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  • 7/30/2019 Elec 3202 Chap 8

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    UAE University Department of Electrical Engineering Dr.Hazem N.Nounou

    Chapter 8

    AC SteadyState Analysis

    -So far , we have discussed the response of circuits due to DC source.

    -Here , we discuss circuits with sinusoidal sources.

    * Sinusoidal

    - Here we study sinusoidal functions :Consider the sinusoidal function :

    ( ) ( )tsinXtx M=Where :

    XM = amplitude

    = radian ( angular ) frequency

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    XM

    - X M2

    2

    32

    t

    t)X(

    T

    Note , X (t) repeats itself every (2 )

    radius.

    DEF : Period , TTime it takes the signal to repeat itself

    f

    1T =

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    Where: f frequency in Hertz ( Hz )

    since T = 2

    = 2 / T = 2 f

    Consider the two signals x 1 (t) & x 2 (t)

    )t(sinX(t)X

    )t(sinX(t)X

    M22

    M11

    +=

    +=

    Subtract from both signals

    t)(sinX(t)X)-t(sinX(t)X

    M22

    M11

    = +=

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    If = X1(t) & X2 (t) are in phase

    If X1(t) leads X2 (t) by Or X2(t) lags X1 (t) by

    t

    t)(sinX(t)X M22 =)-t(sinX(t)X M11 +=

    {

    )t(sinXX(t) M +=Q

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    ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )sinsincoscoscos

    cos

    sin

    cos

    sin

    sin==m

    mm

    ( ) ( ) ( ) ( )[ ]( ) ( ) ( ) ( )

    ( ) ( )tcosBtsinAX(t)

    tcossinXcostsinXX(t)

    tcossincostsinXX(t)

    MM

    M

    +=

    +=

    +=

    From trigonometry , we have

    Where :

    A = XM cos ()

    B = XM sin ()Also : tan () = sin () / cos () = B / A

    = tan -1 ( B / A )

    B

    A

    X M

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    ( )

    ( )

    ( )

    ++=

    +=

    +=

    A

    BtantsinBAX(t)

    tsinXX(t)

    BAX

    122

    M

    22

    M

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    Example :

    30)t(50cos20(t)X60)t(50sin10(t)X

    2

    1

    +=+=

    120)t(50sin20(t)X

    90)30t(50sin2030)t(50cos20(t)X

    2

    2

    +=

    ++=+=QFind the frequency , phase angle between X1 and X2

    Subtract 120 from both sides

    t)(50sin20(t)X'60)t(50sin10(t)X'

    2

    1

    = =

    Let

    Note :

    cos (x) = sin ( x + 90 o)

    sin (x) = cos ( x 90o

    )

    Frequency is50 rad/sec.

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    X 1 (t) leads X 2 (t) by -60 ( X2 leads by 60 )

    X 2 (t) lags X 1 (t) by -60 ( X1 lags by 60 )

    * Relationship between sinusoidal and complex numbers

    We know that if the complex number

    z = x + j y = r e j

    Where :

    =

    +=

    x

    ytan

    yxr

    1

    22

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    Example :

    Consider the circuit :

    V(t) = VM cos ( t)

    Find i (t) ?

    V (t)+

    -

    R

    Li (t)

    ( ) (1)dtdi(t)L(t)iRtcosV

    dt

    di(t)L(t)iRV(t)

    M LLL+=

    +=Using KVL :

    Assume that the solution is a sinusoidal function

    ( )( ) ( )

    )sin(A-A2),cos(AA1where

    (2)tsinAtcosAi(t)

    tcosAi(t)

    21

    ==

    +=+=

    LL

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    Now find A1 and A2

    Plug (2) in (1)

    ( ) ( ) ( )

    ( ) ( )[ ][ ] ( ) [ ] ( )tsinALARtcosALAR

    tcosAtsinA-L

    tsinARtcosARtcosV

    1221

    21

    21M

    ++=++

    +=

    ALAR0

    ALARV

    12

    21M

    =+=

    Solving for A1 and A2 , we find

    222

    M2222

    M1

    LR

    VLA,

    LR

    VRA

    +=

    +=

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    ( )

    2

    2

    2

    1

    1

    M

    M

    1

    2

    AAA

    RLtan

    R

    L

    VR

    VL

    A

    Atan

    )sin(A-A2and)cos(AA1since

    +=

    =

    =

    =

    =

    ==

    ( ) ( )( )

    ( ) 2222

    M

    2222

    2

    M

    222

    2222

    2

    M

    22

    2222

    2

    M

    22

    2

    2

    1

    LR

    V

    LR

    VLR

    LR

    VL

    LR

    VRAA

    +=

    +

    +=

    ++

    +=+

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    ( )

