SINUSOIDAL STEADY STATE ANALYSISocw.nthu.edu.tw/ocw/upload/12/241/09handout.pdf · SINUSOIDAL...

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SINUSOIDAL STEADY SINUSOIDAL STEADY

STATE ANALYSIS STATE ANALYSIS

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9.1 Sinusoidal Source9.1 Sinusoidal Source

9.2 Phasors9.2 Phasors

9.3 Kirchhoff9.3 Kirchhoff’’s Laws in the Frequency Domains Laws in the Frequency Domain

9.4 Component Models in Phasor Domain9.4 Component Models in Phasor Domain

9.5 Series, Parallel, and Delta9.5 Series, Parallel, and Delta--toto--Wye Wye simplificationssimplifications

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9.6 The Node9.6 The Node--Voltage MethodVoltage Method

9.7 The Mesh9.7 The Mesh--Current MethodCurrent Method

9.8 Circuit Theorems9.8 Circuit Theorems

9.9 The Coupling Inductors and Ideal Transformer9.9 The Coupling Inductors and Ideal Transformer

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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.


9.1 Sinusoidal Source 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.



Consider the sinusoidal voltage



( ) sin( )mv t V tω θ= +

Vm : the amplitude of the sinusoid ω : the angular frequency , rad/sθ : phase angle , rad

ω =2πff : frequency , Hz


9.1 Sinusoidal Source 9.1 Sinusoidal Source 1 2Tf

πω

= = , period , in sec.


9.1 Sinusoidal Source 9.1 Sinusoidal Source ( ) ( )

( )( )

( )

sin

sin 2

sin

m

m

m

v t T V t T

V t

V t

v t

ω ω θ

ω π θ

ω θ

+ = + +

= + +

= +

=

( ) ( )

( ) ( )( )( )

cos

sin 90

cos 180

sin 270

m

m

m

m

If i t I tthen

i t I t

I t

I t

ω θ

ω θ

ω θ

ω θ

= +

= + + °

= − + + °

= − + + °

cos

-sin

-cos

sin

i(t)

θ

ψ



cos

-sin

-cos

sin

i(t)

θ

ψsin

cos

sin

cos

om

om

om

om

also , i(t)= -I (wt -ψ) , ψ= 90 -θ = -I (wt -ψ - 90 ) = I (wt -ψ -180 ) = I (wt -ψ - 270 )



-1

2 2

cos sin cos

tan

A wt + B wt = C (wt -θ)B θ=A

C A B= + θ


9.2 Phasors 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 : cos sin

1

je j

j

θ θ θ= +

= −


9.2 Phasors 9.2 Phasors

( ( )) , cos sin , ,

m

jm

m m

m

Let v(t)=V cos(wt +θ)The phasor transform P is defined as

P v t V e exponential formV jV rectangular formV polar form

θ

θ θ

θ

=

= +

= ∠

, V phasor domain , complex (frequency domain)

ur@


9.2 Phasors 9.2 Phasors

It is assumed that the angular frequency is known.

1 ( ) Re[ ] Re[ ] cos( ) ( ) ,

jwt

j jwtm

m

The inverse phasor transform is defined as

P V VeV e e

V wtv t time domain , real

θ

θ

− =

=

= +

=

ur ur

value

100 30oV V= ∠ur

50 20I A°= ∠ −r


9.2 Phasors 9.2 Phasors Example :( ) 100cos( 30 )V( ) 50sin( 70 )A

The phasor diagram is as follows :

o

o

v t wti t wt

= +

= +

o

o

Phasor leads by 50

Phasor lags by 50

V I

I V

uur r

r uur

A convention used in power systems is as follows:

The electric load is lagging, which means the load current phasor lags the voltage phasor.



