Solutions Chap11
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CHAPTER 11 SOLUTIONS 1. 50 (− j80) 106 = − j80 Ω, = 42.40∠ − 32.01°Ω j 500 × 25 50 − j80 ∴ V = 84.80∠ − 32.01° V, I R = 1.696∠ − 32.01° A Zc =
I c = 1.0600∠57.99° A ps (π / 2ms) = 84.80 cos (45° − 32.01°) 2 cos 45° = 116.85 W pR = 50 × 1.696 2 cos 2 (45° − 32.01°) = 136.55 W pc = 84.80 cos (45° − 32.01°) = 1.060 cos (45° + 57.99°) = −19.69 W
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 2. (a)
1 2 1 Li = × 4 (4t 4 − 4t 2 + 1) 2 2 4 2 4 2 ∴ wL = 8t − 8t + 2 ∴ wL (3) − wL (1) = 8 × 3 − 8 × 3 + 2 − 8 ×1 + 8 ×1 − 2 = 576 J 4H : i = 2t 2 − 1∴ v = Li′ = 4 (4t ) = 16t , wL =
t
(b)
1 t 2 2 2 2 0.2 F : vc = (2t − 1) dt + 2 = 5 t 3 − t + 2 = 5 t 3 − t − 5 − 1 + 2 ∫ 0.2 1 3 1 3 3 10 10 61 61 ∴ vc (2) = × 8 − 10 − + 5 + 2 = V ∴ Pc (2) = × 7 = 142.33 W 3 3 3 3
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS R 1 = 2, ω o2 = = 3, s1,2 = −2 ± 1 = −1, − 3 2L LC
3.
vc (0) = −2V, i (0) = 4A, α =
(a)
1 i = Ae− t + Be−3t ∴ A + B = 4; i (0+ ) = vL (0+ ) = (−4 × 4 × +2) = −14 1 ∴− A − 38 = −14 ∴ B = 5, A = −1, i = −e− t + 5e −3t A t
∴+ vc = 3∫ (−e − t + 5e −3t ) dt − 2 = 3(e− t − 5e −3t ) to − 2 = e − t − 3 − 5e −3t + 5 − 2 o
−t
∴ vc = 3e − 5e −3t ∴ Pc (0 + ) = (3 − 5) (−1 + 5) = −8 W (b)
Pc (0.2) = (3e −0.2 − 5e −0.6 ) (−e0.2 + 5e −0.6 ) = −0.5542 W
(c)
Pc (0.4) = (3e −0.4 − 5e −1.2 ) (5e −1.2 − e −0.4 ) = 0.4220 W
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 4.
We assume the circuit has already reached sinusoidal steady state by t = 0. 2.5 kΩ → 2.5 kΩ, 1 H → j1000 Ω, 4 µF → -j250 Ω, 10 kΩ → 10 kΩ Zeq = j1000 || -j250 || 10000 = 11.10 – j333.0 Ω (20∠30)(11.10 − j 333.0) = 2.631∠50.54o V 2500 + 11.10 − j 333.0 Veq Veq I10k = = 0.2631 ∠ - 50.54o mA I1 H = = 2.631 ∠ - 140.5o mA 10000 j1000 Veq (20∠30)(2500) I4 µF = = 10.52 ∠39.46o mA V2.5k = = 19.74∠37.55o V − j 250 2500 + 11.10 − j 333.0 Veq =
[19.74 cos 37.55 ] P2.5k =
o 2
Thus,
=
2500
[
97.97 mW
][
]
P1 H = 2.631cos(− 50.54 ) 2.631 × 10-3 cos(−140.5o ) = - 3.395 mW o
[ ( )][ [2.631cos(− 50.54 )] = P2.5k =
]
P4 µF = 2.631cos − 50.54o 10.52 × 10-3 cos(39.46o ) = 13.58 mW o
10000
2
279.6 µW
FREQ IM(V_PRINT1) IP(V_PRINT1) 1.592E+02 7.896E-03 3.755E+01
FREQ VM(L,0) 1.592E+02 2.629E+00
FREQ VM(R2_5k,$N_0002)VP(R2_5k,$N_0002) 1.592E+02 1.974E+01 3.755E+01
FREQ IM(V_PRINT11) IP(V_PRINT11) 1.592E+02 1.052E-02 3.946E+01
FREQ IM(V_PRINT2) IP(V_PRINT2) 1.592E+02 2.628E-03 -1.405E+02
FREQ IM(V_PRINT12) IP(V_PRINT12) 1.592E+02 2.629E-04 -5.054E+01
Engineering Circuit Analysis, 6th Edition
VP(L,0) -5.054E+01
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 5. is → 5∠0° A, C → − j 4 Ω, Z in = 8 (3 − j 4) =
40∠ − 53.13° 11 − j 4
= 3.417∠ − 33.15°∴ Vs = 17.087∠ − 33.15°, vs = 17.087 cos (25t − 33.15°) V ∴ Ps , abs (0.1) = −17.087 cos (2.5rad − 33.147°) × 5cos 2.5rad = −23.51 W 17.087 cos (25t − 33.15°) ∴ 8 i8 (0.1) = 2.136 cos (2.5rad − 33.15°) = −0.7338 A i8 =
∴ P8, abs = 0.73382 × 8 = 4.307 W; I3 =
17.087∠ − 33.15° = 3.417∠19.98° A 3 − j4
∴ i3 (0.1) = 3.417 cos (2.5rad + 19.98°) = −3.272 A ∴ P3, abc = 3.2722 × 3 = 32.12 W Vc = − j 4 (3.417∠19.983°) = 13.67∠ − 70.02°, vc (0.1) = 13.670 cos (2.5rad − 70.02°) = 3.946 V ∴ Pc , abc = 3.946 (−3.272) = −12.911 W
Engineering Circuit Analysis, 6th Edition
(Σ = 0)
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 6. j 5(10 − j 5) = 4 + 2.5 + j 5 = 6.5 + j 5 Ω 10 100 ∴ Is = = 12.194∠ − 37.57° A 6.5 + j 5 1 ∴ Ps ,abs = − × 100 × 12.194 cos 37.57° = −483.3W 2 1 P4, abs = (12.194) 2 4 = 297.4 W, 2 Pcabs = 0 Zin = 4 +
100 j5 = 6.097∠52.43° so 6.5 + j 5 10 1 P10,abs = (6.097) 2 × 10 = 185.87 W 2 PL = 0 (Σ = 0) I10 =
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 7. V = (10 + j10)
40∠30° = 52.44∠69.18° V 5∠50° + 8∠ − 20°
1 × 10 × 52.44 cos 69.18° = 93.20 W 2 1 = × 10 × 52.44 cos (90° − 69.18°) = 245.08 W 2
P10, gen = Pj10, gen
2
P8 ∠− 20 abs
1 52.44 = 8cos (−20°) = 161.51 W 2 8
Engineering Circuit Analysis, 6th Edition
(Σ gen = Σ abs )
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 8. ZR = 3 +
1 = 3 + 1 + j3 = 4 + j3 Ω 0.1 − j 0.3
Ignore 30° on Vs , I R = 5
2 + j5 5 29 , IR = 6 + j8 10
2
1 5 29 = × 3 = 10.875 W 2 10
(a)
P3 Ω
(b)
Vs = 5∠0° ∴ Ps , gen =
(2 + j 5) (4 + j 3) = 13.463∠51.94° V 6 + j8
1 × 13.463 × 5cos 51.94° = 20.75 W 2
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 9.
Pj10 = P− j 5 = 0, V10 − 50 V10 V10 − j 50 + + =0 j10 10 − j5 ∴ V10 (− j 0.1 + 0.1 + j 0.2) + j 5 + 10 = 0 ∴ V10 = 79.06∠16.57° V 1 79.06 2 = 312.5 W; 2 10 79.06∠161.57° − 50 I 50 = = 12.75∠78.69° A j10 1 ∴ P50V = × 50 × 12.748cos 78.69° = 62.50 W 2 79.06∠161.57° − j 50 = 15.811∠ − 7.57° : I j 50 = − j5 1 Pj 50 = × 50 × 15.811cos (90° + 71.57°) = −375.0 W 2 P10 Ω =
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 10. Vx − 20 Vx − Vc + = 2Vc , 2 3 3Vx − 60 + 2Vx − 2Vc = 12Vc Vc − Vx V + c =0 3 − j2 ∴ 2Vc − 2Vx + j3Vc = 0, − 2Vx + (2 + j3) Vc = 0 ∴ 5Vx − 14Vc = 60,
60 −14 0 2 + j3 120 + j180 Vx = = = 9.233∠ − 83.88° V 5 −14 10 + j15 − 28 −2 2 + j 3
Pgen
5 60 −2 0 Vc = = 5.122∠ − 140.9° V ∴ −18 + j15 1 = × 9.233 × 2 × 5.122 cos ( −83.88° + 140.19°) = 26.23 W 2
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 11. (a)
X in = 0 ∴ Z L = R th + j 0
(b)
R L , X L independent∴ Z L = Zth∗ = R th − jXth
(c)
Vth 1 R L fixed∴ PL = × R L ∴ Z L = R L − jXth 2 (R th + R L )2 + (X th + X L )2
(d)
X L fixed, Let X L + Xth = a ∴ f =
2
2PL Vth
2
=
RL (R th + R L ) 2 + a 2
R + R 2L + a 2 − 2R L (R th + R L ) df = th =0 2 dRL (R th + R L ) 2 + a 2 R th2 + 2R th R L + R 2L + a 2 − 2R th R L = 2R 2L = 0 ∴ R L = R th2 + a 2 = (e)
R th2 + (Xth + X L )2
X L = 0 ∴ R L = R th2 + X th2 = Zth
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 12.
−10 = 107.33∠ − 116.57° V 10 + j 5 − j10 (10 + j15) Zth = = 8 − j14 Ω 10 + j 5 Vth = 120
(a)
∴ Z L = 8 + j15 Ω
(b)
IL =
107.33∠ − 116.57° ∴ 16 2
PL ,max =
1 107.33 × 8 = 180 W 2 16
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 13.
