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

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Electric Circuit Theory

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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

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Electrical Circuit Analysis

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Branch current method (1)

R

1

R

2

R

3

R

4

E

1

E

2

J

+

–

+

–

i

1

i

2

i

3

i

4

+

v

1

–

+

v

3

–

+

v

2

–

a

b

c

+

v

4

–

Node a: i

1

+ i

2

– i

3

= 0

Node b: i

3

– i

4

+ J = 0

A

B

Loop A: – E

1

+ R

1

i

1

– R

2

i

2

+ E

2

= 0

Loop B: – E

2

+ R

2

i

2

+ R

3

i

3

+ R

4

i

4

= 0

i

1

+ i

2

– i

3

= 0

i

3

– i

4

= –J

R

1

i

1

– R

2

i

2

= E

1

– E

2

R

2

i

2

+ R

3

i

3

+ R

4

i

4

= E

2

i

1

i

2

i

3

i

4

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Branch current method (2)

A

B

i

1

+ i

2

– i

3

= 0

i

3

– i

4

= –J

R

1

i

1

– R

2

i

2

= E

1

– E

2

R

2

i

2

+ R

3

i

3

+ R

4

i

4

= E

2

n

KCL

= n – 1

n

KVL

= b – n + 1

Ex. 1

R

1 R

2

R

3

R

4

E

1

E

2

J

+

–

+

–

i

1

i

2

i

3

i

4

+

v

1

–

+

v

3

–

+

v

2

–

a

b

c

+

v

4

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Branch current method (3)

A

B

i

1

+ i

2

– i

3

= 0

i

3

– i

4

= –J

R

1

i

1

– R

2

i

2

= E

1

– E

2

R

2

i

2

+ R

3

i

3

+ R

4

i

4

= E

2

1. Find:

n

KCL

= n – 1, and

n

KVL

= b – n + 1

2. Apply KCL at n

KCL

nodes

3. Apply KVL at n

KVL

loops

4. Solve simultaneous equations

Ex. 1

R

1 R

2

R

3

R

4

E

1

E

2

J

+

–

+

–

i

1

i

2

i

3

i

4

+

v

1

–

+

v

3

–

+

v

2

–

a

b

c

+

v

4

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Branch current method (4)

1. Find:

n

KCL

= n – 1, and

n

KVL

= b – n + 1

2. Apply KCL at n

KCL

nodes

3. Apply KVL at n

KVL

nodes

4. Solve simultaneous equations

n

KCL

= 4 – 1 = 3;

n

KVL

= 6 – 4 + 1 = 3

A: R

1

i

1

+ R

5

i

5

+

R

2

i

2

– E

1

= 0

B: R

3

i

3

+ R

5

i

5

– R

4

i

4

= 0

C: R

2

i

2

+ R

6

i

6

+ R

4

i

4

– E

6

= 0

a: – i

1

+ i

2

–

i

6

= 0

b: i

1

– i

5

+

i

3

+ J

= 0

c: – i

3

– i

4

+ i

6

–

J = 0

A

B

C

Ex. 2

+

–

+

–

R

1

R

3

R

2

R

4

R

5

R

6

E

1

E

6

i

1

i

2

i

3

J

i

4

i

5

+

–

a

b

c

d

i

6

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Branch current method (5)

A

B

C

2

3

6 :

0 b

i

+ + =

i

4

3

5 :

0 c

i

− + =

i

1

i

5

i

2

i

6

i

3

i

4

1

4 :

0 d

− − − =

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

R

1

R

2

R

3

R

4

R

5

R

6

E

6

E

3

E

1

J

d

a

b

c

+

–

+

–

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Branch current method (6)

+

–

+

–

R

1

E

1

R

2

J

R

3

E

3

i

1

i

2

i

3

Ex. 4

R

1

= 10

Ω

, R

2

= 20

Ω

, R

3

= 15

Ω

, E

1

= 30V,

E

3

= 45V, J = 2A. Find currents?

1 2 3

0 i

+ − + =

i

J

1 1 2 2 1

R i

−

R i

=

E

2 2 3 3 3

R i

+

R i

=

E

1 2 3

1 2

2 3

2

10

20

30

20

15

45 i

+

− = −





→



−

=



+

=



1 2 3

;

i

=

∆

i

=

∆

i

=

∆

1

10

20 0 ;

0

20

15 −

∆ =

−

1

2

1

30

20 0 ;

45

20

15 −

∆ =

−

2

1

2

1

10

30 0 ;

0

45

15 −

∆ =

3

1

2

10

20

30

0

20

45 −

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Branch current method (7)

+

–

+

–

R

1

E

1

R

2

J

R

3

E

3

i

1

i

2

i

3

R

1

= 10

Ω

, R

2

= 20

Ω

, R

3

= 15

Ω

, E

1

= 30V,

E

3

= 45V, J = 2A. Find currents?

1 2 3

0 i

+ − + =

i

J

1 1 2 2 1

R i

−

R i

=

E

2 2 3 3 3

R i

+

R i

=

E

1 2 3

1 2

2 3

2

10

20

30

20

15

45 i

+

− = −





→



−

=



+

=



1

10

20

0

20

15 −

∆ =

−

1

20

0

10

1

0

1

20

15 20 15

20

0 −

=

−

+

−

1( 20 15 20 0) 10[1 15 20( 1)] 0[1 0 ( 20)( 1)]

= − × −

× −

− +

× − −

−

= −

650

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

<b>Electric Circuit Theory</b>

Contents

I.

Basic Elements Of Electrical Circuits

II.

Basic Laws

<b>III. Electrical Circuit Analysis</b>

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

Electrical Circuit Analysis

Branch current method (1)

<i>R</i>

<sub>1</sub>

<i>R</i>

<sub>2</sub>

<i>R</i>

<sub>3</sub>

<i>R</i>

<sub>4</sub>

<i>E</i>

<sub>1</sub>

<i>E</i>

<sub>2</sub>

<i>J</i>

<b>+</b>

<b>–</b>

<b>+</b>

<b>–</b>

<i>i</i>

<sub>1</sub>

<i>i</i>

<sub>2</sub>

<i>i</i>

<sub>3</sub>

<i>i</i>

<sub>4</sub>

+

<i><sub>v</sub></i>

<sub>1</sub>

–

+

<i>v</i>

<sub>3</sub>

–

+

<i>v</i>

<sub>2</sub>

–

<i>a</i>

<i>b</i>

<i>c</i>

+

<i>v</i>

<sub>4</sub>

–

<i>Node a: i</i>

<i>+ i</i>

<i>– i</i>

= 0

<i>Node b: i</i>

<i>– i</i>

<i>+ J = 0</i>

<i>A</i>

<i>B</i>

<i>Loop A: – E</i>

<i>+ R</i>

<i>i</i>

<i>– R</i>

<i>i</i>

<i>+ E</i>

= 0