1

1

( ) 0 ,

cos( ) 0

n

KK

n

mK KK

KCL

i t for any node

For sinusoidal circuits

I wt θ

=

=

=

+ =

∑

∑

1

1

0 0

0 ,

n

mK KK

n

KK

Take P transform on both sides

I j

or I for any node

θ=

=

∠ = +

=

∑

∑uur



n

kk=1

m

mk Kk=1

KVL

v (t)= 0 , for any loop

For sinusoidal voltages

V cos(wt + )= 0φ

∑

∑

1

1

0 0

0 ,

m

mk kkm

KK

Take P transform

V j

or V for any loop

φ=

=

∠ = +

=

∑

∑uur



R

( ) cos( ) ( ) cos( )

( ( )) =

=

R m

m

R m

R m

R

R

Given i t I wtthen v t RI wt

Take P transform

Given I Ithen P v t RI

RI

V

θ

θ

θθ

= +

= +

= ∠∠

=

r

r

ur

RRV and I are in phase.ur r

RVur

RIr



VR

IR

R

=

, 1

RR

R

R R R

V RI

Similarity given V one can find

I V GVR

∴

= =

ur r

ur

r ur ur

R : resistance , G : conductance symbol



( 90)

90

( ) cos( )

, ( ) - sin( )

cos( 90 )

( ( ))

L m

L m

L m

mj

L mj j

m

Given i t I t

then I IdiAlso v t L LI tdt

LI t

v t LI e

LI e e

LI

θ

θ

ω θ

θ

ω ω θ

ω ω θ

Ρ ω

ω

ω

+

= +

= ∠

= = +

= + + °

=

=

=

r

r 90jLe



LVr

LIr

, 1

, , ( )

1 - , ,

( )

L L L L

L

L L L L

L

L

L L

V j LI j I

similarly given V one can find

I V j Vj L

L inductive reactance in

inductive susceptance in SL

and are frequenc

ω

ωω

ω

∴ = Χ

= Β

Χ = Ω

Β =

Χ Β

r r r@r

r r r@

電感性電抗

電感性電納

.y dependent

symbol



Re

Im

LVr

LIr

0

90 .

90 .L L

L L

I lags V by

V leads I by

°

°

r r

r r



( 90 )

( ) cos( ) ( ( ))

( ) ( )

- sin( ) cos( 90 )

( ( ))

C m

C m

CC

m

mj

C mj

m

Given v t V tv t V

dv tthen i t CdtCV t

CV t

i t CV e

CV e

θ

ω θ

Ρ θ

ω ω θ

ω ω θ

Ρ ω

ω

+ °

= +

= ∠

=

= +

= + + °

=

= 90

90

j

jC

C

e

CV e

j CV

θ

ω

ω

=

=

r

r



C

C

C

,

1 XC

, , ( )

1 - , ,

( )

C C C C

C C C

C C

I j CV j Vsimilarly one can get

V I j Ij

C capacitive susceptance in S

capacitive reactance inC

and are frequency dependent

ω

ωω

ω

∴ = Β

=

Β =

Χ = Ω

Β Χ

r r r@

r r r@

電容性電納

電容性電抗

.

90 .

90 .C C

C C

I leads V by

V lags I by

°

°

r r

r r

Re

Im

CIr

CVr


Summary of the component modelsElement Time domain Frequency domain

R

L

C

v Rii Gv

==

0

1(0)t

div Ldt

i i vdtL

=

= + ∫

01(0)

tv v idt

Cdvi Cdt

= +

=

∫

V RII GV

=

=

r rr r

j1

j L

V LI

I V

ω

ω

=

=

r r

r r

1V Ij C

I j CVω

ω

=

=

r r

r r


From the above phasor domain models of the R, L, C elements, the ratio of the phasor voltage to the phasor current.

, ,

1 ,

R for resistorV j L for inductorI

j for capacitorC

ω

ω

= −

rr

, , VLet Z impedance inI

Ωrr @

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.


