Bài giải phần giải mạch P10

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Bài giải phần giải mạch P10

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Chapter 10, Solution 1. ω=1 10 cos( t − 45°)  → 10∠ - 45° 5 sin( t + 30°)  → 5∠ - 60° 1H  → 1F  → jωL = j 1 = -j jωC The circuit becomes as shown below. 3Ω Vo jΩ 10∠-45° V + − 2 Io + − 5∠-60° V Applying nodal analysis, (10∠ - 45°) − Vo (5∠ - 60°) − Vo Vo + = 3 j -j j10∠ - 45° + 15∠ - 60° = j Vo Vo = 10 ∠ - 45° + 15∠ - 150° = 15.73∠247.9° Therefore, v o ( t ) = 15.73 cos(t + 247.9°) V Chapter 10, Solution...

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  1. Chapter 10, Solution 1. ω=1 10 cos( t − 45°)  → 10∠ - 45° 5 sin( t + 30°)  → 5∠ - 60° 1H  → jωL = j 1 1F  → = -j jωC The circuit becomes as shown below. 3Ω Vo jΩ + + 10∠-45° V 2 Io 5∠-60° V − − Applying nodal analysis, (10∠ - 45°) − Vo (5∠ - 60°) − Vo Vo + = 3 j -j j10∠ - 45° + 15∠ - 60° = j Vo Vo = 10 ∠ - 45° + 15∠ - 150° = 15.73∠247.9° Therefore, v o ( t ) = 15.73 cos(t + 247.9°) V Chapter 10, Solution 2. ω = 10 4 cos(10t − π 4)  → 4∠ - 45° 20 sin(10 t + π 3)  → 20 ∠ - 150° 1H  → jωL = j10 1 1 0.02 F  → = = - j5 jωC j 0.2
  2. The circuit becomes that shown below. 10 Ω Vo Io + 20∠-150° V j10 Ω -j5 Ω 4∠-45° A − Applying nodal analysis, (20∠ - 150°) − Vo Vo Vo + 4∠ - 45° = + 10 j10 - j5 20 ∠ - 150° + 4∠ - 45° = 0.1(1 + j) Vo Vo 2 ∠ - 150° + 4 ∠ - 45° Io = = = 2.816 ∠150.98° j10 j (1 + j) Therefore, i o ( t ) = 2.816 cos(10t + 150.98°) A Chapter 10, Solution 3. ω= 4 2 cos(4t )  → 2∠0° 16 sin(4 t )  → 16∠ - 90° = -j16 2H  → jωL = j8 1 1 1 12 F  → = = - j3 jωC j (4)(1 12) The circuit is shown below. 4Ω -j3 Ω Vo j8 Ω 6Ω + -j16 V 1Ω 2∠0° A −
  3. Applying nodal analysis, - j16 − Vo Vo Vo +2= + 4 − j3 1 6 + j8 - j16  1 1  + 2 = 1 + + V 4 − j3  4 − j3 6 + j8  o 3.92 − j2.56 4.682∠ - 33.15° Vo = = = 3.835∠ - 35.02° 1.22 + j0.04 1.2207 ∠1.88° Therefore, v o ( t ) = 3.835 cos(4t – 35.02°) V Chapter 10, Solution 4. 16 sin(4 t − 10°)  → 16∠ - 10°, ω = 4 1H  → jωL = j4 1 1 0.25 F  → = = -j jωC j (4)(1 4) Ix j4 Ω V1 -j Ω + + 16∠-10° V 0.5 Ix 1Ω Vo − − (16∠ - 10°) − V1 1 V + Ix = 1 j4 2 1− j But (16∠ - 10°) − V1 Ix = j4 3 ((16∠ - 10°) − V1 ) V So, = 1 j8 1− j
