Solucionario introducción al análisis de circuitos boylestad 11ed

Instructor’s Resource Manual to accompany Introductory Circuit Analysis Eleventh Edition Robert L Boylestad Upper Saddle River, New Jersey Columbus, Ohio www.elsolucionario.net http://www.elsolucionario.net LIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE MUCHOS DE ESTOS LIBROS LOS SOLUCIONARIOS CONTIENEN TODOS LOS EJERCICIOS DEL LIBRO RESUELTOS Y EXPLICADOS DE FORMA CLARA VISITANOS PARA DESARGALOS GRATIS www.elsolucionario.net Contents CHAPTER 1 CHAPTER CHAPTER 13 CHAPTER 22 CHAPTER 29 CHAPTER 39 CHAPTER 52 CHAPTER 65 CHAPTER 86 CHAPTER 10 106 CHAPTER 11 124 CHAPTER 12 143 CHAPTER 13 150 CHAPTER 14 157 CHAPTER 15 170 CHAPTER 16 195 CHAPTER 17 202 CHAPTER 18 222 CHAPTER 19 253 CHAPTER 20 266 CHAPTER 21 280 CHAPTER 22 311 CHAPTER 23 318 CHAPTER 24 333 CHAPTER 25 342 TEST ITEM FILE 353 www.elsolucionario.net iii Chapter 1 − − − υ= d 20,000 ft ⎡ mi ⎤ ⎡ 60 s ⎤ ⎡ 60 ⎤ = = 1363.64 mph t 10 s ⎢⎣ 5,280 ft ⎥⎦ ⎢⎣1 ⎥⎦ ⎢⎣ h ⎥⎦ ⎡ 1h ⎤ ⎢ ⎥ = 0.067 h ⎣ 60 ⎦ d 31 mi υ= = = 29.05 mph t 1.067 h a b c 95 mi ⎡ 5,280 ft ⎤ ⎡ h ⎤ ⎡1 ⎤ = 139.33 ft/s h ⎢⎣ mi ⎥⎦ ⎢⎣ 60 ⎥⎦ ⎢⎣ 60 s ⎥⎦ 60 ft d t= = = 0.431 s υ 139.33 ft/s d 60 ft ⎡ 60 s ⎤ ⎡ 60 ⎤ ⎡ mi ⎤ = 40.91 mph υ= = t s ⎢⎣1 ⎥⎦ ⎢⎣ h ⎥⎦ ⎢⎣ 5,280 ft ⎥⎦ − − − 10 MKS, CGS, °C = 11 ⎡ 0.7378 ft - lb ⎤ 1000 J ⎢ ⎥ = 737.8 ft-lbs 1J ⎣ ⎦ 12 ⎡ ft ⎤ ⎡12 in ⎤ ⎡ 2.54 cm ⎤ 0.5 yd ⎢ ⎥⎢ ⎥⎢ ⎥ = 45.72 cm ⎣1 yd ⎦ ⎣ ft ⎦ ⎣ in ⎦ 13 a 104 14 a 15 × 103 b e 4.02 × 10−4 f × 10−10 5 (°F − 32) = (68 − 32) = (36) = 20° 9 SI: K = 273.15 + °C = 273.15 + 20 = 293.15 b 106 c 103 30 × 10−3 d 10−3 c 2.4 × 106 Chapter e 100 f 10−1 d 150 × 103 www.elsolucionario.net 15 a b c d 4.2 × 103 + 48.0 × 103 = 52.2 × 103 = 5.22 × 104 90 × 103 + 360 × 103 = 450 × 103 = 4.50 × 105 50 × 10−5 − × 10−5 = 44 × 10−5 = 4.4 × 10−4 1.2 × 103 + 0.05 × 103 − 0.6 × 103 = 0.65 × 103 = 6.5 × 102 16 a b c d e f (102)(103) = 105 = 100 × 103 (10−2)(103) = 101 = 10 (103)(106) = × 109 (102)(10−5) = × 10−3 (10−6)(10 × 106) = 10 (104)(10−8)(1028) = × 1024 17 a b c d (50 × 103)(3 × 10−4) = 150 × 10−1 = 1.5 × 101 (2.2 × 103)(2 × 10−3) = 4.4 × 100 = 4.4 (82 × 106)(2.8 × 10−6) = 229.6 = 2.296 × 102 (30 × 10−4)(4 × 10−3)(7 × 108) = 840 × 101 = 8.40 × 103 18 a b c d e f 102/104 = 10−2 = 10 × 10−3 10−2/103 = 10−5 = 10 × 10−6 104/10−3 = 107 = 10 × 106 10−7/102 = 1.0 × 10−9 1038/10−4 = 1.0 × 1042 100 / 10 −2 = 101/10−2 = × 103 19 a