Bài giải phần giải mạch P9

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Bài giải phần giải mạch P9

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Chapter 9, Solution 1. (a) (b) angular frequency frequency f = ω = 103 rad/s ω = 159.2 Hz 2π (c) period T = 1 = 6.283 ms f (d) Since sin(A) = cos(A – 90°), vs = 12 sin(103t + 24°) = 12 cos(103t + 24° – 90°) vs in cosine form is vs = 12 cos(103t – 66°) V vs(2.5 ms) = 12 sin((10 3 )(2.5 × 10 -3 ) + 24°) = 12 sin(2.5 + 24°) = 12 sin(143.24° + 24°) = 2.65 V (e) Chapter 9, Solution 2. (a) (b) (c) (d) amplitude = 8 A ω = 500π = 1570.8 rad/s f = ω = 250 Hz 2π Is =...

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  1. Chapter 9, Solution 1. (a) angular frequency ω = 103 rad/s ω (b) frequency f = = 159.2 Hz 2π 1 (c) period T = = 6.283 ms f (d) Since sin(A) = cos(A – 90°), vs = 12 sin(103t + 24°) = 12 cos(103t + 24° – 90°) vs in cosine form is vs = 12 cos(103t – 66°) V (e) vs(2.5 ms) = 12 sin((10 3 )(2.5 × 10 -3 ) + 24°) = 12 sin(2.5 + 24°) = 12 sin(143.24° + 24°) = 2.65 V Chapter 9, Solution 2. (a) amplitude = 8 A (b) ω = 500π = 1570.8 rad/s ω (c) f = = 250 Hz 2π (d) Is = 8∠-25° A Is(2 ms) = 8 cos((500π )(2 × 10 -3 ) − 25°) = 8 cos(π − 25°) = 8 cos(155°) = -7.25 A Chapter 9, Solution 3. (a) 4 sin(ωt – 30°) = 4 cos(ωt – 30° – 90°) = 4 cos(ωt – 120°) (b) -2 sin(6t) = 2 cos(6t + 90°) (c) -10 sin(ωt + 20°) = 10 cos(ωt + 20° + 90°) = 10 cos(ωt + 110°)
  2. Chapter 9, Solution 4. (a) v = 8 cos(7t + 15°) = 8 sin(7t + 15° + 90°) = 8 sin(7t + 105°) (b) i = -10 sin(3t – 85°) = 10 cos(3t – 85° + 90°) = 10 cos(3t + 5°) Chapter 9, Solution 5. v1 = 20 sin(ωt + 60°) = 20 cos(ωt + 60° − 90°) = 20 cos(ωt − 30°) v2 = 60 cos(ωt − 10°) This indicates that the phase angle between the two signals is 20° and that v1 lags v2. Chapter 9, Solution 6. (a) v(t) = 10 cos(4t – 60°) i(t) = 4 sin(4t + 50°) = 4 cos(4t + 50° – 90°) = 4 cos(4t – 40°) Thus, i(t) leads v(t) by 20°. (b) v1(t) = 4 cos(377t + 10°) v2(t) = -20 cos(377t) = 20 cos(377t + 180°) Thus, v2(t) leads v1(t) by 170°. (c) x(t) = 13 cos(2t) + 5 sin(2t) = 13 cos(2t) + 5 cos(2t – 90°) X = 13∠0° + 5∠-90° = 13 – j5 = 13.928∠-21.04° x(t) = 13.928 cos(2t – 21.04°) y(t) = 15 cos(2t – 11.8°) phase difference = -11.8° + 21.04° = 9.24° Thus, y(t) leads x(t) by 9.24°. Chapter 9, Solution 7. If f(φ) = cosφ + j sinφ, df = -sinφ + j cos φ = j (cos φ + j sin φ) = j f (φ ) dφ df = j dφ f
