Magnetically Coupled Circuits XIII.Frequency Response. XIV.The Laplace Transform.[r]
(1)Electric Circuit Theory
(2)Contents
I. Basic Elements Of Electrical Circuits II. Basic Laws
III Electrical Circuit Analysis
IV Circuit Theorems
V Active Circuits
VI Capacitor And Inductor VII First Order Circuits VIII.Second Order Circuits
IX Sinusoidal Steady State Analysis X AC Power Analysis
XI Three-phase Circuits
XII Magnetically Coupled Circuits XIII.Frequency Response
(3)Circuit Theorems
1 Source Transformation
2 Linearity and Superposition
(4)Source Transformation (1)
+
–
E
R
J R
E = Ri + v i E v
R R
= −
E J
R
=
(5)Source Transformation (2)
J R1
a
b R2 i2
+
– E
R1
R2 i2
a
b
Ex 1
J = A; R1 = Ω; R2 = Ω; find i2?
+
– E
R
J R
E J
R
=
E = RJ
1 4 2 8 V
E = R J = × =
2
1 2
8
0.8 A
4 6
E i
R R
= = =
(6)Source Transformation (3)
R1
R2
R3
R4
E1
E2
J
+
–
+
–
i3
a b
c
Ex 2
(7)Circuit Theorems
1 Source Transformation
2 Linearity and Superposition
(8)Linearity and Superposition (1)
+ – E
J R1 R2
i
+
–
+ –
R2 R1J
E R1
i
1
1 2
R J E R J E
i
R R R R R R
+
= = +
+ + +
0
E J
i = i =
= +
J R1 R2
0
E
i = – +
E
R1 R2
0
(9)Linearity and Superposition (2)
1 1 2 2 n n
y = a x + a x + + a x
1 2 n
y y y
= + + +
0 ,
k
i x
(10)Linear and Superposition (3)
+
–
deactivated
deactivated
+
–
v = 0
(11)Linear and Superposition (4)
R1
R2
R3
R4
E1 E2
J
+
–
+
– i2
a b
c
R1
R2
R3
R4
E2
J
+
–
i2
a b
(12)Linear and Superposition (5)
R1
R2
R3
R4
E1 E2
J
+
–
+
– i2
a b
c
To deactivate E2?
R1
R2
R3
R4
E
J
+
– i2
(13)Linear and Superposition (6)
R1
R2
R3
R4
E1 E2
J
+
–
+
– i2
a b
c
To deactivate J?
R1
R2
R3
R4
E1 E
2
J
+
–
+
– i2
(14)Linear and Superposition (7) E J R R + – Ex 1
R1 = 10Ω, R2 = 20Ω, E = 30V, J = 2A, find the current of R2?
1 Deactivate J, find i2|E
E R R +
– 2 E
i 2 30 1A 10 20 E E i R R = + = = +
2 Deactivate E, find i2|J
J R R 2 J i R R + – 2 J i 2 E R J i R R = + 10 2 10 20 0.67 A × = + =
3 Find i2|E2 + i2|J
2 3
1 0.67 1.67 A
j
(15)Linear and Superposition (8)
R1
R2
R3
R4
E1
E2
J
+
–
+
–
i2
a b
c
E1 = 16 V; E2 = V;
J = A; R1 = Ω; R2 = Ω;
R3 = Ω; R4 = 10 Ω; find i2?
Ex 2
(16)Linear and Superposition (9) R1 R2 R3 R4 E1 E2 J + – + – i2 a b c
1 Deactivate E2 & J, find i2|E1
Ω = + + + = + + + = 4 10 2 6 ) 10 2 ( 6 ) ( 4 234 R R R R R R R 4 2
R i ×
1 1 1234 16 2 A 8 E E i R = = = Ω = + = +
= 1 234 4 4 8 1234 R R
R
E1 = 16 V; E2 = V;
J = A; R1 = Ω; R2 = Ω;
R3 = Ω; R4 = 10 Ω; find i2?
R1
R2
R3
R4
E1 + –
i2
(17)Linear and Superposition (10) R1 R2 R3 R4 E1 E2 J + – + – i2 a b c
2 Deactivate E1 & J, find i2|E2
E1 = 16 V; E2 = V;
J = A; R1 = Ω; R2 = Ω;
R3 = Ω; R4 = 10 Ω; find i2?
R1 R3
i2|E2
R2 R
4 + – i2 Ω = + + + = + + + = 3 10 2 4 ) 10 2 ( 4 ) ( 4 134 R R R R R R R 2 2 2134 9 1A 9 E E i R = = = Ω = + = +
= 2 134 6 3 9
2134 R R
R
(18)Linear and Superposition (11) R1 R2 R3 R4 E1 E2 J + – + – i2 a b c
3 Deactivate E1 & E2, find i2|J
E1 = 16 V; E2 = V;
J = A; R1 = Ω; R2 = Ω;
R3 = Ω; R4 = 10 Ω; find i2?
R1
R2
R3
R4 J
i2
i |
4 10 2 20 V
E = R J = × =
4
12
20
1.39 A 2.4 10
J
E i
R R R
(19)Linear and Superposition (12)
R1
R2
R3
R4
E1
E2
J
+
–
+
–
i2
a b
c
4 Find i2 = i2|E1 + i2|E2 + i2|J E1 = 16 V; E2 = V;
J = A; R1 = Ω; R2 = Ω;
R3 = Ω; R4 = 10 Ω; find i2?
2 J 0.56A
i = −
2 E2 1A
i =
2 E1 1.33A
i = −
→ i2 = –1.33 + – 0.56 = – 0.89A
(20)Circuit Theorems
1 Source Transformation
2 Linearity and Superposition
3 Thévenin Equivalent Subcircuits