Bài giải phần giải mạch P11

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Chapter 11, Solution 1. v( t ) = 160 cos(50t ) i( t ) = -20 sin(50t − 30°) = 2 cos(50t − 30° + 180° − 90°) i( t ) = 20 cos(50t + 60°) p( t ) = v( t ) i( t ) = (160)(20) cos(50t ) cos(50t + 60°) p( t ) = 1600 [ cos(100 t + 60°) + cos(60°) ] W p( t ) = 800 + 1600 cos(100t + 60°) W P= 1 1 Vm I m cos(θ v − θi ) = (160)(20) cos(60°) 2 2 P = 800 W Chapter 11, Solution 2. First, transform the...

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Chapter 11, Solution 1. v( t ) = 160 cos(50t ) i( t ) = -20 sin(50t − 30°) = 2 cos(50t − 30° + 180° − 90°) i( t ) = 20 cos(50t + 60°) p( t ) = v( t ) i( t ) = (160)(20) cos(50t ) cos(50t + 60°) p( t ) = 1600 [ cos(100 t + 60°) + cos(60°) ] W p( t ) = 800 + 1600 cos(100t + 60°) W 1 1 P= Vm I m cos(θ v − θi ) = (160)(20) cos(60°) 2 2 P = 800 W Chapter 11, Solution 2. First, transform the circuit to the frequency domain. 30 cos(500t )  → 30 ∠0° , ω = 500 0.3 H  → jωL = j150 1 -j 20µF  → = = - j100 jωC (500)(20)(10 -6 ) I I2 -j100 Ω I1 + 30∠0° V j150 Ω 200 Ω − 30∠0° I1 = = 0.2∠ − 90° = - j0.2 j150 i1 ( t ) = 0.2 cos(500 t − 90°) = 0.2 sin(500 t ) 30∠0° 0.3 I2 = = = 0.1342∠26.56° = 0.12 + j0.06 200 − j100 2 − j
i 2 ( t ) = 0.1342 cos(500 t + 25.56°) I = I 1 + I 2 = 0.12 − j0.14 = 0.1844 ∠ - 49.4° i( t ) = 0.1844 cos(500t − 35°) For the voltage source, p( t ) = v( t ) i( t ) = [ 30 cos(500t ) ] × [ 0.1844 cos(500t − 35°) ] At t = 2 s , p = 5.532 cos(1000) cos(1000 − 35°) p = (5.532)(0.5624)(0.935) p = 2.91 W For the inductor, p( t ) = v( t ) i( t ) = [ 30 cos(500t ) ] × [ 0.2 sin(500t ) ] At t = 2 s , p = 6 cos(1000) sin(1000) p = (6)(0.5624)(0.8269) p = 2.79 W For the capacitor, Vc = I 2 (- j100) = 13.42∠ - 63.44° p( t ) = v( t ) i( t ) = [13.42 cos(500 − 63.44°) ] × [ 0.1342 cos(500t + 25.56°) At t = 2 s , p = 18 cos(1000 − 63.44°) cos(1000 + 26.56°) p = (18)(0.991)(0.1329) p = 2.37 W For the resistor, VR = 200 I 2 = 26.84 ∠25.56° p( t ) = v( t ) i( t ) = [ 26.84 cos(500t + 26.56°) ] × [ 0.1342 cos(500t + 26.56°) ] At t = 2 s , p = 3.602 cos 2 (1000 + 25.56°) p = (3.602)(0.1329 2 p = 0.0636 W