  4. 48∠ - 10° V1 = - 1 + j4 Using voltage division, 1 48∠ - 10° Vo = V1 = = 8.232∠ - 69.04° 1− j (1 - j)(-1 + j4) Therefore, v o ( t ) = 8.232 sin(4t – 69.04°) V Chapter 10, Solution 5. Let the voltage across the capacitor and the inductor be Vx and we get: Vx − 0.5I x − 10∠30° Vx Vx + + =0 4 − j2 j3 Vx (3 + j6 − j4)Vx − 1.5I x = 30∠30° but I x = = j0.5Vx − j2 Combining these equations we get: 30∠30° (3 + j2 − j0.75)Vx = 30∠30° or Vx = 3 + j1.25 30∠30° I x = j0.5 = 4.615∠97.38° A 3 + j1.25 Chapter 10, Solution 6. Let Vo be the voltage across the current source. Using nodal analysis we get: Vo − 4Vx Vo 20 −3+ = 0 where Vx = Vo 20 20 + j10 20 + j10 Combining these we get: Vo 4Vo Vo − −3+ = 0 → (1 + j0.5 − 3)Vo = 60 + j30 20 20 + j10 20 + j10 60 + j30 20(3) Vo = or Vx = = 29.11∠–166˚ V. − 2 + j0.5 − 2 + j0.5
  5. Chapter 10, Solution 7. At the main node, 120∠ − 15 o − V V V 115.91 − j31.058 = 6∠30 o + +  → − 5.196 − j3 = 40 + j20 − j30 50 40 + j20  1 j 1 V  40 + j20 + 30 + 50     − 3.1885 − j4.7805 V= = 124.08∠ − 154 o V 0.04 + j0.0233 Chapter 10, Solution 8. ω = 200, 100mH  → jωL = j200x 0.1 = j20 1 1 50µF  → = = − j100 jωC j200x 50x10 − 6 The frequency-domain version of the circuit is shown below. 0.1 Vo 40 Ω V1 Io V2 + -j100 Ω 6∠15 o 20 Ω Vo j20 Ω - At node 1, V V1 V − V2 6∠15 o + 0.1V1 = 1 + + 1 20 − j100 40 or 5.7955 + j1.5529 = (−0.025 + j 0.01)V1 − 0.025V2 (1)
  6. At node 2, V1 − V2 V = 0.1V1 + 2 → 0 = 3V1 + (1 − j2)V2 (2) 40 j20 From (1) and (2), (−0.025 + j0.01) − 0.025 V1   (5.7955 + j1.5529)    =  or AV = B  3 (1 − j2)  V2      0   Using MATLAB, V = inv(A)*B leads to V1 = −70.63 − j127.23, V2 = −110.3 + j161.09 V − V2 Io = 1 = 7.276∠ − 82.17 o 40 Thus, i o ( t ) = 7.276 cos(200 t − 82.17 o ) A Chapter 10, Solution 9. 10 cos(10 3 t )  → 10 ∠0°, ω = 10 3 10 mH  → jωL = j10 1 1 50 µF  → = = - j20 jωC j (10 )(50 × 10 -6 ) 3 Consider the circuit shown below. 20 Ω V1 -j20 Ω V2 j10 Ω Io + + 10∠0° V 20 Ω 4 Io 30 Ω Vo − − At node 1, 10 − V1 V1 V1 − V2 = + 20 20 - j20 10 = (2 + j) V1 − jV2 (1)
  7. At node 2, V1 − V2 V V2 V = (4) 1 + , where I o = 1 has been substituted. - j20 20 30 + j10 20 (-4 + j) V1 = (0.6 + j0.8) V2 0.6 + j0.8 V1 = V2 (2) -4+ j Substituting (2) into (1) (2 + j)(0.6 + j0.8) 10 = V2 − jV2 -4+ j 170 or V2 = 0.6 − j26.2 30 3 170 Vo = V2 = ⋅ = 6.154 ∠70.26° 30 + j10 3 + j 0.6 − j26.2 Therefore, v o ( t ) = 6.154 cos(103 t + 70.26°) V Chapter 10, Solution 10. 50 mH  → jωL = j2000x50 x10 − 3 = j100, ω = 2000 1 1 2µF  → = = − j250 jωC j2000 x 2x10 − 6 Consider the frequency-domain equivalent circuit below. V1 -j250 V2 36
