SINUSOIDAL STEADY
STATE ANALYSIS
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9.1 Sinusoidal Source
9.2 Phasors
9.3 Kirchhoff’s Laws in the Frequency Domain
9.4 Component Models in Phasor Domain
9.5 Series, Parallel, and Delta-to-Wye
simplifications
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9.6 The Node-Voltage Method
9.7 The Mesh-Current Method
9.8 Circuit Theorems
9.9 The Coupling Inductors and Ideal Transformer
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9.1 Sinusoidal Source
A sinusoidal current is usually referred to as
alternating current (ac).
Circuits driven by sinusoidal current or voltage
sources are called ac circuits.
(a)A sinusoidal signal is easy to generate and
transmit.
It is the dominant form of signal in electric
power industries and communication.
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9.1 Sinusoidal Source
(b) Nature itself is characteristically sinusoidal.
(c) Through Fourier analysis, any practical
periodic signal cab be represented by a sum
of sinusoids.
(d) A sinusoid is easy to handle mathematically.
The derivative and integral of a sinusoid are
themselves sinusoids.
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9.1 Sinusoidal Source
Consider the sinusoidal voltage
v(t ) = Vm sin(ωt + θ )
Vm : the amplitude of the sinusoid
ω : the angular frequency , rad/s
θ : phase angle , rad
ω =2πf
f : frequency , Hz
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9.1 Sinusoidal Source
T=
1 2π
=
f
ω
, period , in sec.
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9.1 Sinusoidal Source
v (t + T
If
) = V m sin (ω t + ω T + θ )
= V m sin (ω t + 2 π + θ )
= V m sin (ω t + θ )
= v (t )
i ( t ) = I m cos (ω t + θ )
-sin
ψ
θ
-cos
i(t)
cos
then
i ( t ) = I m sin (ω t + θ + 90 ° )
= − I m cos (ω t + θ + 180 ° )
sin
= − I m sin (ω t + θ + 270 ° )
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9.1 Sinusoidal Source
also , i(t)= -I msin(wt - ψ) , ψ = 90 o - θ
-sin
ψ
= -I m cos(wt - ψ - 90 )
o
= I msin(wt - ψ - 180 o )
= I m cos(wt - ψ - 270 o )
i(t)
θ
-cos
cos
sin
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9.1 Sinusoidal Source
A coswt + B sinwt = C cos(wt - θ)
θ = tan -1
B
A
C = A2 + B 2
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θ
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9.2 Phasors
A phasor is a complex number that represents the
amplitude and phase of a sinusoid.
Introduced by Charles Proteus Steinmetz, a
German-Austrian mathematician and engineer, in
1893.
The idea of phasor representation is based on
Euler’s identity : jθ
e = cos θ + j sin θ
j = −1
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9.2 Phasors
Let v(t)= Vm cos(wt +θ)
The phasor transform P is defined as
P (v(t )) = Vm e jθ , exponential form
= Vm cos θ + jVm sin θ , rectangular form
= Vm ∠θ , polar form
ur
@ V , phasor domain , complex
(frequency domain)
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9.2 Phasors
The inverse phasor transform is defined as
ur
ur jwt
−1
P (V ) = Re[Ve ]
= Re[Vm e jθ e jwt ]
= Vm cos(wt + θ )
= v (t ) , time domain , real value
It is assumed that the angular frequency is known.
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9.2 Phasors
Example :
v(t ) = 100 cos( wt + 30o )V
i (t ) = 50sin( wt + 70 )A
ur
V = 100∠30o V
o
The phasor diagram is as follows :
uur
r
Phasor V leads I by 50o
r
uur
Phasor I lags V by 50o