Chapter 11, Solution 3. 10 cos(2t + 30°)  → 10∠30° , ω= 2 1H  → jωL = j2 1 0.25 F  → = -j2 jωC I 4Ω I1 2Ω I2 + 10∠30° V j2 Ω -j2 Ω − ( j2)(2 − j2) j2 || (2 − j2) = = 2 + j2 2 10 ∠30° I= = 1.581∠11.565° 4 + 2 + j2 j2 I1 = I = j I = 1.581∠101.565° 2 2 − j2 I2 = I = 2.236 ∠56.565° 2 For the source, 1 S = V I* = (10∠30°)(1.581∠ - 11.565°) 2 S = 7.905∠18.43° = 7.5 + j2.5 The average power supplied by the source = 7.5 W For the 4-Ω resistor, the average power absorbed is 1 2 1 P = I R = (1.581) 2 (4) = 5 W 2 2 For the inductor, 1 2 1 S = I 2 Z L = (2.236) 2 ( j2) = j5 2 2 The average power absorbed by the inductor = 0 W
For the 2-Ω resistor, the average power absorbed is 1 2 1 P = I 1 R = (1.581) 2 (2) = 2.5 W 2 2 For the capacitor, 1 2 1 S= I 1 Z c = (1.581) 2 (- j2) = - j2.5 2 2 The average power absorbed by the capacitor = 0 W Chapter 11, Solution 4. 20 Ω 10 Ω + I1 I2 50 V -j10 Ω j5 Ω − For mesh 1, 50 = (20 − j10) I 1 + j10 I 2 5 = (2 − j) I 1 + j I 2 (1) For mesh 2, 0 = (10 + j5 − j10) I 2 + j10 I 1 0 = (2 − j) I 2 + j2 I 1 (2) In matrix form, 5  2 − j j  I 1   0 =  j2 2 − j I      2  ∆ = 5 − j4 , ∆ 1 = 5 (2 − j) , ∆ 2 = -j10 ∆ 1 5 (2 − j) I1 = = = 1.746∠12.1° ∆ 5 − j4 ∆ 2 - j10 I2 = = = 1.562 ∠128.66° ∆ 5 - j4 For the source, 1 S= V I 1 = 43.65∠ - 12.1° * 2
The average power supplied = 43.65 cos(12.1°) = 42.68 W For the 20-Ω resistor, 1 2 P = I 1 R = 30.48 W 2 For the inductor and capacitor, P=0W For the 10-Ω resistor, 1 2 P = I 2 R = 12.2 W 2 Chapter 11, Solution 5. Converting the circuit into the frequency domain, we get: 1Ω 2Ω + –j2 8∠–40˚ − j6 8∠ − 40° I1Ω = = 1.6828∠ − 25.38° j6(2 − j2) 1+ j6 + 2 − j2 1.6828 2 P1Ω = 1 = 1.4159 W 2 P3H = P0.25F = 0 j6 I 2Ω = 1.6828∠ − 25.38° = 2.258 j6 + 2 − j2 2.258 2 P2Ω = 2 = 5.097 W 2
Chapter 11, Solution 6. 20 Ω 10 Ω + I1 I2 50 V -j10 Ω j5 Ω − For mesh 1, (4 + j2) I 1 − j2 (4 ∠60°) + 4 Vo = 0 (1) Vo = 2 (4 ∠60° − I 2 ) (2) For mesh 2, (2 − j) I 2 − 2 (4∠60°) − 4Vo = 0 (3) Substituting (2) into (3), (2 − j) I 2 − 8∠60° − 8 (4 ∠60° − I 2 ) = 0 40∠60° I2 = 10 − j Hence,  40∠60°  - j8∠60° Vo = 2  4 ∠60° − =  10 − j  10 − j Substituting this into (1), j32 ∠60° 14 − j  (4 + j2) I 1 = j8∠60° + = ( j8∠60°)   10 − j 10 − j  (4∠60°)(1 + j14) I1 = = 2.498∠125.06° 21 + j8 1 2 1 P4 = I 1 R = (2.498) 2 (4) = 12.48 W 2 2 Chapter 11, Solution 7. 20 Ω 10 Ω + I1 I2 50 V -j10 Ω j5 Ω −
