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Chapter 10, Solution 1.
ω
1=
°∠→°− 45-10)45tcos(10
°∠→°+ 60-5)30tsin(5
jLjH1 =ω→
j-
Cj
1
F1 =
ω
→
The circuit becomes as shown below.
V
o
3
Ω
+
−
10∠-45° V
+
−
2 I
o
j
Ω
5
∠
-60° V
Applying nodal analysis,
j-j
)60-5(
3
)45-10(
ooo
VVV
=
−
°
∠
+
−
°∠
o
j60-1545-10j V
=
°
∠
+
°∠
°
∠
=
°
∠
+
°∠= 9.24715.73150-1545-10
o
V
Therefore,
=
)t(v
o
15.73 cos(t + 247.9°) V
Chapter 10, Solution 2.
ω 10=
°∠→π− 45-4)4t10cos(4
°∠→π+ 150-20)3t10sin(20
10jLjH1 =ω→
5j-
2.0j
1
Cj
1
F02.0 ==
ω
→
The circuit becomes that shown below.
I
o
V
o
10 Ω
j
10
Ω
-
j
5
Ω
20∠-150° V
+
−
4∠-45° A
Applying nodal analysis,
5j-10j
45-4
10
)150-20(
ooo
VVV
+=°∠+
−°∠
o
)j1(1.045-4150-20 V
+
=
°∠+°∠
°∠=
+
°∠+°∠
== 98.150816.2
)j1(j
45-4150-2
10j
o
o
V
I
Therefore,
=
)t(i
o
2.816 cos(10t + 150.98°) A
Chapter 10, Solution 3.
ω 4=
°∠→ 02)t4cos(2
-j1690-16)t4sin(16 =°∠→
8jLjH2 =ω→
3j-
)121)(4(j
1
Cj
1
F121 ==
ω
→
The circuit is shown below.
V
o
+
−
2
∠
0° A
-j16 V
4 Ω -
j
3
Ω
6 Ω
1
Ω
j
8
Ω
Applying nodal analysis,
8j61
2
3j4
16j-
ooo
+
+=+
−
−
VVV
o
8j6
1
3j4
1
12
3j4
16j-
V
+
+
−
+=+
−
°∠=
°∠
°
∠
=
+
−
= 02.35-835.3
88.12207.1
15.33-682.4
04.0j22.1
56.2j92.3
o
V
Therefore,
=)t(v
o
3.835 cos(4t – 35.02°) V
Chapter 10, Solution 4.
16 4,10-16)10t4sin( =ω°∠→°−
4jLjH1 =ω→
j-
)41)(4(j
1
Cj
1
F25.0 ==
ω
→
j
4
Ω
1
Ω
I
x
-
j
Ω
16∠-10° V
+
−
V
1
0.5 I
x
+
V
o
−
j12
1
4j
)10-16(
1
x
1
−
=+
−°∠ V
I
V
But
4j
)10-16(
1
x
V
I
−°∠
=
So,
j18j
))10-16((3
11
−
=
−°∠ VV
4j1-
10-48
1
+
°∠
=V
Using voltage division,
°∠=
+
°
∠
=
−
= 04.69-232.8
)4j1j)( (1
10-48
j1
1
1o
VV
Therefore,
=
)t(v
o
8.232 sin(4t – 69.04°) V
Chapter 10, Solution 5.
Let the voltage across the capacitor and the inductor be V
x
and we get:
0
3j
V
2j
V
4
3010I5.0V
xxxx
=+
−
+
°∠−−
x
x
xxx
V5.0j
2j
V
Ibut3030I5.1V)4j6j3( =
−
=°∠=−−+
Combining these equations we get:
A38.97615.4
25.1j3
3030
5.0jI
25.1j3
3030
Vor3030V)75.0j2j3(
x
xx
°∠=
+
°∠
=
+
°
∠
=°∠=−+
Chapter 10, Solution 6.
Let V
o
be the voltage across the current source. Using nodal analysis we get:
0
10j20
V
3
20
V4V
oxo
=
+
+−
−
where
ox
V
10j20
20
V
+
=
Combining these we get:
30j60V)35.0j1(0
10j20
V
3
10j20
V4
20
V
o
ooo
+=−+→=
+
+−
+
−
=
+−
=
+−
+
=
5.0j2
)3(20
Vor
5.0j2
30j60
V
xo
29.11∠–166˚ V.
Chapter 10, Solution 7.
At the main node,
++
+
=−−
+
−
→+
−
+∠=
+
−−∠
50
1
30
j
20j40
1
V
3j196.5
20j40
058.31j91.115
50
V
30j
V
306
20j40
V15120
o
o
V 15408.124
0233.0j04.0
7805.4j1885.3
V
o
−∠=
+
−
−
=
Chapter 10, Solution 8.
,200=ω
20j1.0x200jLjmH100 ==ω→
100j
10x50x200j
1
Cj
1
F50
6
−==
ω
→µ
−
The frequency-domain version of the circuit is shown below.
