Lecture Electric circuit theory: Sinusoidal steady-state analysis - Nguyễn Công Phương

Nguyễn Công Phương

Electric Circuit Theory
Sinusoidal Steady-State Analysis


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. Sinusoid and Phasors
X. Sinusoidal Steady State Analysis
XI. AC Power Analysis
XII. Three-phase Circuits
XIII.Magnetically Coupled Circuits
XIV.Frequency Response
XV. The Laplace Transform
XVI.Two-port Networks

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Sinusoidal Steady-State Analysis
1.

2.
3.
4.
5.
6.
7.
8.
9.
10.
11.

Sinusoidal Steady-State Analysis
Ohm’s Law
Kirchhoff’s Laws
Impedance Combinations
Branch Current Method
Node Voltage Method
Mesh Current Method
Superposition Theorem
Source Transformation
Thévenin & Norton Equivalent Circuits
Op Amp AC Circuits
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Sinusoidal Steady-State Analysis (1)
10sin5t V
–+

20i  6

20Ω

di
1

idt  10sin 5t
dt 0.02 

i

di
d 2i
i
 20  6 2 
 50 cos 5t
dt
dt
0.02
i  I m sin(5t   )

6H

0.02F

10 0o

20

–+

 100 I m cos(5t   )  150 I m sin(5t   ) 

I

 50 I m sin(5t   )  50 cos 5t

10 0o

 2 2 I m sin(5t    135 )  sin(5t  90 )
o

 2 2 I m  1
 I m  0.35


o
o
o
  135  90
  45

j 30

o

I

20  j 30 

1
j 0.1

1
j 0.1

 0.35  45o A

 i  0.35sin(5t  45o ) A

 i  0.35sin(5t  45o ) A
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Sinusoidal Steady-State Analysis (2)
10sin5t V
–+

20Ω

i

1. Transform to phasor
domain
2. Solve the problem using
dc circuit analysis
3. Transform the resulting

phasor to the timedomain.

6H

0.02F

10 0o

j 30

20

–+

1
j 0.1

I
I

10 0o
20  j 30 

1
j 0.1

 0.35  45o A

 i  0.35sin(5t  45o ) A

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Sinusoidal Steady-State Analysis
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.

Sinusoidal Steady-State Analysis
Ohm’s Law
Kirchhoff’s Laws
Impedance Combinations
Branch Current Method
Node Voltage Method
Mesh Current Method
Superposition Theorem
Source Transformation
Thévenin & Norton Equivalent Circuits
Op Amp AC Circuits
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Ohm’s Law (1)

VR

R
I

VR  RI

VL
 j L
VL  j LI 
I

VC 

I
jC

VC
1


jC
I

V
 Z  V  ZI
I

Z: impedance (Ω)

1
Admittance (S): Y 
Z
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Ohm’s Law (2)
V
Z
I

VR
R
I

 ZR  R

VL
 j L  Z L  j L
I

1

j
VC
1

 ZC 

jC
I
jC C

1
YR 
R
1
j
YL 

j L  L
YC  jC

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Ohm’s Law (3)

 0

 

Z L  j L

j
ZC 
C

ZL  0

ZC  

Short circuit

Open circuit

ZL  

ZC  0

Open circuit

Short circuit

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Ohm’s Law (4)
I

+

Z

V

–

Z  R  jX
R: resistance
X: reactance

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Ohm’s Law (5)
L

C
If

L
C

If

j L 

Z  j L 

1
jC

1
 LC  1
0
 0 
jC
jC
2

1
LC

1
L/C
jC
Z

1
1
j L 
j L 
jC
jC

Z0

j L

1
 2 LC  1
j L 
0
 0 
jC
jC

Z
1
LC

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Sinusoidal Steady-State Analysis
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.

11.

Sinusoidal Steady-State Analysis
Ohm’s Law
Kirchhoff’s Laws
Impedance Combinations
Branch Current Method
Node Voltage Method
Mesh Current Method
Superposition Theorem
Source Transformation
Thévenin & Norton Equivalent Circuits
Op Amp AC Circuits
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Kirchhoff’s Law (1)
v1  v2  ...  vn  0
 Vm1 sin(t  1 )  Vm 2 sin(t  2 )  ...  Vmn sin(t  n )  0

 V1  V2  ...  Vn  0

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Kirchhoff’s Law (2)

i1  i2  ...  in  0
 I m1 sin(t  1 )  I m 2 sin(t  2 )  ...  I mn sin(t  n )  0
 I1  I 2  ...  I n  0

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14


Sinusoidal Steady-State Analysis
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.