Applying KVL to the left-hand side of the circuit, 8∠20° = 4 I o + 0.1Vo (1) Applying KCL to the right side of the circuit, V V1 8Io + 1 + =0 j5 10 − j5 10 10 − j5 But, Vo = V  → V1 = Vo 10 − j5 1 10 10 − j5 Vo Hence, 8Io + Vo + =0 j50 10 I o = j0.025 Vo (2) Substituting (2) into (1), 8∠20° = 0.1 Vo (1 + j) 80∠20° Vo = 1+ j Vo 10 I1 = = ∠ - 25° 10 2 1 2  1 100  P= I 1 R =   (10) = 250 W 2  2  2  Chapter 11, Solution 8. We apply nodal analysis to the following circuit. V1 Io -j20 Ω V2 I2 6∠0° A j10 Ω 0.5 Io 40 Ω At node 1, V1 V1 − V2 6= + V1 = j120 − V2 (1) j10 - j20 At node 2,
V2 0 .5 I o + I o = 40 V1 − V2 But, Io = - j20 1.5 (V1 − V2 ) V2 Hence, = - j20 40 3V1 = (3 − j) V2 (2) Substituting (1) into (2), j360 − 3V2 − 3V2 + j V2 = 0 j360 360 V2 = = (-1 + j6) 6 − j 37 V2 9 I2 = = (-1 + j6) 40 37 2 1 2 1 9  P = I2 R =   (40) = 43.78 W 2 2  37  Chapter 11, Solution 9.  6 Vo = 1 +  Vs = (4)(2) = 8 V rms  2 Vo2 64 P10 = = mW = 6.4 mW R 10 The current through the 2 -kΩ resistor is Vs = 1 mA 2k P2 = I 2 R = 2 mW Similarly, P6 = I 2 R = 6 mW
Chapter 11, Solution 10. No current flows through each of the resistors. Hence, for each resistor, P = 0 W. Chapter 11, Solution 11. ω = 377 , R = 10 4 , C = 200 × 10 -9 ωRC = (377)(10 4 )(200 × 10 -9 ) = 0.754 tan -1 (ωRC) = 37.02° 10k Z ab = ∠ - 37.02° = 6.375∠ - 37.02° kΩ 1 + (0.754) 2 i( t ) = 2 sin(377 t + 22°) = 2 cos(377 t − 68°) mA I = 2 ∠ - 68° 2  2 × 10 -3  S= I 2 Z ab =   (6.375∠ - 37.02°) × 10 3 rms  2  S = 12.751∠ - 37.02° mVA P = S cos(37.02) = 10.181 mW Chapter 11, Solution 12. (a) We find Z Th using the circuit in Fig. (a). Zth 8Ω -j2 Ω (a) (8)(-j2) 8 Z Th = 8 || -j2 = = (1 − j4) = 0.471 − j1.882 8 − j2 17 Z L = Z * = 0.471 + j1.882 Ω Th
We find VTh using the circuit in Fig. (b). Io + 8Ω Vth -j2 Ω 4∠0° A − (b) - j2 Io = (4∠0°) 8 − j2 - j64 VTh = 8 I o = 8 − j2 2  64  2   VTh  68  Pmax = = = 15.99 W 8RL (8)(0.471) (b) We obtain Z Th from the circuit in Fig. (c). 5Ω -j3 Ω j2 Ω Zth 4Ω (c) (5)(4 − j3) Z Th = j2 + 5 || (4 − j3) = j2 + = 2.5 + j1.167 9 − j3 Z L = Z * = 2.5 − j1.167 Ω Th
Chapter 11, Solution 13. (a) We find Z Th at the load terminals using the circuit in Fig. (a). j100 Ω Zth 80 Ω -j40 Ω (a) (-j40)(80 + j100) Z Th = -j40 || (80 + j100) = = 51.2 − j1.6 80 + j60 Z L = Z * = 51.2 + j1.6 Ω Th (b) We find VTh at the load terminals using Fig. (b). Io j100 Ω + 3∠20° A 80 Ω -j40 Ω Vth − (b) 80 (8)(3∠20°) Io = (3∠20°) = 80 + j100 − j40 8 + j6 (- j40)(24∠20°) VTh = - j40 I o = 8 + j6 2  40  2  ⋅ 24 VTh  10  Pmax = = = 22.5 W 8RL (8)(51.2)
