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Sinusoidal Source Phasors Kirchhoff Laws in the

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					                        SINUSOIDAL STEADY
                        STATE ANALYSIS




C.T. Pan                                       1




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
C.T. Pan                                       2
9.6 The Node-Voltage Method
9.7 The Mesh-Current Method
9.8 Circuit Theorems
9.9 The Coupling Inductors and Ideal Transformer



C.T. Pan                                            3




           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.
C.T. Pan                                            4
              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.

C.T. Pan                                                5




              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
C.T. Pan                                                6
                   9.1 Sinusoidal Source
                  1 2π
           T=       =           , period , in sec.
                  f   ω




C.T. Pan                                                                        7




                   9.1 Sinusoidal Source
v (t + T       ) = V m sin (ω t + ω T + θ )             -sin
                = V m sin (ω t + 2 π + θ )                     ψ   i(t)
                = V m sin (ω t + θ )
                = v (t )                         -cos
                                                               θ
                                                                          cos
If     i ( t ) = I m cos (ω t + θ )
then
           i ( t ) = I m sin (ω t + θ + 90 ° )          sin
                = − I m cos (ω t + θ + 180 ° )
                = − I m sin (ω t + θ + 270 ° )
C.T. Pan                                                                        8
                  9.1 Sinusoidal Source
   also , i(t)= -I msin(wt - ψ) , ψ = 90 o - θ          -sin
                                                               ψ       i(t)
                  = -I m cos(wt - ψ - 90 )
                                        o


                  = I msin(wt - ψ - 180 o )
                                                                   θ
                  = I m cos(wt - ψ - 270 o )     -cos                         cos




                                                        sin



 C.T. Pan                                                                      9




                  9.1 Sinusoidal Source


A coswt + B sinwt = C cos(wt - θ)
                         B
            θ = tan -1
                         A
            C = A2 + B 2                                   θ




 C.T. Pan                                                                     10
                      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
C.T. Pan                                                  11




                      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)

C.T. Pan                                                  12
                         9.2 Phasors
        The inverse phasor transform is defined as
                  −1
                     r
                     u       r
                             u jwt
                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.


 C.T. Pan                                                                   13




                         9.2 Phasors
Example :
v(t ) = 100 cos( wt + 30o )V                               ur
                                                           V = 100∠30o V

i (t ) = 50sin( wt + 70 )A
                     o


The phasor diagram is as follows :
       uur      r                                         r
                                                          I = 50∠ − 20° A
Phasor V leads I by 50o
       r       r
              uu
Phasor I lags V by 50o

 A convention used in power systems is as follows:
 The electric load is lagging, which means the load current
 phasor lags the voltage phasor.
 C.T. Pan                                                                   14
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
                                 mK   cos( wt + θ K ) = 0

            Take P transform on both sides
                         n

                     ∑I
                     K =1
                                  mK      ∠θ K = 0 + j 0
                         n        r
                                 uu
             or     ∑I
                     K =1
                                  K       = 0 , for any node
 C.T. Pan                                                            15




9.3 Kirchhoff’s Laws in the Frequency Domain
             KVL
                             n

                         ∑ v (t) = 0
                         k=1
                                      k             , for any loop

                   For sinusoidal voltages
                             m

                         ∑V
                         k=1
                                      mk   cos(wt + φ K ) = 0

               Take P transform
                              m

                             ∑V
                             k =1
                                          mk   ∠ φk = 0 + j 0
                              muur
              or             ∑ VK = 0 , for any loop
                             K =1
 C.T. Pan                                                            16
9.4 Component Models in Phasor Domain

                                         Given iR (t ) = I m cos( wt + θ )
                                         then vR (t ) = RI m cos( wt + θ )
                                       Take P transform
                                               r
                                         Given I R = I m ∠θ
                        r
                        u
                                         then P(vR (t )) = RI m ∠θ
                        VR

                                                             r
                                                         = RI R
                  r                                        r
                                                           u
                                                         =V R
                  IR




           r
           u       r
C.T. Pan   V R and I R are in phase.                                   17