r
I = 50∠ − 20° A
A convention used in power systems is as follows:
The electric load is lagging, which means the load current
phasor lags the voltage phasor.
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9.3 Kirchhoff’s Laws in the Frequency Domain
KCL
n
∑i
K =1
K
(t ) = 0 , for any node
For sinusoidal circuits
n
∑I
K =1
cos( wt + θ K ) = 0
mK
Take P transform on both sides
n
∑I
K =1
n
uur
∑I
or
K =1
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mK
∠θ K = 0 + j 0
= 0 , for any node
K
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9.3 Kirchhoff’s Laws in the Frequency Domain
KVL
n
∑ v (t) = 0
k
, for any loop
k=1
For sinusoidal voltages
m
∑V
mk
cos(wt + φ K ) = 0
k=1
Take P transform
m
∑V
k =1
mk
∠ φk = 0 + j 0
uur
V
∑ K = 0 , for any loop
m
or
K =1
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9.4 Component Models in Phasor Domain
Given iR (t ) = I m cos( wt + θ )
then vR (t ) = RI m cos( wt + θ )
ur
VR
r
IR
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Take P transform
r
Given I R = I m ∠θ
then P(vR (t )) = RI m ∠θ
r
= RI R
ur
=V R
ur
r
V R and I R are in phase.
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9.4 Component Models in Phasor Domain
ur
r
∴V R =RI R
ur
Similarity given V R , one can find
r
ur
1 ur
I R = V R = GV R
R
IR
VR
R
R : resistance , G : conductance
symbol
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9.4 Component Models in Phasor Domain
Given iL (t ) = I m cos(ωt + θ )
r
then I L = I m ∠θ
Also, vL (t ) = L
di
= -ω LI m sin(ωt + θ )
dt
= ω LI m cos(ωt + θ + 90°)
Ρ (vL (t )) = ω LI m e j (θ + 90)
= ω LI m e jθ e j 90
r
= ω LI L e j 90
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9.4 Component Models in Phasor Domain
r
IL
r
VL
symbol
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r
r
r
∴ VL = jω LI L @ j Χ L I L
r
similarly, given VL one can find
r
r
1 r
IL =
VL @ jΒ LVL
jω L
Χ L = ω L , inductive reactance , in Ω
(電感性電抗 )
1
, inductive susceptance , in S
ωL
(電感性電納)
Χ L and Β L are frequency dependent.
ΒL = -
20
9.4 Component Models in Phasor Domain
r
VL
Im
r
IL
Re
0
r
r
I L lags VL by 90°.
r
r
VL leads I L by 90°.
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9.4 Component Models in Phasor Domain
Given vC (t ) = Vm cos(ω t + θ )
Ρ (vC (t )) = Vm ∠θ
then
dvC (t )
dt
= - ω CVm sin(ω t + θ )
iC (t ) = C
= ω CVm cos(ω t + θ + 90°)
Ρ (iC (t )) = ω CVm e j (θ + 90°)
= ω CVm e jθ e j 90
r
= ω CVC e j 90
r
= jω CVC
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9.4 Component Models in Phasor Domain
r
r
r
∴ IC = jωCVC @ jΒCVC
similarly, one can get
r
r
1 r
VC =
I C @ jX C I C
jω C
ΒC = ωC , capacitive susceptance , in S
(電容性電納)
ΧC = -
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, capacitive reactance , in Ω
ωC
(電容性電抗 )
ΒC and ΧC are frequency dependent.
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9.4 Component Models in Phasor Domain
r
IC
Im
r
VC
Re
r
r
IC leads VC by 90°.
r
r
VC lags IC by 90°.
9.4 Component Models in Phasor Domain
Summary of the component models
Element
R
Time domain
v = Ri
i = Gv
v=L
L
C
di
dt
i = i(0) +
1 t
vdt
L ∫0
v = v(0) +
1 t
idt
C ∫0
i =C
dv
dt
Frequency domain
r
r
V = RI
r
r
I = GV
r
r
V = jω LI
r
1 r
I=
V
jωL
r
1 r
V=
I
jωC
r
r
I = jωCV
9.4 Component Models in Phasor Domain
From the above phasor domain models of the R, L, C
elements, the ratio of the phasor voltage to the phasor
current.