  3. Integrating both sides ln f = jφ + ln A f = Aejφ = cosφ + j sinφ f(0) = A = 1 i.e. f(φ) = ejφ = cosφ + j sinφ Chapter 9, Solution 8. 15∠45° 15∠45° (a) + j2 = + j2 3 − j4 5∠ - 53.13° = 3∠98.13° + j2 = -0.4245 + j2.97 + j2 = -0.4243 + j4.97 (b) (2 + j)(3 – j4) = 6 – j8 + j3 + 4 = 10 – j5 = 11.18∠-26.57° 8∠ - 20° 10 8∠ - 20° (-5 − j12)(10) + = + (2 + j)(3 - j4) - 5 + j12 11.18∠ - 26.57° 25 + 144 = 0.7156∠6.57° − 0.2958 − j0.71 = 0.7109 + j0.08188 − 0.2958 − j0.71 = 0.4151 − j0.6281 (c) 10 + (8∠50°)(13∠-68.38°) = 10+104∠-17.38° = 109.25 – j31.07 Chapter 9, Solution 9. 3 + j4 (3 + j4)(5 + j8) (a) 2+ = 2+ 5 − j8 25 + 64 15 + j24 + j20 − 32 = 2+ 89 = 1.809 + j0.4944 1 − j2 2.236 ∠ - 63.43° (b) 4∠-10° + = 4∠-10° + 3∠6° 3∠6°
  4. = 4∠-10° + 0.7453∠-69.43° = 3.939 – j0.6946 + 0.2619 – j0.6978 = 4.201 – j1.392 8∠10° + 6 ∠ - 20° 7.879 + j1.3892 + 5.638 − j2.052 (c) = 9∠80° − 4∠50° 1.5628 + j8.863 − 2.571 − j3.064 13.517 − j0.6629 13.533∠ - 2.81° = = − 1.0083 + j5.799 5.886∠99.86° = 2.299∠-102.67° = -0.5043 – j2.243 Chapter 9, Solution 10. (a) z1 = 6 − j8, z 2 = 8.66 − j 5, and z 3 = −4 − j 6.9282 z1 + z 2 + z 3 = 10.66 − j19.93 z1 z 2 (b) = 9.999 + j 7.499 z3 Chapter 9, Solution 11. (a) z 1 z 2 = (-3 + j4)(12 + j5) = -36 – j15 + j48 – 20 = -56 + j33 z1 - 3 + j4 (-3 + j4)(12 + j5) (b) ∗ = = = -0.3314 + j0.1953 z2 12 − j5 144 + 25 (c) z 1 + z 2 = (-3 + j4) + (12 + j5) = 9 + j9 z 1 − z 2 = (-3 + j4) – (12 + j5) = -15 – j z1 + z 2 9 (1 + j) - 9 (1 + j)(15 - j) - 9 (16 + j14) = = = z1 − z 2 - (15 + j) 15 2 − 12 226 = -0.6372 – j0.5575
  5. Chapter 9, Solution 12. (a) z 1 z 2 = (-3 + j4)(12 + j5) = -36 – j15 + j48 – 20 = -56 + j33 z1 - 3 + j4 (-3 + j4)(12 + j5) (b) ∗ = = = -0.3314 + j0.1953 z2 12 − j5 144 + 25 (c) z 1 + z 2 = (-3 + j4) + (12 + j5) = 9 + j9 z 1 − z 2 = (-3 + j4) – (12 + j5) = -15 – j z1 + z 2 9 (1 + j) - 9 (1 + j)(15 - j) - 9 (16 + j14) = = = z1 − z 2 - (15 + j) 15 − 1 2 2 226 = -0.6372 – j0.5575 Chapter 9, Solution 13. (a) (−0.4324 + j 0.4054)+ (−0.8425 − j 0.2534) = − 1.2749 + j 0.1520 50∠ − 30 o (b) = − 2.0833 24∠150 o (c) (2+j3)(8-j5) –(-4) = 35 +j14 Chapter 9, Solution 14. 3 − j14 (a) = − 0.5751 + j 0.5116 − 15 + j11 (62.116 + j 231.82 + 138.56 − j80)(60 − j80) 24186 − 6944.9 (b) = = − 1.922 − j11.55 (67 + j84)(16.96 + j10.5983) 246.06 + j 2134.7 (c) (− 2 + j 4 ) 2 (260 − j120) = − 256.4 − j 200.89