9.4 Component Models in Phasor Domain

  ur    r
 ∴V R =RI R                                                  IR
                   ur
  Similarity given V R , one can find
   r    1 u r     ur
   I R = V R = GV R                               VR              R
        R
  R : resistance , G : conductance
                                                 symbol

C.T. Pan                                                               18
9.4 Component Models in Phasor Domain
           Given iL (t ) = I m cos(ωt + θ )
                r
           then I L = I m ∠θ
                                 di
           Also, vL (t ) = L        = -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
C.T. Pan                                                   19




9.4 Component Models in Phasor Domain
                r          r         r
      r     ∴ VL = jω LI L @ j Χ L I L
      IL                      r
            similarly, given VL one can find
                r        1 r           r
r              IL =          VL @ jΒ LVL
VL                      jω L
            Χ L = ω L , inductive reactance , in Ω
                             (電感性電抗 )
symbol               1
            ΒL = -       , inductive susceptance , in S
                    ωL
                           (電感性電納)
            Χ L and Β L are frequency dependent.
C.T. Pan                                                   20
9.4 Component Models in Phasor Domain
                 Im
           r
           VL
                                  r
                                  IL

                                       Re
                0

           r        r
           I L lags VL by 90°.
            r        r
           VL leads I L by 90°.
C.T. Pan                                                  21




9.4 Component Models in Phasor Domain
                Given vC (t ) = Vm cos(ω t + θ )
                       Ρ (vC (t )) = Vm ∠θ
                                    dvC (t )
                then    iC (t ) = C
                                      dt
                                = - ω CVm sin(ω t + θ )
                                = 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
C.T. Pan                                                  22
9.4 Component Models in Phasor Domain
              r           r           r
           ∴ IC = jωCVC @ jΒCVC
           similarly, one can get
                r       1 r             r
              VC =          I C @ jX C I C
                       jω C
           ΒC = ωC , capacitive susceptance , in S
                          (電容性電納)
                     1
           ΧC = -      , capacitive reactance , in Ω
                    ωC
                       (電容性電抗 )
C.T. Pan   ΒC and ΧC are frequency dependent.          23




9.4 Component Models in Phasor Domain

                           Im
                     r
                     IC

                                       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        Time domain             Frequency domain
                                          r     r
                v = Ri                    V = RI
    R                                     r     r
                i = Gv                    I = GV
                v=L
                      di                  r       r
                      dt                  V = jω LI
    L                        1 t          r    1 r
                                          I=
                             L ∫0
                i = i(0) +        vdt             V
                                              jωL
                             1 t          r     1 r
                             C ∫0
               v = v(0) +         idt     V=       I
                                               jωC
    C                 dv                  r        r
               i =C                       I = jωCV
                      dt




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        R , for resistor
           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
                θ = tan -1
                         R
                R= Z cos θ , X= Z sin θ
                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

        1     1   R − jX
 QY =     =     =          = G + jB
        Z R + jX R 2 + X 2
            R
 ∴G =
        R2 + X 2
          −X
   B=
       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        r
                             uu                 uur
         ur       + V1 -   + V2 -             + VN -
         V
         −




                               N     r
                                    uu
                         ur    ∑    Vk  N
                         V
                  Z eq @ r =      r = ∑ Zk
                               k =1

                         I        I    k =1
9.5 Series, Parallel, and Delta-to-Wye
    Simplifications
 special case N=2

         r                                                  ur      Z1 u   r
         I                                                  V1 =          V
                                                                 Z1 + Z 2
    +          ur            +r
    ur       + V1 -          uu                              r
                                                            uu      Z2 u   r
    V                        V2                             V2 =          V
    −                        -                                   Z1 + Z 2

