Magnetically Coupled Circuits XIII.Frequency Response. XIV.The Laplace Transform XV.[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)Electrical Circuit Analysis
(4)Branch current method (1)
R1
R2
R3
R4 E1
E2
J
+
–
+
–
i1
i2 i3
i4
+ v1 – + v3 –
+
v2
–
a b
c
+
v4
–
Node a: i1 + i2 – i3 = 0 Node b: i3 – i4 + J = 0
A B
Loop A: – E1 + R1i1 – R2i2 + E2 = 0 Loop B: – E2 + R2i2 + R3i3 + R4i4 = 0
i1 + i2 – i3 = 0 i3 – i4 = –J
R1i1 – R2i2 = E1 – E2 R2i2 + R3i3 + R4i4 = E2
i1 i2 i3 i4
(5)Branch current method (2)
A B
i1 + i2 – i3 = 0 i3 – i4 = –J
R1i1 – R2i2 = E1 – E2 R2i2 + R3i3 + R4i4 = E2
nKCL = n – 1
nKVL = b – n + 1
Ex 1 R1
R2
R3
R4 E1
E2
J
+
–
+
–
i1
i2 i3
i4
+ v1 – + v3 –
+
v2
–
a b
c
+
v4
(6)Branch current method (3)
A B
i1 + i2 – i3 = 0 i3 – i4 = –J
R1i1 – R2i2 = E1 – E2 R2i2 + R3i3 + R4i4 = E2
1. Find:
nKCL = n – 1, and
nKVL = b – n + 1
2. Apply KCL at nKCL nodes 3. Apply KVL at nKVL loops
4. Solve simultaneous equations
Ex 1 R1
R2
R3
R4 E1
E2
J
+
–
+
–
i1
i2 i3
i4
+ v1 – + v3 –
+
v2
–
a b
c
+
v4
(7)Branch current method (4)
1. Find:
nKCL = n – 1, and
nKVL = b – n + 1
2. Apply KCL at nKCL nodes 3. Apply KVL at nKVL nodes 4. Solve simultaneous equations
nKCL = – = 3; nKVL = – + = 3
A: R1i1 + R5i5 + R2i2 – E1 = 0 B: R3i3 + R5i5 – R4i4 = 0
C: R2i2 + R6i6 + R4i4 – E6 = 0 a: – i1 + i2 – i6 = 0
b: i1 – i5 + i3 + J = 0 c: – i3 – i4 + i6 – J = 0
A B
C
Ex 2
+ –
+
–
R1 R3
R2 R4
R5
R6 E1
E6 i1
i2
i3 J
i4 i5
+ +
+ +
+
– –
– –
–
a
b
c
d
i6
(8)Branch current method (5)
A
B
C
2 3 6
: 0
b i + + =i i
4 3 5
: 0
c i − + =i i
i1
i5
i2
i6
i3 i4
1 4
: 0
d − − − =i i J
1 1 5 5 4 4 1
:
A R i + R i − R i = E
3 3 6 6 5 5 3 6
:
B R i − R i + R i = E − E
6 6 2 2 6
:
C R i − R i = E
R1 R2
R3 R4
R5
R6 E6
E3
E1 J
d
a
b c
+
–
+
–
(9)Branch current method (6) + – + – R1 E1
R2 J R3
E3 i1
i2
i3
Ex 4
R1 = 10Ω, R2 = 20Ω, R3 = 15Ω, E1 = 30V, E3 = 45V, J = 2A Find currents?
1 0
i + − + =i i J
1 2
R i − R i = E
2 3
R i + R i = E
1
1
2
2
10 20 30
20 15 45
i i i i i i i + − = − → − = + =
1
1 ; ;
i = ∆ i = ∆ i = ∆
∆ ∆ ∆
1 1 1
10 20 0 ; 0 20 15
−
∆ = − 1
2 1 1
30 20 0 ; 45 20 15
− −
∆ = − 2
1 2 1
10 30 0 ; 0 45 15
− −
∆ = 3
1 1 2
10 20 30 0 20 45
−
(10)Branch current method (7)
+
–
+ –
R1 E1
R2 J R3
E3 i1
i2
i3
R1 = 10Ω, R2 = 20Ω, R3 = 15Ω, E1 = 30V, E3 = 45V, J = 2A Find currents?
1 0
i + − + =i i J
1 2
R i − R i = E
2 3
R i + R i = E
1
1
2
2
10 20 30
20 15 45
i i i i i
i i
+ − = −
→ − =
+ =
1 1 1
10 20 0 0 20 15
−
∆ = − 1 20 0 10 1 1 0 1 1
20 15 20 15 20 0
− − −
= − +
−
1( 20 15 20 0) 10[1 15 20( 1)] 0[1 ( 20)( 1)]
= − × − × − × − − + × − − − = −650