The impedance can be expressed in rectangular form or polar form as

2 2

-1

tan

R= cos , X= sin

Z R jX Z

Z R XXR

Z Z

θ

θ

θ θ

+ ∠

= +

=

@ A


R電阻 , X電抗 ,Z 阻抗

Definition

1

Re , , Im , ,

: ,

IY G jBZV

G Y conductance SB Y susceptance SY admittance in S

= = +

==

rr@

電導電納導納G : , B : , Y :


,'

V Z I I YVcomplex ohm s law

= =ur r r ur

2 2

2 2

2 2

1 1

1 0

:

R jXY G jBZ R jX R X

RGR X

XBR X

G if XR

immittance refer to either impedance or admittance

−= = = = +

+ +

∴ =+

−=

+

∴ ≠ ≠

Q


(a) Series connection

Ir

V

+

−

ur1 -V+

ur2 -V+

uur -NV+uur

1

1

N

k Nk

eq kk

VVZ ZI I

=

=

= =∑

∑

uururr r@

9.5 Series, Parallel, and Delta9.5 Series, Parallel, and Delta--toto--Wye Wye SimplificationsSimplifications

special case N=2

Ir

V

+

−

ur1 -V+

ur

2-V+uur

11

1 2

22

1 2

ZV VZ Z

ZV VZ Z

=+

=+

ur ur

uur ur

voltage division principle


(b) Parallel connection

1Iur

2Iuur

NIuur

V

+

−

ur

Ir

1

1

1

1 1

N

k Nk

keq k

N

eq kk

II

Z ZV V

or Y Y

=

=

=

= =

=

∑∑

∑

uurrur ur@


special case N=2

Z1 Z2

1Iur

2Iuur

V

+

−

urIr

21

1 2

12

1 2

ZI IZ Z

ZI IZ Z

=+

=+

ur r

uur r

1 2

1 2eq

Z ZZZ Z

=+

current division principle


(c) wye-delta tronsformation CY onversion− ∆

1

2

3

a

b

c

ZZ

ZZ

ZZ

∆=

∆=

∆=

1 2 2 3 3 1Z Z Z Z Z Z∆ + +@

Special case Z1=Z2=Z3 then Za=Zb=Zc=3Z1


1

2

3

b c

a b c

c a

a b c

a b

a b c

Z ZZZ Z Z

Z ZZZ Z Z

Z ZZZ Z Z

=+ +

=+ +

=+ +

Special case Za=Zb=Zc then Z1=Z2=Z3=1 Za3

CY onversion∆ −a b

c

Z1 Z2

Z3

Zc

Zb Zan


Example1. Determine vo(t)

3

4 /1 110 25

4 10 105 4 5 2020 cos(4 15 ) 20 15o o

rad s

mF jj C j

H j L j jt

ω

ωω

−

=

⇒ = = − Ω× ×

⇒ = × = Ω

− ⇒ ∠ −


20 15o∠ − 25j− Ω 20j ΩOV

uur

1

2 5 2 02 5 2 0 1 0 02 5 2 0

1 0 0 ( 2 0 1 5 )6 0 1 0 0

1 7 .1 5 1 5 .9 6

( ) 1 7 .1 5 c o s ( 4 1 5 .9 6 ) ( )

oO

o

O

o

j jj j jj j

jVj

p V t Vv t

−

− ×− Ω Ω = = Ω

− +Ω

∴ = ∠ −+ Ω

= ∠

= +

=

uur

uur

P


Example 2Example 2：：

Ir

50 0∠ °

Find I in the circuit.r


ab c Y4 ( 2 - 4 ) 1 6 8 1 .6 0 .8

4 ( 2 4 ) 8 1 04 (8 ) 3 .21 0

8( 2 - 4 ) 1 .6 3 .21 0

j j jZ a n jj jjZ b n j

jZ cn j

∆ →+

= = = + Ω+ − +

= = Ω

= = − Ω

Ir

50 0∠ °


50 0∠ °

Ir

12 ( 3) //( 8 6) 13.64 4.204

an bn cnZ Z Z j Z j∴ = Ω + + − + +

= ∠ ° Ω

50 0 3.666 4.20413.64 4.204

VI AZ

∠ °∴ = = = ∠ − °

∠ °

rr



9.6 The Node9.6 The Node--Voltage MethodVoltage Methodnn The same analysis techniques for DC resistive The same analysis techniques for DC resistive

circuits can be extended to AC circuit analysis.circuits can be extended to AC circuit analysis.nn For example For example

voltage divider and Yvoltage divider and Y-- transformation transformation Nodal analysis Nodal analysis Mesh analysis Mesh analysis Superposition theoremSuperposition theoremSource transformationSource transformationThevenin and Norton equivalentsThevenin and Norton equivalents

