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Overview
• Phasors Transform
• Phasor Mathematics
• Circuit Analysis Using Phasors
02-Phasors
ECEGR 450
Electromechanical Energy Conversion
Dr. Louie
Questions
Phasors
Why are AC circuits solved in the phasor domain?
• Analysis of circuits with R, L and/or C
components requires solving differential
equations
• We will consider only sinusoidal steady-state
voltages and currents of the same frequency
• We can then analyze the circuits in the phasor
domain much easier than in the time domain
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Dr. Louie
Phasor Transform
• Used for steady-state calculations
• Contains amplitude and phase angle information
 Assumed that frequency is known
v(t)  vmax cos  t  v 
time-domain representation
e jx  cos  x   j sin  x 
Euler’s Identity

v(t)  vmax Re e
• Relies on Euler’s Identity
Phasor
Transform
j t v 

using Euler’s Identity
implied, so we suppress this
v(t)  vmax Re e jt e jv
regrouped


Vmax v
v(t)  vmax ejv  vmax v
V
vmax
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Phasor Transform
• Shorthand for writing sinusoidal functions
v(t)  vmax cos  t  v 
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e jv  Vrmse jv
transformed
in power we divide by 2 , also known
as the effective phasor
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Phasor Transform
Notation
• We use the effective phasor because
• A bug in the notation
• Book uses a tilde to indicate that a variable is a
phasor, as in V, I
P  vrmsirms cos()
 So we can then write
P  VI cos()
• Unless otherwise specified, assume that voltages
and currents are given in RMS and all phasors are
“effective phasors”
• Also note:
 In other words, V, I are understood to have a
magnitude and phase component
• Book uses V, I to indicate the magnitude of the
phasor
e j90  cos(90o )  jsin(90o )  jsin(90o )   j
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Notation
Phasors
• Lecture slides use bold uppercase variables (e.g.
V, I) for phasors and other vectors
• Capital letters (e.g. V, I) or absolute values of
phasors (|V|, |I|) are used to indicate the
magnitude of the phasor
• write the phasor
representation of
v1(t)  1.41 cos  377t  0 
V  V
• Lowercase variables (e.g. v, i) are preferred to
represent scalars not associated with phasors and
vectors
 Notable exceptions P, Q for real and imaginary
power
• write the phasor
representation of
v2 (t)  2.12 cos  377t  45
• write the phasor
representation of
v3 (t)  1.41 cos  t  0 
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Phasors
Phasors
Phasors have a direct geometric interpretation
• write the phasor
representation of
solution
• write the phasor
representation of
solution
v1(t)  1.41 cos  377t  0 
 Polar form
V1  10
v2 (t)  2.12 cos  377t  45
V1  145
V1  45o
V2  1.545
• write the phasor
representation of
V2  30
Horizontal is reference
solution
v3 (t)  1.41 cos  t  0 
V3  10
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Phasors
Addition of Phasors
• Another way of specifying phasors is in
rectangular form
• Addition and subtraction of phasors are simple
using rectangular form
 Let the Y-axis be the imaginary (j) axis
 Let the X-axis be the real axis
 Simply add/subtract the real values and
add/subtract the imaginary values
V3  V1  V2  3.707  j0.707
• Resolving into real and imaginary components
V1  145  0.707  j0.707
Addition is “tip to tail”
Subtraction is “tail to tip”
V2  30  3  j0
imag
imag
V1  145
V1
V3
real
V2  30
real
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Phasors
Phasors
• To convert from rectangular to polar:
• What is V3 in Phasor form?
V3  3.707  j.707