  8. At node 1, V1 V V − V2 36 = + 1 + 1  → 36 = (0.0005 − j0.006)V1 − j0.004V2 (1) 2000 j100 − j250 At node 2, V1 − V2 V = 0.1V1 + 2 → 0 = (0.1 − j0.004)V1 + (0.00025 + j0.004)V2 (2) − j250 4000 Solving (1) and (2) gives Vo = V2 = −535.6 + j893.5 = 8951.1∠93.43o vo (t) = 8.951 sin(2000t +93.43o) kV Chapter 10, Solution 11. cos(2t )  → 1∠0°, ω = 2 8 sin( 2t + 30°)  → 8∠ - 60° 1 1 1H  → jωL = j2 12F  → = = -j jωC j (2)(1 2) 1 1 2H  → jωL = j4 14F  → = = - j2 jωC j (2)(1 4) Consider the circuit below. 2 Io 2 Io 2 -j Ω 2 Io 2 2 2 Io 2 2 Io I -j Ω
  9. At node 1, (8∠ - 60°) − V1 V1 V1 − V2 = + 2 -j j2 8∠ - 60° = (1 + j) V1 + j V2 (1) At node 2, V1 − V2 (8∠ - 60°) − V2 1+ + =0 j2 j4 − j2 V2 = 4 ∠ - 60° + j + 0.5 V1 (2) Substituting (2) into (1), 1 + 8∠ - 60° − 4 ∠30° = (1 + j1.5) V1 1 + 8∠ - 60° − 4∠30° V1 = 1 + j1.5 V1 1 + 8∠ - 60° − 4 ∠30° Io = = = 5.024∠ - 46.55° -j 1.5 − j Therefore, i o ( t ) = 5.024 cos(2t – 46.55°) Chapter 10, Solution 12. 20 sin(1000t )  → 20 ∠0°, ω = 1000 10 mH  → jωL = j10 1 1 50 µF  → = = - j20 jωC j (10 )(50 × 10 -6 ) 3 The frequency-domain equivalent circuit is shown below. 2 Io V1 10 Ω V2 Io 20∠0° A 20 Ω -j20 Ω j10 Ω
  10. At node 1, V1 V1 − V2 20 = 2 I o + + , 20 10 where V2 Io = j10 2V2 V1 V1 − V2 20 = + + j10 20 10 400 = 3V1 − (2 + j4) V2 (1) At node 2, 2V2 V1 − V2 V V + = 2 + 2 j10 10 - j20 j10 j2 V1 = (-3 + j2) V2 or V1 = (1 + j1.5) V2 (2) Substituting (2) into (1), 400 = (3 + j4.5) V2 − (2 + j4) V2 = (1 + j0.5) V2 400 V2 = 1 + j0.5 V2 40 Io = = = 35.74 ∠ - 116.6° j10 j (1 + j0.5) Therefore, i o ( t ) = 35.74 sin(1000t – 116.6°) A Chapter 10, Solution 13. Nodal analysis is the best approach to use on this problem. We can make our work easier by doing a source transformation on the right hand side of the circuit. –j2 Ω 18 Ω j6 Ω + + + 40∠30º V − Vx 3Ω 50∠0º V − −
  11. Vx − 40∠30° Vx Vx − 50 + + =0 − j2 3 18 + j6 which leads to Vx = 29.36∠62.88˚ A. Chapter 10, Solution 14. At node 1, 0 − V1 0 − V1 V2 − V1 + + = 20∠30° - j2 10 j4 - (1 + j2.5) V1 − j2.5 V2 = 173.2 + j100 (1) At node 2, V2 V2 V2 − V1 + + = 20∠30° j2 - j5 j4 - j5.5 V2 + j2.5 V1 = 173.2 + j100 (2) Equations (1) and (2) can be cast into matrix form as 1 + j2.5 j2.5  V1   - 200 ∠30° =  j2.5  - j5.5 V2   200 ∠30°      1 + j2.5 j2.5 ∆= = 20 − j5.5 = 20.74∠ - 15.38° j2.5 - j5.5 - 200 ∠30° j2.5 ∆1 = = j3 (200∠30°) = 600∠120° 200 ∠30° - j5.5 1 + j2.5 - 200∠30° ∆2 = = (200 ∠30°)(1 + j5) = 1020∠108.7° j2.5 200∠30° ∆1 V1 = = 28.93∠135.38° ∆ ∆2 V2 = = 49.18∠124.08° ∆