Steps to Analysis of AC CircuitsSteps to Analysis of AC Circuitsnn step1 : Transform the circuit to the phasor step1 : Transform the circuit to the phasor

(or frequency ) domain.(or frequency ) domain.nn step2 : Solve the problem step2 : Solve the problem

Complex Algebraic Equations ( Not realComplex Algebraic Equations ( Not realalgebraic equations as in DC circuit algebraic equations as in DC circuit analysis).analysis).

nn step3 : Transform the answer back to the time step3 : Transform the answer back to the time domain. domain.

C.T. PanC.T. Pan 4242


n Example 9.6.1: Find ix using nodal analysis

n Step1: Phasor Domain

20cos 4 vt 2 xi

10Ω

020 0∠

4j L jω = Ω1V

ur2V

uur

1 2.5jj Cω

= − Ω 2 xIuur

2j ΩxIuur




n Step2 : Choose datum node & , After source transformation

0

1

2

1 1 10.4 20 010 4 4

101 1 1

24 4 2 x

jVj j

V Ij j j

+ + − ∠ = − +

ur

uur uur

9.6 The Node9.6 The Node--Voltage MethodVoltage Method1V

ur2V

uur

xIuur

20 010∠ °

10Ω1 2.5j

j Cω= − Ω

4j L jω = Ω1V

ur2V

uur

2 xIuur

2j Ω



1 10.4 , 2 0.8x xI V j I j V= × = ×uur ur uur ur

1

2

1 1.5 2.5 200.8 0.25 (0.75) 0

Vj jj j j V

+ = − + −

ur

uur

xIuur

20 010∠ °

10Ω1 2.5j

j Cω= − Ω

4j L jω = Ω1V

ur2V

uur

2 xIuur

2j Ω


1

2

1 1.5 2.5 2011 15 0

Vj j

V

+ =

ur

uur

01

01

18.97 18.43 (V)

0.4 7.59 108.4 (A)x

V

I j V

= ∠

= × = ∠

ur

uur ur

20j×


xIuur

20 010∠ °

10Ω1 2.5j

j Cω= − Ω

4j L jω = Ω1V

ur2V

uur

2 xIuur

2j Ω



Step3 : 07.59cos(4 108.4 ) (A)xi t= +

20cos 4 vt 2 xi


9.7 The Mesh9.7 The Mesh--Current Method Current Method Example 9.7.1: Determine using mesh analysisoI

uur

20 90oV∠

0I5 0o A∠ 2j− Ω

2j− Ω

10j Ω

8Ω

3Iur

1Iur

2Iuur

4Ω

1 2 3(8 10 2) ( 2) 10 0 0j j I j I j I j+ − − − − = +ur uur ur

n Mesh1 :

9.7 The Mesh9.7 The Mesh--Current Method Current Method


n Mesh2 :One current source, only two unknowns

02 1 3(4 2 2) ( 2) ( 2) 20 90j j I j I j I− − − − − − = − ∠

uur ur ur

20 90oV∠

0I5 0o A∠ 2j− Ω

2j− Ω

10j Ω

8Ω

3Iur

1Iur

2Iuur

4Ω

03 5 0 (A )I = ∠

ur

9.7 The Mesh9.7 The Mesh--Current Method Current Method


20 90oV∠

0I5 0o A∠ 2j− Ω

2j− Ω

10j Ω

8Ω

3Iur

1Iur

2Iuur

4Ω

1

2

8 8 2 502 4 4 30

Ij j jj j jI

+ = − −

ur

uur

02

00 2

6.12 35.22 A

6.12 144.78 A

I

I I

∴ = ∠−

∴ = − = ∠

uur

uur uur


9.8 Circuit Theorems9.8 Circuit Theoremsn 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 inthe frequency domain.Take Example 9.7.1 as an illustration.