  6. Chapter 9, Solution 15. 10 + j6 2 − j3 (a) = -10 – j6 + j10 – 6 + 10 – j15 -5 -1 + j = -6 – j11 20∠ − 30° - 4∠ - 10° (b) = 60∠15° + 64∠-10° 16∠0° 3∠45° = 57.96 + j15.529 + 63.03 – j11.114 = 120.99 – j4.415 1− j − j 0 j 1 −j (c) 1 j 1+ j = 1 + 1 + 0 − 1 − 0 + j2 (1 − j) + j2 (1 + j) 1− j − j 0 j 1 −j = 1 − 1 (1 − j + 1 + j) = 1 – 2 = -1 Chapter 9, Solution 16. (a) -10 cos(4t + 75°) = 10 cos(4t + 75° − 180°) = 10 cos(4t − 105°) The phasor form is 10∠-105° (b) 5 sin(20t – 10°) = 5 cos(20t – 10° – 90°) = 5 cos(20t – 100°) The phasor form is 5∠-100° (c) 4 cos(2t) + 3 sin(2t) = 4 cos(2t) + 3 cos(2t – 90°) The phasor form is 4∠0° + 3∠-90° = 4 – j3 = 5∠-36.87° Chapter 9, Solution 17. (a) Let A = 8∠-30° + 6∠0° = 12.928 – j4 = 13.533∠-17.19° a(t) = 13.533 cos(5t + 342.81°)
  7. (b) We know that -sinα = cos(α + 90°). Let B = 20∠45° + 30∠(20° + 90°) = 14.142 + j14.142 – 10.261 + j28.19 = 3.881 + j42.33 = 42.51∠84.76° b(t) = 42.51 cos(120πt + 84.76°) (c) Let C = 4∠-90° + 3∠(-10° – 90°) = -j4 – 0.5209 – j2.954 = 6.974∠265.72° c(t) = 6.974 cos(8t + 265.72°) Chapter 9, Solution 18. (a) v1 ( t ) = 60 cos(t + 15°) (b) V2 = 6 + j8 = 10∠53.13° v 2 ( t ) = 10 cos(40t + 53.13°) (c) i1 ( t ) = 2.8 cos(377t – π/3) (d) I 2 = -0.5 – j1.2 = 1.3∠247.4° i 2 ( t ) = 1.3 cos(103t + 247.4°) Chapter 9, Solution 19. (a) 3∠10° − 5∠-30° = 2.954 + j0.5209 – 4.33 + j2.5 = -1.376 + j3.021 = 3.32∠114.49° Therefore, 3 cos(20t + 10°) – 5 cos(20t – 30°) = 3.32 cos(20t + 114.49°) (b) 4∠-90° + 3∠-45° = -j40 + 21.21 – j21.21 = 21.21 – j61.21 = 64.78∠-70.89° Therefore, 40 sin(50t) + 30 cos(50t – 45°) = 64.78 cos(50t – 70.89°) (c) Using sinα = cos(α − 90°), 20∠-90° + 10∠60° − 5∠-110° = -j20 + 5 + j8.66 + 1.7101 + j4.699 = 6.7101 – j6.641 = 9.44∠-44.7° Therefore, 20 sin(400t) + 10 cos(400t + 60°) – 5 sin(400t – 20°) = 9.44 cos(400t – 44.7°)
  8. Chapter 9, Solution 20. (a) V = 4∠− 60 o − 90 o − 5∠40 o = −3.464 − j 2 − 3.83 − j 3.2139 = 8.966∠ − 4.399 o Hence, v = 8.966 cos(377t − 4.399 o ) (b) I = 10∠0 o + jω 8∠20 o − 90 o , ω = 5 , i.e. I = 10 + 40∠20 o = 49.51∠16.04 o i = 49.51 cos(5t + 16.04 o ) Chapter 9, Solution 21. (a) F = 5∠15 o − 4∠− 30 o − 90 o = 6.8296 + j 4.758 = 8.3236∠34.86 o f (t ) = 8.324 cos(30t + 34.86 o ) (b) G = 8∠ − 90 o + 4∠50 o = 2.571 − j 4.9358 = 5.565∠ − 62.49 o g (t ) = 5.565 cos(t − 