R L = Zth ∴ R L = 82 + 142 = 16.125 Ω PL =
1 107.332 × 16.125 = 119.38 W 2 (8 + 16.125)2 + 142
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 14. − j 9.6 = −4.8 I x − j1.92 I x − +4.8I x 9.6 =5 1.92 ∴ V = (0.6 × 5)8 = 24 V 1 ∴ Po = × 24 × 1.6 × 5 = 96 W ( gen) 2 ∴ Ix =
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 15. j 480 80 − j 60 80 + j 60 80 − j 60 = 28.8 + j 38.4 Ω ∴ Z L max = 28.8 − j 38.4 Ω
(a)
Zth = 80 j 60 =
(b)
Vth = 5(28.8 + j 38.4) = 144 + j192 V, 144 + j192 2 × 28.8 1 1442 + 1922 and PL ,max = × 28.8 = 250 W 2 4 × 28.82
∴ IL =
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 16.
Zeq = (6 – j8) || (12 + j9) = 8.321 ∠ -19.44o W Veq = (5 ∠-30o) (8.321 ∠ -19.44o) = 41.61 ∠ -49.44o V Ptotal = ½ (41.61)(5) cos (-19.44o) = 98.09 W I6-j8 = Veq / (6 – j8) = 4.161 ∠ 3.69o A I4+j2 = I8+j7 = Veq/ 12+j9 = 2.774 ∠ -86.31o A P6-j8 = ½ (41.61)(4.161) cos (-49.44o – 3.69o) = 51.94 W P4+j2 = ½ (2.774)2 (4) = 15.39 W P8+j7 = ½ (2.774)2 (8) = 30.78 W Check: Σ = 98.11 W (okay)
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 17. Vth = 100
j10 (20) j10 = 20 + j 40, Z th = = 4 + j8 Ω 20 + j10 20 + j10
∴ R L = Zth ∴ R L = 8.944 Ω ∴ PL ,max =
1 202 + 402 × 8.944 = 38.63 W 2 (4 + 8.944)2 + 64
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 18.
We may write a single mesh equation: 170 ∠0o = (30 + j10) I1 – (10 – j50)(-λI1) Solving, 170∠0 o I1 = 30 + j10 + 10λ − j 50λ o 170∠0 (a) λ = 0, so I1 = = 5.376∠ - 18.43 o A and, with the same current flowing 30 + j10 through both resistors in this case, P20 = ½ (5.376)2 (20) = 289.0 W P10 = ½ (5.376)2 (10) = 144.5 W 170∠0 o = 3.005∠45 o A 40 − j 40 P20 = ½ (3.005)2 (20) = 90.30 W The current through the 10-Ω resistor is I1 + λI1 = 2 I1 = 6.01 ∠ 45o so
(b) λ = 1, so I1 =
P10 = ½ (6.01)2 (10) = 180.6 W (c)
(a) FREQ IM(V_PRINT3) 6.000E+01 5.375E+00
IP(V_PRINT3) -1.846E+01
FREQ IM(V_PRINT4) 6.000E+01 5.375E+00
IP(V_PRINT4) -1.846E+01
(b) FREQ IM(V_PRINT3) 6.000E+01 6.011E+00
IP(V_PRINT3) 4.499E+01
FREQ IM(V_PRINT4) 6.000E+01 3.006E+00
IP(V_PRINT4) 4.499E+01
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS
19.
(10)(1) + (−5)(1) + 0(1) = 1.667 A 3 1 (20)(1) + 0(1) Waveform (b): Iavg = 2 = 5A 2
(a) Waveform (a): Iavg =
Waveform (c): 1 Iavg = 1 × 10 −3 =−
(b)
∫
2πt dt = - 8 × 10 3 8sin −3 4 × 10
(
10 − 3 0
)
4 × 10 −3 πt cos −3 2π 2 × 10
10 −3
0
16 (0 − 1) = 16 A π π
(100)(1) + (25)(1) + (0)(1) = 41.67 A 2 3 Waveform (b): i(t) = -20×103 t + 20 i2(t) = 4×108 t2 – 8×105 t + 400 10 -3 1 2 I avg = 4 × 10 8 t 2 - 8 × 10 5 t + 400 dt -3 ∫ 0 2 × 10 5 4 × 10 8 2 1 0.1333 − 3 3 8 × 10 = 66.67 A 2 = 10 10 −3 + 400 10 −3 = -3 -3 2 × 2 × 10 3 2 10 Waveform (c):
2 Waveform (a): I avg =
(
)
(
2 I avg
1 = 1 × 10 −3
∫
Engineering Circuit Analysis, 6th Edition
)
(
0
)
(
(
)
2πt sin π × 10 3 t 3 t 64sin 2 dt = 64 × 10 4 × 10 −3 2π × 10 3 2
10 − 3
10 = 64 × 10 3 2
(
)
−3
−
)
10 −3
0
sin π = 32 A 2 3 2π × 10
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 20.
At ω = 120π, 1 H → j377 Ω, and 4 µF → -j663.1 Ω Define Zeff = j377 || -j663.1 || 10 000 = 870.5 ∠ 85.01o Ω
(400
)
2∠ − 9 o 2500 V2.5k = = 520.4 ∠ - 27.61o V o 2500 + 870.5 ∠85.01 400 2∠ − 9 o 870.5 ∠85.01o V10k = = 181.2 ∠57.40 o V o 2500 + 870.5 ∠85.01
(
)(
)
Thus, P2.5k = ½ (520.4)2 / 2 500 P10k = ½ (181.2)2 / 10 000 P1H P4µF
= = = =
54.16 W 1.642 W 0 0 (A total absorbed power of 55.80 W.)
To check, the average power delivered by the source: Isource =
400 2∠ − 9 o = 0.2081 ∠ - 27.61o A 2500 + 870.5∠85.01o
and Psource = ½ ( 400 2 )(0.2081) cos (-9o + 27.61o) = 55.78 W (checks out).
FREQ IM(V_PRINT1) 6.000E+01 2.081E-01
IP(V_PRINT1) -2.760E+01
FREQ VM(L,0) 6.000E+01 1.812E+02
VP(L,0) 5.740E+01
FREQ VM(R2_5k,$N_0002) VP(R2_5k,$N_0002) 6.000E+01 5.204E+02 -2.760E+01
FREQ IM(V_PRINT11) 6.000E+01 2.732E-01
IP(V_PRINT11) 1.474E+02
FREQ IM(V_PRINT2) 6.000E+01 4.805E-01
FREQ IM(V_PRINT12) 6.000E+01 1.812E-02
IP(V_PRINT12) 5.740E+01
IP(V_PRINT2) -3.260E+01
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 21. (a)
v = 10 + 9 cos100t + 6 sin100t 1 1 ∴ Veff = 100 + × 81 + × 36 = 158.5 = 12.590 V 2 2
(b)
Feff =
1 2 (10 + 20 2 + 102 ) = 150 = 12.247 4
(c)
Favg =
(10)(1) + (20)(1) + (10)(1) 40 = = 10 4 4
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 22. (a)
g(t) = 2 + 3cos100t + 4cos(100t – 120o) 3 ∠0 + 4∠-120 = 3.606 ∠-73.90 so Geff = o
(b)
o
3.6062 4+ = 3.240 2
h (t ) = 2 + 3cos100t + 4 cos (101t − 120°) 1 1 ∴ H eff = 2 + 32 + 4 2 = 16.5 = 4.062 2 2
(c)
f (t ) = 100t , 0 < t < 0.1∴ Feff = =
1 0.1 6 2 10 t dt 0.3 ∫0
10 1 × 106 × × 10−3 = 33.33 3 3
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 23.
f (t ) = (2 − 3cos100t ) 2
(a)
f (t ) = 4 − 12 cos100t + 9 cos 2 100t ∴ f (t ) = 4 − 12 cos100t + 4.5 + 4.5cos 200t ∴ Fav = 4 + 4.5 = 8.5
(b)
Feff = 8.52 +
1 1 × 122 + × 4.52 = 12.43 2 2
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS
24.
(a)
ieff
(
)
1 = 10 2 + (−5) 2 + 0 3
1
2
= 6.455 A
(b) ieff
1 1 = ∫ [− 20t + 20] dt + 0 2 0
(c) ieff
1 1 2π t dt = ∫ 8sin 0 4 1
Engineering Circuit Analysis, 6th Edition
1
2
1
2
=
5 = 2.236 A
1
=
2 πt - 8 π cos 2 = 2.257 A 0
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 25. (a)
A = B = 10V, C = D = 0 ∴10∠0° + 10∠ − 45° = 18.48∠ − 22.50o 1 1 ∴ P = × × 18.482 = 42.68 W 2 4
(b)
A = C = 10V, B = D = 0, vs = 10cos10t + 10 cos 40t , P=
(c)
1 102 1 102 + = 25 W 2 4 2 4
vs = 10 cos10t − 10sin (10t + 45°) → 10 − 10∠ − 45° = 7.654∠67.50o 1 7.6542 ∴P = = 7.322 W 2 4
(d)
v = 10 cos10t + 10sin (10t + 45°) + 10 cos 40t ; 10∠0° + 10∠ − 45° = 18.48∠ − 22.50o ∴P =
(e)
1 1 1 1 × 18.482 × + × 102 × = 55.18 W 2 4 2 4
102 // + 10dc ∴ Pav = 55.18 + = 80.18 W 4
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS
26.
Zeq = R || j0.3ω =
j 0.3Rω . By voltage division, then, we write: R + j 0.3Rω
j 0.1ω - 0.03ω 2 + j 0.1ωR V100mH = 120∠0 = 120∠0 j 0.3Rω − 0.03ω 2 + j 0.4 Rω j 0.1ω + R + j 0.3ω j 0.3Rω j 36 Rω R + j 0.3ω V300mH = 120∠0 = 120∠0 j 0.3Rω − 0.03ω2 + j 0.4 Rω j 0.1ω + R + j 0.3ω (a) We’re interested in the value of R that would lead to equal voltage magnitudes, or j 36 Rω Thus, 36Rω =
=
(
(120) - 0.03ω 2 + j 0.1ωR
)
12.96ω 4 + 144ω 2 R 2 or R = 0.1061 ω
(b) Substituting into the expression for V100mH, we find that V100mH = 73.47 V, independent of frequency. To verify with PSpice, simulate the circuit at 60 Hz, or ω = 120π rad/s, so R = 40 Ω. We also include a miniscule (1 pΩ) resistor to avoid inductor loop warnings. We see from the simulation results that the two voltage magnitudes are indeed the same.