R , for resistor
r

V
r =  jω L , for inductor
I

1
− j
, for capacitor
 ωC
r
V
Let r @ Z , impedance, in Ω
I
The impedance represents the opposition that the
circuit exhibits to the flow of sinusoidal current.
It is a complex quantity. It is not a phasor.
9.4 Component Models in Phasor Domain
The impedance can be expressed in rectangular
form or polar form as
Z @ R + jX A Z ∠θ
Z = R2 + X 2
X
R
R= Z cos θ , X= Z sin θ
θ = tan -1
R電阻 , X電抗 ,Z 阻抗
9.4 Component Models in Phasor Domain
Definition
r
I 1
Y @ r = = G + jB
V Z
G = Re{Y } , conductance , S
B = Im{Y } , susceptance , S
Y : admittance , in S
G : 電導 , B : 電納, Y : 導納
ur
r r
ur
V = Z I , I = YV
complex ohm ' s law
9.4 Component Models in Phasor Domain
QY =
∴G =
B=
1
1
R − jX
=
=
= G + jB
Z R + jX R 2 + X 2
R
R2 + X 2
−X
R2 + X 2
1
∴G ≠
if X ≠ 0
R
immittance : refer to either impedance or admittance
9.5 Series, Parallel, and Delta-to-Wye
Simplifications
(a) Series connection
r
I
+
ur
V
ur
+ V1 -
uur
+ V2 -
−
ur
V
Z eq @ r =
I
uur
V
∑
k
N
k =1
r = ∑ Zk
I
k =1
N
uur
+ VN -
9.5 Series, Parallel, and Delta-to-Wye
Simplifications
special case N=2
r
I
+
ur
V
−
ur
V1 =
ur
+ V1 -
Z1 ur
V
Z1 + Z 2
uur
Z 2 ur
V2 =
V
Z1 + Z 2
+
uur
V2
-
voltage division principle
9.5 Series, Parallel, and Delta-to-Wye
Simplifications
(b) Parallel connection
r
I
ur
I1
+
ur
V
−
uur
I2
r
1
I
@ ur =
Z eq V
or
Yeq =
N
uur
∑I
ur
V
k =1
N
∑Y
k =1
uur
IN
k
k
=
N
1
∑Z
k =1
k
9.5 Series, Parallel, and Delta-to-Wye
Simplifications
special case N=2
r
I
+
ur
V
−
Z eq =
ur
I1
Z1
uur
I2
ur
I1 =
Z2
Z2 r
I
Z1 + Z 2
uur
Z1 r
I2 =
I
Z1 + Z 2
Z1Z 2
Z1 + Z 2
current division principle
9.5 Series, Parallel, and Delta-to-Wye
Simplifications
(c) wye-delta tronsformation
Y − ∆ C o n versio n
Za =
∆
Z1
Zb =
∆
Z2
Zc =
∆
Z3
∆ @ Z 1Z 2 + Z 2 Z 3 + Z 3 Z 1
Special case Z1=Z2=Z3 then Za=Zb=Zc=3Z1
9.5 Series, Parallel, and Delta-to-Wye
Simplifications
Zc
a
b
Z1
Zb
∆ − Y Conversion
Z2
n
Za
Z1 =
Zb Zc
Z a + Zb + Zc
Z2 =
Zc Za
Z a + Zb + Zc
Z3 =
Z a Zb
Z a + Zb + Zc
Z3
c
1
Special case Za=Zb=Zc then Z1=Z2=Z3= 3 Za
9.5 Series, Parallel, and Delta-to-Wye
