电路理论

5
i1
i4
+
i6
0
− i3 − i5 + i6 = 0
→ KVL: -u1 + u2 + u4 = 0
− 5(1+ i1) + (3i2 + 3) + (4i4 − 2) = 0
− (4i4 − 2) + 2i5 + 6i6 = 0 8(i3 − 2) − 2i5 − (3i2 + 3) = 0
2.1 支路电流分析法 Branch Current Analysis
1Ω
4Ω 6Ω
3Ω 4Ω
+
10V
-
6Ω
5Ω I
3Ω 2Ω ，
+
10V
-
3Ω
5Ω
2Ω
2Ω
2Ω
2Ω
2Ω
2Ω
6Ω
6Ω
4Ω
4Ω
6Ω
6Ω
4Ω
4Ω
6Ω
6Ω
4Ω
4Ω
10V
10V
10V
6Ω
6Ω
6Ω
6Ω
6Ω
5Ω I
5Ω
10Ω
I
10V 6Ω
10Ω
2.2 等效变换 Equivalent Transformation 5.Bridge Circuits (3-3)
i
i
us u
当 us = Ris 时，电路等效。
R u = us + Ri
Thevenin model Two models of real DC source
u
is
R
u = Ris + Ri Norton model
-
a
5V
+
10Ω
0.5A
→
10Ω
b
2
2.2 等效变换 Equivalent Transformation
4.电桥 Bridge CS ircuAi2tsN' 5.星N 形－三角形互主换讲教Wy师e-Delta Transformation 6.电源转移华中So科ur技ce 大Tr学ansfe颜r 秋容
2.1 支路电流分析法 Branch Current Analysis 2A
1.电路的基本方程 －2b方程
Problem1: Write the equations of branch current analysis.
VCR: u1 = 5(1+ i1) u3 = 8(i3 − 2)
u2 = 3i2 + 3 u4 = 4i4 − 2
u5 = 2i5
u6 = 6i6
i3
2A
3Ω + 3V -
2Ω
i2
i5
2.1 支路电流分析法 Branch Current Analysis
Problem2: 计算独立电源的功率。 KCL：i1 = i + i2
KVL： 5 = 5i1 + 5i 5i = 10i2 +10i
5Ω
i1 5V l1
10Ω i2
i
5Ω l2
10i
i
=
2 3
A，i1
=
1 3
A，i2
=
−
1 3
(c) Practice Find the equivalent circuit.
5Ω ix
10ix
+-
a
2Ω 8Ω
b
+
+
2Ω
a
2U1
-
U1 2Ω 2A
4Ω
-
b
Answer:
a 4.6Ω
b
a
1A
b
3.5Ω a
+
1V
-
b
2.2 等效变换 Equivalent Transformation
(c) Practice
2A
3.支路电流分析法 VCR
i3 8Ω
u1 = 5(1+ i1) u3 = 8(i3 − 2) u2 = 3i2 + 3 u4 = 4i4 − 2
3Ω + 3V -
2
2Ω
1
1A
i2
4Ω
i5
3
u5 = 2i5
u6 = 6i6
5Ω
-
6Ω
2V
KCL: i1 + i2 + i3 = 0 − i2 + i4 + i5 = 0
Short circuit: Rsc = 0 Gsc = ∞
? Req = ? i, u
Geq = ?
i
u
R
i
u
R
2.2 等效变换 Equivalent Transformation
3.Interconnections of Resistor and Independent Source
i
i
i
u
i1
5Ω ix 2Ω 8Ω
Find the equivalent circuit.
10ix
+
-
a
b
2.2 等效变换 Equivalent Transformation
(c) Practice
Find the equivalent circuit.
+
+
2Ω
a
2U1
-
U1 2Ω 2A
4Ω
-
b
3
2.2 等效变换 Equivalent Transformation (c) Practice
i
i
u
Equivalent networks u
f (u,i) = 0
u-i 关系相同
f (u,i) = 0
1
2.2 等效变换 Equivalent Transformation
2.Series and Parallel Connections of Resistors
Series connection：
扩展电桥:
R1
R4
R7
R3
R6
R2
R5
R8
R1
R4
R7
R2
R5
R8
R3
R6
2.2 等效变换 Equivalent Transformation 5.Bridge Circuits (3-4)
Symmetric circuit
4
2.2 等效变换 Equivalent Transformation
a6.WR1 ye-DeltaR→2Tranasformation
Current
division : ik
=
Gk Geq
i
2.2 等效变换 Equivalent Transformation
Problem1: Find the equivalent resistance for the circuit.
4Ω
a 4Ω c
Rab = 2.5Ω
5Ω 6Ω
b
4Ω
4Ω
4Ω
d
i
i
+
+
ri
-
u
+
+
αu
-
u
R
-
R
-
Req = R + r i
Req
=
R 1−α
i
i1 +
Ru
βi
-
i1 Ru βi1
Req = (1− β )R
Req
=
R 1+ β
i
+
+
αu1
+-
u
u1 -
R
-
Req = (1+ α )R
i
i1 Ru gu
Req
=
R 1+ gR
2.2 等效变换 Equivalent Transformation
(a) Transformation of independent sources
Find the current I .