                 voltage division principle




9.5 Series, Parallel, and Delta-to-Wye
    Simplifications
  (b) Parallel connection
             r
             I
                        ur                       r
                                                uu                  r
                                                                   uu
     +                  I1                      I2                 IN
     ur
     V
     −
                                    N    r
                                        uu
                         r      ∑I          k         N
                   1     I                                  1
                      @ ur =
                  Z eq V
                                 k =1
                                      ur
                                      V
                                                 =   ∑Z
                                                     k =1   k
                                N
                  or   Yeq =   ∑Y
                               k =1
                                        k
 9.5 Series, Parallel, and Delta-to-Wye
     Simplifications
   special case N=2
                        ur        r
                                 uu         ur
            +                                      Z2 r
     r                                      I1 =
                        I1       I2
            ur                                   Z1 + Z 2
                                                          I
     I      V            Z1      Z2          r
                                            uu      Z1 r
                                            I2 =          I
            −                                    Z1 + Z 2

                       Z1Z 2
             Z eq =
                      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
a
                   Zc
                                          b
                                              ∆ − Y Conversion

           Z1                   Z2
                                                          Zb Zc
                                               Z1 =
                                                      Z a + Zb + Zc
                   n
      Zb                             Za                   Zc Za
                                               Z2 =
                           Z3                         Z a + Zb + Zc
                                                          Z a Zb
                                               Z3 =
                                                      Z a + Zb + Zc
                       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
                                                               r
                                                              uu
          20∠ − 15o                                   j 20Ω   VO




                                    − j25 × j20
        − j25Ω P j20Ω =                         = j1 0 0 Ω
                                    − j25 + j20
          uur           j1 0 0 Ω
        ∴ VO =                     (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
           Zan =                        =           = 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




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    The same analysis techniques for DC resistive
       circuits can be extended to AC circuit analysis.
  n    For example
           voltage divider and Y-△ transformation
           Nodal analysis
           Mesh analysis
           Superposition theorem
           Source transformation
           Thevenin and Norton equivalents
 C.T. Pan                                                 41




       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.
 C.T. Pan                                                 42
            9.6 The Node-Voltage Method
    n       Example 9.6.1: Find ix using nodal analysis


              20 cos 4t v                                  2ix



    n       Step1: Phasor Domain
                                         ur                     r
                                                               uu
                              10Ω        V1 jω L = j 4Ω        V2
                                     r
                                    uu
                                    Ix      1                         r
                                                                     uu
               20∠00                           = − j 2.5Ω           2I x    j 2Ω
                                           jωC


                                                                                          43
 C.T. Pan




            9.6 The Node-Voltage Method
                                                          ur         r
                                                                    uu
    n       Step2 : Choose datum node & V1 , V2
                    After source transformation
                                           ur                  r
                                                              uu
                                           V1     jω L = j 4Ω V
                                       r
                                      uu
                                                               2


              20∠0°                   Ix         1                           r
                                                                            uu
                                                    = − j 2.5Ω             2I x    j 2Ω
                            10Ω                 jωC
                10



                  1            1             1  ur
                  10 + j 0.4 + j 4         −
                                              j 4  V1  
                                                               20∠ 0 0 
                                                    uu  =           
                                                        r      10 
                                                1              r
                                                                  uu
                                             +  V2   2 I x 
                           1               1
                       −                                              
C.T. Pan                  j4              j4 j2                                          44
           9.6 The Node-Voltage Method
                                      ur                  r
                                                         uu
                                      V1     jω L = j 4Ω V
                                 r
                                uu
                                                          2


           20∠0°                Ix          1                    r
                                                                uu
                                               = − j 2.5Ω      2I x   j 2Ω
                        10Ω                jωC
             10



                     r
                    uu ur               r
                                       uu         ur
                    I x = V1 × j0.4 , 2I x = j0.8×V1
                                               ur
              1 + j1.5            j 2.5    V1   20
                                                r
              − j 0.8 + j 0.25 − j (0.75)   uu  =  0 
                                           V2   
                                              
C.T. Pan                                                                     45




           9.6 The Node-Voltage Method
                                      ur                   r
                                                          uu
                                      V1      jω L = j 4Ω V
                                  r
                                 uu
                                                           2


            20∠0°                Ix         1                    r
                                                                uu
                                               = − j 2.5Ω      2I x   j 2Ω
                         10Ω               jωC
              10


