Ah = t = 5 A
25. Ah = (0.8 A)(75 h) = 60.0 Ah
26. t(hours) = A 1.28 Ah 32 rating Ah = I = 25 h
27. 40 Ah(for 1 h): W1 = VQ = V I t = (12 V)(40 A)(1 h)
min 1 s 60 h 1 min 60 = 1.728 × 106 J
60 Ah(for 1 h): W2 = (12 V)(60 A)(1 h)
Ratio W2/W1 = 1.5 or 50% more energy available with 60 Ah rating.
For 60 s discharge: 40 Ah = It = I []
I(16.67 × 10 3 h) and I = h 10 16.67 Ah 40 -3 × = 2400 A 60 Ah = It = I []
⎣
min 60 h 1 s 60 min 1 s 60 = I(16.67 × 10 3 h) and I = h 10 16.67 Ah 60 -3 × = 3600 A
I2/I1 = 1.5 or 50 % more starting current available at 60 Ah
28. I = 3 Ah 6.0 h = 500 mA
Q = It = (500 mA)(6 h) ⎥ ⎦
⎡ min 1 s 60 h 1 min 60 = 10.80 kC
W = QV = (10.8 kC)(12 V) ≅ 129.6 kJ
4 min
⎦ ⎤ ⎢ ⎣ ⎡ min 1 s 60 = 240 s
Q = It = (2.5 A)(240 s) = 600 C
36. Q = It = (10 × 10 3 A)(20 s) = 200 mC
W = VQ = (12.5 V)(200 × 10 3 C) = 2.5 J
Chapter 3
1. a. 0.5 in. = 500 mils
b. 0.02 in. ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ in. 1 mils 1000 = 20 mils
c. 4 1 in. = 0.25 in. ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ in. 1 mils 1000 = 250 mils
d. 1 in. = 1000 mils
e. 0.02 ft ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ in. 1 mils 10 ft 1 in. 12 3 = 240 mils
f. 0.1 cm ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ in. 1 mils 1000 cm 2.54 in. 1 = 39.37 mils
2. a. ACM = (30 mils)2 = 900 CM
b. 0.016 in. = 16 mils, ACM = (16 mils)2 = 256 CM
c. 1 8 = 0.125" = 125 mils, ACM = (125 mils)2 = 15.63 × 103 CM
d. 1 cm ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ in. 1 mils 1000 cm 2.54 in. 1 = 393.7 mils, ACM = (393.7 mils)2 = 155 × 103 CM
e. 0.02 ft ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ in. 1 mils 1000 ft 1 in. 12 = 240 mils, ACM = (240 mils)2 = 57.60 × 103 CM
f. 0.0042 m ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ m 1 in. 39.37 = 0.1654 in. = 165.4 mils, ACM = (165.4 mils)2 = 27.36 × 103 CM
3. ACM = (dmils )2 → dmils = CMA
a. d = CM 1600 = 40 mils = 0.04 in.
b. d = 820 CM = 28.64 mils = 0.029 in.
c. d = CM 40,000 = 200 mils = 0.2 in. "
d. d = CM 625 = 25 mils = 0.025 in.
e. d = CM 6.25 = 2.5 mils = 0.0025 in.
f. d = 100 CM = 10 mils = 0.01 in.
4. 0.01 in. = 10 mils, ACM = (10 mils)2 = 100 CM R = A l ρ = CM 100 ) (200 (10.37) ′ = 20.74 Ω
5. ACM = (4 mils)2 = 16 CM, R = A l ρ = CM 16 ft) (150 9) (9 = 92.81 Ω
6. a. A = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ Ω ′ = 5 2 80 17 R l ρ = 544 CM b. d = CM 544 CM = A = 23.32 mils = 23.3 × 10 3 in. 7. 32 1 = 0.03125" = 31.25 mils, ACM = (31.25 mils)2 = 976.56 CM R = R l ρ ⇒ l = 600 CM) )(976.56 (2.2 Ω = ρ RA = 3.58 ft
8. a. ACM = A l ρ = (10.37)(300) 3.3 ′ Ω = 942.73 CM d = 942.73 CM = 30.70 mils = 30.7 × 10 3 in.