8V 8Ω
a
12V 2A 4Ω
I 12Ω 4Ω
a I
1A 6Ω
4Ω
b
b
I=0.6A
2.2 等效变换 Equivalent Transformation
(b) Transformation of dependent sources
i R1 R2
Rn
i
u
u
Req
n
∑ Equivalent resistance: Req = Rk k =1
Parallel connection： i
Voltage
division:uk
=
Rk Req
u
i
u
G1
G2
Gn
u
Geq
n
∑ Equivalent conductance : Geq = Gk k =1
i
i
u1 u
u
i1
u2
+
us
Ｎ
−
u2
us = u1 − u2
Find the current i. 1Ω
1A 10V
i u 2Ω
us = u2
u1
R
us = u1
3Ω
i=?
i=5A
i
u
ㄨu1
u2
i
+
→
u -
+ us
-
2.2 等效变换 Equivalent Transformation
4.Source transformation
? R4
R2 R4
b
Balanced bridge
R2 R3
R4
R5
The voltage source supplies 150W. Find the resistance R.
2Ω
5Ω 1Ω 10Ω
+
30V
-
R 2Ω 4Ω
5.Bridge Circuits (3-2)
2Ω
2Ω
Find the current I in the circuit.
Find the equivalent circuit.
R2 i
R2 i
R2
i
+ + u2 -+
+
+
gu2
+ u2 - +
R1
u→
-
( gR1 )u2
-
R1
u
→ -
(−gR1R2 )i
-
u
R1
-
i
+
+
(−gR1R2 )i
→-
u
→
R1 + R2
-
− gR1R2 R1 + R2
i
+
u
→ -
i
+
u R1 + R2 − gR1R2
R1
R1
R1
R1
R1
R1
Req = ? R2
R2
Req = ? R1
R2
R2
R2
R2 = 2R1
R2 R1
2.2 等效变换 Equivalent Transformation
Problem2: Find the equivalent resistance.
Open circuit: Roc = ∞ Goc = 0
Find the equivalent circuit.
2.2 等效变换 Equivalent Transformation
电压源转移 电流源转移
2.2 等效变换 Equivalent Transformation
5.Bridge Circuits (3-1)
a
R1
R2
R1 a
b
R5 R3
R1 = R3
R12
i2 2
2.2 等效变换 Equivalent Transformation Practice Find the voltage U in the circuit.
U
I1Fra Baidu bibliotek4Ω c 4Ω
a 5A
I 2 6Ω 2Ω d
2Ω b 24Ω
I1
=
I2
=
5 2
A
(Current division)
U = 4I1 − 2I2 = 5V （KVL）
i2
i3
is = i1 − i2 + i3
is Ｎ
u1
u
i1
uㄨ
i2 is = i2
i2
i i
i1
is
u R
→u
is = i1
2.2 等效变换 Equivalent Transformation
3.Interconnections of Resistor and Independent Source
A
p1
=
5i1＝
5 3
W
? •Supplying
•Absorbing
2.2 等效变换 Equivalent Transformation
1.等效的概念 Concept of Equivalence
a
3Ω
i
i
5V
4Ω 4Ω
5V
5Ω
b
i = 1A
Rab = 5Ω Equivalent resistance or input resistance
i3
电路分析以获得支路电压和
8Ω
电流为最终目标，对于b 条支 路、n 结点的电路，基本方程 为：
KCL: n－1 个 KVL: b－n+1 个 支路的u-i 关系(VCR): b个
2.电路分析方程
3Ω
3V
+-
2
2Ω
1
1A
i2
4Ω
i5
3
5Ω
-
6Ω
2V
i1
i4
+
i6
0
支路电流分析方程: b equations of i1～ ib
a
R2
c
R5
dc
dc
d
b R3
R4
b R3
R4 b
R4
三角形→星形
星形→三角形
R3 i3 3
1 i1 R1 R2
uij = fk (i1, i2, i3)
R12
=
R1R2
+
R2 R3 R3
+
R3 R1
R31
i2
R1
=
R12
R12 R31 + R23 +
R31
2
i3 3
u-i 关系相同
1 i1 R23
第2章 简单电阻电路分析
*
2.1 支路A 电流W分析法IA IA1 *
2.2 等效变换方法 B
IA2 A1
Z1
IR
Z1
1.等效变换的概念 Concept of EqZ1uivalence
2.串C 联与并联 Series and Parallel ConnectionRs
3.电源变换 SoZu2 rcZe2 trZa2nsformation
Method2:
u − i 关系 : u = R2i + R1(gu2 + i) = (R1 + R2 − gR1R2 )i
2.2 等效变换 Equivalent Transformation
(b) Transformation of dependent sources
Find the equivalent resistance.
消元 支路电压分析方程: b equations of u1～ub
基本方程
结点电位分析方程: n-1 equations of
nodel voltage un1 ~ un(n-1) 网孔电流分析方程: b − n + 1 equations of mesh current
2.1 支路电流分析法 Branch Current Analysis
1A
-
5Ω
2V
+
KCL: i1 + i2 + i3 = 0
i1
i4
i6
− i2 + i4 + i5 = 0
− i3 − i5 + i6 = 0
i3 = 2A
KVL:
→ − 5(1+ i1) + (3i2 + 3) + ((4−i4 2− 2) ) = 0
− (42i4 − 2) + 2i5 + 6i6 = 0 8v(3i3 − 2) − 2i5 − (3i2 + 3) = 0