  12. Chapter 10, Solution 15. We apply nodal analysis to the circuit shown below. 5A 2Ω V1 jΩ V2 I + -j20 V -j2 Ω 2I 4Ω − At node 1, - j20 − V1 V V − V2 = 5+ 1 + 1 2 - j2 j - 5 − j10 = (0.5 − j0.5) V1 + j V2 (1) At node 2, V1 − V2 V2 5 + 2I + = , j 4 V1 where I = - j2 5 V2 = V1 0.25 − j (2) Substituting (2) into (1), j5 - 5 − j10 − = 0.5 (1 − j) V1 0.25 − j j40 (1 − j) V1 = -10 − j20 − 1 − j4 160 j40 ( 2 ∠ - 45°) V1 = -10 − j20 + − 17 17 V1 = 15.81∠313.5°
  13. V1 I= = (0.5∠90°)(15.81∠313.5°) - j2 I = 7.906∠43.49° A Chapter 10, Solution 16. At node 1, V1 V1 − V2 V1 − V2 j2 = + + 20 10 - j5 j40 = (3 + j4) V1 − (2 + j4) V2 At node 2, V1 − V2 V1 − V2 V + +1+ j = 2 10 - j5 j10 10 (1 + j) = - (1 + j2) V1 + (1 + j) V2 Thus,  j40   3 + j4 - 2 (1 + j2)  V1  = 10 (1 +  j)   - (1 + j2)   1 + j  V2    3 + j4 - 2 (1 + j2) ∆= = 5 − j = 5.099 ∠ - 11.31° - (1 + j2) 1+ j j40 - 2 (1 + j2) ∆1 = = −60 + j100 = 116.62 ∠120.96° 10 (1 + j) 1+ j 3 + j4 j40 ∆2 = = -90 + j110 = 142.13∠129.29° - (1 + j2) 10 (1 + j) ∆1 V1 = = 22.87∠132.27° V ∆ ∆2 V2 = = 27.87∠140.6° V ∆
  14. Chapter 10, Solution 17. Consider the circuit below. j4 Ω 1Ω Io 2Ω + 100∠20° V V1 V2 − 3Ω -j2 Ω At node 1, 100∠20° − V1 V1 V1 − V2 = + j4 3 2 V1 100 ∠20° = (3 + j10) − j2 V2 (1) 3 At node 2, 100∠20° − V2 V1 − V2 V2 + = 1 2 - j2 100 ∠20° = -0.5 V1 + (1.5 + j0.5) V2 (2) From (1) and (2), 100∠20°  - 0.5 0.5 (3 + j)  V1  = 100∠20° 1 + j10 3    - j2  V2    - 0.5 1.5 + j0.5 ∆= = 0.1667 − j4.5 1 + j10 3 - j2 100∠20° 1.5 + j0.5 ∆1 = = -55.45 − j286.2 100∠20° - j2 - 0.5 100∠20° ∆2 = = -26.95 − j364.5 1 + j10 3 100∠20°
  15. ∆1 V1 = = 64.74 ∠ - 13.08° ∆ ∆2 V2 = = 81.17 ∠ - 6.35° ∆ V1 − V2 ∆ 1 − ∆ 2 - 28.5 + j78.31 Io = = = 2 2∆ 0.3333 − j 9 I o = 9.25∠-162.12° Chapter 10, Solution 18. Consider the circuit shown below. V1 8Ω j6 Ω V 4Ω j5 Ω 2 + + 4∠45° A 2Ω Vx 2 Vx -j Ω -j2 Ω Vo − − At node 1, V1 V1 − V2 4∠45° = + 2 8 + j6 200 ∠45° = (29 − j3) V1 − (4 − j3) V2 (1) At node 2, V1 − V2 V V2 + 2Vx = 2 + , where Vx = V1 8 + j6 - j 4 + j5 − j2 (104 − j3) V1 = (12 + j41) V2 12 + j41 V1 = V (2) 104 − j3 2 Substituting (2) into (1), (12 + j41) 200∠45° = (29 − j3) V − (4 − j3) V2 104 − j3 2
  16. 200 ∠45° = (14.21∠89.17°) V2 200∠45° V2 = 14.21∠89.17° - j2 - j2 - 6 − j8 Vo = V2 = V2 = V2 4 + j5 − j2 4 + j3 25 10∠233.13° 200∠45° Vo = ⋅ 25 14.21∠89.17° Vo = 5.63∠189° V Chapter 10, Solution 19. We have a supernode as shown in the circuit below. j2 Ω V1 V2 4Ω V3 + 2Ω Vo -j4 Ω 0.2 Vo − Notice that Vo = V1 . At the supernode, V3 − V2 V2 V1 V1 − V3 = + + 4 - j4 2 j2 0 = (2 − j2) V1 + (1 + j) V2 + (-1 + j2) V3 (1) At node 3, V1 − V3 V3 − V2 0.2V1 + = j2 4 (0.8 − j2) V1 + V2 + (-1 + j2) V3 = 0 (2) Subtracting (2) from (1),