b. larger
c. smaller
9. a. Rsilver > Rcopper > Raluminum b. Silver: R = A l ρ = (9.9)(10 ft) 1 CM = 99 Ω { ACM = (1 mil)2 = 1 CM Copper: R = A l ρ = (10.37)(50 ft) 100 CM = 5.19 Ω { ACM = (10 mils)2 = 100 CM
10. ρ = 1000 CM) )(94 (500 ′ Ω = l RA = 47 ⇒ nickel "
Aluminum: R = A l ρ = (17)(200 ft) 2500 CM = 1.36 Ω { ACM = (50 mils)2 = 2500 CM
11. a. 3" = 3000 mils, 1/2" = 0.5 in. = 500 mils
Area = (3 × 103 mils)(5 × 102 mils) = 15 × 105 sq. mils 15 × 105 sq mils
(17)(4
b.
=
Aluminum bus-bar has almost 64% higher resistance.
c. increases
d. decreases
12.
13.
15. a. #8: R = 1800 ft ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ Ω ft 1000 0.6282 = 1.13 Ω #18: R = 1800 ft ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ Ω ft 1000 6.385 = 11.49 Ω
b. #18:#8 = 11.49 Ω:1.13 Ω = 10.17:1 ≅ 10:1
c. #18:#8 = 1624.3 CM:16,509 CM = 1:10.16 ≅ 1:10
16. a. A = 3 10 6 CM 1 311 m 6 ) 37)(30 (10 × = Ω ′ = R l ρ = 51,850 CM ⇒ #3 but 110 A ⇒ #2
b. A = 3 10 3 CM 311.1 m 3 ) (10.37)(30 × = Ω ′ = R l ρ = 103,700 CM ⇒ #0
17. a. A/CM = 230 A/211,600 CM = 1.09 mA/CM
b. 1.09 mA1 CM1000 mils1000 mils CM1 in.1 in. sq mils 4 π
c. 5 kA 2 1 in. 1.39 kA ⎡⎤ ⎢⎥ ⎣⎦ = 3.6 in.2
18. 1 12 in. = 0.083 in. 2.54 cm 1 in. ⎛⎞ ⎜⎟ ⎝⎠ = 0.21 cm A = 22 (3.14)(0.21 cm) 44 dπ = = 0.035 cm2 l = 2 6 (2 )(0.035
cm
= 1.39 kA/in.2
c. increases
d. decreases
20. Rs = d ρ = 100 ⇒ d = 100 ρ = 6 25010 100 × = 2.5 μcm
21. R = Rs l w ⇒ w = (150 )(1/2 in.) 500 Rsl = R Ω Ω = 0.15 in.
22. a. d = 1 in. = 1000 mils A
25.
26. 2 40 5 234 76 0 30 5 234 R = Ω + R2 = 264.5 ) (194.5)(0.76 Ω = 0.56 Ω
27. 2 243(30)2430 0.04 R +−+ = Ω R2 = (243)(40 m) 213 Ω = 46 mΩ
28. a. 68°F = 20°C, 32°F = 0°C
=
+
m
+
=
(334.5)(2 m)
m
Ω
2.63 m
30. a. K = 273.15 + °C 50 = 273.15 + °C °C = 223.15° 2 15 223 5 234 10 20 5 234 R = Ω + R2 = 254.5 11.35 (10 Ω) = 0.446 Ω b. K = 273.15 + °C 38.65 = 273.15 + °C °C = 234.5° 2 5 234 5 234 10 20 5 234 R = Ω + R2 = 254.5 (0)10 Ω = 0 Ω
Recall: 234.5° = Inferred absolute zero R = 0 Ω
c. F = 5 9 32 C 5 9 = + ° ( 273.15°) + 32 = 459.67°
31. a. α20 = 5 254 1 20 5 234 1 C 20 1 = + = ° + Ti = 0.003929 ≅ 0.00393
b. R = R20[1 + α20(t 20°C)] 1 Ω = 0.8 Ω[1 + 0.00393(t 20°)]
1.25 = 1 + 0.00393t 0.0786 1.25 0.9214 = 0.00393t 0.3286 = 0.00393t t = 0.00393 3286 0 = 83.61°C
32. R = R20[1 + α20(t 20°C)] = 0.4 Ω[1 + 0.00393(16 20)] = 0.4 Ω[1 0.01572] = 0.39 Ω
33. Table: 1000′ of #12 copper wire = 1.588 Ω @ 20°C C° = 9 5 (F° 32) = 9 5 (115 32) = 46.11°C
R = R20[1 + α20(t 20°C)] = 1.588 Ω[1 + 0.00393(46.11 20)] = 1.75 Ω
34. ΔR = 6 nominal 10 R (PPM)( ΔT) = 106 22 Ω (200)(65° 20°) = 0.198 Ω R = Rnominal + ΔR = 22.198 Ω
35. ΔR = 6 nominal 10 R (PPM)( ΔT) = 6 100 10 Ω (100)(50° 20°) = 0.30 Ω R = Rnominal + ΔR = 100 Ω + 0.30 Ω = 100.30 Ω
36.