62.49 o ) (c) H = 1 jω (10∠0 o + 5∠ − 90 o ), ω = 40 i.e. H = 0.25∠ − 90 o + 0.125∠ − 180 o = − j 0.25 − 0.125 = 0.2795∠ − 116.6 o h(t ) = 0.2795 cos(40t − 116.6 o ) Chapter 9, Solution 22. t dv Let f(t) = 10v(t ) + 4 − 2 ∫ v(t )dt dt −∞ 2V F = 10V + jω 4V − , ω = 5, V = 20∠ − 30 o jω F = 10V + j 20V − j 0.4V = (10 − j19.6)(17.32 − j10) = 440.1∠ − 92.97 o f (t ) = 440.1 cos(5t − 92.97 o )
  9. Chapter 9, Solution 23. (a) v(t) = 40 cos(ωt – 60°) (b) V = -30∠10° + 50∠60° = -4.54 + j38.09 = 38.36∠96.8° v(t) = 38.36 cos(ωt + 96.8°) (c) I = j6∠-10° = 6∠(90° − 10°) = 6∠80° i(t) = 6 cos(ωt + 80°) 2 (d) I = + 10∠-45° = -j2 + 7.071 – j7.071 j = 11.5∠-52.06° i(t) = 11.5 cos(ωt – 52.06°) Chapter 9, Solution 24. (a) V V+ = 10∠0°, ω = 1 jω V (1 − j) = 10 10 V= = 5 + j5 = 7.071∠45° 1− j Therefore, v(t) = 7.071 cos(t + 45°) (b) 4V jωV + 5V + = 20∠(10° − 90°), ω = 4 jω  4 V  j4 + 5 +  = 20 ∠ - 80°  j4  20∠ - 80° V= = 3.43∠ - 110.96° 5 + j3 Therefore, v(t) = 3.43 cos(4t – 110.96°)
  10. Chapter 9, Solution 25. (a) 2jωI + 3I = 4∠ - 45°, ω = 2 I (3 + j4) = 4∠ - 45° 4∠ - 45° 4∠ - 45° I= = = 0.8∠ - 98.13° 3 + j4 5∠53.13° Therefore, i(t) = 0.8 cos(2t – 98.13°) (b) I 10 + jωI + 6I = 5∠22°, ω = 5 jω (- j2 + j5 + 6) I = 5∠22° 5∠22° 5∠22° I= = = 0.745∠ - 4.56° 6 + j3 6.708∠26.56° Therefore, i(t) = 0.745 cos(5t – 4.56°) Chapter 9, Solution 26. I jωI + 2I + = 1∠0°, ω = 2 jω  1 I  j2 + 2 +  = 1  j2  1 I= = 0.4∠ - 36.87° 2 + j1.5 Therefore, i(t) = 0.4 cos(2t – 36.87°) Chapter 9, Solution 27. V jωV + 50V + 100 = 110∠ - 10°, ω = 377 jω  j100  V  j377 + 50 −  = 110∠ - 10°  377  V (380.6∠82.45°) = 110∠ - 10° V = 0.289 ∠ - 92.45° Therefore, v(t) = 0.289 cos(377t – 92.45°).
  11. Chapter 9, Solution 28. v s ( t ) 110 cos(377 t ) i( t ) = = = 13.75 cos(377t) A. R 8 Chapter 9, Solution 29. 1 1 Z= = = - j 0.5 jωC j (10 )(2 × 10 -6 ) 6 V = IZ = (4∠25°)(0.5∠ - 90°) = 2 ∠ - 65° Therefore v(t) = 2 sin(106t – 65°) V. Chapter 9, Solution 30. Z = jωL = j (500)(4 × 10 -3 ) = j2 V 60 ∠ - 65° I= = = 30∠ - 155° Z 2∠90° Therefore, i(t) = 30 cos(500t – 155°) A. Chapter 9, Solution 31. i(t) = 10 sin(ωt + 30°) = 10 cos(ωt + 30° − 90°) = 10 cos(ωt − 60°) Thus, I = 10∠-60° v(t) = -65 cos(ωt + 120°) = 65 cos(ωt + 120° − 180°) = 65 cos(ωt − 60°) Thus, V = 65∠-60° V 65∠ - 60° Z= = = 6.5 Ω I 10∠ - 60° Since V and I are in phase, the element is a resistor with R = 6.5 Ω.