 V = a + jb
V  V 
V2
b
a  b  tan  
 a
2
1
2
imag
V1
V2
V3
real
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Phasors
Phasors
• What is V3 in Phasor form?
• What is V3 in the time domain?
V3  3.707  j.707
V  V 
imag
 0.707 
3.707 2  0.707 2  tan1 
  3.7710.8
 3.707 
V1
imag
V2
V3
V1
V2
V3
real
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real
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Phasors
Multiplication of Phasors
• What is V3 in the time domain?
• Multiplication and division are easier in polar form
 For multiplication: multiply magnitudes, add angles
 For division: divide magnitudes, subtract angles
v(t)  3.77 2 cos(t  10.8 )
V4  V1V2  (145 )(30 )
V4  345
add angles
V4
multiply magnitudes
imag
V1
V2
imag
V3
real
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real
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Multiplication of Phasors
Phasor Analysis of Inductors
Using Matlab
• For inductors
>> V1=1*exp(j*45*pi/180)
V1 =
0.7071 + 0.7071i
>> V2=3*exp(j*0*pi/180)
V2 =
3
>> V3=V1+V2
V3 =
3.7071 + 0.7071i
>> Vmag=abs(V3)
Vmag =
3.7739
>> Vanglerad=angle(V3)
Vanglerad =
0.1885
>> Vangledeg=Vanglerad*180/pi
Vangledeg =
10.7991
i(t)  imax cos(t  i )
di
dt
v(t)  Limax sin(t  i )  Limax cos(t  i  90o )
v(t)  L
• Transforming into phasor form:
v(t)  Limax cos(t  i  90o )
o
 V  LIe ji e j90
V  jLIe ji  jLI
j 90
 j
using e
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Phasor Analysis
i(t)
v(t)
L
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Phasor Analysis
V  jLI
• We can rewrite Ohm’s Law to include complex
impedances
• V = IZ
• Define XL = L (inductive reactance)
• Therefore
 Z: complex impedance (Ohms)
 V = jXLI
• Z = R + jXL + jXC (if in series)
• 1/Z = 1/R + j/XL + j/XC (if in parallel)
• A similar derivation for capacitors yields
 XC = -1/(C) (capacitive reactance)
 V = jXCI
Dr. Louie
• Z will have a magnitude and phase associated
with it Z  Zz
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Phasor Analysis
Phasor Analysis
• Z can be found by adding the R, XL and XC if in
series
• Parallel:
 1/Z = 1/R + 1/(-jXC) + 1/(jXL)
 Z = R + jXC + jXL
• Example
• Example
 Z = 10 + -j2 + j3
 Z = 10 + j (rectangular coordinates)
R=10
 1/Z = 1/10 + 1/(j3) = 0.1 -0.333j
 Z = 2.87 73.3o
jXC =-j2
Note:
Re{Z} = R
Im{Z} = Xc + XL
jXL =j3
R=10
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Phasor Analysis
Phasor Analysis
• Note:
• Find the current out of the source, the power out
of the source, and power consumed by the
resistor assuming:
 |V| = Vrms (magnitude of the phasor)
 |I| = Irms
 Vs = 120 Volts at 60 Hz
 L = 0.01 Henry
 R = 10 Ohms
• Using phasors
 V = IZ
 P = |V||I|cos()
 P = |I|2R =|V|2/R
I
L
R
Vs
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jXL =j3
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Phasor Analysis
Phasor Analysis
• Vs = 120 Volts (RMS) at 60 Hz
• L = 0.01 Henry
Z  (3.77 j  10) 
 jXL = jL = j(60 x 2p)x.01 = j3.77
 3.77 
3.77 2  102  tan1 

 10 
Z  10.6920.7 
• R = 10 Ohms
Vs  IZ
I
1200
 11.22  20.7 A
10.6920.7
phasor diagram
I
L
Vs
I
R
Vs
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Phasor Analysis
Summary
• Power from the source
• Phasor Domain: allows steady state AC circuits to be easily
analyzed (also graphical interpretation)
P | Vs || I | cos()  (120)(11.22) cos(20.7 )  1.26kW
• Power consumed by the load resistor
• Phasors: magnitude and phase info; freq. is assumed
v
V  max e jv  Vrmse jv
2
• Effective phasors are used in power systems
P | I |2 R  11.222  10  1.26kW
• The inductor does not consume any power P
I
Vs  IZ
L
• Ohm’s Law in phasor domain:
R
Vs
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