37.
38. #12: Area = 6529 CM d = CM 6529 = 80.8 mils = 0.0808 in. ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ in. 1 cm 2.54 = 0.205 cm A = 4 cm) (0.205 4 2 2 π π = d = 0.033 cm2 I = ] cm 033 [0 cm MA 1 2 2 = 33 kA >>> 20 A
39.
40. a. 2 times larger b. 4 times larger
41. 10 kΩ 3.5 kΩ = 6.5 kΩ
42. 6.25 kΩ and 18.75 kΩ
43.
44. a. 560 kΩ ± 5%, 560 kΩ ± 28 kΩ, 532 kΩ ↔ 588 kΩ
b. 220 Ω ± 10%, 220 Ω ± 22 Ω, 198 Ω ↔ 242 Ω
c. 100 Ω ± 20%, 100 Ω ± 20 Ω, 80 Ω ↔ 120 Ω
45.
a. 120 Ω = Brown, Red, Brown, Silver
b. 8.2 Ω = Gray, Red, Gold, Silver
c. 6.8 kΩ = Blue, Gray, Red, Silver
d. 3.3 MΩ = Orange, Orange, Green, Silver
46.1020%812 1520%1218 Ω±⇒Ω−Ω ⎫ ⎬ Ω±⇒Ω−Ω ⎭ no overlap, continuance
47.1010%101 911 1510%151.5 13.516.5
48. a. 621 = 62 × 101 Ω = 620 Ω = 0.62 kΩ
b. 333 = 33 × 103 Ω = 33 kΩ
c. Q2 = 3.9 × 102 Ω = 390 Ω
d. C6 = 1.2 × 106 Ω = 1.2 MΩ
49. a. G = 11 120 R = Ω = 8.33 mS
b. G = Ωk 4 1 = 0.25 mS
c. G = ΩM 2 2 1 = 0.46 μS
Ga > Gb > Gc vs. Rc > Rb > Ra
⎭ No overlap
50. a. Table 3.2, Ω/1000′ = 1.588 Ω G = 11 1.588 R = Ω = 629.72 mS or G = 6529.9 CM (Table 3.2) (10.37)(1000) A lρ = ′ = 629.69 mS (Cu)
b. G = 6529.9 CM (17)(1000) ′ = 384.11 mS (Al)
c. G = 6529.9 CM (74)(1000) ′ = 88.24 mS (Fe)
51. A2 = 2 1 3 A1 = 5 3 A1, l2 = 1 1 2 1 33 l l ⎛⎞ = ⎜⎟
G2 = 5G1 = 5(100 S) = 500 S
52. 53. 54.