    ++=

    +=+

    =+=

    R

    Ltantcos

    LR

    Vi(t)

    tcosAi(t)LR

    VAAA

    1222

    M

    222

    M2

    2

    2

    1

    It is clear that it is very complicated to find the solution usingsinusoidal function

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    [ ]

    [ ][ ]

    VVwhereeVv(t)

    eVRe

    eeVRe

    eVRe)t(cosVv(t)

    )t(sinVj)t(cosVeV

    x)sin(j(x)cose

    formulasEulerUsing

    )tw(cosV(t)v

    thatassumedEarlier we

    M

    tj

    tj

    M

    jtj

    M

    )t(j

    MM

    MM

    )t(j

    M

    jx

    M

    ==

    ==

    =+=+++=

    +=

    +=

    +

    +

    Dropping e jt since it

    exists in all termsPhasor form

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    Phasor representation :

    ( )

    ( ) )90(AtsinA

    AtcosAformPhsordomainTime

    o

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    Example :

    V (t)+

    -

    R

    L

    o0VVisformphasorThe

    t)(cosVv(t)

    M

    M

    =

    =

    Assume :

    tj

    M

    tj

    M

    eIi(t)

    eIi(t)

    )t(cosIi(t)

    o==

    +=

    We know that

    tjo

    M

    tj

    eVtv

    Vetv

    0)(

    )(

    =

    =

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    KVL :

    [ ] [ ] [ ]

    jILIRV

    ejILeIReV

    eIdt

    dLeIReV

    dt

    di(t)L(t)iRv(t)

    tjtjtj

    tjtjtj

    +=+=

    +=

    +=

    [ ]

    +

    =

    +=+=

    222 LR

    jLRVI

    jLRVIjLRIV

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    ( )

    +

    +=

    +

    =

    R

    LtanLR

    LR

    V

    LR

    jLR0VI

    1222

    222

    M

    222M

    o

    Example :

    Convert from time domain to phasor form

    ( )( )

    Aooo

    o

    o

    o

    3012)90(12012I

    V)(-4524V

    120t337sin12i(t)

    45t337cos24v(t)

    ==

    =

    +==

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    Example :

    Convert from phasor form to time domain

    zHK1fif)(-7510I

    V12016V

    o

    o

    ==

    =

    ( ) 200010002f2 ===

    ( )

    ( )o

    o

    75t2000cos10i(t)

    120t2000cos16v(t)

    =

    +=

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    * Phasor relationships for circuit elements

    1. Resistor :

    V (t)

    +

    -R

    i (t)

    Assume :

    iv

    tj

    iM

    tj

    vM

    )t(j

    M

    )t(j

    M

    )t(j

    M

    )t(j

    M

    &

    IRV

    eIReV

    eIReV

    i(t)Rv(t)

    eVv(t)&

    eIi(t)

    iv

    v

    i

    =

    =

    =

    =

    =

    =

    =

    ++

    +

    +

    In phasor form

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    t

    i (t)

    v (t)

    )t(j

    M

    )t(jM

    i

    v

    eIi(t)

    eVv(t)+

    +

    ==

    2. Inductor :

    V (t)

    +

    -

    L

    i (t)Assume :

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    [ ])t(jM ieIdt

    dL

    dt

    di(t)Lv(t)

    +==

    o

    iv

    tj

    iM

    tj

    vM

    )t(j

    M

    )t(j

    M

    90&

    ILjVeILjeV

    eILjeV iv

    +=

    = =

    = ++

    t

    i(t)

    v(t)

    o90

    {

    The voltage leads the

    Current by 90o

    .

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    3. Capacitori (t)

    V (t)

    +

    -

    )t(j

    M

    )t(j

    M

    i

    v

    eIi(t)eVv(t) +

    +

    ==

    [ ])t(j

    Mv

    eVdt

    dCdt

    dv(t)Ci(t)

    +

    ==

    Assume :

    o

    iv

    tjvM

    tjiM

    )t(j

    M

    )t(j

    M

    90

    ICj

    1Vor

    VCjI

    eVCjeI

    eVCjeI vi

    =

    =

    ==

    = ++

    t

    i(t) v(t)

    o90

    {

    The current leadsThe voltage by 90o.

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    Example :

    If the voltage across the 20 mH inductor is

    v (t) = 12 cos (377 t + 20o ) find i (t) ?

    ( ) ( )

    ( ) ( )

    ( )o

    o

    o

    o

    o

    70t377cos1.59i(t)

    )07(1.5990m20377

    2012I

    Im20377j2012

    ILjV

    =

    =

    =

    =

    =

    L

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    Example :

    If the voltage across the capacitor is

    v (t) = 100 cos (314 t + 15o ) find i (t) ??