                                                ur
           × j 20     1 + j1.5        j 2.5 V1   20 
                       11                       r
                                               uu  =
                                       15  V2   0 
                                               
                      ur
                      V1 = 18.97∠18.430 (V)
                       r
                      uu          ur
C.T. Pan
                      I x = j0.4 ×V1 = 7.59∠108.40 (A)                       46
           9.6 The Node-Voltage Method


               20 cos 4t v                               2ix




        Step3 :        ix = 7.59cos(4t + 108.40 )              (A)




C.T. Pan                                                                  47




                          u
                         ur
           9.7 The Mesh-Current Method
                         I
  Example 9.7.1: Determine               o   using mesh analysis
                                              4Ω

                             ur                          I0
                 5∠0o A      I3          − j 2Ω
                                                r
                                               uu
                                               I2         20∠90o V
                             j10Ω
                               ur
                     8Ω        I1        − j 2Ω


                                    ur               r
                                                    uu         ur
    n      Mesh1 : (8 + j10 − j 2)I − (− j 2)I − j10I = 0 + j0
                                   1          2      3

    C.T. Pan                                                         48
     9.7 The Mesh-Current Method
                                                            4Ω

                                         ur                              I0
                       5∠0o A            I3              − j 2Ω
                                                                r
                                                               uu
                                                               I2         20∠90o V
                                         j10Ω
                                           ur
                             8Ω            I1            − j 2Ω


                                   r
                                  uu              ur                ur
    Mesh2 : (4 − j 2 − j 2) I 2 − ( − j 2) I1 − ( − j 2) I 3 = −20∠90
                                                                      0
n

    One current source, only two unknowns
                  ur
                  I 3 = 5 ∠ 0 0 (A )
C.T. Pan                                                                             49




     9.7 The Mesh-Current Method
                                                 4Ω

                                  ur                        I0
                    5∠0o A        I3          − j 2Ω
                                                     r
                                                    uu
                                                    I2       20∠90o V
                                  j10Ω
                                    ur
                       8Ω           I1        − j 2Ω


                                     ur
                  8 + j8    j2    I1   j50 
                                      r
                   j2 4 − j4 uu  = − j30
                                  I2  
                                              
                     r
                    uu
                  ∴I2 = 6.12∠− 35.220 A
                     r
                    uu uu r
C.T. Pan          ∴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.
C.T. Pan                                                  51




                9.8 Circuit Theorems
                               r
                              uu
           The component I o ′ due to 20∠ 900 V source
                                        4Ω
                                             uu r
                                             I 0′
                                   − j 2Ω

                                               20∠90o V
                       j10Ω
               8Ω                  − j 2Ω


                  r
                 uu ′
                 I o = −2.353 + j 2.353 A
                                    r
                                   uu
C.T. Pan     The component I o ′′ due to 5∠00 A           52
                       9.8 Circuit Theorems
                                                                 r ''
                                                                 Io
                              5A
                                                   -j2Ω

                                    j10Ω
                                                   -j2Ω


                              r
                             uu ′′
                             I o = −2.647 + j1.176 A
By superposition theorem
            r     r
           uu uu ′ uu ′′r
           I 0 = I o + I o = − 5 + j 3.529                              A
                                        =6.12 ∠144.78 0
    C.T. Pan                                                                     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




    C.T. Pan                                                                     54
                   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


                               −1
                     ∴ v1 =        × 5V = −1V
  C.T. Pan
                              1+ 4                                 55




                   9.8 Circuit Theorems
    (2) v2(t)=? , ω2=2 rad/s
             j4Ω     1Ω         4Ω
                       r
                      uu                                  v0
                      V2
10∠0o                          -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
  C.T. Pan     v2 ( t ) = 2.498cos ( 2t − 30.79o ) V               56
                          9.8 Circuit Theorems
      (3) v3(t)=? , ω3=5 rad/s
                     ur                  Z1 = ( − j 2Ω ) P 4Ω = 0.8 − j1.6Ω
                     I1
                          uu
                           r             Current divider
                          V3             r         j10Ω
                                         I1 =               × 2∠ − 90o A
                     2∠ − 90o A                j10 + 1 + Z1
                                         r r
                                         u
                                         V 3 = I 1 ×1Ω = 2.328∠ − 80o V
                               v0        v3 ( t ) = 2.328cos ( 5t − 80o ) V
10cos 2t v