Simplifications
Example1. Determine vo(t)
ω = 4 rad / s
1
1
10 mF ⇒
=
= − j 25 Ω
jω C
j 4 × 10 × 10 − 3
5 H ⇒ jω L = j 4 × 5 = j 20 Ω
20 cos(4 t − 15 o ) ⇒ 20 ∠ − 15 o
9.5 Series, Parallel, and Delta-to-Wye
Simplifications
− j 25Ω
20∠ − 15o
− j25Ω P j20Ω =
uur
∴ VO =
j 20Ω
uur
VO
− j25 × j20
= j1 0 0 Ω
− j25 + j20
j1 0 0 Ω
(20∠ − 15o )
6 0 + j1 0 0 Ω
= 1 7 .1 5 ∠ 1 5 .9 6 o
uur
p − 1 (V O ) = 1 7 .1 5 c o s ( 4 t + 1 5 .9 6 )V
= vo (t )
9.5 Series, Parallel, and Delta-to-Wye
Simplifications
r
Find I in the circuit.
Example 2：
r
I
50∠0°
9.5 Series, Parallel, and Delta-to-Wye
Simplifications
r
I
50∠0°
∆ abc
→
Y
j4(2 - j4)
1 6 + j8
=
= 1 .6 + j 0 .8 Ω
j4 + (2 − j4) + 8
10
j 4 (8 )
Zbn =
= j 3 .2 Ω
10
8(2 - j 4)
Z cn =
= 1 .6 − j 3 .2 Ω
10
Zan =
9.5 Series, Parallel, and Delta-to-Wye
Simplifications
r
I
50∠0°
∴ Z = 12Ω + Z an + ( Zbn − j 3) //( Z cn + 8 + j 6)
= 13.64∠4.204° Ω
r
r
V
50∠0°
∴I =
=
= 3.666∠ − 4.204° A
Z 13.64∠4.204°
9.6 The Node-Voltage Method
n
n
The same analysis techniques for DC resistive
circuits can be extended to AC circuit analysis.
For example
voltage divider and Y-△ transformation
Nodal analysis
Mesh analysis
Superposition theorem
Source transformation
Thevenin and Norton equivalents
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9.6 The Node-Voltage Method
Steps to Analysis of AC Circuits
n step1 : Transform the circuit to the phasor
(or frequency ) domain.
n step2 : Solve the problem
Complex Algebraic Equations ( Not real
algebraic equations as in DC circuit
analysis).
n step3 : Transform the answer back to the time
domain.
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9.6 The Node-Voltage Method
n
Example 9.6.1: Find ix using nodal analysis
20 cos 4t v
n
2ix
Step1: Phasor Domain
10Ω
uur
Ix
20∠00
ur
V1 jω L = j 4Ω
uur
V2
1
= − j 2.5Ω
jωC
uur
2I x
j 2Ω
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9.6 The Node-Voltage Method
n
ur
20∠0°
10
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uur
Step2 : Choose datum node & V1 , V2
After source transformation
uur
Ix
10Ω
1
1
10 + j 0.4 + j 4