b c d (2 × 103)/(8 × 10−5) = 0.25 × 108 = 2.50 × 107 (4 × 10−3)/(60 × 104) = 4/60 × 10−7 = 0.667 × 10−7 = 6.67 × 10−8 (22 × 10−5)/(5 × 10−5) = 22/5 × 100 = 4.4 (78 × 1018)/(4 × 10−6) = 1.95 × 1025 20 a c (102)3 = 1.0 × 106 (104)8 = 100.0 × 1030 21 a b c d 22 a (4 × 102)2 = 16 × 104 = 1.6 × 105 (6 × 10−3)3 = 216 × 10−9 = 2.16 × 10−7 (4 × 10−3)(6 × 102)2 = (4 × 10−3)(36 × 104) = 144 × 101 = 1.44 × 103 ((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 −3 −6 (−10 ) = 1.0 × 10 (10 )(10 −4 ) = 10−2/103 = 1.0 × 10−5 10 (10 −3 ) (10 ) (10 −6 )(10 ) 10 −4 = = = 1.0 × 10−8 10 10 10 (10 )(10 ) = 107/10−4 = 1.0 × 1011 10 − b c d b d (10−4)1/2 = 10.0 × 10−3 (10−7)9 = 1.0 × 10−63 Chapter www.elsolucionario.net e (1 × 10−4)3(102)/106 = (10−12)(102)/106 = 10−10/106 = 1.0 × 10−16 [(10 )(10 )] [(10 ) ][10 ] −2 f 23 a b 2 −3 −3 = 1 = = 1.0 × 10−1 −3 (10 )(10 ) 10 (3 × 10 ) (10 ) = (9 × 104)(102)/(3 × 104) = (9 × 106)/(3 × 104) = × 102 = 300 × 10 (4 × 10 ) 16 × 10 = = × 105 = 200.0 × 103 (20) × 10 c (6 × 104 ) 36 × 108 = = 9.0 × 1012 −2 −4 ( × 10 ) × 10 d (27 × 10 −6 )1 / 3 × 10 −2 = 1.5 × 10−7 = 150.0 × 10−9 = × 10 × 10 e (4 × 10 ) (3 × 10 ) (16 × 10 )(3 × 10 ) 48 × 10 = 24.0 × 1012 = = −4 −4 −4 × 10 × 10 × 10 f (16 × 10−6)1/2(105)5(2 × 10−2) = (4 × 10−3)(1025)(2 × 10−2) = × 1020 = 800.0 × 1018 g 1/ ⎡ (3 × 10−3 )3 ⎤ ⎡1.60 × 102 ⎤ ⎡(2 × 102 )(8 × 10−4 ) ⎤ ⎣ ⎦⎣ ⎦ ⎣ ⎦ (7 × 10−5 ) (27 × 10−9 )(2.56 × 104 )(16 × 10−2 )1/ 49 × 10−10 (69.12 × 10−5 )(4 × 10−1 ) 276.48 × 10−6 = = 49 × 10−10 49 × 10−10 = 5.64 × 104 = 56.4 × 103 = +3 24 a × 103 = 0.006 × 10+6 −3 −3 b × 10−3 = 4000 × 10−6 +3 −2 c +3 +3 50 × 105 = 5000 × 103 = × 106 = 0.005 × 109 +2 −3 −3 Chapter www.elsolucionario.net −3 +5 d −3 30 × 10−8 = 0.0003 × 10−3 = 0.3 × 10−6 = 300 × 10−9 −5 +3 +3 −3 25 a 0.05 × 100 s = 50 × 10−3 s = 50 ms +3 +3 b 2000 × 10−6 s = × 10−3 s = 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 −6 −3 e × 10−3 × 103 m = × 100 m = 4000 × 10−3 m = 4000 mm +3 100 f increase by 260 × 103 × 10−3 m = 0.26 × 103 m = 0.26 km −3 26 a b c d ⎡ 60 s ⎤ 1.5 ⎢ ⎥ = 90 s ⎣1 ⎦ ⎡ 60 ⎤ ⎡ 60 s ⎤ 0.04 h ⎢ ⎥⎢ ⎥ = 144 s ⎣ h ⎦ ⎣1 ⎦ ⎡ μs ⎤ 0.05 s ⎢ −6 ⎥ = 0.05 × 106 μs = 50 × 103 μs ⎣10 s ⎦ ⎡ mm ⎤ 0.16 m ⎢ −3 ⎥ = 0.16 × 103 mm = 160 mm ⎣10 m ⎦ Chapter www.elsolucionario.net 27 28 e ⎡ ns ⎤ 1.2 × 10−7 s ⎢ −9 ⎥ = 1.2 × 102 ns = 120 ns ⎣10 s ⎦ f ⎡1 ⎤ ⎡ h ⎤ ⎡1 day ⎤ 3.62 × 106 s ⎢ ⎥ = 41.90 days ⎥⎢ ⎥⎢ ⎣ 60 s ⎦ ⎣ 60 ⎦ ⎣ 