  12. Chapter 9, Solution 32. V = 180∠10°, I = 12∠-30°, ω = 2 V 180∠10° Z= = = 15∠40° = 11.49 + j 9.642 Ω I 12∠ - 30° One element is a resistor with R = 11.49 Ω. The other element is an inductor with ωL = 9.642 or L = 4.821 H. Chapter 9, Solution 33. 110 = v 2 + v 2 R L v L = 110 2 − v 2 R v L = 110 2 − 85 2 = 69.82 V Chapter 9, Solution 34. 1 1 v o = 0 if ωL =  → ω = ωC LC 1 ω= = 100 rad/s (5 × 10 −3 )(2 × 10 − 3 ) Chapter 9, Solution 35. Vs = 5∠0° jωL = j (2)(1) = j2 1 1 = = - j2 jωC j (2)(0.25) j2 j2 Vo = Vs = 5∠0° = (1∠90°)(5∠0°) = 5∠90° 2 − j2 + j2 2 Thus, v o ( t ) = 5 cos(2t + 90°) = -5 sin(2t) V
  13. Chapter 9, Solution 36. Let Z be the input impedance at the source. 100 mH  → jωL = j 200 x100 x10 −3 = j 20 1 1 10 µF  → = = − j 500 jωC j10 x10 −6 x 200 1000//-j500 = 200 –j400 1000//(j20 + 200 –j400) = 242.62 –j239.84 Z = 2242.62 − j 239.84 = 2255∠ − 6.104 o 60∠ − 10 o I= = 26.61∠ − 3.896 o mA 2255∠ − 6.104 o i = 266.1 cos(200t − 3.896 o ) Chapter 9, Solution 37. jωL = j (5)(1) = j5 1 1 = = -j jωC j (5)(0.2) (2)( j5) j10 Let Z1 = - j , Z 2 = 2 || j5 = = 2 + j5 2 + j5 Z2 Then, Ix = I , where I s = 2∠0° Z1 + Z 2 s j10 2 + j5 j20 Ix = (2) = = 2.12 ∠32° j10 5 + j8 - j+ 2 + j5 Therefore, i x ( t ) = 2.12 sin(5t + 32°) A
  14. Chapter 9, Solution 38. 1 1 1 (a) F  → = = - j2 6 jωC j (3)(1 / 6) - j2 I= (10 ∠45°) = 4.472∠ - 18.43° 4 − j2 Hence, i(t) = 4.472 cos(3t – 18.43°) A V = 4I = (4)(4.472∠ - 18.43°) = 17.89∠ - 18.43° Hence, v(t) = 17.89 cos(3t – 18.43°) V 1 1 1 (b) F  → = = - j3 12 jωC j (4)(1 / 12) 3H  → jωL = j (4)(3) = j12 V 50∠0° I= = = 10∠36.87° Z 4 − j3 Hence, i(t) = 10 cos(4t + 36.87°) A j12 V= (50∠0°) = 41.6 ∠33.69° 8 + j12 Hence, v(t) = 41.6 cos(4t + 33.69°) V Chapter 9, Solution 39. ( j5)(- j10) Z = 8 + j5 || (- j10) = 8 + = 8 + j10 j5 − j10 V 40 ∠0° 20 I= = = = 3.124∠ - 51.34° Z 8 + j10 6.403∠51.34° - j10 I1 = I = 2 I = 6.248∠ - 51.34° j5 − j10 j5 I2 = I = - I = 3.124∠128.66° - j5 Therefore, i1 ( t ) = 6.248 cos(120πt – 51.34°) A i 2 ( t ) = 3.124 cos(120πt + 128.66°) A
  15. Chapter 9, Solution 40. (a) For ω = 1 , 1H  → jωL = j (1)(1) = j 1 1 0.05 F  → = = - j20 jωC j (1)(0.05) - j40 Z = j + 2 || (- j20) = j + = 1.98 + j0.802 2 − j20 V 4 ∠0° 4∠0° Io = = = = 1.872 ∠ - 22.05° Z 1.98 + j0.802 2.136∠22.05° Hence, i o ( t ) = 1.872 cos(t – 22.05°) A (b) For ω = 5 , 1H  → jωL = j (5)(1) = j5 1 1 0.05 F  → = = - j4 jωC j (5)(0.05) - j4 Z = j5 + 2 || (- j4) = j5 + = 1.6 + j4.2 1 − j2 V 4∠0° 4∠0° Io = = = = 0.89∠ - 69.14° Z 1.6 + j4 4.494∠69.14° Hence, i o ( t ) = 0.89 cos(5t – 69.14°) A (c) For ω = 10 , 1H  → jωL = j (10)(1) = j10 1 1 0.05 F  → = = - j2 jωC j (10)(0.05) - j4 Z = j10 + 2 || (- j2) = j10 + = 1 + j9 2 − j2 V 4∠0° 4 ∠0° Io = = = = 0.4417 ∠ - 83.66° Z 1 + j9 9.055∠83.66° Hence, i o ( t ) = 0.4417 cos(10t – 83.66°) A