9.8 Circuit Theorems9.8 Circuit TheoremsThe component due to source

oI ′uur 020 90 V∠

20 90oV∠

0I ′uur

2j− Ω

2j− Ω

10j Ω

8Ω

4Ω

2.353 2.353 AoI j′ = − +uur

C.T. PanC.T. Pan 5252The component due tooI ′′uur 05 0 A∠


5A -j2Ω

-j2Ωj10Ω

''oI

r


2.647 1.176 AoI j′′ = − +uur

By superposition theorem

00

5 3.529 A

=6.12 144.78 Ao oI I I j′ ′′= + = − +

∠

uur uur uur


5454C.T. PanC.T. Pan

n Example 9.8.2: Linear AC circuit containing three source frequencies.Find steady state by using superposition theorem .

0( )v t

4Ω1Ω

0.1F

2H

10cos 2 vt0v

2sin 5 At5V



Three frequencies1 2 30 , 2 /sec , 5 /secrad radω ω ω= = =

0 1 2 3( ) ( ) ( ) ( )v t v t v t v t∴ = + +

1v

11 5 1

1 4v V V−

∴ = × = −+

(1) v1(t) ?


(2) v2(t)=? , ω2=2 rad/s4Ω1Ω

-j5Ω

j4Ω

10 0o∠ 2Vuur


10cos 2 vt0v

2sin 5 At

( ) ( )

2

2

5 4 2.439 1.951 1 10 0 2.498 30.79

1 4

2.498cos 2 30.79

o o

o

Z j jVoltage divider

V Vj Z

v t t V

= − Ω = −

Ω= ∠ = ∠ −

+ +

= −

P

ur


(3) v3(t)=? , ω3=5 rad/s

2 90o A∠ −

3Vuur

1Iur

10cos 2 vt0v

2sin 5 At

( )

( ) ( )( )

( )

1

1

1

13

3

00

2 4 0.8 1.6 10 2 90

10 1

1 2.328 80

2.328cos 5 80

1 2.498cos(2 30.79 )

2.328sin 5 10

o

o

o

o

Z j jCurrent divider

jI Aj Z

V I V

v t t V

v t t

t V

= − Ω Ω = − Ω

Ω= × ∠ −

+ +

= × Ω = ∠ −

= −

∴ = − + −

+ +

P

r

ur r




n Source Transformation

sVuur

sIur

ss sV Z I=uv v s

s

s

VIZ

=uvv


9.8 Circuit Theorems9.8 Circuit Theoremsn Example 9.8.3 : Calculate xV

ur

20 90oV∠ −xV

uur

4 90o A∠ −xV

uur

( )1 5 3 4 2.5 1.25Z j j= Ω + Ω = + ΩP



14 905 10

o Zj V

∠ −= −

xVuur

5.519 28 oxV V by voltage division principle∴ = ∠ −

ur


9.8 Circuit Theorems9.8 Circuit Theoremsnn Thevenin and Norton Equivalent CircuitsThevenin and Norton Equivalent Circuits