24 h ⎦ a ⎡10 −6 F ⎤ ⎡ pF ⎤ −6 12 0.1 μF ⎢ ⎥ ⎢ −12 ⎥ = 0.1 × 10 × 10 pF = 10 pF F μ 10 F ⎦ ⎣ ⎦⎣ b ⎡100 cm ⎤ = 8000 × 10−3 cm = cm 80 × 10−3 m ⎢ ⎥ ⎣ 1m ⎦ c ⎡ m ⎤ ⎡ km ⎤ −5 60 cm ⎢ ⎥⎢ ⎥ = 60 × 10 km ⎣100 cm ⎦ ⎣1000 m ⎦ d ⎡ 60 ⎤ 3.2 h ⎢ ⎥ ⎣ 1h ⎦ e ⎡10−3 m ⎤ ⎡ μ m ⎤ 0.016 mm ⎢ ⎥ ⎢ − ⎥ = 0.016 × 10 μm = 16 μm ⎣ mm ⎦ ⎣10 m ⎦ f ⎡ 1m ⎤⎡ 1m ⎤ −4 60 cm2 ⎢ ⎥ ⎢100 cm ⎥ = 60 × 10 m 100 cm ⎣ ⎦⎣ ⎦ a ⎡ 1m ⎤ 100 in ⎢ ⎥ = 2.54 m ⎣ 39.37 in ⎦ b ⎡12 in ⎤ ⎡ m ⎤ ft ⎢ ⎥ = 1.22 m ⎥⎢ ⎣ ft ⎦ ⎣ 39.37 in ⎦ c ⎡ 4.45 N ⎤ lb ⎢ ⎥ = 26.7 N ⎣ lb ⎦ d ⎡ N ⎤ ⎡ lb ⎤ 60 × 103 dynes ⎢ ⎥⎢ ⎥ = 0.13 lb ⎣10 dynes ⎦ ⎣ 4.45 N ⎦ e ⎡ in ⎤ 150,000 cm ⎢ ⎥ ⎣ 2.54 cm ⎦ f ⎡ 5280 ft ⎤ ⎡12 in ⎤ ⎡ m ⎤ 0.002 mi ⎢ ⎥⎢ ⎥⎢ ⎥ = 3.22 m ⎣ mi ⎦ ⎣ ft ⎦ ⎣ 39.37 in ⎦ ⎡ ms ⎤ ⎢10− s ⎥ = 11.52 × 10 ms ⎦ ⎣ ⎡ 60 s ⎤ ⎢1 ⎥ ⎦ ⎣ ⎡ ft ⎤ ⎢12 in ⎥ = 4921.26 ft ⎣ ⎦ Chapter www.elsolucionario.net 29 ⎡1 yd ⎤ 5280 ft ⎢ ⎥ = 1760 yds ⎣ ft ⎦ ⎡ 12 in ⎤ ⎡ m ⎤ 5280 ft ⎢ ⎥⎢ ⎥ = 1609.35 m, 1.61 km ⎣ ft ⎦ ⎣ 39.37 in ⎦ 5280 ft, m ⎡ 39.37 in ⎤ ⎡ ft ⎤ ⎡ mi ⎤ ⎡ 60 s ⎤ ⎡ 60 ⎤ s ⎢⎣ m ⎥⎦ ⎢⎣12 in ⎥⎦ ⎢⎣ 5280 ft ⎥⎦ ⎢⎣1 ⎥⎦ ⎢⎣ h ⎥⎦ = 670,615,288.1 mph ≅ 670.62 × 106 mph 30 299,792,458 31 ⎡ ft ⎤ 100 yds ⎢ ⎥ ⎣1 yd ⎦ ⎡ mi ⎤ ⎢ 5,280 ft ⎥ = 0.0568 mi ⎣ ⎦ 60 mi ⎡ h ⎤ ⎡1 ⎤ = 0.0167 mi/s h ⎢⎣ 60 ⎥⎦ ⎢⎣ 60 s ⎥⎦ t= 32 33 34 d υ = 0.0568 mi = 3.40 s 0.0167 mi/s 30 mi ⎡ 5280 ft ⎤ ⎡12 in ⎤ ⎡ m ⎤ ⎡ h ⎤ ⎡ ⎤ = 13.41 m/s h ⎢⎣ mi ⎥⎦ ⎢⎣ ft ⎥⎦ ⎢⎣ 39.37 in ⎥⎦ ⎢⎣ 60 ⎥⎦ ⎢⎣ 60 s ⎥⎦ 50 yd ⎡ 60 ⎤ ⎡ ft ⎤ ⎡ mi ⎤ = 1.705 mi/h ⎢ ⎥ ⎢⎣ h ⎥⎦ ⎣1 yd ⎦ ⎢⎣ 5,280 ft ⎥⎦ 3000 mi d ⎡1 day ⎤ t= = = 1760 h ⎢ ⎥ = 73.33 days υ 1.705 mi/h ⎣ 24 h ⎦ ⎡1000 m ⎤ ⎡ 39.37 in ⎤ ⎡ ft ⎤ ⎡ mi ⎤ 10 km ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥ ⎣ km ⎦ ⎣ m ⎦ ⎣12 in ⎦ ⎣ 5280 ft ⎦ mi d 6.214 mi = 40.39 υ= ,t= = mi 6.5 υ 6.5 = 6.214 mi 35 ⎡ ft ⎤ ⎡12 in ⎤ 100 yds ⎢ ⎥⎢ ⎥ = 3600 in ⇒ 3600 quarters ⎣1 yd ⎦ ⎣ ft ⎦ 36 60 mph: 100 mi = 1.67 h = h:40.2 υ 60 mi/h d 100 mi 75 mph: t= = = 1.33 h = 1h:19.98 υ 75 mi/h difference = 20.22 minutes t= d = Chapter www.elsolucionario.net 37 cm ⎤ ⎡ [0.016 h ] ⎡⎢ 60 ⎤⎥ d = υt = ⎢600 ⎥ s ⎦ ⎣ 1h ⎦ ⎣ 38 ⎡ 14 ft ⎤ d = 86 stories ⎢ ⎥ ⎣ story ⎦ υ= 39 40 ⎡ 1m ⎤ ⎢100 cm ⎥ = 345.6 m ⎦ ⎣ ⎡ ⎤ ⎢1 step ⎥ ⎢ ⎥ = 1605 steps ⎢ ft ⎥ ⎣⎢ 12 ⎦⎥ d d 1605 steps ⎡ minute ⎤ ⇒t = = = 802.5 seconds ⎢ ⎥ = 13.38 minutes steps t υ ⎣ 60 seconds ⎦ second ⎡ 14 ft ⎤ ⎡ mile ⎤ d = (86 stories) ⎢ ⎥ = 1204 ft ⎢ ⎥ = 0.228 miles story ⎣ 5,280 ft ⎦ ⎣ ⎦ 10.7833 = = 47.30 min/mile mile 0.228 miles mile ⎡ 5,280 ft ⎤ 1056 ft , = ⇒ mile ⎢⎣ 1mile ⎥⎦ minute υ= 41 ⎡ 60 s ⎤ ⎢1 ⎥ ⎦ ⎣ 1204 ft d d = 1.14 minutes ⇒t = = υ 1056 ft t a ⎡ Btu ⎤ −3 J⎢ ⎥ = 4.74 × 10 Btu ⎣1054.35 J ⎦ b ⎡ gallon ⎤ 24 ounces ⎢ ⎥ ⎣128 ounces ⎦ c d ⎡ 14 ft ⎤ distance = 86 stories ⎢ ⎥ = 1204 ft ⎣ story ⎦ ⎡ ⎤ m3 −4 ⎢ ⎥ = 7.1 × 10 m ⎣ 264.172 gallons ⎦ ⎡ 86,400 s ⎤ 1.4 days ⎢ ⎥ = 1.21 × 10 s day ⎣ ⎦ ⎡ 264.172 gallons ⎤ ⎡ pints ⎤ m3 ⎢ ⎥ ⎢1 gallon ⎥ = 2113.38 pints m3 ⎣ ⎦⎣ ⎦ 42 6(4 + 8) = 72 43 (20 + 32)/4 = 13 (82 + 122 ) = 14.42 44 45 MODE = DEGREES: cos 50° = 0.64 46 MODE = DEGREES: tan−1(3/4) = 36.87° Chapter www.elsolucionario.net 20 τ = RC = (5 kΩ)(0.04 μF) = 0.2 ms (throughout) υC = E(1 − e−t/τ) = 20 V(1 − e−t/0.2 ms) (Starting at t = for each plot) a T= 1 = = ms f 500 Hz T = ms 5τ = ms = b T= T 1 = 10 ms = f 100 Hz T = ms 5τ = ms = c T= 1⎛T ⎞ ⎜ ⎟ 5⎝ ⎠ 1 = 0.2 ms = f Hz T = 0.1 ms ⎛T ⎞ 5τ = ms = 10 ⎜ ⎟ ⎝2⎠ 21 The mathematical expression for iC is the same for each frequency! τ = RC = (5 kΩ)(0.04 μF) = 0.2 ms 20 V −t / 0.2 ms = mAe−t/0.2 ms and iC = e kΩ a T= T = ms, = ms 500 Hz 5τ = 5(0.2 ms) = ms = b T= c T= T T = 10 ms, = ms 100 Hz 1⎛T ⎞ 5τ = ms = ⎜ ⎟ 5⎝ ⎠ T = 0.2 ms, = 0.1 ms 5000 Hz ⎛T ⎞ 5τ = ms = 10 ⎜ ⎟ ⎝2⎠ 338 CHAPTER 24 www.elsolucionario.net 22 τ = 0.2 ms as above T= = ms 500 Hz T 5τ = ms = T → : υC = 20 V(1 − e−t/0.2 ms) T → T: Vi = 20 V, Vf = −20 V υC = Vi + (Vf − Vi)(1 − e−t/τ) = 20 + (−20 − 20)(1 − e−t/0.2 ms) = 20 − 40(1 − e−t/0.2 ms) = 20 − 40 + 40e−t/0.2 ms υC = −20 V+ 40 Ve−t/0.2 ms T→ 23 T : Vi = −20 V, Vf = +20 V υC = Vi + (Vf − Vi)(1 − e−t/τ) = −20 + (20 − (−20))(1 − e−t/τ) = −20 + 40(1 − e−t/τ) = −20 + 40 − 40e−t/τ υC = 20 V − 40 Ve−t/0.2 ms υC = Vi + (Vf − Vi)(1 − e−t/RC) Vi = 20 V, Vf = 20 V υC = 20 + (20 − 20)(1 − e−t/RC) T = 20 V (for → ) For T → T, υi = V and υC = 20 Ve−t/τ τ = RC = 0.2 ms with T T = ms and 5τ = 2 T , υi = 20 V υC = 20 V(1 − e−t/τ) For T → For T → T, υ i = V υC = 20 Ve−t/τ CHAPTER 24 339 www.elsolucionario.net 24 25 τ = RC = 0.2 ms T 5τ = ms = Vi = −10 V, Vf = +20 V T 0→ : υC = Vi + (Vf − Vi)(1 − e−t/τ) = −10 + (20 − (−10))(1 − e−t/τ) = −10 + 30(1 − e−t/τ) = −10 + 30 − 30e−t/τ υC = +20 V − 30 Ve−t/0.2 ms T → T: Vi = 20 V, Vf = V υC = 20 Ve−t/0.2 ms 1 = = 5.31 MΩ 2π fC 2π (10 kHz)(3 pF) (9 M Ω ∠0°)(5.31 M Ω ∠ − 90°) = 4.573 MΩ ∠−59.5° Zp = M Ω − j 5.31 M Ω Zp: XC = Zs: CT = 18 pF + pF = 27 pF 1 XC = = = 0.589 MΩ 2π fC T 2π (10 kHz)(27 pF) Zs = Vscope = (1 M Ω ∠0°)(0.589 