Lecture Electric circuit theory: Circuit theorems - Nguyễn Công Phương

Magnetically Coupled Circuits XIII.Frequency Response. XIV.The Laplace Transform.[r]

Electric Circuit Theory

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

Circuit Theorems

1 Source Transformation

2 Linearity and Superposition

Source Transformation (1)

+

–

E

R

J

R

E

=

Ri

+

v

i

E

v

R

R

= −

E

J

R

=

Source Transformation (2)

J

R

1

a

b

R

2

i

2

+

–

E

R

1

R

2

i

2

a

b

Ex 1

J = A; R

1

=

Ω

; R

2

=

Ω

; find i

2

?

+

–

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

=

=

=

Source Transformation (3)

R

1

R

2

R

3

R

4

E

1

E

2

J

+

–

+

–

i

3

a

b

c

Ex 2

Circuit Theorems

1 Source Transformation

2 Linearity and Superposition

Linearity and Superposition (1)

+

–

E

J

R

1

R

2

i

+

–

+

–

R

2

R

1

J

E

R

1

i

1

1 2

R J

E

R J

E

i

R

R

R

R

R

R

+

=

=

+

+

+

+

0

E J

i

=

i

=

=

+

J

R

1

R

2

0

E

i

=

–

+

E

R

1

R

2

0

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

Linear and Superposition (3)

+

–

deactivated

deactivated

+

–

v = 0

Linear and Superposition (4)

R

1

R

2

R

3

R

4

E

1

E

2

J

+

–

+

–

i

2

a

b

c

R

1

R

2

R

3

R

4

E

2

J

+

–

i

2

a

b

Linear and Superposition (5)

R

1

R

2

R

3

R

4

E

1

E

2

J

+

–

+

–

i

2

a

b

c

To deactivate E

2

?

R

1

R

2

R

3

R

4

E

J

+

–

i

2

Linear and Superposition (6)

R

1

R

2

R

3

R

4

E

1

E

2

J

+

–

+

–

i

2

a

b

c

To deactivate J?

R

1

R

2

R

3

R

4

E

1

E

2

J

+

–

+

–

i

2

Linear and Superposition (7)

E

J

R

R

+

–

Ex 1

R

1

= 10

Ω

, R

2

= 20

Ω

, E = 30V, J = 2A, find the current of R

2

?

1 Deactivate J, find i

2

|

E

E

R

R

+

–

2 E

i

2

30

1A

10

20

E

E

i

R

R

=

+

=

=

+

2 Deactivate E, find i

2

|

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 i

2

|

E2

+ i

2

|

J

2 3

1 0.67

1.67 A

j

Linear and Superposition (8)

R

1

R

2

R

3

R

4

E

1

E

2

J

+

–

+

–

i

2

a

b

c

E

1

= 16 V; E

2

= V;

J = A; R

1

=

Ω

; R

2

=

Ω

;

R

3

=

Ω

; R

4

= 10

Ω

;

find i

2

?

Ex 2

Linear and Superposition (9)

R

1

R

2

R

3

R

4

E

1

E

2

J

+

–

+

–

i

2

a

b

c

1 Deactivate E

2

& J, find i

2

|

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

E

1

= 16 V; E

2

= V;

J = A; R

1

=

Ω

; R

2

=

Ω

;

R

3

=

Ω

; R

4

= 10

Ω

;

find i

2

?

R

1

R

2

R

3

R

4

E

1

+

–

i

2

Linear and Superposition (10)

R

1

R

2

R

3

R

4

E

1

E

2

J

+

–

+

–

i

2

a

b

c

2 Deactivate E

1

& J, find i

2

|

E2

E

1

= 16 V; E

2

= V;

J = A; R

1

=

Ω

; R

2

=

Ω

;

R

3

=

Ω

; R

4

= 10

Ω

;

find i

2

?

R

1

R

3

i

2

|

E2

R

2

R

4

+

–

i

2

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

Linear and Superposition (11)

R

1

R

2

R

3

R

4

E

1

E

2

J

+

–

+

–

i

2

a

b

c

3 Deactivate E

1

& E

2

, find i

2

|

J

E

1

= 16 V; E

2

= V;

J = A; R

1

=

Ω

; R

2

=

Ω

;

R

3

=

Ω

; R

4

= 10

Ω

;

find i

2

?

R

1

R

2

R

3

R

4

J

i

2

i

|

4

10 2

20 V

E

=

R J

= × =

4

12

20

1.39 A

2.4 10

J

E

i

R

R

R

Linear and Superposition (12)

R

1

R

2

R

3

R

4

E

1

E

2

J

+

–

+

–

i

2

a

b

c

4 Find i

2

= i

2

|

E1

+ i

2

|

E2

+ i

2

|

J

E

1

= 16 V; E

2

= V;

J = A; R

1

=

Ω

; R

2

=

Ω

;

R

3

=

Ω

; R

4

= 10

Ω

;

find i

2

?

2 J

0.56A

i

= −

2 E2

1A

i

=

2 E1

1.33A

i

= −

→

i

2

= –1.33 + – 0.56 = – 0.89A

Circuit Theorems

1 Source Transformation

2 Linearity and Superposition

3 Thévenin Equivalent Subcircuits