A linear two terminal circuit under sinusoidal steady state condition

THVur

NIr

THV =V No N

TH N

I ZZ Z

=

=

ur ur r/N s TH TH

N TH

I I V ZZ Z

= ==

r r ur



Example 9.8.4 : Find the Thevenin equivalent

0120 75 V∠

120 75oV∠

( ) ( )6 8 4 12 6.48 2.64

THZ j jj

= − Ω Ω + Ω Ω

= − Ω

P P



8 12 120 75 8 6 4 12

37.95 220.31

oTH ab a b

o

voltage division

jV V V V Vj j

V

∴ = = − = − ∠ − +

= ∠

ur ur ur ur

THVur

ZTHa

b

9.8 Circuit Theorems9.8 Circuit Theoremsn Example 9.8.5 : Obtain using Norton’s theoremoI

r

40 90oV∠

oIr

3 0o A∠

5 N THZ Z with dead independent sources= = Ω

By superposition theorem40 90 3 0 8 3

5

oo

N s v IVI I I I j A∠

= = + = + ∠ = +Ω

r r r r




3 0o A∠II

r

NIr

oIr

( )5 1.465 38.48

5 20 15o

o NI I Aj

= = ∠+ +

r r


1 11

2 2 2

( ) ( )( ) ( )

v t L M i tdv t M L i tdt

=

Under sinusoidal steady state , given

i1(t)= I1mcos(ωt+θ1)i2(t)= I2mcos(ωt+θ2)

9.9 The Coupling Inductors and 9.9 The Coupling Inductors and Ideal TransformerIdeal Transformer


1 1 1

2 2 2

( ) ( )( ) ( )

v t L M i tdv t M L i tdt

=

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°)



Take phasor transform on both sides

1 1 1 2

2 1 2 2

V j L I j M I

V j M I j L I

ω ω

ω ω

= +

= +

uur ur uur

uur ur uur




1j Lω 2j Lω

j Mω

1Iur

2Iuur

Example 9.9.1: Example 9.9.1: Find the input impedance ZFind the input impedance Zabab of the following circuitof the following circuit


ZS R1

VS 1j Lω 2j Lω ZL

R2a

b

c

d

j Mω

1Iur

2Iuur

Use mesh current method:Use mesh current method:1 1 1 2

1 2 2 2

( ) (A)

0 ( ) (B)S S

L

V Z R j L I j M I

j M I R j L Z I

ω ω

ω ω

= + + −

= − + + +

uur ur uur

ur uur

From (B)From (B) 2 12 2 L

j MI IR j L Z

ωω

=+ +

uur ur

From (A)From (A)2 2

11 1 1

2 2

( )S SL

M IV Z R j L IR j L Z

ωω

ω= + + +

+ +

uruur ur



ZS R1

VS 1j Lω 2j Lω ZL

R2a

b

c

d

j Mω

1Iur

2Iuur

2 2

1 12 21

Sab S

L

V MZ Z R j LR j L ZI

ωω

ω∴ = − = + +

+ +

uurur

The reflected impedance due to the secondary coil The reflected impedance due to the secondary coil and load impedanceand load impedance

2 2

2 2r

L

MZR j L Z

ωω

=+ +



An ideal transformer is the limiting case of two lossless An ideal transformer is the limiting case of two lossless coupled inductors where the inductances approach coupled inductors where the inductances approach infinity and the coupling is perfect.infinity and the coupling is perfect.

1Vur

2Vuur

1Iur

2Iuurj Mω

1L 2L1 1 1 2

2 1 2 2

(A)

(B)

V j L I j M I

V j M I j L I

ω ω

ω ω

= +

= +

ur ur uur

uur ur uur



1 21

1

(C)V j M IIj L

ωω

−=

ur uururFrom (A)From (A)

Substitute (C) into (B)Substitute (C) into (B)2

2 2 2 1 21 1

M j MV j L I V IL L

ωω= + −

uur uur ur uur

1 21 2

1 Mk M L LL L

= = ⇒ =Q

22 2 2 1 2 2

1

2 21 1

1 1

LV j L I V j L IL

L NV VL N

ω ω∴ = + −

= =

uur uur ur uur

ur ur



As LAs L11, L, L22, M , M →∞→∞ such that remains the such that remains the same, the coupled coils become an ideal same, the coupled coils become an ideal transformer.transformer.