                                         ∴ v0 ( t ) = −1 + 2.498cos(2t − 30.790 )
                           2sin 5t A



                                                         + 2.328sin ( 5t + 10o ) V
     C.T. Pan                                                                  57




                          9.8 Circuit Theorems
        n       Source Transformation


                uu
                 r                                  ur
                Vs                                  Is



                                                                v
                                                                u
                          uv       v                       v Vs
                          V s = Zs I s                     Is =
                                                                Zs


     C.T. Pan                                                                  58
                  9.8 Circuit Theorems
                            r
                            u
n Example 9.8.3 : Calculate Vx




                                                         r
                                                        uu
                                                        Vx
                20∠ − 90o V




                                                             r
                                                            uu
                                                            Vx
                        4∠ − 90o A



                   Z1 = 5Ω P ( 3 + j 4 ) Ω = 2.5 + j1.25Ω
C.T. Pan                                                              59




                  9.8 Circuit Theorems



           4∠ − 90o Z1                                            r
                                                                 uu
                                                                 Vx
           = 5 − j10V

      ur
     ∴V x = 5.519∠ − 28o V by voltage division principle




C.T. Pan                                                              60
                     9.8 Circuit Theorems
n    Thevenin and Norton Equivalent Circuits
       A linear two terminal circuit under sinusoidal steady
       state condition


         ur                                   r
         V TH                                 IN


               ur    ur r                     r     r u    r
               V TH =V o = I N Z N            I N = I s = V TH / ZTH
               ZTH = Z N                      Z N = ZTH
    C.T. Pan                                                           61




                     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 )
                           = 6.48 − j 2.64Ω
    C.T. Pan                                                           62
                 9.8 Circuit Theorems
         voltage division
           r
           u     ur     ur u   r  8            j12 
         ∴V TH = V ab = V a − V b =         −        120∠75 V
                                                             o

                                     8 − j 6 4 + j12 
                = 37.95∠220.31o V

                                 ZTH
                                               a
                      ur
                      V TH
                                               b

    C.T. Pan                                                      63




                 9.8 Circuit Theorems
                                r
n   Example 9.8.5 : Obtain I o using Norton’s theorem
                                                   r
                                                   Io
                                       3∠0 A
                                          o




                    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
    C.T. Pan                                                      64
               9.8 Circuit Theorems
                                                 r
                                                 II
                                     3∠0o A




                                        r
                                        Io
                      r
                      IN


               r             5         r
               Io =                    I N = 1.465∠38.48o A
C.T. Pan              5 + ( 20 + j15 )                                65




           9.9 The Coupling Inductors and
                 Ideal Transformer

                              v1 (t )   L1   M  d  i1(t ) 
                             v 2(t )  =  M   L 2  dt i 2(t ) 
                                                             


Under sinusoidal steady state , given

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

C.T. Pan                                                              66
           9.9 The Coupling Inductors and
                 Ideal Transformer
                       v1(t )   L1    M  d  i1(t ) 
                      v 2(t )  =  M   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°)
C.T. Pan                                                       67




           9.9 The Coupling Inductors and
                 Ideal Transformer
Take phasor transform on both sides


                              r
                             uu         ur         uu r
                             V 1 = jω L1I 1 + jω M I 2
                             uur         ur          uur
                             V 2 = jω M I 1 + jω L 2 I 2




C.T. Pan                                                       68
            9.9 The Coupling Inductors and
                  Ideal Transformer
Example 9.9.1:
    Find the input impedance Zab of the following circuit

                                               jω M


                         ur                                          r
                                                                    uu
                         I1            jω L1           jω L2
                                                                    I2




 C.T. Pan                                                                                            69