M Ω ∠ − 90°) = 0.507 MΩ ∠−59.5° M Ω − j 0.589 M Ω (0.507 M Ω ∠ − 59.5°)(100 V ∠0°) ZsVi = Z s + Z p (0.257 M Ω − j 0.437 M Ω) + (2.324 M Ω − j 3.939 M Ω) 50.7 × 106 V ∠ − 59.5° (100 V ∠0°) = 10 V ∠0° = 10 5.07 × 10 ∠ − 59.5° = θ Z p = −59.5° = θ Zs 26 Zp: XC = 1 = = 3.333 MΩ ωC (10 rad/s)(3 pF) (9 M Ω ∠0°)(3.333 M Ω) = 3.126 MΩ ∠−69.68° M Ω − j 3.333 M Ω 1 = XC = = 0.370 MΩ ωC (10 rad/s)(27 pF) (1 M Ω ∠0°)(0.370 M Ω ∠ − 90°) = 0.347 MΩ ∠−69.68° Zs = M Ω − j 0.370 M Ω ✓ θ Z p =θ Z s Zp = Zs: 340 CHAPTER 24 www.elsolucionario.net Vscope = (0.347 M Ω ∠ − 69.68°)(100 V ∠0°) ZsVi = Z s + Z p (0.121 M Ω − j 0.325 M Ω) + (1.086 M Ω − j 2.931 M Ω) 34.70 × 106 V ∠ − 69.68° 3.470 × 106 ∠ − 69.68° ≅ 10 V ∠0° = (100 V ∠0°) 10 = CHAPTER 24 341 www.elsolucionario.net Chapter 25 I: a no b no c yes d no e yes II: a yes b yes c yes d yes e no III: a yes b yes c no d yes e yes IV: a no b no c yes d yes e yes b i= c 2I m ⎛ 2 ⎞ ⎜1+ cos(2ωt − 90°) − cos(4ωt − 90°) + cos(6ωt − 90°) + … ⎟ π ⎝ 15 35 ⎠ ⎡ π⎤ ⎢1 − ⎥ π ⎣ ⎦ 2 ⎡ π ⎤ i = I m ⎢1 − + cos(2ωt − 90°) − cos(4ωt − 90°) + cos(6ωt − 90°) + …⎥ π ⎣ 15 35 ⎦ 2Im − Im 2Im = π d 2 ⎡ π ⎤ i = −2 I m ⎢1 − + cos (2ωt − 90°) − cos (4ωt − 90°) + cos (6ωt − 90°) + …⎥ π ⎣ 15 35 ⎦ a υ = −4 + sin α b υ = (sin α)2 342 CHAPTER 25 www.elsolucionario.net c i = − cos α a b CHAPTER 25 343 www.elsolucionario.net a b c 344 CHAPTER 25 www.elsolucionario.net a Vav = 100 V Veff = (50 V) + (25 V) = 107.53 V Iav = A b Ieff = (100 V) + (3 A) + (2 A) + (0.8 A) = 3.36 A a Veff = (20 V) + (15 V) + (10 V) = 19.04 V b Ieff = (6 A) + (2 A) + (1 A) = 4.53 A PT = V0I0 + V1I1 cos θ1 + … + VnIn cos θn (50 V)(2 A) (25 V)(0.8 A) cos 53° + cos 70° = (100 V)(3 A) + 2 = 300 + (50)(0.6018) + (10)(0.3420) = 333.52 W P= 10 a DC: E = 18 V, Io = b Ieff = (20 V)(6 A) (15 V)(2 A) (10 V)(1 A) cos 20° + cos 30° + cos 60° 2 = 60(0.9397) + 15(0.866) + 5(0.5) = 71.87 W E 18 V = 1.5 A = R 12 Ω ω = 400 rad/s: XL = ωL = (400 rad/s)(0.02 H) = Ω Z = 12 Ω + j8 Ω = 14.42 Ω ∠33.69° E 30 V/ ∠0° 2.08 A = ∠−33.69° I= = Z 14.42 Ω ∠33.69° ⎛ 2.08 ⎞ i = 1.5 + ⎜ ⎟ sin(400t − 33.69°) ⎝ ⎠ i = 1.5 + 2.08 sin(400t − 33.69°) (1.5 A) + (2.08 A) = 2.10 A CHAPTER 25 345 www.elsolucionario.net c ⎛ 2.08 A ⎞ DC: υR = E = 18 V, VR = ⎜ ∠ − 33.69° ⎟ (12 Ω ∠0°) ⎝ ⎠ 24.96 V = ∠ − 33.69° ⎛ 24.96 ⎞ υR = 18 + ⎜ ⎟ sin(400t − 33.69°) ⎝ ⎠ υR = 18 + 24.96 sin(400t − 33.69°) d V Reff = (18 V ) + e DC: VL = V (24.96 V ) = 25.21 V ⎛ 2.08 A ⎞ ω = 400 rad/s: VL = ⎜ ∠ − 33.69° ⎟ (8 Ω ∠90°) ⎝ ⎠ 16.64 A ∠56.31° = υL = + 16.64 sin(400t + 56.31°) 11 (16.64 V) = 11.77 V f V Leff = + g R = (2.101 A)2 12 Ω = 52.97 W P = I eff a DC: IDC = b Ieff = c υR = iR = i(12 Ω) = 24 + 24.96 sin(400t − 33.69°) + sin(800t − 53.13°) d Veff = 24 V =2A 12 Ω ω = 400 rad/s: Z = 12 Ω + j(400 rad/s)(0.02 H) = 12 Ω + j8 Ω = 14.422 Ω ∠33.69° 30 V ∠0° = 2.08 A ∠−33.69° (peak values) I= 14.422 Ω ∠33.69° ω = 800 rad/s: Z = 12 Ω + j(800 rad/s)(0.02 H) = 12 Ω + j16 Ω = 20 Ω ∠53.13° 10 V ∠0° I= = 0.5 A ∠−53.13° (peak values) 20 Ω ∠53.13° i = + 2.08 sin(400t − 33.69°) + 0.5 sin(800t − 53.13°) (2 A) + (2.08 A) + (0.5 A ) = 2.51 A (24 V) + (24.96 V) + (6 V) = 30.09 V 346 CHAPTER 25 www.elsolucionario.net 12 e DC: VL = V ω = 400 rad/s: VL = (2.08 A ∠−33.69°)(8 Ω ∠90°) = 16.64 V ∠56.31° ω = 800 rad/s: VL = (0.5 A ∠−53.13°)(16 Ω ∠90°) = V ∠36.87° υL = + 16.64 sin(400t + 56.31°) + sin(800t + 36.87°) f Veff = g PT = I eff R = (2.508 A)2 12 Ω = 75.48 W a (0) + (16.64 V) + (8 V) = 13.06 V 60 V = −5A 12 Ω ω = 300 rad/s: XL = ωL = (300 rad/s)(0.02 H) = Ω Z = 12 Ω + j16 Ω = 13.42 Ω ∠26.57° E = (0.707)(20 V) ∠0° = 14.14 V ∠0° E 14.14 V ∠0° I= = = 1.054 A ∠−26.57° Z 13.42 Ω ∠26.57° ω = 600 rad/s: XL = ωL = (600 rad/s)(0.02 H) = 12 Ω Z = 12 Ω + j12 Ω = 16.97 Ω ∠45° E = −(0.707)(10 V) ∠0° = −7.07 V ∠0° E 7.07 V ∠0° I= =− = −0.417 A ∠−45° Z 16.97 Ω ∠45° i = −5 + (1.414)(1.054)sin(300t − 26.57°) − (1.414)(0.417)sin(600t − 45°) i = −5 + 1.49 sin(300t − 26.57°) − 0.59 sin(600t − 45°) DC: I = − (10 A) + (1.49 A) + (0.59 A) = 10.06 A b Ieff = c DC: V = IR = (−5 A)(12 Ω) = −60 V ω = 300 rad/s: VR = (1.054 A ∠−26.57°)(12 Ω ∠0°) = 12.648 V ∠−26.57° ω = 600 rad/s: VR = (−0.417 A ∠−45°)(12 Ω ∠0°) = −5 V ∠−45° υR = −60 + (1.414)(12.648)sin(300t − 26.57°) − (1.414)(5)sin(600t − 45°) υR = −60 + 17.88 sin(300t − 26.57°) − 7.07 sin(600t − 45°) d V R eff = (60 V ) + (17.88 V ) + (7.07 V ) = 61.52 V CHAPTER 25 347 www.elsolucionario.net e 13 DC: VL = V VL = (1.054 A ∠−26.57°)(6 Ω ∠90°) = 6.324 V ∠63.43° ω = 300 rad/s: VL = (−0.417 A ∠−45°)(6 Ω ∠90°) = −2.502 V ∠45° ω = 600 rad/s: υL = + (1.414)(6.324)sin(300t + 63.43°) − (1.414)(2.502)sin(600t + 45°) υL = 8.94 sin(300t + 63.43°) − 3.54 sin(600t + 45°) f (8.94 V ) + (3.54 V ) = 6.8 V V Leff = g P = I eff R = (10.06 A)2 12 Ω = 1214.44 W a DC: I = A 1 = = 20 Ω ωC (400 rad/s)(125 μ F) Z = 15 Ω − j20 