2

1

NN



To have some understanding of the physical meaning, To have some understanding of the physical meaning, consider the open and short circuit cases as follows:consider the open and short circuit cases as follows:

1 1 1

2 1

V j L I

V j M I

ω

ω

=

=

ur ur

uur ur

1 22 2 2

1 1 1 11

L LV L NML L L NV

∴ = = = =uurur

1Vur

2Vuur

1Iur

2 0I =uurj Mω

1j Lω 2j Lω



2 1 2 2 0V j M I j L Iω ω= + =uur ur uur

1 2 2 2 2

1 12 1 2

I L L L NM L NI L L

− −∴ = = = − = −

uruur

1Vur

2 0V =uur

1Iur

2Iuurj Mω

1j Lω 2j Lω



2 2

11

1 2

12

V NNV

I NNI

=

= −

uurur

uruur

The circuit symbol for ideal transformersThe circuit symbol for ideal transformers

1Vur

2Vuur

1Iur

2Iuur

1 2:N N

when , called an isolation transformerwhen , called an isolation transformer

NN2 2 > N> N11 , a step, a step--up transformerup transformerNN2 2 < N< N11 , a step, a step--down transformerdown transformer

2

1

1NN

=



2 2

11

1 2

12

V NNV

I NNI

=

= −

uurur

uruur

Example for different algebraic signsExample for different algebraic signs

1Vur

2Vuur

1Iur

2Iuur

1 2:N N



2 2

11

1 2

12

V NNV

I NNI

= −

=

uurur

uruur1V

ur2V

uur1I

ur2I

uur1 2:N N



2 2

11

1 2

12

V NNV

I NNI

= −

= −

uurur

uruur1V

ur2V

uur1I

ur2I

uur1 2:N N



Rule 1Rule 1 : If and are both positive at the : If and are both positive at the dotdot--marked terminal, use a plus sign for the marked terminal, use a plus sign for the turn ratio Nturn ratio N22/N/N11..Otherwise, use a negative sign.Otherwise, use a negative sign.

Rule 2Rule 2 : If and are both directed into or out of : If and are both directed into or out of the dotthe dot--marked terminal, use a minus sign for marked terminal, use a minus sign for the turn ratio.the turn ratio.Otherwise, use a plus sign.Otherwise, use a plus sign.

1Vur

2Vuur

1Iur

2Iuur



Example 9.9.2: Find the input impedance Zab for the following circuit


ZS

ZL

a

b

c

d

1: n

1V

+

−

ur

2Iuur

1Iur

2V

+

−

uur

Zab

SVuur


From the ideal transformer model

2 1

1 2

21 2

ab 2 21 2 2

1 1Z = L

V In nV I

VV Vn Zn nI nI I

= =

∴ = = =

uur uruur uur

uuruur uurur uur uur

,


ZS

ZL

a

b

c

d

1: n

1V

+

−

ur

2Iuur

1Iur

2V

+

−

uur

Zab

SVuur


Hence, ideal transformers can be used to raise or lower the impedance level of a load by choosing proper turn ratio n.


1: n

1V

+

−

ur

2Iuur

1Iur

2V

+

−

uurSV

uur



The T-equivalent circuit for magnetically coupled coils

1 21 1

1 22 2

di div L Mdt dtdi div M Ldt dt

= +

= +

Assumption: the voltage between b and d must be zero, i.e. , vbd=0



1 1 21 1

1 2 22 2

( )

( )

di di diV L M M Mdt dt dt

di di diV M M L Mdt dt dt

= − + +

= + + −

L1-M L2-M

M

a

b

c

d

i1 i2

V1 V2

+

-

+

-

time domain

phasor domain

Mathematically, they are equivalent.

L1-M or L2-M could be negative



Modeling of a practical transformer

Ll1,Ll2 : leakage inductancesLm : magnetizing inductanceRc : represents core lossR1,R2 : winding resistancesn : turn ratio of ideal transformer

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.


SummarySummary

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.


SummarySummary


SummarySummaryChapter Problems : 9.11(c)

9.199.239.279.379.489.539.619.749.79

Due within one week.

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