    C = 100 F

    ( ) ( ) ( )

    ( ) ( )( )( )o

    o6-4

    6-

    105t314cos3.14i(t)

    1053.14)9015(1010314I

    1510*100100314j

    VCjI

    +==+=

    =

    =

    oo

    o

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    In summary

    1. Resistor : RR IRV =

    LL ILjV =

    CC ICj

    1V =

    2. Inductor :

    3. Capacitor

    * Impedance and admittance :

    Definition : impedance : ( Z )The ratio of the voltage over the current

    )(I

    V

    I

    V

    I

    VZ

    V

    I

    Z

    1

    y&I

    V

    Z

    iV

    m

    m

    im

    Vm =

    ==

    ===

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    ivz

    m

    m &

    I

    VZwhere ==

    In complex form :

    ( )

    ( )

    =

    +=

    =

    =

    +=

    R

    Xtan

    XRZ

    sinZX

    cosZRwhere

    XjRZ

    1

    z

    22

    z

    z

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    In summary :

    ==

    +==

    +==

    R

    Xtan

    XRZI

    VZ

    XjRZzZZ

    formcomplexformphasor

    1

    zivz

    22

    m

    m

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    Impedance of R,L,C

    C

    1j

    Cj

    10Z)90(

    C

    1ZC

    Lj0Z90LZL

    0jRZ0RZR

    form)(complexZform)(phasorZElement

    C

    o

    C

    L

    o

    L

    R

    o

    R

    =+==

    +==

    +==

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    * Parallel and series connections of ( Z )

    Series : Z 1 Z 2 Z 3 Z n

    Z eq

    Z 1 Z 2 Z n

    Parallel

    Z eq

    n21eq Z

    1

    Z

    1

    Z

    1

    Z

    1

    +++= L

    n21eq ZZZZ +++= L

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    Definition : admittance ( y )

    It is the reciprocal of Z

    BjGy

    xRxj

    xRRy

    xR

    xjR

    xjR

    1

    Z

    1y

    V

    I

    Z

    1y

    2222

    22

    +=

    ++

    +=

    +=

    +==

    == [S] siemens

    Conductance Susceptance

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    * In terms of R ,L ,and C

    o

    CC

    o

    L

    LL

    R

    RR

    90CCjyCj

    1Z-

    )90(L

    1

    Lj

    1

    Z

    1yLjZ-

    GR

    1

    Z

    1yRZ-

    ===

    ====

    ====

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    * Parallel and series connections of ( y )

    Series : y 1 y 2 y 3 y n

    y S

    y 1 y 2 y n

    Parallel

    y P

    n21S y

    11

    y

    1

    y

    1+++= L

    y

    n21P yyyy +++= L

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    Example :

    Find the equivalent impedance

    sec/rad2

    F2C,F1C

    H2L,2R

    H1L,1R

    21

    22

    11

    =

    ==

    ==

    ==

    L 2

    L 1 C 1

    C 2

    R1

    R2

    Z eq

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    ( ) ( )

    4

    1j

    C

    1jZ,

    2

    1j

    C

    1jZ

    4jLjZ,2j12jLjZ

    2RZ,1RZ

    2

    C2

    1

    C1

    2L21L1

    2R21R1

    ====

    =====

    ====

    ( )( ) ( )

    22.33.2430

    37j30

    30

    845j3Z

    2

    3j3

    154j

    2

    3j3

    415j

    1

    22

    1j2j1

    41j4j

    41j4j

    ZZZZZ//ZZ

    o

    eq

    R2C1L1R1C2L2eq

    =+=

    +=

    ++=++=

    +++

    =

    ++++=

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    v (t)

    40 m H

    20+

    - F50

    i (t)

    Example :

    Find the current i (t) in the circuit

    ( )( )( )

    sec/rad377

    V)-30(120V

    30t377cos120v(t)

    9060t377cos120v(t)

    60t377sin120v(t)

    o

    o

    oo

    o

    ==

    =

    +=

    +=

    20RZ

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    V

    Z L

    R+

    -CZ

    I

    ( ) ( )

    ( )( )

    53.05jZ

    503771j

    C1jZ

    15.08jZ,m40377jLjZ

    20RZ

    C

    C

    LL

    R

    =

    ==

    ===

    ==

    Z eq+

    -V

    ( )( ) ( )

    ( )oo

    o

    o

    o

    CLReq

    eq

    39.23t377cos3.87i(t)

    A)39.23(3.879.2330.96

    )30(120I

    9.2330.9653.05j//15.08j20

    Z//ZZZ

    Z

    VI

    =

    =

    =

    =+=+=

    =

    4j

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    Example :

    Find the equivalent impedance :

    4

    Z eq

    2

    2

    2j

    6j

    4j-

    ( ) ( )[ ]

    ( ) ( )[ ]( ) ( )

    ( )17.653.551.076j3.38Z

    2

    4j6

    2j42j2

    22j4//2j2

    24j6j4//2j2Z

    o

    eq

    eq

    =+=

    ++

    ++=

    +++=

    +++=

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    Phasor diagram :