FREQ VM($N_0002,$N_0003)VP($N_0002,$N_0003) 6.000E+01 7.349E+01 -3.525E+01 FREQ VM($N_0001,$N_0002)VP($N_0001,$N_0002) 6.000E+01 7.347E+01 3.527E+01
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 27. (a)
Vav ,1 = 30V 1 Vav ,2 = (10 + 30 + 50) = 30V 3
(b)
Veff ,1 =
1 3 1 1 (20t ) 2 dt = × 400 × × 27 = 1200 = 34.64V ∫ 0 3 3 3
Veff ,2 =
1 2 1 (10 + 30 2 + 50 2 ) = × 3500 = 34.16 V 3 3
(c) PSpice verification for Sawtooth waveform of Fig. 11.40a:
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS
28.
− j106 − jR106 = Zeff = R || 6 3ω 3ωR − j10
(
)
120∠0 120ω 3ωR - j106 = ISRC = 106 R106 − j106 3ωR − j106 − jωR106 −j −j ω 3ωR − j106 R I3µF = ISRC 106 R− j 3ω
(
)
R
= 1 . This is 106 R− j 3ω only true when R = ∞; otherwise, current is shunted through the resistor and the two capacitor currents will be unequal. (b) In this case, the capacitor current is
(a) For the two current magnitudes to be equal, we must have
120∠0
1 6
10 106 −j −j ω 3ω
= j 90ω µA, or
90ω cos(ωt + 90o ) µA
(c) PSpice verification: set f = 60 Hz, simulate a single 0.75-µF capacitor, and include a 100-MΩ resistor in parallel with the capacitor to prevent a floating node. This should resit in a rms current amplitude of 33.93 mA, which it does.
FREQ IM(V_PRINT3) 6.000E+01 3.393E-02
Engineering Circuit Analysis, 6th Edition
IP(V_PRINT3) 9.000E+01
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 29. v(t ) = 10t [u (t ) − u (t − 2)] + 16e −0.5(t −3) [u (t − 3) − u (t − 5)] V Find eff. value separately V1,eff =
1 2 20 100t 2 dt = × 8 = 7.303 ∫ 0 5 3
V2,eff =
1 5 256 3 − t 5 256e − (t −3) dt = e (−e )3 = 6.654 ∫ 3 5 5
∴ Veff = 7.3032 + 6.6542 = 9.879 Veff =
5 1 2 2 100 t dt + 256e3e − t dt ∫ ∫ 0 3 5
=
1 100 × 8 + 256e3 (e −3 − e−5 ) 5 3
=
1 800 + 256 (1 − e−2 ) = 9.879 V OK 5 3
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 30.
The peak instantaneous power is 250 mW. The combination of elements yields Z = 1000 + j1000 Ω = 1414 ∠45o Ω. Arbitrarily designate V = Vm ∠0 , so that I =
Vm ∠0 Vm ∠ − 45o = A. Z 1414
We may write p(t) = ½ Vm Im cos φ + ½ Vm Im cos (2ωt + φ) where φ = the angle of the current (-45o). This function has a maximum value of ½ VmIm cos φ + ½ VmIm. Thus, 0.250 = ½ VmIm (1 + cos φ) = ½ (1414) Im2 (1.707) and Im = 14.39 mA. In terms of rms current, the largest rms current permitted is 14.39 /
Engineering Circuit Analysis, 6th Edition
2 = 10.18 mA rms.
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 31.
I = 4∠35° A rms
(a)
V = 20I + 80∠35° Vrms, Ps , gen = 80 × 10 cos 35° = 655.3 W
(b)
PR = I R = 16 × 20 = 320 W
(c)
PLoad = 655.3 − 320 = 335.3 W
(d)
APs , gen = 80 × 10 = 800 VA
(e)
APR = PR = 320 VA
(f)
I L = 10∠0° − 4∠35° = 7.104∠ − 18.84° A rms
2
∴ APL = 80 × 7.104 = 568.3 VA (g)
PFL = cos θ L = since I L lags V,
PL 335.3 = = 0.599 APL 568.3 PFL is lagging
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 32. (a)
(b) (c)
(d)
120 = 9.214∠ − 26.25° A rms j192 4+ 12 + j16 ∴ PFs = cos 26.25 = 0.8969 lag Is =
Ps = 120 × 9.214 × 0.8969 = 991.7W j 48 1 = 4+ (192 + j144) 3 + j4 25 11.68 − j 5.76 ∴ Z L = 11.68 + j 5.76 Ω, YL = 11.682 + 5.762 j 5.76 , C = 90.09 µ F ∴ j120π C = 11.682 + 5.762 ZL = 4 +
PSpice verification
FREQ VM($N_0003,0) VP($N_0003,0) 6.000E+01 1.200E+02 0.000E+00 FREQ IM(V_PRINT1) IP(V_PRINT1) 6.000E+01 9.215E+00 -2.625E+01
; (a) and (b) are correct
Next, add a 90.09-µF capacitor in parallel with the source: FREQ IM(V_PRINT1) IP(V_PRINT1) 6.000E+01 8.264E+00 -9.774E-05
Engineering Circuit Analysis, 6th Edition
;(c) is correct (-9.8×10-5 degrees is essentially zero, for unity PF).
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 33.
Z A = 5 + j 2 Ω, Z B = 20 − j10 Ω, Z c = 10∠30° Ω = 8.660 + j5 Ω Z D = 10∠ − 60° = 5 − j8.660 Ω 200 −20 + j10 0 33.66 − j13.660 7265∠22.09° I1 = = = 15.11∠3.908° A rms 25 − j8 −20 + j10 480.9∠ − 26.00° −20 + j10 33.66 − j13.660 25 − j8 200 −20 + j10 0 200 (20 − j10) I2 = = = 9.300∠ − 0.5681° A rms 480.9∠ − 26.00° 480.9∠20.00° APA = I1 Z A = 15.1082 29 = 1229 VA 2
APB = I1 − I 2
2
Z B = 5.8812 × 10 5 = 773.5 VA
APC = I 2 2 ZC = 9.32 × 10 = 86.49 VA APD = I 2
2
Z1 = 9.32 × 10 = 864.9 VA
APS = 200 I1 = 200 × 15.108 = 3022 VA
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 34.
Perhaps the easiest approach is to consider the load and the compensation capacitor separately. The load draws a complex power Sload = P + jQ. The capacitor draws a purely reactive complex power SC = -jQC. θload = tan-1(Q/P), or Q = P tan θload QC = SC = Vrms
Vrms 2 2 = ω CVrms = ω CVrms (− j / ω C)
Stotal = Sload + SC = P + j(Q – QC) Q-QC θnew = ang(Stotal) = tan −1 P
, so that Q – QC = P tan θnew
Substituting, we find that QC = P tan θload – P tan θnew or 2 ω CVrms = P (tan θload – tan θnew) Thus, noting that θold = θload, C =
Engineering Circuit Analysis, 6th Edition
P ( tan θ old - tan θ new ) 2 ω Vrms
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CHAPTER 11 SOLUTIONS 35.
Z1 = 30∠15°Ω, Z 2 = 40∠40°Ω
(a)
Ztot = 30∠15° + 40∠40° = 68.37∠29.31°Ω ∴ PF = cos 29.3° = 0.8719 lag
(b)
Ztot = 68.37∠29.31° = 59.62 + j 33.48 PFnew = 0.9 lag ∴θ new = cos −1 0.9 = 25.84° tan 25.84° = 0.4843 = ∴ 33.48 −
X new ∴ X new = 28.88 Ω 59.62
1 = 28.88, 100π C
C = 691.8µ F
Engineering Circuit Analysis, 6th Edition
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CHAPTER 11 SOLUTIONS 36.
θ1 = cos-1(0.92) = 23.07o, θ 2 = cos-1 (0.8) = 36.87o, θ 3 = 0 100 ∠23.07o S1 = = 100 + j 42.59 VA 0.92 250 ∠36.87 o S2 = = 250 + j187.5 VA 0.8 500 ∠0o S3 = = 500 VA 1 Stotal = S1 + S2 + S3 = 500 + j230.1 VA = 550.4 ∠24.71o VA (a) Ieff =
Stotal 550.4 = = 4.786 A rms Veff 115
(b) PF of composite load = cos (24.71o) = 0.9084 lagging
Engineering Circuit Analysis, 6th Edition
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CHAPTER 11 SOLUTIONS 37. APL = 10, 000 VA, PFL = 0.8lag, I L = 40A rms Let I L = 40∠0° A rms; PL = 10, 000 × 0.8 = 8000 W 8000 =5 Ω 402 cosθ L = 0.8lag∴θ L = cos−1 0.8 = 36.87° Let Z L = R L + jX L ∴ R L =
∴ X L = 5 tan 36.87° = 3.75 Ω, Z L = 5 + j 3.75, Ztot = 5.2 + j 3.75 Ω 1 5.2 + j 3.75 = 0.12651 + j (120π C − 0.09124),
∴ Vs = 40 (5.2 + j 3.75) = 256.4∠35.80° V; Ytot = = 0.12651 − j 0.09124S, Ynew
PFnew = 0.9 lag,θ new = 25.84°∴ tan 25.84° = 0.4843 0.09124 − 120π C ∴ 0.12651 C = 79.48µ F =
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 38.
