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
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
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