From Fig.(d), we obtain VTh using the voltage division principle. 5Ω -j3 Ω j2 Ω + 10∠30° V + 4Ω − Vth − (d)  4 − j3   4 − j3  10  VTh =  (10∠30°) =   ∠30°  9 − j3   3 − j  3  2  5 10  2  ⋅  VTh  10 3  Pmax = = = 1.389 W 8RL (8)(2.5) Chapter 11, Solution 14. j24 Ω –j10 Ω I + 16 Ω VTh ZTh 40∠90º A 10 Ω j8 Ω _ (10 + j24)(16 + j8) Z Th = − j10 + = − j10 + 8.245 + j7.7 = 8.245 − j2.3Ω 10 + j24 + 16 + j8 Z = Z∗ = 8.245 + j2.3Ω Th
10 VTh = I(16 + j8) = j40(16 + j8) 10 + j24 + 16 + j8 = 173.55∠65.66° = 71.53 + j158.12 V 2 VTh 2 Pmax = I 2 8.245 = 2 8.245 = 456.6 W rms (2x8.245) 2 Chapter 11, Solution 15. To find Z Th , insert a 1-A current source at the load terminals as shown in Fig. (a). 1Ω 1 -j Ω 2 + Vo jΩ 2 Vo 1A − (a) At node 1, Vo Vo V2 − Vo + =  → Vo = j V2 (1) 1 j -j At node 2, V2 − Vo 1 + 2 Vo =  → 1 = j V2 − (2 + j) Vo (2) -j Substituting (1) into (2), 1 = j V2 − (2 + j)( j) V2 = (1 − j) V2 1 V2 = 1− j V2 1 + j VTh = = = 0.5 + j0.5 1 2 Z L = Z * = 0.5 − j0.5 Ω Th
We now obtain VTh from Fig. (b). 1Ω -j Ω + + + 12∠0° V Vo jΩ 2 Vo Vth − − − (b) 12 − Vo Vo 2 Vo + = 1 j - 12 Vo = 1+ j Vo − (- j × 2 Vo ) + VTh = 0 (12)(1 + j2) VTh = -(1 + j2)Vo = 1+ j 2 12 5  2     VTh  2  Pmax = = = 90 W 8RL (8)(0.5) Chapter 11, Solution 16. 1 1 ω = 4, 1H  → jωL = j 4, 1 / 20F  → = = − j5 jωC j 4 x1 / 20 We find the Thevenin equivalent at the terminals of ZL. To find VTh, we use the circuit shown below. 0.5Vo 2Ω V1 4Ω V2 + + + 10
At node 1, 10 − V1 V V − V2 = 1 + 0.25V1 + 1  → 5 = V1 (1 + j 0.2) − 0.25V2 (1) 2 − j5 4 At node 2, V1 − V2 V + 0.25V1 = 2  → 0 = 0.5V1 + V2 (−0.25 + j 0.25) (2) 4 j4 Solving (1) and (2) leads to VTh = V2 = 6.1947 + j 7.0796 = 9.4072∠48.81o Chapter 11, Solution 17. We find R Th at terminals a-b following Fig. (a). -j10 Ω 30 Ω a b 40 Ω j20 Ω (a) (30)( j20) (40)(-j10) Z Th = 30 || j20 + 40 || (- j10) = + 30 + j20 40 − j10 Z Th = 9.23 + j13.85 + 2.353 − j9.41 Z Th = 11.583 + j4.44 Ω Z L = Z * = 11.583 − j4.44 Ω Th We obtain VTh from Fig. (b). I1 I2 -j10 Ω 30 Ω j5 A + VTh − 40 Ω j20 Ω (b)