Zeff = j100 + j300 || 200 = 237 ∠54.25o. PF = cos 54.25o = 0.5843 lagging. (a) Raise PF to 0.92 lagging with series capacitance Znew = j100 + jXC + j300 || 200 = 138.5 + j(192.3 + XC) Ω 192.3 + X C -1 o tan −1 = cos 0.92 = 23.07 138.5 Solving, we find that XC = -133.3 Ω = -1/ωC, so that C = 7.501 µF (b) Raise PF to 0.92 lagging with parallel capacitance Znew = j100 || jXC +
j300 || 200 =
− 100 X C +138.5 + j92.31 Ω j (100 + X C )
100X C Ω = 138.5 + j 92.31 + 100 + X C 100X C 92.31 + 100 + X C −1 tan = cos-1 0.92 = 23.07 o 138.5 Solving, we find that XC = -25 Ω = -1/ωC, so that C =
40 µF
General circuit for simulations. Results agree with hand calculations With no compensation: With series compensation: With parallel compensation:
FREQ 1.592E+02 1.592E+02 1.592E+02
Engineering Circuit Analysis, 6th Edition
IM(V_PRINT1) 4.853E-01 7.641E-01 7.641E-01
IP(V_PRINT1) -5.825E+01 -2.707E+01 -2.707E+01
θ 54.25o 23.07o 23.07o
PF 0.5843 lag 0.9200 lag 0.9200 lag
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 39. (a) (b)
Ps ,tot = 20 + 25 × 0.8 + 30 × 0.75 = 70 kW 20, 000 = 80∠0° A rms 250 I 2 = 25, 000 / 250 = 100 A rms I1 =
∠I 2 = − cos−1 0.8 = −36.87 ∴ I 2 = 100∠ − 36.87o A rms 30, 000 40, 000 = 40, 000 VA, I 3 = = 160 A rms 0.75 250 ∠ I 3 = − cos−1 0.75 = −41.41° ∴ I 3 = 160∠ − 41.41° A rms AP3 =
∴ I s = 80∠0° + 100∠ − 36.87° + 160∠ − 41.41° = 325.4∠ − 30.64° A rms ∴ APs = 250 × 325.4 = 81, 360 VA (c)
PF3 =
70, 000 = 0.8604 lag 81,360
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 40.
200 kW average power and 280 kVAR reactive result in a power factor of PF = cos (tan-1 (280/200) = 0.5813 lagging, which is pretty low. (a) 0.65 peak = 0.65(200) = 130 kVAR Excess = 280 – 130 = 150 kVAR, for a cost of (12)(0.22)(150) = $396 / year. (b) Target = S = P + j0.65 P θ = tan-1(0.65P/P) = 33.02o, so target PF = cos θ = 0.8385 (c) A single 100-kVAR increment costs $200 to install. The excess kVAR would then be 280 – 100 – 130 = 50 kVAR, for an annual penalty of $332. This would result in a first-year savings of $64. A single 200-kVAR increment costs $395 to install, and would remove the entire excess kVAR. The savings would be $1 (wow) in the first year, but $396 each year thereafter. The single 200-kVAR increment is the most economical choice.
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 41. 20 (1 + j 2) = 10.769 − j 3.846 = 11.435+ ∠ − 19.65° Ω 3 + j2 100 ∴ Is = = 8.745∠19.65° 11.435∠ − 19.654° ∴ S s = − Vs I∗s = −100 × 8.745∠ − 19.65° = −823.5 + j 294.1VA
Zin = − j10 +
I 20 = 8.745∠19.65° ×
10 + j 20 = 5.423∠49.40° 30 + j 20
∴ S 20 = 20 × 5.432 2 = 588.2 + j 0 VA I10 =
20 × 5.423∠49.40 = 4.851∠ − 14.04° 10 + j 20
S10 = 10 × 4.8512 = 235.3 + j 0 VA S j 20 = j 20 × 4.8512 = j 470.6 VA, S − j10 = − j10 × 8.7452 =
− j 764.7 VA,
Engineering Circuit Analysis, 6th Edition
Σ=0
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 42. Vx − 100 V V − j100 + x + x =0 6 + j4 − j10 5 1 100 ∴ Vx + j 0.1 + 0.2 = + j 20 6 + j4 6 + j4 ∴ Vx = 53.35− ∠42.66° V 100 − 53.35− ∠42.66° = 9.806∠ − 64.44° A 6 + j4 1 ∴ S1. gen = × 100 × 9.806∠64.44° = 211.5 + j 4423VA 2 1 S 6,abs = × 6 × 9.8062 = 288.5 + j 0 VA 2 1 S j 4, abs = ( j 4) 9.8062 = 0 + j192.3VA 2 j100 − 53.35− ∠42.66° I2 = = 14.99∠121.6°, 5 1 S5 abs = × 5 × 14.992 = 561.5 + j 0 VA 2 1 S 2, gen = ( j100)14.99∠ − 121.57° = 638.4 − j 392.3VA 2 1 53.35 S − j10,abs = (− j10) = 0 − j142.3VA = 142.3∠0 VA 2 10 ∴ I1 =
Engineering Circuit Analysis, 6th Edition
Σ=0
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 43. (a)
500 VA, PF = 0.75 lead∴ S = 500∠ − cos −1 0.75 = 375 − j 330.7 VA
(b)
500W, PF = 0.75 lead∴ S = 500 −
(c)
500 sin (cos −1 0.75) = 500 − j 441.0 VA j.075
−500 VAR, PF = 0.75 (lead) ∴θ = − cos −1 0.75 = −41.41° ∴ P 500 / tan 41.41° = 566.9W, S = 566.9 − j 500 VA
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 44. (a)
S s = 1600 + j500 VA (gen) 1600 + j 500 = 4 + j1.25 ∴ I s = 4 − j1.25 400 400 Ic = = j 3.333A rms∴ I L = I s − I c = 4 − j1.25 − j 3.333 − j120 ∴ I L = 4 − j 4.583A rms∴ I ∗s =
S L = 400 (4 + j 4.583) = 1600 + j1833 VA (b)
1833.3 + PFL = cos tan −1 = 0.6575 lag 1600
(c)
S s = 1600 + j 500 = 1676∠17.35° VA ∴ PFs = cos17.35° = 0.9545 lag
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 45.
(cos−1 0.8 = 36.87°, cos−1 0.9 = 25.84°)
(a)
Stot = 1200∠36.87° + 1600∠25.84° + 900 = 960 + j 720 + 1440 + j 697.4 + 900 = 3300 + j1417.4 = 3592∠23.25° VA 3591.5 ∴ Is = = 15.62 A rms 230
(b)
PFs = cos 23.245° = 0.9188
(c)
S = 3300 + j1417 VA
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 46.
V = 339 ∠-66o V, ω = 100π rad/ s, connected to Z = 1000 Ω. 339 = 239.7 V rms 2 (b) pmax = 3392 / 1000 = 114.9 W (a) Veff =
(c) pmin = 0 W 339 Veff2 339 2 (d) Apparent power = Veff Ieff = = = 57.46 VA 2 1000 1000 (e) Since the load is purely resistive, it draws zero reactive power. (f) S = 57.46 VA
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 47.
V = 339 ∠-66o V, ω = 100π rad/s to a purely inductive load of 150 mH (j47.12 Ω) V 339∠ - 66o = = 7.194 ∠ - 156o A Z j 47.12 7.194 = 5.087 A rms so Ieff = 2 (b) p(t) = ½ VmIm cos φ + ½ VmIm cos(2ωt + φ) where φ = angle of current – angle of voltage pmax = ½ VmIm cos φ + ½ VmIm = (1 + cos(-90o)) (339)(7.194)/ 2 = 1219 W (a) I =
(c) pmin = ½ VmIm cos φ - ½ VmIm = -1219 W 339 (5.087 ) = 1219 VA 2 (e) reactive power = Q = Veff Ieff sin (θ – φ) = 1219 VA
(d) apparent power = Veff Ieff =
(f) complex power = j1219 VA
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 48.
1 H → j Ω, 4 µF → -j250 kΩ Zeff = j || -j250×103 || 103 Ω = 1 ∠89.99o Ω V10k = (a) pmax
(5∠0) (1 ∠89.99o ) = 0.002 ∠89.97o V 2500 + (1 ∠89.99o ) = (0.002)2 / 10×103 = 400 pW
(b) 0 W (purely resistive elements draw no reactive power) (c) apparent power = VeffIeff = ½ VmIm = ½ (0.002)2 / 10000 = 200 pW 5∠0 = 0.002 ∠ - 0.02292o A o 2500 + 1∠89.99 S = ½ VmIm ∠(89.99o + 0.02292o) = 0.005 ∠90.01o VA (d) Isource =
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 49.
(a) At ω = 400 rad/s, 1 µF → -j2500 Ω, 100 mH → j40 Ω Define Zeff = -j2500 || (250 + j40) = 256 ∠ 3.287o Ω 12000∠0 = 43.48 ∠ - 3.049o A rms 20 + 256∠3.287o Ssource = (12000)(43.48) ∠ 3.049o = 521.8 ∠3.049o kVA
IS =
S20Ω = (43.48)2 (20) ∠0 = 37.81 ∠0 kVA Veff
(12000∠0)(256∠3.287 o ) = = 11130 ∠0.2381o V rms o 20 + 256∠3.287
I1µF =
Veff = 4.452 ∠90.24o A rms - j 2500 so S1µF = (11130)(4.452) ∠-90o = 49.55 ∠-90o kVA
V100mH =
(11130∠0.2381o )( j 40) = 1758 ∠81.15o V rms 250 + j 40
V100mH = 43.96 ∠ - 8.852o A rms j 40 so S100µΗ = (1758)(4.43.96) ∠90o = 77.28 ∠90o kVA
I100mH =
(11130∠0.2381o )(250) = 10990 ∠ − 8.852o V rms 250 + j 40 so S250Ω = (10990)2 / 250 = 483.1 ∠0o kVA
V250Ω =
(b) 37.81 ∠0 + 49.55 ∠-90o +77.28 ∠90o + 483.1 ∠0o = 521.6 ∠3.014o kVA, which is within rounding error of the complex power delivered by the source. (c) The apparent power of the source is 521.8 kVA. The apparent powers of the passive elements sum to 37.81 + 49.55 + 77.28 + 483.1 = 647.7 kVA, so NO! Phase angle is important! (d) P = Veff Ieff cos (ang VS – ang IS) = (12000)(43.48) cos (3.049o) = 521 kW (e) Q = Veff Ieff sin (ang VS – ang IS) = (12000)(43.48) sin (3.049o) = 27.75 kVAR
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 50.
(a) Peak current = 28 2 = 39.6 A (b) θload = cos-1(0.812) = +35.71o (since lagging PF). Assume ang (V) = 0o.