1

−

j4

ur
V1
uur
jω L = j 4Ω V
2
1
= − j 2.5Ω
jωC
uur
2I x
1  ur
 20∠ 0 0 
j 4  V1  
  uur  =
10 

uur
1
1 
+  V2   2 I x 
j4 j2 
j 2Ω
−
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9.6 The Node-Voltage Method
20∠0°
10
uur
Ix
ur
V1
uur
jω L = j 4Ω V
2
1
= − j 2.5Ω
jωC
10Ω
uur
2I x
j 2Ω
uur ur
uur
ur
I x = V1 × j0.4 , 2I x = j0.8×V1
ur

j 2.5  V1   20
 1 + j1.5
 − j 0.8 + j 0.25 − j (0.75)   uur  =  0 

 V2   
45
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9.6 The Node-Voltage Method
20∠0°
10
× j 20
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uur
Ix
10Ω
ur
V1
uur
jω L = j 4Ω V
2
1
= − j 2.5Ω
jωC
uur
2I x
j 2Ω
ur

j 2.5 V1   20 
 uur  =
15  V2   0 
1 + j1.5
 11

ur
V1 = 18.97∠18.430 (V)
uur
ur
I x = j0.4 ×V1 = 7.59∠108.40 (A)
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9.6 The Node-Voltage Method
20 cos 4t v
Step3 :
2ix
ix = 7.59cos(4t + 108.40 )
(A)
47
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uur
9.7 The Mesh-Current
Method
I
Example 9.7.1: Determine
o
using mesh analysis
4Ω
5∠0o A
8Ω
n
ur
I3
I0
− j 2Ω
uur
I2
j10Ω
ur
I1
20∠90o V
− j 2Ω
ur
uur
ur
Mesh1 : (8 + j10 − j 2)I − (− j 2)I − j10I = 0 + j0
1
2
3
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9.7 The Mesh-Current Method
4Ω
ur
I3
5∠0o A
− j 2Ω
uur
I2
j10Ω
ur
I1
8Ω
n
I0
uur
20∠90o V
− j 2Ω
ur
ur
0
Mesh2 : (4 − j 2 − j 2) I 2 − ( − j 2) I1 − ( − j 2) I 3 = −20∠90
One current source, only two unknowns
ur
I 3 = 5 ∠ 0 0 (A )
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9.7 The Mesh-Current Method
4Ω
5∠0o A
8Ω
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ur
I3
j10Ω
ur
I1
I0
− j 2Ω
uur
I2
20∠90o V
− j 2Ω
ur

