chapter 7 sinusoidal steady-state analysis
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Sinusoidal SteadySinusoidal Steady--StateState
AnalysisAnalysis
Chapter 7
Artemio P. MagaboArtemio P. MagaboProfessor of Electrical EngineeringProfessor of Electrical Engineering
Department of Electrical and Electronics EngineeringUniversity of the Philippines - Diliman
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The Sinusoidal Function
T
)tcos(F)t(f m +=where
Fm = amplitude or peak value = angular frequency, rad/sec = phase angle at t=0, rad
The sinusoid is described by the expression
t, rad
-Fm
Fm
2
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The sinusoid is generally plotted in terms oft,expressed either in radians or degrees. Considerthe plot of the sinusoidal function f(t)=Fmcos t.
-Fm
Fm
t, deg180 360
t, rad 2
T
When t=2, t=T. Thus we get T=2 or = .The function may also be written as
T2
tT
2cosFtcosF)t(f mm
==
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The peak value of the voltage is Vm=311 volts. Theangular frequency is =377 rad/sec. The frequencyis f=60 Hz. The period is T=16.67 msec.
Define: The frequency of the sinusoid
T
1f= sec-1 or cycles/sec or Hertz (Hz)
Then, the sinusoid may also be expressed as
ft2cosFtcosF)t(f mm ==
Note:: The nominal voltage in the Philippines is asinusoid described by
V)t377cos(311)t(v +=
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Leading and Lagging Sinusoids
Note: We say either f11(t) leads f2(t) by an angleof 30 or that f2(t) lags f1(t) by an angle of 30.
)30tcos(F)t(f m2 +=
Consider the plot of the sinusoidal functions
and )60tcos(F)t(f m1 +=
30
60
90 180-30
-60
-90t, deg
270 360
)t(f1
)t(f2
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Note: The current is in phase with the voltage.
From Ohms law, we get tcosRIRiv mRR ==
Consider a resistor. Let thecurrent be described by
tcosIi mR =
R
+ -vR
iR
90 180-90t, deg
270 360
vRiR
The Resistor
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The Inductor
Consider an inductor. Let thecurrent be described by
tcosIi mL =
LiL
+ -vL
From vL= , we get tsinLIv mL =dt
diL L
90 180-90t, deg
270 360
iL
vL
Note: The current lags the voltage by 90o.
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The Capacitor
iC
vC+ -
CConsider a capacitor. Let thecurrent be described by
tcosIi mC =
From vC= , we get tsinC
Iv mC
= dtiC1
C
90 180-90t, deg
270 360
iC
Note: The current leads the voltage by 90o.
vC
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Note: We will show later that:
Summary:
In a resistor, iR and vR are in phase.1.
In an inductor, iL lags vL by 90o. In a capacitor,
iC leads vC by 90o (ELI the ICE man).
2.
For an RL network, the current lags thevoltage by an angle between 0 and 90.
1.
For an RC network, the current leads thevoltage by an angle between 0 and 90.
2.
For an RLC network, either 1 or 2 will hold.3.
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Consider a DC (constant) current I and an AC(sinusoidal) current i(t)=Imcos t.
The sinusoidal current i(t) is said to be as effectiveas the constant current I if i(t) dissipates the sameaverage power in the same resistor R.
Since the current I is constant, then PAV,DC = I2R.
Consider R with the DC current I.
The power dissipated by R isRIP 2=
R
I
Effective Value of a Sinusoid
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The average value of any sinusoidal function canbe shown to be equal to zero. Thus
RIP2
m21
AC,AV =
tcosRIRi)t(p 22
m2 ==
R
i(t)
Consider next R with the ACcurrent i(t). The instantaneouspower dissipated by R is
Simplifying, we get
+=
2
t2cos1RI)t(p
2
m
t2cosRIRI2
m212
m21 +=
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Note: The same definition applies to a sinusoidalvoltage v(t)=Vm cos t.
Equating average power, we get
RIRI2
m212 =
orm
m I707.02
II =
Definition: The effective value of a sinusoidalcurrent with an amplitude Im is equal to
2
II mEFF =
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The effective value of a periodic function is also
called the Root-Mean-Square (RMS) value.That is, given a periodic function f(t), we get
==T
0
2RMSEFF dt)t(f
T
1FF
The effective or RMS value of the voltage is
V220)311(707.0V ==
Note:: The nominal voltage in the Philippines is asinusoid described by
V)t377cos(311)t(v +=
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From KVL, we get
tcosVRidt
diL m =+
Network with Sinusoidal Source
Consider the network shown.Let v(t)=Vm cos t where Vmand are constant. Find thesteady-state current i(t).