(
)(
)
p(t) = 2300 2 39.60 2 cos (120πt ) cos (120πt - 35.71o ) at t = 2.5 ms, then, p(t) = 147.9 kW (c) P = Veff Ieff cos θ = (2300)(28) cos (35.71o) = 52.29 kW (d) S = Veff Ieff ∠θ = 64.4 ∠ 35.71o kVA (e) apparent power = |S| = 64.4 kVA (f) |Zload| = |V/ I| = 2300/28 = 82.14 Ω. Thus, Zload = 82.14 ∠ 35.71o Ω (g) Q = Veff Ieff sin θ = 37.59 kVAR
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
I c = 1.0600∠57.99° A ps (π / 2ms) = 84.80 cos (45° − 32.01°) 2 cos 45° = 116.85 W pR = 50 × 1.696 2 cos 2 (45° − 32.01°) = 136.55 W pc = 84.80 cos (45° − 32.01°) = 1.060 cos (45° + 57.99°) = −19.69 W
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 2. (a)
1 2 1 Li = × 4 (4t 4 − 4t 2 + 1) 2 2 4 2 4 2 ∴ wL = 8t − 8t + 2 ∴ wL (3) − wL (1) = 8 × 3 − 8 × 3 + 2 − 8 ×1 + 8 ×1 − 2 = 576 J 4H : i = 2t 2 − 1∴ v = Li′ = 4 (4t ) = 16t , wL =
t
(b)
1 t 2 2 2 2 0.2 F : vc = (2t − 1) dt + 2 = 5 t 3 − t + 2 = 5 t 3 − t − 5 − 1 + 2 ∫ 0.2 1 3 1 3 3 10 10 61 61 ∴ vc (2) = × 8 − 10 − + 5 + 2 = V ∴ Pc (2) = × 7 = 142.33 W 3 3 3 3
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS R 1 = 2, ω o2 = = 3, s1,2 = −2 ± 1 = −1, − 3 2L LC
3.
vc (0) = −2V, i (0) = 4A, α =
(a)
1 i = Ae− t + Be−3t ∴ A + B = 4; i (0+ ) = vL (0+ ) = (−4 × 4 × +2) = −14 1 ∴− A − 38 = −14 ∴ B = 5, A = −1, i = −e− t + 5e −3t A t
∴+ vc = 3∫ (−e − t + 5e −3t ) dt − 2 = 3(e− t − 5e −3t ) to − 2 = e − t − 3 − 5e −3t + 5 − 2 o
−t
∴ vc = 3e − 5e −3t ∴ Pc (0 + ) = (3 − 5) (−1 + 5) = −8 W (b)
Pc (0.2) = (3e −0.2 − 5e −0.6 ) (−e0.2 + 5e −0.6 ) = −0.5542 W
(c)
Pc (0.4) = (3e −0.4 − 5e −1.2 ) (5e −1.2 − e −0.4 ) = 0.4220 W
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 4.
We assume the circuit has already reached sinusoidal steady state by t = 0. 2.5 kΩ → 2.5 kΩ, 1 H → j1000 Ω, 4 µF → -j250 Ω, 10 kΩ → 10 kΩ Zeq = j1000 || -j250 || 10000 = 11.10 – j333.0 Ω (20∠30)(11.10 − j 333.0) = 2.631∠50.54o V 2500 + 11.10 − j 333.0 Veq Veq I10k = = 0.2631 ∠ - 50.54o mA I1 H = = 2.631 ∠ - 140.5o mA 10000 j1000 Veq (20∠30)(2500) I4 µF = = 10.52 ∠39.46o mA V2.5k = = 19.74∠37.55o V − j 250 2500 + 11.10 − j 333.0 Veq =
[19.74 cos 37.55 ] P2.5k =
o 2
Thus,
=
2500
[
97.97 mW
][
]
P1 H = 2.631cos(− 50.54 ) 2.631 × 10-3 cos(−140.5o ) = - 3.395 mW o
[ ( )][ [2.631cos(− 50.54 )] = P2.5k =
]
P4 µF = 2.631cos − 50.54o 10.52 × 10-3 cos(39.46o ) = 13.58 mW o
10000
2
279.6 µW
FREQ IM(V_PRINT1) IP(V_PRINT1) 1.592E+02 7.896E-03 3.755E+01
FREQ VM(L,0) 1.592E+02 2.629E+00
FREQ VM(R2_5k,$N_0002)VP(R2_5k,$N_0002) 1.592E+02 1.974E+01 3.755E+01
FREQ IM(V_PRINT11) IP(V_PRINT11) 1.592E+02 1.052E-02 3.946E+01
FREQ IM(V_PRINT2) IP(V_PRINT2) 1.592E+02 2.628E-03 -1.405E+02
FREQ IM(V_PRINT12) IP(V_PRINT12) 1.592E+02 2.629E-04 -5.054E+01
Engineering Circuit Analysis, 6th Edition
VP(L,0) -5.054E+01
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 5. is → 5∠0° A, C → − j 4 Ω, Z in = 8 (3 − j 4) =
40∠ − 53.13° 11 − j 4
= 3.417∠ − 33.15°∴ Vs = 17.087∠ − 33.15°, vs = 17.087 cos (25t − 33.15°) V ∴ Ps , abs (0.1) = −17.087 cos (2.5rad − 33.147°) × 5cos 2.5rad = −23.51 W 17.087 cos (25t − 33.15°) ∴ 8 i8 (0.1) = 2.136 cos (2.5rad − 33.15°) = −0.7338 A i8 =
∴ P8, abs = 0.73382 × 8 = 4.307 W; I3 =
17.087∠ − 33.15° = 3.417∠19.98° A 3 − j4
∴ i3 (0.1) = 3.417 cos (2.5rad + 19.98°) = −3.272 A ∴ P3, abc = 3.2722 × 3 = 32.12 W Vc = − j 4 (3.417∠19.983°) = 13.67∠ − 70.02°, vc (0.1) = 13.670 cos (2.5rad − 70.02°) = 3.946 V ∴ Pc , abc = 3.946 (−3.272) = −12.911 W
Engineering Circuit Analysis, 6th Edition
(Σ = 0)
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 6. j 5(10 − j 5) = 4 + 2.5 + j 5 = 6.5 + j 5 Ω 10 100 ∴ Is = = 12.194∠ − 37.57° A 6.5 + j 5 1 ∴ Ps ,abs = − × 100 × 12.194 cos 37.57° = −483.3W 2 1 P4, abs = (12.194) 2 4 = 297.4 W, 2 Pcabs = 0 Zin = 4 +
100 j5 = 6.097∠52.43° so 6.5 + j 5 10 1 P10,abs = (6.097) 2 × 10 = 185.87 W 2 PL = 0 (Σ = 0) I10 =
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 7. V = (10 + j10)
40∠30° = 52.44∠69.18° V 5∠50° + 8∠ − 20°
1 × 10 × 52.44 cos 69.18° = 93.20 W 2 1 = × 10 × 52.44 cos (90° − 69.18°) = 245.08 W 2
P10, gen = Pj10, gen
2
P8 ∠− 20 abs
1 52.44 = 8cos (−20°) = 161.51 W 2 8
Engineering Circuit Analysis, 6th Edition
(Σ gen = Σ abs )
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 8. ZR = 3 +
1 = 3 + 1 + j3 = 4 + j3 Ω 0.1 − j 0.3
Ignore 30° on Vs , I R = 5
2 + j5 5 29 , IR = 6 + j8 10
2
1 5 29 = × 3 = 10.875 W 2 10
(a)
P3 Ω
(b)
Vs = 5∠0° ∴ Ps , gen =
(2 + j 5) (4 + j 3) = 13.463∠51.94° V 6 + j8
1 × 13.463 × 5cos 51.94° = 20.75 W 2
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 9.
Pj10 = P− j 5 = 0, V10 − 50 V10 V10 − j 50 + + =0 j10 10 − j5 ∴ V10 (− j 0.1 + 0.1 + j 0.2) + j 5 + 10 = 0 ∴ V10 = 79.06∠16.57° V 1 79.06 2 = 312.5 W; 2 10 79.06∠161.57° − 50 I 50 = = 12.75∠78.69° A j10 1 ∴ P50V = × 50 × 12.748cos 78.69° = 62.50 W 2 79.06∠161.57° − j 50 = 15.811∠ − 7.57° : I j 50 = − j5 1 Pj 50 = × 50 × 15.811cos (90° + 71.57°) = −375.0 W 2 P10 Ω =
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 10. Vx − 20 Vx − Vc + = 2Vc , 2 3 3Vx − 60 + 2Vx − 2Vc = 12Vc Vc − Vx V + c =0 3 − j2 ∴ 2Vc − 2Vx + j3Vc = 0, − 2Vx + (2 + j3) Vc = 0 ∴ 5Vx − 14Vc = 60,
60 −14 0 2 + j3 120 + j180 Vx = = = 9.233∠ − 83.88° V 5 −14 10 + j15 − 28 −2 2 + j 3
Pgen
5 60 −2 0 Vc = = 5.122∠ − 140.9° V ∴ −18 + j15 1 = × 9.233 × 2 × 5.122 cos ( −83.88° + 140.19°) = 26.23 W 2
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 11. (a)
X in = 0 ∴ Z L = R th + j 0
(b)
R L , X L independent∴ Z L = Zth∗ = R th − jXth
(c)
Vth 1 R L fixed∴ PL = × R L ∴ Z L = R L − jXth 2 (R th + R L )2 + (X th + X L )2
(d)
X L fixed, Let X L + Xth = a ∴ f =
2
2PL Vth
2
=
RL (R th + R L ) 2 + a 2
R + R 2L + a 2 − 2R L (R th + R L ) df = th =0 2 dRL (R th + R L ) 2 + a 2 R th2 + 2R th R L + R 2L + a 2 − 2R th R L = 2R 2L = 0 ∴ R L = R th2 + a 2 = (e)
R th2 + (Xth + X L )2
X L = 0 ∴ R L = R th2 + X th2 = Zth
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 12.
−10 = 107.33∠ − 116.57° V 10 + j 5 − j10 (10 + j15) Zth = = 8 − j14 Ω 10 + j 5 Vth = 120
(a)
∴ Z L = 8 + j15 Ω
(b)
IL =
107.33∠ − 116.57° ∴ 16 2
PL ,max =
1 107.33 × 8 = 180 W 2 16
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 13.