Using current division, 30 + j20 I1 = ( j5) = -1.1 + j2.3 70 + j10 40 − j10 I2 = ( j5) = 1.1 + j2.7 70 + j10 VTh = 30 I 2 + j10 I 1 = 10 + j70 2 VTh 5000 Pmax = = = 53.96 W 8RL (8)(11.583) Chapter 11, Solution 18. We find Z Th at terminals a-b as shown in the figure below. 40 Ω -j10 Ω 40 Ω 80 Ω a Zth j20 Ω b (80)(-j10) Z Th = j20 + 40 || 40 + 80 || (-j10) = j20 + 20 + 80 − j10 Z Th = 21.23 + j10.154 Z L = Z * = 21.23 − j10.15 Ω Th Chapter 11, Solution 19. At the load terminals, (6)(3 + j) Z Th = - j2 + 6 || (3 + j) = -j2 + 9+ j Z Th = 2.049 − j1.561 R L = Z Th = 2.576 Ω
To get VTh , let Z = 6 || (3 + j) = 2.049 + j0.439 . By transforming the current sources, we obtain VTh = (4 ∠0°) Z = 8.196 + j1.756 2 VTh 70.258 Pmax = = = 3.409 W 8RL 20.608 Chapter 11, Solution 20. Combine j20 Ω and -j10 Ω to get j20 || -j10 = -j20 To find Z Th , insert a 1-A current source at the terminals of R L , as shown in Fig. (a). 4 Io Io 40 Ω V1 V2 + − -j20 Ω -j10 Ω 1A (a) At the supernode, V1 V V 1= + 1 + 2 40 - j20 - j10 40 = (1 + j2) V1 + j4 V2 (1) - V1 Also, V1 = V2 + 4 I o , where I o = 40 V2 1.1 V1 = V2  → V1 = (2) 1 .1 Substituting (2) into (1), V  40 = (1 + j2)  2  + j4 V2  1 .1 
44 V2 = 1 + j6.4 V2 Z Th = = 1.05 − j6.71 Ω 1 R L = Z Th = 6.792 Ω To find VTh , consider the circuit in Fig. (b). 4 Io Io 40 Ω V1 V2 + − + + 120∠0° V -j20 Ω -j10 Ω Vth − − (b) At the supernode, 120 − V1 V V = 1 + 2 40 - j20 - j10 120 = (1 + j2) V1 + j4 V2 (3) 120 − V1 Also, V1 = V2 + 4 I o , where I o = 40 V2 + 12 V1 = (4) 1 .1 Substituting (4) into (3), 109.09 − j21.82 = (0.9091 + j5.818) V2 109.09 − j21.82 VTh = V2 = = 18.893∠ - 92.43° 0.9091 + j5.818 2 VTh (18.893) 2 Pmax = = = 6.569 W 8RL (8)(6.792)
Chapter 11, Solution 21. We find Z Th at terminals a-b, as shown in the figure below. 100 Ω -j10 Ω a 40 Ω Zth 50 Ω j30 Ω b Z Th = 50 || [ - j10 + 100 || (40 + j30) ] (100)(40 + j30) where 100 || (40 + j30) = = 31.707 + j14.634 140 + j30 (50)(31.707 + j4.634) Z Th = 50 || (31.707 + j4.634) = 81.707 + j4.634 Z Th = 19.5 + j1.73 R L = Z Th = 19.58 Ω Chapter 11, Solution 22. i (t ) = 4 sin t , 0
Chapter 11, Solution 23. 15, 0 < t < 2 v( t ) =   5, 2 < t < 6 Vrms = 2 1 6 [ ∫ 15 0 2 2 dt + ∫2 5 2 dt = 6 ] 550 6 Vrms = 9.574 V Chapter 11, Solution 24.  5, 0 < t < 1 T = 2, v( t ) =  - 5, 1 < t < 2 Vrms = 2 1 2 [∫ 5 0 1 2 dt + ∫1 (-5) 2 dt = 2 ] 25 2 [1 + 1] = 25 Vrms = 5 V Chapter 11, Solution 25. 2 1 T 2 f rms = T 1 1 2 [2 3 2 ∫ 0 f ( t )dt = 3 ∫ 0 (−4) dt + ∫ 1 0dt + ∫2 4 dt ] 1 32 = [16 + 0 + 16] = 3 3 32 f rms = = 3.266 3