Sinusoidal Steady-State Analysis
Ohm’s Law
Kirchhoff’s Laws
Impedance Combinations
Branch Current Method
Node Voltage Method
Mesh Current Method
Superposition Theorem

Source Transformation
Thévenin & Norton Equivalent Circuits
Op Amp AC Circuits
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Impedance Combinations (1)
a
b

Z1

Z2

a
b

Z eq  Z1  Z 2

Z1
V1 
Vab
Z1  Z 2
Z eq  Z1  Z 2  ...  Z n

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Impedance Combinations (2)
a

a
Z1

b

Z2
b

Z1Z 2
Z eq 
Z1  Z 2

Z2
I1 
I ab
Z1  Z 2

1
1
1
1


 ... 
Z eq Z1 Z 2

Zn
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Impedance Combinations (3)
Zc

a

Z1

Z2

Zb

Z3

Zb Z c
Z1 
Z a  Zb  Z c
Zc Za
Z2 
Z a  Zb  Z c
Z a Zb
Z3 
Z a  Zb  Z c

b

c

Za
Z1Z 2  Z 2 Z3  Z3 Z1
Za 
Z1

Z1Z 2  Z 2 Z3  Z3 Z1
Zb 
Z2
Z1Z 2  Z 2 Z3  Z3 Z1
Zc 
Z3

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Sinusoidal Steady-State Analysis
1.
2.
3.
4.
5.
6.
7.
8.
9.

10.
11.

Sinusoidal Steady-State Analysis
Ohm’s Law
Kirchhoff’s Laws
Impedance Combinations
Branch Current Method
Node Voltage Method
Mesh Current Method
Superposition Theorem
Source Transformation
Thévenin & Norton Equivalent Circuits
Op Amp AC Circuits
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Branch Current Method (1)
I1

Ex. 1

+
–

E1

Z1

a I 3 Z3

+ V1 –
–
V1
A +
+

1.

Find:
nKCL = n – 1, and
nKVL = b – n + 1

2.
3.
4.

I4
Z4

J

–

E2
c

I2+ V3 –

+
Z2
B V4
–

b

Apply KCL at nKCL nodes
Apply KVL at nKVL loops
Solve simultaneous equations

I1 + I2 – I3 = 0
I3 – I4 = –J
Z1I1 – Z2I2 = E1 – E2
Z2I2 + Z3I3 + Z4I4 = E2

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Branch Current Method (2)
+–

b : Ic  I 2  I3  0
c : I1  I 3  J  0
A : Z1I1  Z3I 3  Z 2 I 2  E  0

I c   I1

E

Z2

Ex. 2

I2

a
Z1

βI1
Ic

I1

I3

b
Z3

J c

  I1  I 2  I 3  0

 I1  I 3  J  0
Z I  Z I  Z I  E  0
3 3
2 2
 11

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Branch Current Method (3)

Ex. 3

Z1  10; Z 2  j 20; Z 3  5  j10;

Z1

E1  30 V; E3  45 15 V; J  2  30 A;
o

o

E1

I3

I1
I2

– +

E3

+

Find currents?

–

I1  I 2  I 3  J  0

Z2

J

Z3

Z1I1  Z 2I 2  E1
Z 2I 2  Z3I3  E3
 I1
 I2
 I3  2  30o

 10I1  j 20I 2
 30

o
15
j
20
I

(5


j
10)
I

45
2
3




I1  1 ; I 2  2 ; I3  3



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Branch Current Method (4)

Ex. 3

Z1  10; Z 2  j 20; Z 3  5  j10;

Z1

E1  30 V; E3  45 15 V; J  2  30 A;
o

o

I2

E3

–

 I1
 I2
 I3  2  30o

 30
10I1  j 20I 2

o
j
20
I

(5

j
10)
I

45
15
2

3

1
2
3
I1  ; I 2 
; I3 



1

– +

+

Find currents?

1

E1

I3

I1

1

  10  j 20
0

j 20 5  j10
0

Z2

J

Z3

 j 20
0
1
1
1
1
1
 10
0
j 20 5  j10
j 20 5  j10
 j 20 0

 250  j 200
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Branch Current Method (5)

Ex. 3

Z1  10; Z 2  j 20; Z 3  5  j10;

Z1

E1  30 V; E3  45 15 V; J  2  30 A;
o

o

E1

I2

– +

E3

+

Find currents?

–

 I1
 I2
 I3  2  30o

 30

10I1  j 20I 2

o
j
20
I

(5

j
10)
I

45
15
2
3


I1 

I3

I1

2  30o
30

1
 j 20

1
0

45 15o

j 20

5  j10

250  j 200

Z2

I1 

J

Z3

1


; I2  2 ; I3  3




 1.04  j 3.95  4.09 75.2o A

 i1  4.09sin(t  75.2o ) A
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Branch Current Method (6)

Ex. 3

Z1  10; Z 2  j 20; Z 3  5  j10;

Z1

E1  30 V; E3  45 15 V; J  2  30 A;
o

o

E1

I2

– +

E3

+

Find currents?

–

 I1
 I2
 I3  2  30o

 30
10I1  j 20I 2

o
j
20
I

(5

j
10)
I

45
15
2
3


I2 

I3

I1

1

2  30o

1

10

30

0

0

45 15o

5  j10

250  j 200

Z2

I1 

J

Z3

1


; I2  2 ; I3  3




 1.98  j 0.98  2.20 26.4o A

 i2  2.20sin(t  26.4o ) A
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