j2  I1   j50 
8 + j8
 j2 4 − j4 uur = − j30

 I2  

uur
∴I2 = 6.12∠− 35.220 A
uur uur
∴I0 = −I2 = 6.12∠144.780 A
50
9.8 Circuit Theorems
n
Superposition theorem is important for a linear AC
circuit containing different source frequencies.
n
The total response is obtained by adding the individual
responses in the time domain (not frequency domain).
Example 9.8.1. Linear circuit with only one frequency.
The individual responses can be added in
the frequency domain.
Take Example 9.7.1 as an illustration.
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9.8 Circuit Theorems
uur
The component I o ′ due to 20∠ 900 V source
4Ω
− j 2Ω
uur
I 0′
20∠90o V
j10Ω
8Ω
− j 2Ω
uur′
I o = −2.353 + j 2.353 A
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uur
The component I o ′′ due to 5∠00 A
52
9.8 Circuit Theorems
r ''
Io
5A
-j2Ω
j10Ω
-j2Ω
uur′′
I o = −2.647 + j1.176 A
By superposition theorem
uur uur ′ uur ′′
I 0 = I o + I o = − 5 + j 3.529
=6.12 ∠144.78 0
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A
53
A
9.8 Circuit Theorems
n
Example 9.8.2:
Linear AC circuit containing three source frequencies.
Find steady state v0 (t ) by using superposition
theorem .
2H
1Ω
4Ω
v0
10 cos 2t v
0.1F
5V
2sin 5t A
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9.8 Circuit Theorems
Three frequencies ω1 = 0 , ω2 = 2rad /sec , ω3 = 5rad /sec
∴v0 (t) = v1(t) + v2 (t) + v3 (t)
(1) v1(t) ?
v1
∴ v1 =
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−1
× 5V = −1V
1+ 4
55
9.8 Circuit Theorems
(2) v2(t)=? , ω2=2 rad/s
j4Ω
10∠0o
1Ω
uur
V2
4Ω
v0
-j5Ω
10cos 2t v
2sin 5t A
Z = − j 5Ω P 4 = 2.439 − j1.951
Voltage divider
ur
1Ω
V2 =
10∠0o = 2.498∠ − 30.79o V
1+ j4 + Z
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v2 ( t ) = 2.498cos ( 2t − 30.79o ) V
56
9.8 Circuit Theorems
(3) v3(t)=? , ω3=5 rad/s
ur
I1
Z1 = ( − j 2Ω ) P 4Ω = 0.8 − j1.6Ω
uur
V3
2∠ − 90o A
v0
10cos 2t v
2sin 5t A
Current divider
r
j10Ω
× 2∠ − 90o A
I1 =
j10 + 1 + Z1
ur r
V 3 = I 1 ×1Ω = 2.328∠ − 80o V
v3 ( t ) = 2.328cos ( 5t − 80o ) V
∴ v0 ( t ) = −1 + 2.498cos(2t − 30.790 )
+ 2.328sin ( 5t + 10o ) V
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9.8 Circuit Theorems
n
Source Transformation
ur
Is
uur
Vs
uv
v
V s = Zs I s
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uv
v Vs
Is =
Zs
58
9.8 Circuit Theorems
ur
n Example 9.8.3 : Calculate Vx
uur
Vx
20∠ − 90o V
uur
Vx
4∠ − 90o A
Z1 = 5Ω P ( 3 + j 4 ) Ω = 2.5 + j1.25Ω
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9.8 Circuit Theorems
4∠ − 90o Z1
= 5 − j10V
uur
Vx
ur
∴V x = 5.519∠ − 28o V by voltage division principle
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9.8 Circuit Theorems
n
Thevenin and Norton Equivalent Circuits
A linear two terminal circuit under sinusoidal steady
state condition
r
IN
ur
V TH
ur
ur r
V TH =V o = I N Z N
r
r ur
I N = I s = V TH / ZTH
ZTH = Z N
Z N = ZTH
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9.8 Circuit Theorems
Example 9.8.4 : Find the Thevenin equivalent
120∠750 V
120∠75o V
ZTH = ( − j 6Ω P 8Ω ) + ( 4Ω P j12Ω )
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= 6.48 − j 2.64Ω
62
9.8 Circuit Theorems
voltage division
ur
ur
ur ur  8
j12 
o
∴V TH = V ab = V a − V b = 
−
120∠75 V
 8 − j 6 4 + j12 
= 37.95∠220.31o V
ZTH
a
ur
V TH
b
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9.8 Circuit Theorems
n
r
Example 9.8.5 : Obtain I o using Norton’s theorem
3∠0 A
o
r
Io
40∠90o V
Z N = ZTH = 5Ω with dead independent sources
By superposition theorem
r
r r r
40∠90o V
IN = Is = Iv + II =
+ 3∠0o = j8 + 3 A
5Ω
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9.8 Circuit Theorems
r
II
3∠0o A
r
Io
r
IN
r
Io =
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r
5
I N = 1.465∠38.48o A
5 + ( 20 + j15 )
65
9.9 The Coupling Inductors and
Ideal Transformer
 v1 (t )   L1
v 2(t )  =  M

 
M  d  i1(t ) 
L 2  dt i 2(t ) 
Under sinusoidal steady state , given
i1(t)= I1mcos(ωt+θ1)
i2(t)= I2mcos(ωt+θ2)
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9.9 The Coupling Inductors and
Ideal Transformer
 v1(t )   L1
v 2(t )  =  M

