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内容
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开始
1
节点分析
2
网孔分析
切换目录
电路理论/双电源激励/节点和网孔
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外观
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隐藏
来自维基教科书,开放的书籍,开放的世界
<
电路理论
|
双电源激励
第一步是将所有内容转换为相量和阻抗,如果可能,以符号形式表示
V
1
=
5
∗
(
3
+
j
)
{\displaystyle \mathbb {V} _{1}=5*({\sqrt {3}}+j)}
I
1
=
1
−
j
2
∗
2
{\displaystyle \mathbb {I} _{1}={\frac {1-j}{2*{\sqrt {2}}}}}
Z
C
1
=
1
j
ω
C
1
=
−
10
j
{\displaystyle Z_{C1}={\frac {1}{j\omega C_{1}}}=-10j}
Z
C
2
=
1
j
ω
C
2
=
−
5
j
{\displaystyle Z_{C2}={\frac {1}{j\omega C_{2}}}=-5j}
Z
L
=
j
ω
L
1
=
L
2
=
1
j
{\displaystyle Z_{L}=j\omega L_{1}=L_{2}=1j}
节点分析
[
编辑
|
编辑源代码
]
标记用于节点分析
V
1
−
V
a
R
1
+
j
ω
L
1
+
I
1
−
V
a
−
V
b
1
j
ω
C
1
=
0
{\displaystyle {\frac {\mathbb {V} _{1}-\mathbb {V} _{a}}{R_{1}+j\omega L_{1}}}+\mathbb {I} _{1}-{\frac {\mathbb {V} _{a}-\mathbb {V} _{b}}{\frac {1}{j\omega C_{1}}}}=0}
V
a
−
V
b
1
j
ω
C
1
−
V
b
R
3
−
V
b
−
V
c
1
j
ω
C
2
=
0
{\displaystyle {\frac {\mathbb {V} _{a}-\mathbb {V} _{b}}{\frac {1}{j\omega C_{1}}}}-{\frac {\mathbb {V} _{b}}{R_{3}}}-{\frac {\mathbb {V} _{b}-\mathbb {V} _{c}}{\frac {1}{j\omega C_{2}}}}=0}
V
b
−
V
c
1
j
ω
C
2
−
I
1
−
V
c
j
ω
L
2
=
0
{\displaystyle {\frac {\mathbb {V} _{b}-\mathbb {V} _{c}}{\frac {1}{j\omega C_{2}}}}-\mathbb {I} _{1}-{\frac {\mathbb {V} _{c}}{j\omega L_{2}}}=0}
使用
matlab
的结果
V
a
=
−
6.5
−
7
i
⇒
V
a
(
t
)
=
9.55
c
o
s
(
1000
t
−
2.32
)
{\displaystyle \mathbb {V} _{a}=-6.5-7i\Rightarrow V_{a}(t)=9.55cos(1000t-2.32)}
V
b
=
−
3.08
−
3.31
i
⇒
V
b
(
t
)
=
4.52
c
o
s
(
1000
t
−
2.32
)
{\displaystyle \mathbb {V} _{b}=-3.08-3.31i\Rightarrow V_{b}(t)=4.52cos(1000t-2.32)}
V
c
=
0.327
+
0.385
i
⇒
V
c
(
t
)
=
0.506
c
o
s
(
1000
t
+
0.866
)
{\displaystyle \mathbb {V} _{c}=0.327+0.385i\Rightarrow V_{c}(t)=0.506cos(1000t+0.866)}
网格分析
[
编辑
|
编辑源代码
]
为网格分析标记
I
1
R
2
+
I
1
+
I
2
j
ω
C
1
+
I
1
+
I
3
j
ω
C
2
−
V
I
s
=
0
{\displaystyle \mathbb {I} _{1}R_{2}+{\frac {\mathbb {I} _{1}+\mathbb {I} _{2}}{j\omega C_{1}}}+{\frac {\mathbb {I} _{1}+\mathbb {I} _{3}}{j\omega C_{2}}}-\mathbb {V} _{Is}=0}
V
1
+
I
2
R
1
+
I
2
+
I
1
j
ω
C
1
+
(
I
2
−
I
3
)
R
3
+
I
2
j
ω
L
1
=
0
{\displaystyle \mathbb {V} _{1}+\mathbb {I} _{2}R_{1}+{\frac {\mathbb {I} _{2}+\mathbb {I} _{1}}{j\omega C_{1}}}+(\mathbb {I} _{2}-\mathbb {I} _{3})R_{3}+\mathbb {I} _{2}j\omega L_{1}=0}
I
3
+
I
1
j
ω
C
2
+
I
3
j
ω
L
2
+
(
I
3
−
I
2
)
R
3
=
0
{\displaystyle {\frac {\mathbb {I} _{3}+\mathbb {I} _{1}}{j\omega C_{2}}}+\mathbb {I} _{3}j\omega L_{2}+(\mathbb {I} _{3}-\mathbb {I} _{2})R_{3}=0}
使用
Matlab
的结果
I
2
=
−
0.239
+
0.234
i
⇒
I
2
(
t
)
=
0.334
c
o
s
(
1000
t
+
2.37
)
{\displaystyle \mathbb {I} _{2}=-0.239+0.234i\Rightarrow I_{2}(t)=0.334cos(1000t+2.37)}
I
3
=
−
0.238
+
0.235
i
⇒
I
3
(
t
)
=
0.334
c
o
s
(
1000
t
+
2.36
)
{\displaystyle \mathbb {I} _{3}=-0.238+0.235i\Rightarrow I_{3}(t)=0.334cos(1000t+2.36)}
V
I
s
=
175
−
179
i
⇒
V
I
s
(
t
)
=
250
c
o
s
(
1000
t
−
0.795
)
{\displaystyle \mathbb {V} _{Is}=175-179i\Rightarrow V_{Is}(t)=250cos(1000t-0.795)}
类别
:
书:电路理论
华夏公益教科书