R L = Zth ∴ R L = 82 + 142 = 16.125 Ω PL =
1 107.332 × 16.125 = 119.38 W 2 (8 + 16.125)2 + 142
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 14. − j 9.6 = −4.8 I x − j1.92 I x − +4.8I x 9.6 =5 1.92 ∴ V = (0.6 × 5)8 = 24 V 1 ∴ Po = × 24 × 1.6 × 5 = 96 W ( gen) 2 ∴ Ix =
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 15. j 480 80 − j 60 80 + j 60 80 − j 60 = 28.8 + j 38.4 Ω ∴ Z L max = 28.8 − j 38.4 Ω
(a)
Zth = 80 j 60 =
(b)
Vth = 5(28.8 + j 38.4) = 144 + j192 V, 144 + j192 2 × 28.8 1 1442 + 1922 and PL ,max = × 28.8 = 250 W 2 4 × 28.82
∴ IL =
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 16.
Zeq = (6 – j8) || (12 + j9) = 8.321 ∠ -19.44o W Veq = (5 ∠-30o) (8.321 ∠ -19.44o) = 41.61 ∠ -49.44o V Ptotal = ½ (41.61)(5) cos (-19.44o) = 98.09 W I6-j8 = Veq / (6 – j8) = 4.161 ∠ 3.69o A I4+j2 = I8+j7 = Veq/ 12+j9 = 2.774 ∠ -86.31o A P6-j8 = ½ (41.61)(4.161) cos (-49.44o – 3.69o) = 51.94 W P4+j2 = ½ (2.774)2 (4) = 15.39 W P8+j7 = ½ (2.774)2 (8) = 30.78 W Check: Σ = 98.11 W (okay)
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 17. Vth = 100
j10 (20) j10 = 20 + j 40, Z th = = 4 + j8 Ω 20 + j10 20 + j10
∴ R L = Zth ∴ R L = 8.944 Ω ∴ PL ,max =
1 202 + 402 × 8.944 = 38.63 W 2 (4 + 8.944)2 + 64
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 18.
We may write a single mesh equation: 170 ∠0o = (30 + j10) I1 – (10 – j50)(-λI1) Solving, 170∠0 o I1 = 30 + j10 + 10λ − j 50λ o 170∠0 (a) λ = 0, so I1 = = 5.376∠ - 18.43 o A and, with the same current flowing 30 + j10 through both resistors in this case, P20 = ½ (5.376)2 (20) = 289.0 W P10 = ½ (5.376)2 (10) = 144.5 W 170∠0 o = 3.005∠45 o A 40 − j 40 P20 = ½ (3.005)2 (20) = 90.30 W The current through the 10-Ω resistor is I1 + λI1 = 2 I1 = 6.01 ∠ 45o so
(b) λ = 1, so I1 =
P10 = ½ (6.01)2 (10) = 180.6 W (c)
(a) FREQ IM(V_PRINT3) 6.000E+01 5.375E+00
IP(V_PRINT3) -1.846E+01
FREQ IM(V_PRINT4) 6.000E+01 5.375E+00
IP(V_PRINT4) -1.846E+01
(b) FREQ IM(V_PRINT3) 6.000E+01 6.011E+00
IP(V_PRINT3) 4.499E+01
FREQ IM(V_PRINT4) 6.000E+01 3.006E+00
IP(V_PRINT4) 4.499E+01
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS
19.
(10)(1) + (−5)(1) + 0(1) = 1.667 A 3 1 (20)(1) + 0(1) Waveform (b): Iavg = 2 = 5A 2
(a) Waveform (a): Iavg =
Waveform (c): 1 Iavg = 1 × 10 −3 =−
(b)
∫
2πt dt = - 8 × 10 3 8sin −3 4 × 10
(
10 − 3 0
)
4 × 10 −3 πt cos −3 2π 2 × 10
10 −3
0
16 (0 − 1) = 16 A π π
(100)(1) + (25)(1) + (0)(1) = 41.67 A 2 3 Waveform (b): i(t) = -20×103 t + 20 i2(t) = 4×108 t2 – 8×105 t + 400 10 -3 1 2 I avg = 4 × 10 8 t 2 - 8 × 10 5 t + 400 dt -3 ∫ 0 2 × 10 5 4 × 10 8 2 1 0.1333 − 3 3 8 × 10 = 66.67 A 2 = 10 10 −3 + 400 10 −3 = -3 -3 2 × 2 × 10 3 2 10 Waveform (c):
2 Waveform (a): I avg =
(
)
(
2 I avg
1 = 1 × 10 −3
∫
Engineering Circuit Analysis, 6th Edition
)
(
0
)
(
(
)
2πt sin π × 10 3 t 3 t 64sin 2 dt = 64 × 10 4 × 10 −3 2π × 10 3 2
10 − 3
10 = 64 × 10 3 2
(
)
−3
−
)
10 −3
0
sin π = 32 A 2 3 2π × 10
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 20.
At ω = 120π, 1 H → j377 Ω, and 4 µF → -j663.1 Ω Define Zeff = j377 || -j663.1 || 10 000 = 870.5 ∠ 85.01o Ω
(400
)
2∠ − 9 o 2500 V2.5k = = 520.4 ∠ - 27.61o V o 2500 + 870.5 ∠85.01 400 2∠ − 9 o 870.5 ∠85.01o V10k = = 181.2 ∠57.40 o V o 2500 + 870.5 ∠85.01
(
)(
)
Thus, P2.5k = ½ (520.4)2 / 2 500 P10k = ½ (181.2)2 / 10 000 P1H P4µF
= = = =
54.16 W 1.642 W 0 0 (A total absorbed power of 55.80 W.)
To check, the average power delivered by the source: Isource =
400 2∠ − 9 o = 0.2081 ∠ - 27.61o A 2500 + 870.5∠85.01o
and Psource = ½ ( 400 2 )(0.2081) cos (-9o + 27.61o) = 55.78 W (checks out).
FREQ IM(V_PRINT1) 6.000E+01 2.081E-01
IP(V_PRINT1) -2.760E+01
FREQ VM(L,0) 6.000E+01 1.812E+02
VP(L,0) 5.740E+01
FREQ VM(R2_5k,$N_0002) VP(R2_5k,$N_0002) 6.000E+01 5.204E+02 -2.760E+01
FREQ IM(V_PRINT11) 6.000E+01 2.732E-01
IP(V_PRINT11) 1.474E+02
FREQ IM(V_PRINT2) 6.000E+01 4.805E-01
FREQ IM(V_PRINT12) 6.000E+01 1.812E-02
IP(V_PRINT12) 5.740E+01
IP(V_PRINT2) -3.260E+01
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 21. (a)
v = 10 + 9 cos100t + 6 sin100t 1 1 ∴ Veff = 100 + × 81 + × 36 = 158.5 = 12.590 V 2 2
(b)
Feff =
1 2 (10 + 20 2 + 102 ) = 150 = 12.247 4
(c)
Favg =
(10)(1) + (20)(1) + (10)(1) 40 = = 10 4 4
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 22. (a)
g(t) = 2 + 3cos100t + 4cos(100t – 120o) 3 ∠0 + 4∠-120 = 3.606 ∠-73.90 so Geff = o
(b)
o
3.6062 4+ = 3.240 2
h (t ) = 2 + 3cos100t + 4 cos (101t − 120°) 1 1 ∴ H eff = 2 + 32 + 4 2 = 16.5 = 4.062 2 2
(c)
f (t ) = 100t , 0 < t < 0.1∴ Feff = =
1 0.1 6 2 10 t dt 0.3 ∫0
10 1 × 106 × × 10−3 = 33.33 3 3
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 23.
f (t ) = (2 − 3cos100t ) 2
(a)
f (t ) = 4 − 12 cos100t + 9 cos 2 100t ∴ f (t ) = 4 − 12 cos100t + 4.5 + 4.5cos 200t ∴ Fav = 4 + 4.5 = 8.5
(b)
Feff = 8.52 +
1 1 × 122 + × 4.52 = 12.43 2 2
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS
24.
(a)
ieff
(
)
1 = 10 2 + (−5) 2 + 0 3
1
2
= 6.455 A
(b) ieff
1 1 = ∫ [− 20t + 20] dt + 0 2 0
(c) ieff
1 1 2π t dt = ∫ 8sin 0 4 1
Engineering Circuit Analysis, 6th Edition
1
2
1
2
=
5 = 2.236 A
1
=
2 πt - 8 π cos 2 = 2.257 A 0
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 25. (a)
A = B = 10V, C = D = 0 ∴10∠0° + 10∠ − 45° = 18.48∠ − 22.50o 1 1 ∴ P = × × 18.482 = 42.68 W 2 4
(b)
A = C = 10V, B = D = 0, vs = 10cos10t + 10 cos 40t , P=
(c)
1 102 1 102 + = 25 W 2 4 2 4
vs = 10 cos10t − 10sin (10t + 45°) → 10 − 10∠ − 45° = 7.654∠67.50o 1 7.6542 ∴P = = 7.322 W 2 4
(d)
v = 10 cos10t + 10sin (10t + 45°) + 10 cos 40t ; 10∠0° + 10∠ − 45° = 18.48∠ − 22.50o ∴P =
(e)
1 1 1 1 × 18.482 × + × 102 × = 55.18 W 2 4 2 4
102 // + 10dc ∴ Pav = 55.18 + = 80.18 W 4
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS
26.
Zeq = R || j0.3ω =
j 0.3Rω . By voltage division, then, we write: R + j 0.3Rω
j 0.1ω - 0.03ω 2 + j 0.1ωR V100mH = 120∠0 = 120∠0 j 0.3Rω − 0.03ω 2 + j 0.4 Rω j 0.1ω + R + j 0.3ω j 0.3Rω j 36 Rω R + j 0.3ω V300mH = 120∠0 = 120∠0 j 0.3Rω − 0.03ω2 + j 0.4 Rω j 0.1ω + R + j 0.3ω (a) We’re interested in the value of R that would lead to equal voltage magnitudes, or j 36 Rω Thus, 36Rω =
=
(
(120) - 0.03ω 2 + j 0.1ωR
)
12.96ω 4 + 144ω 2 R 2 or R = 0.1061 ω
(b) Substituting into the expression for V100mH, we find that V100mH = 73.47 V, independent of frequency. To verify with PSpice, simulate the circuit at 60 Hz, or ω = 120π rad/s, so R = 40 Ω. We also include a miniscule (1 pΩ) resistor to avoid inductor loop warnings. We see from the simulation results that the two voltage magnitudes are indeed the same.