Ω = 25 Ω ∠−53.13° E = (0.707)(30 V) ∠0° = 21.21 V ∠0° E 21.21 V ∠0° I= = = 0.848 A ∠53.13° Z 25 Ω ∠ − 53.13° i = + (1.414)(0.848)sin(400t + 53.13°) i = 1.2 sin(400t + 53.13°) ω = 400 rad/s: XC = (1.2 A) = 0.85 A as above b Ieff = c DC: VR = V ω = 400 rad/s: VR = (0.848 A ∠53.13°)(15 Ω ∠0°) = 12.72 V ∠53.13° υR = + (1.414)(12.72)sin(400t + 53.13°) υR = 18 sin(400t + 53.13°) d (18 V ) = 12.73 V V R eff = e DC: VC = 18 V ω = 400 rad/s: VC = (0.848 A ∠53.13°)(20 Ω ∠−90°) = 16.96 V ∠−36.87° υC = 18 + (1.414)(16.96)sin(400t − 36.87°) υC = 18 + 23.98 sin(400t − 36.87°) f V C eff = (18 V ) + g R = (0.848 A)2 15 Ω = 10.79 W P = I eff (23.98 V ) = 24.73 V 348 CHAPTER 25 www.elsolucionario.net 14 a 400 400 cos 2ωt − cos 4ωt π 3π 15 π = 63.69 + 42.46 sin(2ωt + 90°) − 8.49 sin(4ωt + 90°) ω = 377 rad/s: e = 63.69 + 42.46 sin(754t + 90°) − 8.49 sin(1508t + 90°) DC: XL = ∴ VL = V 1 ω = 754 rad/s: XC = = = 1330 Ω ωC (754 rad/s)(1 μ F) e= 200 + XL = ωL = (754 rad/s)(0.1 H) = 75.4 Ω Z′ = (1 kΩ ∠0°) || 75.4 Ω ∠90° = 75.19 Ω ∠85.69° E = (0.707)(42.46 V) ∠90° = 30.02 V ∠90° Z′(E) (75.19 Ω ∠85.69°)(30.02 V ∠90°) Vo = = 1.799 V ∠−94.57° = Z′ + ZC 75.19 Ω ∠85.69° + 1330 Ω ∠ − 90° ω = 1508 rad/s: XC = 1 = = 6631.13 Ω ωC (1508 rad/s)(1 μ F) XL = ωL = (1508 rad/s)(0.1 H) = 150.8 Ω Z′ = (1 kΩ ∠0°) || 150.8 Ω ∠90° = 149.12 Ω ∠81.42° E = (0.707)(8.49 V) ∠90° = V ∠90° Z′(E) (149.12 Ω ∠81.42°)(6 V ∠90°) = Vo = ′ Z + ZC 149.12 Ω ∠81.42° + 6631.13 Ω ∠ − 90° = 1.73 V ∠−101.1° υo = + 1.414(1.799)sin(754t − 94.57°) − 1.414(1.73)sin(1508t − 101.1°) υo = 2.54 sin(754t − 94.57°) − 2.45 sin(1508t − 101.1°) 15 b Voeff = c P= (2.54 V ) + (2.45 V ) = 2.50 V ( V eff ) (2.50 V ) = 6.25 mW = R 1kΩ i = 0.318Im + 0.500 Im sin ωt − 0.212Im cos 2ωt − 0.0424Im cos 4ωt + … (Im = 10 mA) i = 3.18 × 10−3 + × 10−3 sin ωt − 2.12 × 10−3 sin(2ωt + 90°) − 0.424 × 10−3 sin(4ωt + 90°) + … i ≅ 3.18 × 10−3 + × 10−3 sin ωt − 2.12 × 10−3 sin(2ωt + 90°) DC: Io = A, Vo = V XL = ωL = (377 rad/s)(1.2 mH) = 0.452 Ω ω = 377 rad/s; 1 XC = = = 13.26 Ω ωC ( 377 rad/s ) (200 μ F) Z′ = 200 Ω − j13.26 Ω = 200.44 Ω ∠−3.79° I = (0.707)(5 × 10−3)A ∠0° = 3.54 mA ∠0° ZLI (0.452 Ω ∠90°)(3.54 mA ∠0°) = Io = = 7.98 μA ∠93.66° ′ j 0.452 Ω + 200 Ω − j13.26 Ω ZL + Z CHAPTER 25 349 www.elsolucionario.net Vo = (7.98 μA ∠93.66°)(200 Ω ∠0°) = 1.596 mV ∠93.66° ω = 754 rad/s: XL = ωL = (754 rad/s)(1.2 mH) = 0.905 Ω 1 XC = = = 6.63 Ω ωC (754 rad/s)(200 