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Lecture Electric circuit theory: Electrical circuit analysis - Nguyễn Công Phương

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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 iR i = E

3 3 6 6 5 5 3 6

:

B R iR i + R i = EE

6 6 2 2 6

:

C R iR 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 iR 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 iR 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

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