0.1
V
o
40
Ω
V
1
I
o
V
2
+ -j100
Ω
6
o
15∠
20
Ω
V
o
j20
Ω
-
At node 1,
40
VV
100j
V
20
V
V1.0156
2111
1
o
−
+
−
+=+∠
or
21
025.0)01.0025.0(5529.17955.5 VVjj
−
+
−
=
+ (1)
At node 2,
21
2
1
21
V)2j1(V30
20j
V
V1.0
40
VV
−+=→+=
−
(2)
From (1) and (2),
BAVor
0
)5529.1j7955.5(
V
V
)2j1(3
025.0)01.0j025.0(
2
1
=
+
=
−
−+−
Using MATLAB,
V = inv(A)*B
leads to
V 09.1613.110,23.12763.70
21
jVj
+
−
=
−−=
o
21
o
17.82276.7
40
VV
I −∠=
−
=
Thus,
A )17.82t200cos(276.7)t(i
o
o
−=
Chapter 10, Solution 9.
10
33
10,010)t10cos( =ω°∠→
10jLjmH10 =ω→
20j-
)1050)(10(j
1
Cj
1
F50
6-3
=
×
=
ω
→µ
Consider the circuit shown below.
V
1
20 Ω
V
2
-
j
20
Ω
20
Ω
I
o
30
Ω
+
V
o
−
4 I
o
+
−
10∠0° V
j
10
Ω
At node 1,
20j-2020
10
2111
VVVV −
+=
−
21
j)j2(10 VV −+=
(1)
At node 2,
10j3020
)4(
20j-
2121
+
+=
− VVVV
, where
20
1
o
V
I =
has been substituted.
21
)8.0j6.0()j4-( VV +=+
21
j4-
8.0j6.0
VV
+
+
= (2)
Substituting (2) into (1)
22
j
j4-
)8.0j6.0)(j2(
10 VV −
+
++
=
or
2.26j6.0
170
2
−
=
V
°∠=
−
⋅
+
=
+
= 26.70154.6
2.26j6.0
170
j3
3
10j30
30
2o
VV
Therefore,
=)t(v
o
6.154 cos(10
3
t + 70.26°) V
Chapter 10, Solution 10.
2000,100j10x50x2000jLj mH 50
3
=ω==ω→
−
250j
10x2x2000j
1
Cj
1
F2
6
−==
ω
→µ
−
Consider the frequency-domain equivalent circuit below.
V
1
-j250 V
2
36<0
o
2k
Ω
j100 0.1V
1
4k
Ω
At node 1,
21
2111
V004.0jV)006.0j0005.0(36
250j
VV
100j
V
2000
V
36 −−=→
−
−
++= (1)
At node 2,
21
2
1
21
V)004.0j00025.0(V)004.0j1.0(0
4000
V
V1.0
250j
VV
++−=→+=
−
−
(2)
Solving (1) and (2) gives
o
2o
43.931.89515.893j6.535VV ∠=+−==
v
o
(t) = 8.951 sin(2000t +93.43
o
) kV
Chapter 10, Solution 11.
cos( 2,01)t2 =ω°∠→
°∠→°+ 60-8)30t2sin(8
2jLjH1 =ω→
j-
)21)(2(j
1
Cj
1
F2 ==
ω
→1
4jLjH2 =ω→
2j-
)41)(2(j
1
Cj
1
F4 ==
ω
→1
Consider the circuit below.
-
j
Ω
-
j
Ω
2 I
o
2 I
o
2 I
o
2 I
o
2
I
2
2
2 I
o
2
At node 1,
2jj-2
)60-8(
2111
VVVV
−
+=
−°∠
21
j)j1(60-8 VV
+
+=°∠ (1)
At node 2,
0
2j4j
)60-8(
2j
1
221
=
−
−
°∠
+
−
+
VVV
12
5.0j60-4 VV
+
+°∠= (2)
Substituting (2) into (1),
1
)5.1j1(30460-81 V
+
=
°∠−°∠+
5.1j1
30460-81
1
+
°
∠
−
°∠+
=
V
°∠=
−
°∠−°∠+
== 46.55-024.5
j5.1
30460-81
j-
1
o
V
I
Therefore,
=)t(i
o
5.024 cos(2t – 46.55°)
Chapter 10, Solution 12.
20 1000,020)t1000sin( =ω°∠→
10jLjmH10 =ω→
20j-
)1050)(10(j
1
Cj
1
F50
6-3
=
×
=
ω
→µ
The frequency-domain equivalent circuit is shown below.
2 I
o
-
j
20
Ω
20
Ω
V
2
V
1
20∠0° A
10
Ω
I
o
j
10 Ω
At node 1,
1020
220
211
o
VVV
I
−
++=
,
where
10j
2
o
V
I
=
102010j
2
20
2112
VVVV
−
++=
21
)4j2(3400 VV +−= (1)
At node 2,
10j20j-1010j
2
22212
VVVVV
+=
−
+
21
)2j3-(2j VV +=
or
(2)
21
)5.1j1( VV +=
Substituting (2) into (1),
222
)5.0j1()4j2()5.4j3(400 VVV
+
=
+
−
+=
5.0j1
400
2
+
=
V
°∠=
+
== 6.116-74.35
)5.0j1(j
40
10j
2
o
V
I
Therefore,
=)t(i
o
35.74 sin(1000t – 116.6°) A
Chapter 10, Solution 13.
Nodal analysis is the best approach to use on this problem. We can make our work easier
by doing a source transformation on the right hand side of the circuit.
–
j
2 Ω
18
Ω
j
6
Ω
+
−
50
∠
0º V
+
−
+
V
x
−
3
Ω
40∠30º V