            9.9 The Coupling Inductors and
                  Ideal Transformer
    Use mesh current method:
                r
               uu                        ur         r
                                                   uu
               VS = ( Z S + R1 + jω L1 ) I1 − jω M I 2              (A)
                             ur                        r
                                                      uu
                0 = − jω M I1 + ( R2 + jω L2 + Z L ) I 2            (B)
                     r
                    uu         jω M        ur
     From (B) I 2 =                        I1
                          R2 + jω L2 + Z L
                                                                 ur
                     r
                    uu                        ur         ω 2 M 2 I1
     From (A)       VS = ( Z S + R1 + jω L1 ) I1 +
                                                      R2 + jω L2 + Z L
                                                                            jω M
                                                 ZS     a      R1                          R2 c

                                                        ur                                   r
                                                                                            uu
                                        VS              I1          jω L1          jω L2                  ZL
                                                                                            I2
 C.T. Pan                                                                                            70

                                                        b                                        d
           9.9 The Coupling Inductors and
                 Ideal Transformer
                r
               uu
               VS                          ω2M 2
      ∴ Z ab = ur − Z S = R1 + jω L1 +
               I1                      R2 + jω L2 + Z L

      The reflected impedance due to the secondary coil
      and load impedance
                                       ω 2M 2
                        Zr =
                                   R2 + jω L2 + Z L
                                                                         jω M
                                                  ZS   a    R1                          R2 c

                                                       ur                                 r
                                                                                         uu
                                            VS         I1        jω L1          jω L2                  ZL
                                                                                         I2
C.T. Pan                                                                                          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        jω M           r
                                     uu
             I1                      I2          ur         ur        uur
     ur                                          V1 = jω L1 I1 + jω M I 2                     (A)
                                           r
                                          uu      r
                                                 uu          ur        uur
     V1           L1          L2          V2     V2 = jω M I1 + jω L2 I 2                     (B)




C.T. Pan                                                                                          72
           9.9 The Coupling Inductors and
                 Ideal Transformer
                     ur     r
                           uu
                         ur   V1 − jω M I 2
           From (A) I1 =                        (C)
                                  jω L1
           Substitute (C) into (B)
                      r
                     uu          r              r
                                uu M ur jω M 2 uu
                     V2 = jω L2 I 2 + V1 −     I2
                                     L1    L1
                    M
           Qk =           = 1 ⇒ M = L1 L2
                    L1 L2
             r
            uu          r
                       uu    L ur          r
                                          uu
           ∴V2 = jω L2 I 2 + 2 V1 − jω L2 I 2
                             L1
                    L2 ur N 2 ur
C.T. Pan
                =      V1 =    V1                        73
                    L1      N1




           9.9 The Coupling Inductors and
                 Ideal Transformer
                                                 N2
           As L1, L2, M →∞ such that       remains the
                                                 N1
           same, the coupled coils become an ideal
           transformer.




C.T. Pan                                                 74
           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           jω M            r
                                          uu
              I1                          I2 = 0              ur         ur
                                                              V1 = jω L1 I1
   ur                                               r
                                                   uu          r
                                                              uu          ur
   V1              jω L1                           V2
                                  jω L2                       V2 = jω M I1

                       r
                      uu
                     V2 M                       L1 L2        L2 N 2
                    ∴ ur =  =                         =        =
                      V1 L1                     L1           L1 N1
C.T. Pan                                                                               75




           9.9 The Coupling Inductors and
                 Ideal Transformer
              ur           jω M             r
                                           uu
              I1                           I2
                                                               r
                                                              uu        ur          r
                                                                                   uu
    ur                                               r
                                                    uu        V2 = jω M I1 + jω L2 I 2 = 0
    V1             jω L1          jω L2             V2 = 0



                ur
                I1 − L   − L2     L   N
                 r
              ∴ uu = 2 =       =− 2 =− 2
                I2  M    L1 L2    L1  N1