μ F) Z′ = 200 Ω − j6.63 Ω = 200.11 Ω ∠−1.9° I = (0.707)(2.12 mA) ∠90° = 1.5 mA ∠90° Z LI (0.905 Ω ∠90°)(1.5 mA ∠90°) Io = = 6.8 μA ∠181.64° = j 0.905 Ω + 200 Ω − j 6.63 Ω Z L + Z′ Vo = (6.8 μA ∠181.64°)(200 Ω ∠0°) = 1.36 mA ∠181.64° υo = + (1.414)(1.596 × 10−3)sin(377t + 93.66°) − (1.414)(1.360 × 10−3)sin(754t + 181.64°) −3 υo = 2.26 × 10 sin(377t + 93.66°) + 1.92 × 10−3 sin(754t + 1.64°) 16 17 a 60 + 70 sin ωt + 20 sin(2ωt + 90°) + 10 sin(3ωt + 60°) +20 + 30 sin ωt − 20 sin(2ωt + 90°) + sin(3ωt + 90°) DC: 60 + 20 = 80 ω: 70 + 30 = 100 ⇒ 100 sin ωt 2ω: 3ω: 10 ∠60° + 5∠90° = + j8.66 + j5 = + j13.66 = 14.55 ∠69.9° Sum = 80 + 100 sin ωt + 14.55 sin(3ωt + 69.9°) b 20 + 60 sin α + 10 sin(2α − 180°) + sin(3α + 180°) −5 + 10 sin α + − sin(3α − 30°) DC: 20 − = 15 α: 60 + 10 = 70 ⇒ 70 sin α 2α: 10 sin(2α − 180°) 3α: ∠180° − ∠−30° = −5 − [3.46 − j2] = −8.46 + j2 = 8.69 ∠166.7° Sum = 15 + 70 sin α + 10 sin(2α − 180°) + 8.69 sin(3α + 166.7°) iT = i1 + i2 = 10 + 30 sin 20t − 0.5 sin(40t+ 90°) +20 + sin(20t + 90°) + 0.5 sin(40t+ 30°) DC: 10 A + 20 A = 30 A ω = 20 rad/s: 30 A ∠0° + A ∠90° = 30 A + j4 A = 30.27 A ∠7.59° ω = 40 rad/s: −0.5 A ∠90° + 0.5 A ∠30° = −j0.5 A + 0.433 A + j0.25 A = 0.433 A − j0.25 A = 0.5 A ∠−30° iT = 30 + 30.27 sin(20t + 7.59°) + 0.5 sin(40t − 30°) 350 CHAPTER 25 www.elsolucionario.net 18 e = υ1 + υ2 = 20 − 200 sin 600t + 100 sin(1200t + 90°) + 75 sin 1800t −10 + 150 sin(600t + 30°) +0 + 50 sin(1800t + 60°) DC: 20 V − 10 V = 10 V ω: 600 rad/s: −200 V ∠0° + 150 V ∠30° = 102.66 V ∠133.07° ω = 1200 rad/s: 100 sin(1200t + 90°) ω = 1800 rad/s: 75 V ∠0° + 50 V ∠60° = 108.97 V ∠23.41° e = 10 + 102.66 sin(600t + 133.07°) + 100 sin(1200t + 90°) + 108.97 sin(1800t + 23.41°) CHAPTER 25 351 www.elsolucionario.net 352 CHAPTER 25 www.elsolucionario.net ... UNIVERISTARIOS Y SOLUCIONARIOS DE MUCHOS DE ESTOS LIBROS LOS SOLUCIONARIOS CONTIENEN TODOS LOS EJERCICIOS DEL LIBRO RESUELTOS Y EXPLICADOS DE FORMA CLARA VISITANOS PARA DESARGALOS GRATIS www.elsolucionario.net... ⎡ 264.172 gallons ⎤ ⎡ pints ⎤ m3 ⎢ ⎥ ⎢1 gallon ⎥ = 2113.38 pints m3 ⎣ ⎦⎣ ⎦ 42 6(4 + 8) = 72 43 (20 + 32)/4 = 13 (82 + 122 ) = 14.42 44 45 MODE = DEGREES: cos 50° = 0.64 46 MODE = DEGREES: tan−1(3/4)... resistance ≅ 105 Ω-cm 50°C specific resistance ≅ 500 Ω-cm 200°C specific resistance ≅ Ω-cm b negative c No d ρ= a Log scale: 10 fc ⇒ kΩ 100 fc ⇒ 0.4 kΩ b negative c no—log scales imply linearity