  16. Chapter 9, Solution 41. ω = 1, 1H  → jωL = j (1)(1) = j 1 1 1F  → = = -j jωC j (1)(1) - j+1 Z = 1 + (1 + j) || (- j) = 1 + = 2− j 1 Vs 10 I= = , I c = (1 + j) I Z 2− j (1 − j)(10) V = (- j)(1 + j) I = (1 − j) I = = 6.325∠ - 18.43° 2− j Thus, v(t) = 6.325 cos(t – 18.43°) V Chapter 9, Solution 42. ω = 200 1 1 50 µF  → = = - j100 jωC j (200)(50 × 10 -6 ) 0.1 H  → jωL = j (200)(0.1) = j20 (50)(-j100) - j100 50 || -j100 = = = 40 − j20 50 − j100 1 - j2 j20 j20 Vo = (60∠0°) = (60∠0°) = 17.14 ∠90° j20 + 30 + 40 − j20 70 Thus, v o ( t ) = 17.14 sin(200t + 90°) V or v o ( t ) = 17.14 cos(200t) V
  17. Chapter 9, Solution 43. ω= 2 1H  → jωL = j (2)(1) = j2 1 1 1F  → = = - j0.5 jωC j (2)(1) j2 − j0.5 j1.5 Io = I= 4∠0° = 3.328∠33.69° j2 − j0.5 + 1 1 + j1.5 Thus, i o ( t ) = 3.328 cos(2t + 33.69°) A Chapter 9, Solution 44. ω = 200 10 mH → jωL = j (200)(10 × 10 -3 ) = j2 1 1 5 mF  → = = -j jωC j (200)(5 × 10 -3 ) 1 1 1 3+ j Y= + + = 0.25 − j0.5 + = 0.55 − j0.4 4 j2 3 − j 10 1 1 Z= = = 1.1892 + j0.865 Y 0.55 − j0.4 6∠0° 6∠0° I= = = 0.96 ∠ - 7.956° 5 + Z 6.1892 + j0.865 Thus, i(t) = 0.96 cos(200t – 7.956°) A
  18. Chapter 9, Solution 45. We obtain I o by applying the principle of current division twice. I I2 I2 Io Z1 Z2 -j2 Ω 2Ω (a) (b) - j4 Z 1 = - j2 , Z 2 = j4 + (-j2) || 2 = j4 + = 1 + j3 2 - j2 Z1 - j2 - j10 I2 = I= (5∠0°) = Z1 + Z 2 - j2 + 1 + j3 1+ j - j2  - j  - j10  - 10 Io = I2 =   = = -5 A 2 - j2 1 - j  1 + j  1 + 1 Chapter 9, Solution 46. i s = 5 cos(10 t + 40°)  → I s = 5∠40° 1 1 0.1 F  → = = -j jωC j (10)(0.1) 0.2 H  → jωL = j (10)(0.2) = j2 j8 Let Z1 = 4 || j2 = = 0.8 + j1.6 , Z2 = 3 − j 4 + j2 Z1 0.8 + j1.6 Io = Is = (5∠40°) Z1 + Z 2 3.8 + j0.6 (1.789∠63.43°)(5∠40°) Io = = 2.325∠94.46° 3.847 ∠8.97° Thus, i o ( t ) = 2.325 cos(10t + 94.46°) A
  19. Chapter 9, Solution 47. First, we convert the circuit into the frequency domain. Ix 2Ω j4 + 5∠0˚ -j10 20 Ω − 5 5 5 Ix = = = = 0.4607∠52.63° − j10(20 + j4) 2 + 4.588 − j8.626 10.854∠ − 52.63° 2+ − j10 + 20 + j4 is(t) = 0.4607cos(2000t +52.63˚) A Chapter 9, Solution 48. Converting the circuit to the frequency domain, we get: 10 Ω V1 30 Ω Ix + -j20 20∠-40˚ − j20 We can solve this using nodal analysis.
  20. V1 − 20∠ − 40° V1 − 0 V −0 + + 1 =0 10 j20 30 − j20 V1(0.1 − j0.05 + 0.02307 + j0.01538) = 2∠ − 40° 2∠40° V1 = = 15.643∠ − 24.29° 0.12307 − j0.03462 15.643∠ − 24.29° Ix = = 0.4338∠9.4° 30 − j20 ix = 0.4338 sin(100 t + 9.4°) A Chapter 9, Solution 49. ( j2)(1 − j) Z T = 2 + j2 || (1 − j) = 2 + =4 1+ j I Ix 1Ω j2 Ω -j Ω j2 j2 1 Ix = I= I, where I x = 0.5∠0° = j2 + 1 − j 1+ j 2 1+ j 1+ j I= Ix = j2 j4 1+ j 1+ j Vs = I Z T = (4) = = 1 − j = 1.414∠ - 45° j4 j v s ( t ) = 1.414 sin(200t – 45°) V Chapter 9, Solution 50. Since ω = 100, the inductor = j100x0.1 = j10 Ω and the capacitor = 1/(j100x10-3) = -j10Ω. j10 Ix + 5∠40˚ -j10 20 Ω vx −
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