    A diagram that shows the magnitude and phase of variousvoltage and currents in the circuit

    For example :

    o

    1

    o2

    o

    1

    1307I

    605V

    4510V

    ==

    =

    V 1 V 1I 1

    60o

    45o

    130o10

    57

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    Example :

    Draw the phasor diagram of all currents and voltage

    I 1 I 2

    j42 =Z= 21Z

    o

    454I =

    +

    -

    V

    A18.433.574j2

    4j454

    ZZ

    ZII

    454I

    oo

    21

    21

    o

    =

    =

    +

    =

    =

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    ( ) V18.437.1418.433.572ZIV

    A108.431.78

    4j2

    2454

    ZZ

    ZII

    oo

    11

    oo

    21

    12

    ===

    =

    =

    +

    =

    | I | = 4|V | = 7.14

    | I2

    | = 1.79

    45o108.4o

    | I1 | = 3.57

    18.43o

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    Circuit Analysis Technique

    Note :All analysis techniques discussed before can be used in AC steady-

    state analysis

    Example :Find Io using nodal analysis ?

    I 1I 2

    1j

    V2+-

    A901 o

    I o

    V11j

    1j

    o01

    1

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    Here , we have a supernode :

    Applying KCL at supernode

    ( ) ( ) (1)901V450.707V450.7071j1

    1

    1j

    1V

    1j1

    V901

    1j1

    V

    1j

    V

    1j1

    V901

    III901

    o

    2

    o

    1

    o

    21o

    221o

    2o1

    o

    LL=+

    ++

    =

    ++=

    ++=

    )2(01VV

    01VV

    o

    21

    o12

    LL=

    +=

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    A)18.44(1.58

    1j

    VI

    V71.561.58V

    V108.41.58V

    o2o

    o2

    o

    1

    ==

    =

    =

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    Example :

    use mesh analysis to find Vo ?

    I 2

    +

    -

    2

    +

    -

    2j

    2j

    2

    A902

    o

    o024V o

    I 1

    Supermesh

    KVL around supermesh( )

    ( ) (1)024I2j2I2

    0I2I2jI2024

    o

    21

    221

    o

    LL=+

    =++

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    UAE University Department of Electrical Engineering Dr.Hazem N.Nounou

    V36.0210.88I2V

    A36.025.44I

    A15.254.56I

    (2)902II

    902II

    o

    20

    o2

    o

    1

    o21

    o

    12

    ===

    =

    =+

    =

    LL

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    UAE University Department of Electrical Engineering Dr.Hazem N.Nounou

    Example :

    find Vo ?

    I 2

    +

    -

    2 1j

    V012

    o

    A06 o

    V o+

    -

    V 1 V 2

    2j 1

    2

    I 3I 1

    03

    V

    2j

    V

    1j

    VV

    0III

    2221

    321

    =

    =

    V012V o1 =

    KCL at node V2

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    UAE University Department of Electrical Engineering Dr.Hazem N.Nounou

    ( )

    V33.76.65V3

    1

    21

    1VV

    V33.719.969V

    j2

    13

    19012

    2j1

    31

    9012V

    90122j

    21

    3

    1V

    9012j

    012

    3

    1

    j2

    1

    j

    1V

    03

    V

    2j

    V

    1j

    V012

    o

    22o

    o

    2

    oo

    2

    o

    2

    oo

    2

    222

    o

    ==

    +=

    =

    +=

    =

    =

    ++

    =

    =

    ++

    =

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    UAE University Department of Electrical Engineering Dr.Hazem N.Nounou

    Example :

    Use Thevenin Theorm to find Vo ?

    +

    -

    V012 o

    1 2 1

    1j1j

    Vo

    +

    -

    First find VOC

    +

    -

    V012 o

    1 2 1

    j 1j Voc

    +

    -

    I1

    I3

    I2

    Vx

    KCL around Vx

    0III

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    UAE University Department of Electrical Engineering Dr.Hazem N.Nounou

    0III 321 =

    V33.693.32j2

    j1VV

    V)-29.7(7.44V3.69j6.460.8j1.4

    12

    V

    5

    11j

    5

    21V012

    14

    j2j1V012

    j2

    1j1V012

    0j2

    V

    j

    V

    1

    V012

    o

    xoc

    o

    xx

    x

    o

    x

    o

    x

    o

    xxx

    o

    =

    +=

    ==+=

    +

    +=

    +

    ++=

    +

    ++=

    =+

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    1 Vo

    +

    -

    +

    -

    ZTH

    Voc

    V12.541.3V

    33.71.661

    1

    33.693.32

    Z11VV

    o

    o

    o

    o

    TH

    OCo

    =

    +=

    +=

    Find ZTH:

    33.71.661}1||]2)||1{[( oTH =++= jJZ