 
M  d  i1(t ) 
L 2  dt i 2(t ) 
then
v1 =-ωL1I1msin(ωt+θ1)- ωM I2msin(ωt+θ2)
= ωL1I1mcos(ωt+θ1+90°)+ ωM I2mcos(ωt+θ2+90°)
v2 =-ωM I1msin(ωt+θ1)- ωL2I2msin(ωt+θ2)
= ωM I1mcos(ωt+θ1+90°)+ ωL2I2mcos(ωt+θ2+90°)
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9.9 The Coupling Inductors and
Ideal Transformer
Take phasor transform on both sides
uur
ur
uur
V 1 = jω L1I 1 + jω M I 2
uur
ur
uur
V 2 = jω M I 1 + jω L 2 I 2
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9.9 The Coupling Inductors and
Ideal Transformer
Example 9.9.1:
Find the input impedance Zab of the following circuit
jω M
ur
I1
uur
I2
jω L2
jω L1
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9.9 The Coupling Inductors and
Ideal Transformer
Use mesh current method:
uur
ur
uur
VS = ( Z S + R1 + jω L1 ) I1 − jω M I 2
ur
uur
0 = − jω M I1 + ( R2 + jω L2 + Z L ) I 2
uur
From (B) I 2 =
From (A)
ur
jω M
I1
R2 + jω L2 + Z L
uur
ur
VS = ( Z S + R1 + jω L1 ) I1 +
ZS
VS
(A)
(B)
ur
ω 2 M 2 I1
R2 + jω L2 + Z L
a
ur
I1
jω M
R1
jω L1
R2 c
jω L2
uur
I2
ZL
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70
b
d
9.9 The Coupling Inductors and
Ideal Transformer
uur
VS
ω2M 2
ur
∴ Z ab =
− Z S = R1 + jω L1 +
R2 + jω L2 + Z L
I1
The reflected impedance due to the secondary coil
and load impedance
Zr =
ω 2M 2
R2 + jω L2 + Z L
ZS
VS
a
ur
I1
jω M
R1
jω L1
R2 c
jω L2
uur
I2
ZL
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71
b
d
9.9 The Coupling Inductors and
Ideal Transformer
An ideal transformer is the limiting case of two lossless
coupled inductors where the inductances approach
infinity and the coupling is perfect.
ur
I1
ur
V1
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uur
I2
jω M
L1
L2
uur
V2
ur
ur
uur
V1 = jω L1 I1 + jω M I 2
uur
ur
uur
V2 = jω M I1 + jω L2 I 2
(A)
(B)
72
9.9 The Coupling Inductors and
IdealurTransformer
uur
ur
From (A) I1 =
V1 − jω M I 2
jω L1
(C)
Substitute (C) into (B)
uur
uur M ur jω M 2 uur
V2 = jω L2 I 2 + V1 −
I2
L1
L1
Qk =
M
= 1 ⇒ M = L1 L2
L1 L2
uur
uur
uur
L ur
∴V2 = jω L2 I 2 + 2 V1 − jω L2 I 2
L1
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=
L2 ur N 2 ur
V1 =
V1
L1
N1
73
9.9 The Coupling Inductors and
Ideal Transformer
N2
N1
As L1, L2, M →∞ such that
remains the
same, the coupled coils become an ideal
transformer.
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9.9 The Coupling Inductors and
Ideal Transformer
To have some understanding of the physical meaning,
consider the open and short circuit cases as follows:
ur
I1
ur
V1
uur
I2 = 0
jω M
jω L1
uur
V2
jω L2
uur
V2 M
∴ ur =
=
V1 L1
L1 L2
=
L1
ur
ur
V1 = jω L1 I1
uur
ur
V2 = jω M I1
L2 N 2
=
L1 N1
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9.9 The Coupling Inductors and
Ideal Transformer
ur
I1
ur
V1
uur
I2
jω M
jω L1
jω L2
uur
V2 = 0
uur
ur
uur
V2 = jω M I1 + jω L2 I 2 = 0
ur
I
−L
− L2
L
N
∴ uur1 = 2 =
=− 2 =− 2
M
L1
N1
I2
L1 L2
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9.9 The Coupling Inductors and
Ideal Transformer
The circuit symbol for ideal transformers
ur
I1
N1 : N 2
uur
I2
ur
V1
when
uur
V2
uur
V2 N 2
ur =
V1 N1
ur
I
N
uur1 = − 2
N1
I2
N2
= 1 , called an isolation transformer
N1
N2 > N1 , a step-up transformer
N2 < N1 , a step-down transformer
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9.9 The Coupling Inductors and
Ideal Transformer
Example for different algebraic signs
ur
I1
ur
V1
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N1 : N 2