55. a. 50°C specific resistance ≅ 105 Ω-cm 50°C specific resistance ≅ 500 Ω-cm 200°C specific resistance ≅ 7 Ω-cm
b. negative
c. No
d. ρ = cm30030270 cm 1255075 C T
56. a. Log scale: 10 fc ⇒ 3 kΩ 100 fc ⇒ 0.4 kΩ
negative
c. no—log scales imply linearity d. 1 kΩ ⇒ ≅ 30 fc 10 kΩ ⇒ ≅ 2 fc R fc Δ Δ = 10
57. a. @ 0.5 mA, V ≅ 195 V @ 1 mA, V ≅ 200 V @ 5 mA, V ≅ 215 V
b. ΔVtotal = 215 V 195 V = 20 V
c. 5 mA:0.5 mA = 10:1 compared to 215 V: 200 V = 1.08:1
Ω/fc
Chapter 4
1. V = IR = (2.5 A)(47 Ω) = 117.5 V
2. I = 12 V 6.8 V R = Ω = 1.76 A
3. R = 6 V 1.5 mA V = I = 4 kΩ
4. I = 3 12 V 4010 V R = ×Ω = 300 A
5. V = IR = (3.6 μA)(0.02 MΩ) = 0.072 V = 72 mV
6. I = 62 V 15 k V R = Ω = 4.13 mA
7. R = 120 V 2.2 A V = I = 54.55 Ω
8. I = 120 V 7.5 k V R = Ω = 16 mA
9. R = 120 V 4.2 A V = I = 28.57 Ω
10. R = 4.5 V 125 mA V = I = 36 Ω
11. R = 24 mV 20 A V = I μ = 1.2 kΩ
12. V = IR = (15 A)(0.5 Ω) = 7.5 V
13. a. R = 120 V 9.5 A V = I = 12.63 Ω b. t = 1 h 60 min60 s 1 h1 min ⎡ ⎤⎡⎤ ⎢ ⎥⎢⎥ ⎣ ⎦⎣⎦ = 3600 s W = Pt = VIt = (120 V)(9.5 A)(3600 s) = 4.1 × 106 J
14. V = IR = (2.4 μA)(3.3 MΩ) = 7.92 V 15.
16. b. (0.13 mA)(500 h) = 65 mAh
17. 18. 19.
20. P = 420 J420 J 60 s 240 s 4 min 1 min W t == ⎡⎤ ⎢⎥ ⎣⎦ = 1.75 W
21. t = 640 J 40 J/s W P = = 16 s
22. a. 60 min60 s 8 h 1 h1 min ⎡⎤⎡⎤ ⎢⎥⎢⎥ ⎣⎦ ⎣⎦ = 28,800 s W = Pt = (2 W)(28,000 s) = 57.6 kJ
b. kWh = (2 W)(8 h) 1000 = 16 × 10 3 kWh
23. I = 300 C1 min 1 min60 s Q t ⎡⎤ = ⎢⎥ ⎣⎦ = 5 C/s = 5 A
P = I2R = (5 A)2 10 Ω = 250 W
24. P = VI = (3 V)(1.4 A) = 4.20 W t = 12 J 4.2 W W P = = 2.86 s
25. I = 48 C1 min min60 s ⎡⎤ ⎢⎥ ⎣⎦ = 0.8 A P = EI = (6 V)(0.8 A) = 4.8 W
26. P = I2R = (7.2 mA)2 4 kΩ = 207.36 mW
27. P = I2R ⇒ I = 240 mW 2.2 k P R = Ω = 10.44 mA
28. I = 2 W 120 P R = Ω = 129.10 mA
V = IR = (129.10 mA)(120 Ω) = 15.49 V
29. I = 12 V 5.6 k E R = Ω = 2.14 mA P = I2R = (2.14 mA)2 5.6 kΩ = 25.65 mW
W = P ⋅ t = (25.65 mW) 60 min60 s 1 h 1 h1 min
30. E = 324 W 2.7 A P I = = 120 V
31. I = 1 W 4.7 M P R = Ω = 461.27 μA no
= 92.34 J
32. V = (42 mW)(2.2 k)92.40 PR =Ω= = 9.61 V
33. P = EI = (9 V)(45 mA) = 405 mW
34. P = VI, I = 100 W 120 V P V = = 0.833 A R = 120 V 0.833 A V I = = 144.06 Ω
35. V = 450 W 3.75 A P I = = 120 V R = 120 V 3.75 A V I = = 32 Ω
36. a. P = EI and I = 3 0.410 W 3 V P E × = = 0.13 mA b. Ah rating = (0.13 mA)(500 h) = 66.5 mAh
37. I = 3 100 W 510 20 k P R ==× Ω = 70.71 mA V = (100 W)(20 k) PR =Ω = 1.42 kV
38. a. W = Pt = 2 2 12 V 60 s 10 V t R ⎛⎞ ⎛⎞ = ⎜⎟ ⎜⎟ Ω ⎝⎠ ⎝⎠ = 864 J
b. Energy doubles, power the same