FREQ VM($N_0002,$N_0003)VP($N_0002,$N_0003) 6.000E+01 7.349E+01 -3.525E+01 FREQ VM($N_0001,$N_0002)VP($N_0001,$N_0002) 6.000E+01 7.347E+01 3.527E+01
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 27. (a)
Vav ,1 = 30V 1 Vav ,2 = (10 + 30 + 50) = 30V 3
(b)
Veff ,1 =
1 3 1 1 (20t ) 2 dt = × 400 × × 27 = 1200 = 34.64V ∫ 0 3 3 3
Veff ,2 =
1 2 1 (10 + 30 2 + 50 2 ) = × 3500 = 34.16 V 3 3
(c) PSpice verification for Sawtooth waveform of Fig. 11.40a:
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS
28.
− j106 − jR106 = Zeff = R || 6 3ω 3ωR − j10
(
)
120∠0 120ω 3ωR - j106 = ISRC = 106 R106 − j106 3ωR − j106 − jωR106 −j −j ω 3ωR − j106 R I3µF = ISRC 106 R− j 3ω
(
)
R
= 1 . This is 106 R− j 3ω only true when R = ∞; otherwise, current is shunted through the resistor and the two capacitor currents will be unequal. (b) In this case, the capacitor current is
(a) For the two current magnitudes to be equal, we must have
120∠0
1 6
10 106 −j −j ω 3ω
= j 90ω µA, or
90ω cos(ωt + 90o ) µA
(c) PSpice verification: set f = 60 Hz, simulate a single 0.75-µF capacitor, and include a 100-MΩ resistor in parallel with the capacitor to prevent a floating node. This should resit in a rms current amplitude of 33.93 mA, which it does.
FREQ IM(V_PRINT3) 6.000E+01 3.393E-02
Engineering Circuit Analysis, 6th Edition
IP(V_PRINT3) 9.000E+01
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 29. v(t ) = 10t [u (t ) − u (t − 2)] + 16e −0.5(t −3) [u (t − 3) − u (t − 5)] V Find eff. value separately V1,eff =
1 2 20 100t 2 dt = × 8 = 7.303 ∫ 0 5 3
V2,eff =
1 5 256 3 − t 5 256e − (t −3) dt = e (−e )3 = 6.654 ∫ 3 5 5
∴ Veff = 7.3032 + 6.6542 = 9.879 Veff =
5 1 2 2 100 t dt + 256e3e − t dt ∫ ∫ 0 3 5
=
1 100 × 8 + 256e3 (e −3 − e−5 ) 5 3
=
1 800 + 256 (1 − e−2 ) = 9.879 V OK 5 3
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 30.
The peak instantaneous power is 250 mW. The combination of elements yields Z = 1000 + j1000 Ω = 1414 ∠45o Ω. Arbitrarily designate V = Vm ∠0 , so that I =
Vm ∠0 Vm ∠ − 45o = A. Z 1414
We may write p(t) = ½ Vm Im cos φ + ½ Vm Im cos (2ωt + φ) where φ = the angle of the current (-45o). This function has a maximum value of ½ VmIm cos φ + ½ VmIm. Thus, 0.250 = ½ VmIm (1 + cos φ) = ½ (1414) Im2 (1.707) and Im = 14.39 mA. In terms of rms current, the largest rms current permitted is 14.39 /
Engineering Circuit Analysis, 6th Edition
2 = 10.18 mA rms.
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 31.
I = 4∠35° A rms
(a)
V = 20I + 80∠35° Vrms, Ps , gen = 80 × 10 cos 35° = 655.3 W
(b)
PR = I R = 16 × 20 = 320 W
(c)
PLoad = 655.3 − 320 = 335.3 W
(d)
APs , gen = 80 × 10 = 800 VA
(e)
APR = PR = 320 VA
(f)
I L = 10∠0° − 4∠35° = 7.104∠ − 18.84° A rms
2
∴ APL = 80 × 7.104 = 568.3 VA (g)
PFL = cos θ L = since I L lags V,
PL 335.3 = = 0.599 APL 568.3 PFL is lagging
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 32. (a)
(b) (c)
(d)
120 = 9.214∠ − 26.25° A rms j192 4+ 12 + j16 ∴ PFs = cos 26.25 = 0.8969 lag Is =
Ps = 120 × 9.214 × 0.8969 = 991.7W j 48 1 = 4+ (192 + j144) 3 + j4 25 11.68 − j 5.76 ∴ Z L = 11.68 + j 5.76 Ω, YL = 11.682 + 5.762 j 5.76 , C = 90.09 µ F ∴ j120π C = 11.682 + 5.762 ZL = 4 +
PSpice verification
FREQ VM($N_0003,0) VP($N_0003,0) 6.000E+01 1.200E+02 0.000E+00 FREQ IM(V_PRINT1) IP(V_PRINT1) 6.000E+01 9.215E+00 -2.625E+01
; (a) and (b) are correct
Next, add a 90.09-µF capacitor in parallel with the source: FREQ IM(V_PRINT1) IP(V_PRINT1) 6.000E+01 8.264E+00 -9.774E-05
Engineering Circuit Analysis, 6th Edition
;(c) is correct (-9.8×10-5 degrees is essentially zero, for unity PF).
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 33.
Z A = 5 + j 2 Ω, Z B = 20 − j10 Ω, Z c = 10∠30° Ω = 8.660 + j5 Ω Z D = 10∠ − 60° = 5 − j8.660 Ω 200 −20 + j10 0 33.66 − j13.660 7265∠22.09° I1 = = = 15.11∠3.908° A rms 25 − j8 −20 + j10 480.9∠ − 26.00° −20 + j10 33.66 − j13.660 25 − j8 200 −20 + j10 0 200 (20 − j10) I2 = = = 9.300∠ − 0.5681° A rms 480.9∠ − 26.00° 480.9∠20.00° APA = I1 Z A = 15.1082 29 = 1229 VA 2
APB = I1 − I 2
2
Z B = 5.8812 × 10 5 = 773.5 VA
APC = I 2 2 ZC = 9.32 × 10 = 86.49 VA APD = I 2
2
Z1 = 9.32 × 10 = 864.9 VA
APS = 200 I1 = 200 × 15.108 = 3022 VA
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 34.
Perhaps the easiest approach is to consider the load and the compensation capacitor separately. The load draws a complex power Sload = P + jQ. The capacitor draws a purely reactive complex power SC = -jQC. θload = tan-1(Q/P), or Q = P tan θload QC = SC = Vrms
Vrms 2 2 = ω CVrms = ω CVrms (− j / ω C)
Stotal = Sload + SC = P + j(Q – QC) Q-QC θnew = ang(Stotal) = tan −1 P
, so that Q – QC = P tan θnew
Substituting, we find that QC = P tan θload – P tan θnew or 2 ω CVrms = P (tan θload – tan θnew) Thus, noting that θold = θload, C =
Engineering Circuit Analysis, 6th Edition
P ( tan θ old - tan θ new ) 2 ω Vrms
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 35.
Z1 = 30∠15°Ω, Z 2 = 40∠40°Ω
(a)
Ztot = 30∠15° + 40∠40° = 68.37∠29.31°Ω ∴ PF = cos 29.3° = 0.8719 lag
(b)
Ztot = 68.37∠29.31° = 59.62 + j 33.48 PFnew = 0.9 lag ∴θ new = cos −1 0.9 = 25.84° tan 25.84° = 0.4843 = ∴ 33.48 −
X new ∴ X new = 28.88 Ω 59.62
1 = 28.88, 100π C
C = 691.8µ F
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 36.
θ1 = cos-1(0.92) = 23.07o, θ 2 = cos-1 (0.8) = 36.87o, θ 3 = 0 100 ∠23.07o S1 = = 100 + j 42.59 VA 0.92 250 ∠36.87 o S2 = = 250 + j187.5 VA 0.8 500 ∠0o S3 = = 500 VA 1 Stotal = S1 + S2 + S3 = 500 + j230.1 VA = 550.4 ∠24.71o VA (a) Ieff =
Stotal 550.4 = = 4.786 A rms Veff 115
(b) PF of composite load = cos (24.71o) = 0.9084 lagging
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 37. APL = 10, 000 VA, PFL = 0.8lag, I L = 40A rms Let I L = 40∠0° A rms; PL = 10, 000 × 0.8 = 8000 W 8000 =5 Ω 402 cosθ L = 0.8lag∴θ L = cos−1 0.8 = 36.87° Let Z L = R L + jX L ∴ R L =
∴ X L = 5 tan 36.87° = 3.75 Ω, Z L = 5 + j 3.75, Ztot = 5.2 + j 3.75 Ω 1 5.2 + j 3.75 = 0.12651 + j (120π C − 0.09124),
∴ Vs = 40 (5.2 + j 3.75) = 256.4∠35.80° V; Ytot = = 0.12651 − j 0.09124S, Ynew
PFnew = 0.9 lag,θ new = 25.84°∴ tan 25.84° = 0.4843 0.09124 − 120π C ∴ 0.12651 C = 79.48µ F =
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 38.
Zeff = j100 + j300 || 200 = 237 ∠54.25o. PF = cos 54.25o = 0.5843 lagging. (a) Raise PF to 0.92 lagging with series capacitance Znew = j100 + jXC + j300 || 200 = 138.5 + j(192.3 + XC) Ω 192.3 + X C -1 o tan −1 = cos 0.92 = 23.07 138.5 Solving, we find that XC = -133.3 Ω = -1/ωC, so that C = 7.501 µF (b) Raise PF to 0.92 lagging with parallel capacitance Znew = j100 || jXC +
j300 || 200 =
− 100 X C +138.5 + j92.31 Ω j (100 + X C )
100X C Ω = 138.5 + j 92.31 + 100 + X C 100X C 92.31 + 100 + X C −1 tan = cos-1 0.92 = 23.07 o 138.5 Solving, we find that XC = -25 Ω = -1/ωC, so that C =
40 µF
General circuit for simulations. Results agree with hand calculations With no compensation: With series compensation: With parallel compensation:
FREQ 1.592E+02 1.592E+02 1.592E+02
Engineering Circuit Analysis, 6th Edition
IM(V_PRINT1) 4.853E-01 7.641E-01 7.641E-01
IP(V_PRINT1) -5.825E+01 -2.707E+01 -2.707E+01
θ 54.25o 23.07o 23.07o
PF 0.5843 lag 0.9200 lag 0.9200 lag
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 39. (a) (b)
Ps ,tot = 20 + 25 × 0.8 + 30 × 0.75 = 70 kW 20, 000 = 80∠0° A rms 250 I 2 = 25, 000 / 250 = 100 A rms I1 =
∠I 2 = − cos−1 0.8 = −36.87 ∴ I 2 = 100∠ − 36.87o A rms 30, 000 40, 000 = 40, 000 VA, I 3 = = 160 A rms 0.75 250 ∠ I 3 = − cos−1 0.75 = −41.41° ∴ I 3 = 160∠ − 41.41° A rms AP3 =
∴ I s = 80∠0° + 100∠ − 36.87° + 160∠ − 41.41° = 325.4∠ − 30.64° A rms ∴ APs = 250 × 325.4 = 81, 360 VA (c)
PF3 =
70, 000 = 0.8604 lag 81,360
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 40.