C.T. Pan                                                                               76
           9.9 The Coupling Inductors and
                 Ideal Transformer
   The circuit symbol for ideal transformers
             ur               uu
                               r           r
                                          uu
             I1    N1 : N 2   I2          V2 N 2
                                          ur =
      ur                            r
                                   uu     V1 N1
      V1                           V2     ur
                                          I1    N
                                           r
                                          uu = − 2
                                          I2    N1
              N2
     when        = 1 , called an isolation transformer
              N1

              N2 > N1 , a step-up transformer
              N2 < N1 , a step-down transformer
C.T. Pan                                                   77




           9.9 The Coupling Inductors and
                 Ideal Transformer
  Example for different algebraic signs
              ur                    uu
                                     r           r
                                                uu
              I1       N1 : N 2     I2          V2 N 2
                                                ur =
   ur                                    uu
                                          r     V1 N1
                                                ur
   V1                                    V2     I1    N
                                                 r
                                                uu = − 2
                                                I2    N1



C.T. Pan                                                   78
           9.9 The Coupling Inductors and
                 Ideal Transformer
              ur               r
                              uu        uur
              I1   N1 : N 2   I2        V2    N
                                        ur = − 2
   ur                              uu
                                    r   V1
                                        ur
                                              N1
   V1                              V2   I1 N
                                         r
                                        uu = 2
                                        I 2 N1




C.T. Pan                                           79




           9.9 The Coupling Inductors and
                 Ideal Transformer
              ur              uu
                               r         r
                                        uu
              I1   N1 : N 2   I2        V2    N
                                        ur = − 2
   ur                              uu
                                    r   V1
                                        ur
                                              N1
   V1                              V2   I1    N
                                         r
                                        uu = − 2
                                        I2    N1




C.T. Pan                                           80
              9.9 The Coupling Inductors and
                    Ideal Transformer
                    ur         r
                              uu
      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       r
                           uu
      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.

   C.T. Pan                                                   81




              9.9 The Coupling Inductors and
                    Ideal Transformer
Example 9.9.2: Find the input impedance Zab for the
               following circuit
                                   ur           r
                                               uu
                   ZS    a         I1   1: n   I2   c
                         +                          +
               r
              uu         ur                          r
                                                    uu
              VS         V1                         V2   ZL
                         −                          −

                         b                          d
                   Zab


   C.T. Pan                                                   82
           9.9 The Coupling Inductors and
                 Ideal Transformer
             From the ideal transformer model
                r
               uu        ur
               V2        I1
                r         r
               uu = n , uu = n
               V1        I2
                             r
                            uu
                       r
                      uu V 2             r
                                        uu
                     V1          1      V2 1
                               r
               ∴ Zab= ur = uu = 2
                               n         r
                                        uu = 2 ZL
                      I 1 nI 2 n        I2 n
                                                         ur                     r
                                                                               uu
                                         ZS   a          I1        1: n        I2   c
                                              +                                     +
                                    r
                                   uu         ur                                     r
                                                                                    uu
                                   VS         V1                                    V2        ZL
                                              −                                     −
                                              b                                     d
C.T. Pan                                                                                 83
                                        Zab




           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                     r
                                                              uu
                                        I1        1: n        I2
                              +                                           +
                       r
                      uu      ur                                           r
                                                                          uu
                      VS      V1                                          V2
                              −                                           −




C.T. Pan                                                                                 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
C.T. Pan                                                                                 85




           9.9 The Coupling Inductors and
                 Ideal Transformer
                                                     L1-M            L2-M
                                        a                                            c
               di        di       di
V1 = ( L1 − M ) 1 + M 1 + M 2               +
                                                i1                          i2
                                                                                 +
               dt        dt       dt
                                        V1                       M               V2
          di1     di2              di
V2 = M        +M      + ( L2 − M ) 2        -                                    -
          dt      dt               dt
                                        b                                            d
                                                            time domain
L1-M or L2-M could be
negative

Mathematically, they are
equivalent.
                                                        phasor domain
C.T. Pan                                                                                 86
               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
    C.T. Pan                                                    87




                            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.
    C.T. Pan                                                    88
                        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.




    C.T. Pan                                                  89




                        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.
    C.T. Pan                                                  90

				
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