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

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

Electric Circuit Theory

Electrical Circuit Analysis

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

XIV.The Laplace Transform

XV Two-port Networks

Electrical Circuit Analysis

1 Branch current method

2 Node voltage method

3 Mesh current method

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

–

c

+

v 4

–

Node a: i1 + i2 – i3 = 0

Node b: i3 – i4 + J = 0

Loop A: – E1 + R1i1 – R2i2 + E2 = 0

Loop B: – E + R i + R i + R i = 0

i1 + i2 – i3 = 0

i3 – i4 = –J

R1i1 – R2i2 = E1 – E2

R2i2 + R3i3 + R4i4 = E2

i1

i2

i3

i4

Ex 1

Branch current method (2)

i1 + i2 – i3 = 0

i3 – i4 = –J

R1i1 – R2i2 = E1 – E2

R2i2 + R3i3 + R4i4 = E2

nKCL = n – 1

nKVL = 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

–

c

+

v 4

–

Branch current method (3)

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 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

–

c

+

v 4

–

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 = 4 – 1 = 3; nKVL = 6 – 4 + 1 = 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

C

Ex 2

+ –

–

R5

R6

E1

E6

i1

i2

i3 J

i4

i5

+

–

a

b

c

d

i6

+ –

Branch current method (5)

A

B

C

2 3 6

4 3 5

i1

i5

i2

i6

i3

i4

1 4

1 1 5 5 4 4 1

:

3 3 6 6 5 5 3 6

:

6 6 2 2 6

:

R3

R4

R5

R6

E6

E3

d

a

b c

–

+

Ex 3

Branch current method (6)

–

+ –

R1

E1

E3

i1

i2

i3

Ex 4

R1 = 10Ω, R2 = 20Ω, R3 = 15Ω, E1 = 30V,

E3 = 45V, J = 2A Find currents?

i + − + = 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

2 3

2

20 15 45

i i i

i i

i i







i = ∆ i = ∆ i = ∆

10 20 0 ;

0 20 15

−

30 20 0 ;

45 20 15

10 30 0 ;

0 45 15

10 20 30

0 20 45

−

Branch current method (7)

–

+ –

R1

E1

E3

i1

i2

i3

R1 = 10Ω, R2 = 20Ω, R3 = 15Ω, E1 = 30V,

E3 = 45V, J = 2A Find currents?

i + − + = 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

2 3

2

20 15 45

i i i

i i

i i







10 20 0

0 20 15

−

20 15 20 15 20 0

−

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

Ex 4