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