  17. 0 = 1.2V1 + j V2 (3) But at the supernode, V1 = 12 ∠0° + V2 or V2 = V1 − 12 (4) Substituting (4) into (3), 0 = 1.2V1 + j (V1 − 12) j12 V1 = = Vo 1.2 + j 12∠90° Vo = 1.562∠39.81° Vo = 7.682∠50.19° V Chapter 10, Solution 20. The circuit is converted to its frequency-domain equivalent circuit as shown below. R + + 1 Vm∠0° jωL Vo − jωC − L 1 C jωL Let Z = jωL || = = jωC 1 1 − ω2 LC jωL + jωC jωL Z 1 − ω2 LC jωL Vo = Vm = Vm = V R+Z jωL R (1 − ω2 LC) + jωL m R+ 1 − ω2 LC ωL Vm  ωL  Vo = ∠90° − tan -1  R 2 (1 − ω2 LC) 2 + ω2 L2  R (1 − ω LC)  2
  18. If Vo = A∠φ , then ωL Vm A= R 2 (1 − ω 2 LC) 2 + ω 2 L2 ωL and φ = 90° − tan -1 R (1 − ω 2 LC) Chapter 10, Solution 21. 1 Vo jωC 1 (a) = = Vi 1 1 − ω LC + jωRC 2 R + jωL + jωC Vo 1 At ω = 0 , = = 1 Vi 1 Vo As ω → ∞ , = 0 Vi 1 Vo 1 -j L At ω = , = = LC Vi 1 R C jRC ⋅ LC Vo jωL − ω2 LC (b) = = Vi 1 1 − ω2 LC + jωRC R + jωL + jωC Vo At ω = 0 , = 0 Vi Vo 1 As ω → ∞ , = = 1 Vi 1 1 Vo −1 j L At ω = , = = LC Vi 1 R C jRC ⋅ LC
  19. Chapter 10, Solution 22. Consider the circuit in the frequency domain as shown below. R1 R2 + + 1 Vs Vo − jωC jωL − 1 Let Z = (R 2 + jωL) || jωC 1 (R + jωL) jωC 2 R 2 + jωL Z= = 1 1 + jωR 2 − ω2 LC R 2 + jωL + jωC R 2 + jωL Vo Z 1 − ω2 LC + jωR 2 C = = Vs Z + R 1 R 2 + jωL R1 + 1 − ω2 LC + jωR 2 C Vo R 2 + jωL = Vs R 1 + R 2 − ω LCR 1 + jω (L + R 1 R 2 C) 2 Chapter 10, Solution 23. V − Vs V + + jωCV = 0 R 1 jωL + jω C jωRCV V+ + jωRCV = Vs − ω2LC + 1  1 − ω2LC + jωRC + jωRC − jω3RLC2    V = Vs  1 − ω2LC   
  20. (1 − ω2 LC)Vs V= 1 − ω2LC + jωRC(2 − ω2LC) Chapter 10, Solution 24. For mesh 1,  1 1  1 Vs =  +  I1 − I (1)  jωC1 jωC 2  jωC 2 2 For mesh 2, −1  1  0= I 1 +  R + jωL + I (2) jωC 2  jωC 2  2 Putting (1) and (2) into matrix form,  1 1 −1  Vs   jωC + jωC jωC 2   I1  = 1 2  0    −1 1  I 2    R + jωL +   jω C 2 jωC 2    1 1  1  1 ∆ = +  R + jωL + + 2  jωC1 jωC 2  jωC 2  ω C1C 2  1  Vs ∆ 1 = Vs  R + jωL +  and ∆2 =  jωC 2  jωC 2  1  Vs  R + jωL +  ∆1  jωC 2  I1 = = ∆  1 1  1  1  +  R + jωL + + 2  jωC 1 jωC 2  jωC 2  ω C1 C 2 Vs ∆2 jωC 2 I2 = = ∆  1 1  1  1  +  R + jωL + + 2  jωC 1 jωC 2  jωC 2  ω C1 C 2 Chapter 10, Solution 25. ω= 2 10 cos(2t )  → 10∠0°
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