200 kW average power and 280 kVAR reactive result in a power factor of PF = cos (tan-1 (280/200) = 0.5813 lagging, which is pretty low. (a) 0.65 peak = 0.65(200) = 130 kVAR Excess = 280 – 130 = 150 kVAR, for a cost of (12)(0.22)(150) = $396 / year. (b) Target = S = P + j0.65 P θ = tan-1(0.65P/P) = 33.02o, so target PF = cos θ = 0.8385 (c) A single 100-kVAR increment costs $200 to install. The excess kVAR would then be 280 – 100 – 130 = 50 kVAR, for an annual penalty of $332. This would result in a first-year savings of $64. A single 200-kVAR increment costs $395 to install, and would remove the entire excess kVAR. The savings would be $1 (wow) in the first year, but $396 each year thereafter. The single 200-kVAR increment is the most economical choice.
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 41. 20 (1 + j 2) = 10.769 − j 3.846 = 11.435+ ∠ − 19.65° Ω 3 + j2 100 ∴ Is = = 8.745∠19.65° 11.435∠ − 19.654° ∴ S s = − Vs I∗s = −100 × 8.745∠ − 19.65° = −823.5 + j 294.1VA
Zin = − j10 +
I 20 = 8.745∠19.65° ×
10 + j 20 = 5.423∠49.40° 30 + j 20
∴ S 20 = 20 × 5.432 2 = 588.2 + j 0 VA I10 =
20 × 5.423∠49.40 = 4.851∠ − 14.04° 10 + j 20
S10 = 10 × 4.8512 = 235.3 + j 0 VA S j 20 = j 20 × 4.8512 = j 470.6 VA, S − j10 = − j10 × 8.7452 =
− j 764.7 VA,
Engineering Circuit Analysis, 6th Edition
Σ=0
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 42. Vx − 100 V V − j100 + x + x =0 6 + j4 − j10 5 1 100 ∴ Vx + j 0.1 + 0.2 = + j 20 6 + j4 6 + j4 ∴ Vx = 53.35− ∠42.66° V 100 − 53.35− ∠42.66° = 9.806∠ − 64.44° A 6 + j4 1 ∴ S1. gen = × 100 × 9.806∠64.44° = 211.5 + j 4423VA 2 1 S 6,abs = × 6 × 9.8062 = 288.5 + j 0 VA 2 1 S j 4, abs = ( j 4) 9.8062 = 0 + j192.3VA 2 j100 − 53.35− ∠42.66° I2 = = 14.99∠121.6°, 5 1 S5 abs = × 5 × 14.992 = 561.5 + j 0 VA 2 1 S 2, gen = ( j100)14.99∠ − 121.57° = 638.4 − j 392.3VA 2 1 53.35 S − j10,abs = (− j10) = 0 − j142.3VA = 142.3∠0 VA 2 10 ∴ I1 =
Engineering Circuit Analysis, 6th Edition
Σ=0
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 43. (a)
500 VA, PF = 0.75 lead∴ S = 500∠ − cos −1 0.75 = 375 − j 330.7 VA
(b)
500W, PF = 0.75 lead∴ S = 500 −
(c)
500 sin (cos −1 0.75) = 500 − j 441.0 VA j.075
−500 VAR, PF = 0.75 (lead) ∴θ = − cos −1 0.75 = −41.41° ∴ P 500 / tan 41.41° = 566.9W, S = 566.9 − j 500 VA
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 44. (a)
S s = 1600 + j500 VA (gen) 1600 + j 500 = 4 + j1.25 ∴ I s = 4 − j1.25 400 400 Ic = = j 3.333A rms∴ I L = I s − I c = 4 − j1.25 − j 3.333 − j120 ∴ I L = 4 − j 4.583A rms∴ I ∗s =
S L = 400 (4 + j 4.583) = 1600 + j1833 VA (b)
1833.3 + PFL = cos tan −1 = 0.6575 lag 1600
(c)
S s = 1600 + j 500 = 1676∠17.35° VA ∴ PFs = cos17.35° = 0.9545 lag
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 45.
(cos−1 0.8 = 36.87°, cos−1 0.9 = 25.84°)
(a)
Stot = 1200∠36.87° + 1600∠25.84° + 900 = 960 + j 720 + 1440 + j 697.4 + 900 = 3300 + j1417.4 = 3592∠23.25° VA 3591.5 ∴ Is = = 15.62 A rms 230
(b)
PFs = cos 23.245° = 0.9188
(c)
S = 3300 + j1417 VA
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 46.
V = 339 ∠-66o V, ω = 100π rad/ s, connected to Z = 1000 Ω. 339 = 239.7 V rms 2 (b) pmax = 3392 / 1000 = 114.9 W (a) Veff =
(c) pmin = 0 W 339 Veff2 339 2 (d) Apparent power = Veff Ieff = = = 57.46 VA 2 1000 1000 (e) Since the load is purely resistive, it draws zero reactive power. (f) S = 57.46 VA
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 47.
V = 339 ∠-66o V, ω = 100π rad/s to a purely inductive load of 150 mH (j47.12 Ω) V 339∠ - 66o = = 7.194 ∠ - 156o A Z j 47.12 7.194 = 5.087 A rms so Ieff = 2 (b) p(t) = ½ VmIm cos φ + ½ VmIm cos(2ωt + φ) where φ = angle of current – angle of voltage pmax = ½ VmIm cos φ + ½ VmIm = (1 + cos(-90o)) (339)(7.194)/ 2 = 1219 W (a) I =
(c) pmin = ½ VmIm cos φ - ½ VmIm = -1219 W 339 (5.087 ) = 1219 VA 2 (e) reactive power = Q = Veff Ieff sin (θ – φ) = 1219 VA
(d) apparent power = Veff Ieff =
(f) complex power = j1219 VA
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 48.
1 H → j Ω, 4 µF → -j250 kΩ Zeff = j || -j250×103 || 103 Ω = 1 ∠89.99o Ω V10k = (a) pmax
(5∠0) (1 ∠89.99o ) = 0.002 ∠89.97o V 2500 + (1 ∠89.99o ) = (0.002)2 / 10×103 = 400 pW
(b) 0 W (purely resistive elements draw no reactive power) (c) apparent power = VeffIeff = ½ VmIm = ½ (0.002)2 / 10000 = 200 pW 5∠0 = 0.002 ∠ - 0.02292o A o 2500 + 1∠89.99 S = ½ VmIm ∠(89.99o + 0.02292o) = 0.005 ∠90.01o VA (d) Isource =
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 49.
(a) At ω = 400 rad/s, 1 µF → -j2500 Ω, 100 mH → j40 Ω Define Zeff = -j2500 || (250 + j40) = 256 ∠ 3.287o Ω 12000∠0 = 43.48 ∠ - 3.049o A rms 20 + 256∠3.287o Ssource = (12000)(43.48) ∠ 3.049o = 521.8 ∠3.049o kVA
IS =
S20Ω = (43.48)2 (20) ∠0 = 37.81 ∠0 kVA Veff
(12000∠0)(256∠3.287 o ) = = 11130 ∠0.2381o V rms o 20 + 256∠3.287
I1µF =
Veff = 4.452 ∠90.24o A rms - j 2500 so S1µF = (11130)(4.452) ∠-90o = 49.55 ∠-90o kVA
V100mH =
(11130∠0.2381o )( j 40) = 1758 ∠81.15o V rms 250 + j 40
V100mH = 43.96 ∠ - 8.852o A rms j 40 so S100µΗ = (1758)(4.43.96) ∠90o = 77.28 ∠90o kVA
I100mH =
(11130∠0.2381o )(250) = 10990 ∠ − 8.852o V rms 250 + j 40 so S250Ω = (10990)2 / 250 = 483.1 ∠0o kVA
V250Ω =
(b) 37.81 ∠0 + 49.55 ∠-90o +77.28 ∠90o + 483.1 ∠0o = 521.6 ∠3.014o kVA, which is within rounding error of the complex power delivered by the source. (c) The apparent power of the source is 521.8 kVA. The apparent powers of the passive elements sum to 37.81 + 49.55 + 77.28 + 483.1 = 647.7 kVA, so NO! Phase angle is important! (d) P = Veff Ieff cos (ang VS – ang IS) = (12000)(43.48) cos (3.049o) = 521 kW (e) Q = Veff Ieff sin (ang VS – ang IS) = (12000)(43.48) sin (3.049o) = 27.75 kVAR
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
CHAPTER 11 SOLUTIONS 50.
(a) Peak current = 28 2 = 39.6 A (b) θload = cos-1(0.812) = +35.71o (since lagging PF). Assume ang (V) = 0o.
(
)(
)
p(t) = 2300 2 39.60 2 cos (120πt ) cos (120πt - 35.71o ) at t = 2.5 ms, then, p(t) = 147.9 kW (c) P = Veff Ieff cos θ = (2300)(28) cos (35.71o) = 52.29 kW (d) S = Veff Ieff ∠θ = 64.4 ∠ 35.71o kVA (e) apparent power = |S| = 64.4 kVA (f) |Zload| = |V/ I| = 2300/28 = 82.14 Ω. Thus, Zload = 82.14 ∠ 35.71o Ω (g) Q = Veff Ieff sin θ = 37.59 kVAR
Engineering Circuit Analysis, 6th Edition
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
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