uur
I2
uur
V2
uur
V2 N 2
ur =
V1 N1
ur
I
N
uur1 = − 2
N1
I2
78
9.9 The Coupling Inductors and
Ideal Transformer
ur
I1
N1 : N 2
uur
I2
ur
V1
uur
V2
uur
V2
N
ur = − 2
N1
V1
ur
I
N
uur1 = 2
I 2 N1
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9.9 The Coupling Inductors and
Ideal Transformer
ur
I1
ur
V1
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N1 : N 2
uur
I2
uur
V2
uur
V2
N
ur = − 2
N1
V1
ur
I
N
uur1 = − 2
N1
I2
80
9.9 The Coupling Inductors and
Ideal Transformer
ur
uur
Rule 1 : If V1 and V2 are both positive at the
dot-marked terminal, use a plus sign for the
turn ratio N2/N1.
Otherwise, use a negative sign.
ur
uur
Rule 2 : If I1 and I 2 are both directed into or out of
the dot-marked terminal, use a minus sign for
the turn ratio.
Otherwise, use a plus sign.
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9.9 The Coupling Inductors and
Ideal Transformer
Example 9.9.2: Find the input impedance Zab for the
following circuit
ZS
uur
VS
a
ur
I1
1: n
uur
I2
c
+
ur
V1
+
uur
V2
−
−
b
d
ZL
Zab
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9.9 The Coupling Inductors and
Ideal Transformer
From the ideal transformer model
uur
ur
V2
I1
uur = n , uur = n
V1
I2
uur
uur V 2
V1
1
∴ Zab= ur = uunr = 2
I 1 nI 2 n
uur
V2 1
uur = 2 ZL
I2 n
ZS
uur
VS
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ur
I1
a
uur
I2
1: n
c
+
ur
V1
+
uur
V2
−
−
b
d
Zab
ZL
83
9.9 The Coupling Inductors and
Ideal Transformer
Hence, ideal transformers can be used to raise or
lower the impedance level of a load by choosing
proper turn ratio n.
ur
I1
uur
VS
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1: n
uur
I2
+
ur
V1
+
uur
V2
−
−
84
9.9 The Coupling Inductors and
Ideal Transformer
The T-equivalent circuit for magnetically coupled
coils
di
di
v1 = L1 1 + M 2
dt
dt
di
di
v2 = M 1 + L2 2
dt
dt
Assumption:
the voltage between b and d must be zero, i.e. , vbd=0
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9.9 The Coupling Inductors and
Ideal Transformer
di
di
di
V1 = ( L1 − M ) 1 + M 1 + M 2
dt
dt
dt
di1
di2
di
V2 = M
+M
+ ( L2 − M ) 2
dt
dt
dt
L1-M
a
+
V1
L2-M
i1
c
i2
M
-
+
V2
-
b
d
time domain
L1-M or L2-M could be
negative
Mathematically, they are
equivalent.
phasor domain
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9.9 The Coupling Inductors and
Ideal Transformer
Modeling of a practical transformer
Ll1,Ll2 : leakage inductances
Lm : magnetizing inductance
Rc : represents core loss
R1,R2 : winding resistances
n : turn ratio of ideal transformer
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Summary
n
Objective 1 : Be able to perform a phasor transform and an
inverse phasor transform.
n
Objective 2 : Be able to transform a time-domain circuit
with a sinusoidal source into frequency domain
using phasor concepts.
n
Objective 3 : Be able to solve a linear AC circuit under
sinusoidal steady state by extending the
analysis techniques for DC resistive circuits.
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Summary
n
Objective 4 : Be able to analyze AC circuits containing
coupling inductors using phasor methods.
n
Objective 5 : Be able to analyze AC circuits containing ideal
transformers using phasor methods.
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Summary
Chapter Problems : 9.11(c)
9.19
9.23
9.27
9.37
9.48
9.53
9.61
9.74
9.79
Due within one week.
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