R
Lv(t)
+
-
i
tcosBtsinAdt
di+=
tsinBtcosAi +=Let
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Substitution gives
tsinLAtcosLB
tsinRBtcosRAtcosVm
+
+=
Comparing coefficients, we get
LARB0
LBRAVm
=
+=
Solving simultaneously, we get
m222V
LR
RA
+= m222 VLR
LB
+
=and
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Thus, the steady-state current is
tsinVLR
LtcosV
LR
Ri m222m222 +
+
+=
Trigonometric Identity:
)tcos(KtsinBtcosA =+where
22 BAK += andA
Btan 1=
+=
R
Ltantcos
LR
Vi 1
222
m
or
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( )= tcosK
= cosKA = sinKBProof: Let and
Substitution gives
)sintsincost(cosKtsinBtcosA +=+
)sin(cosKBA 22222 +=+
From the definitions of A and B, we get
22 BAK +=or
= tanA
B
A
Btan 1=
andor
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8
0.6Hv(t)
+
-
i
Example: Find the currenti(t) and the average powerdissipated by the resistor.
Assume v(t)=100cos 10t V.
+=
R
Ltantcos
LR
Vi 1
222
m
Earlier we got the steady-state current as
( )
( )
+=
8
6.010tant10cos
6.0108
100i 1
222
Substitution gives
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W400RIP2
RMSAV,R ==
The average power dissipated by R is
( ) A87.36t10cos10i o=
Simplifying, we get
A07.72
10IRMS ==
The RMS value of the current is
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Definition: A complex number consists of a realpart and an imaginary part. For example, given
jbaA +=
A is a complex number with real part equal to aand an imaginary part equal to b. Note: j= .1
Example: The following complex numbers areexpressed in the rectangular-coordinate form.
3j5.0C = 25.4j6D +=
4j3A += 5.3j5.2B =
Algebra of Complex Numbers
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The Complex Plane
Definition: The complex plane is a Cartesiancoordinate system where the abscissa is for realnumbers and the ordinate is for imaginary numbers.
ImaginaryAxis
Real Axis2 4 6 8-2-4-6-8
j4
j2
-j2
-j4
A=3+j4
D=0+j2
C=4+j0
B=2.5-j3.5F=-3-j3
E=-4+j3
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Definition: In the polar-coordinate form, themagnitude and angle of the complex number isspecified.
Polar-Coordinate Form
Consider the complex number A=a+jb.
Imag
+ Reala
jb A
A
From the figure, we get
22 baA +=
abtan 1=
Thus,
=+= AjbaA
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Trigonometric Form
Consider the complex number .=+= AjbaA
Imag
+ Reala
jb A
A
From the figure, we get
= cosAa
= sinAb
Thus, we can also write
)sinj(cosAA +=
For example, A=1036.87 can be expressed as
6j8)87.36sinj87.36(cos10A +=+=
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Given A = a+jb and B = c+jd, then
( ) ( )dbjcaBA +=
Addition or Subtraction
For example, given A=8+j6 and B=4+j10
16j12)106(j)48(BA +=+++=+
4j4)106(j)48(BA =+=
Addition or subtraction of complex numbers canonly be done in the rectangular-coordinate form.
( ) ( )dbjcaBA +++=+
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Given A = a+jb =||||A||||A and B = c+jd =||||B||||B, thenin the rectangular-coordinate form, we get
)jdc(jb)jdc(a +++=
Multiplication of complex numbers can be doneusing the rectangular-coordinate or polar form.
)jdc)(jba(AB ++=
)bcad(j)bdac(AB ++=
bdjjbcjadac 2+++=
Since j2=-1, the product is
Multiplication
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Given A = a+jb =||||A||||A and B = c+jd =||||B||||B, then
)B)(A(AB BA =
)(BAAB BA +=
The rule is multiply magnitude and add angles.We get
For example, given A=3+j4=553.13o andB=4+j3=536.87o
o2 902525j12j16j9j12
)3j4)(4j3(AB
==+++=
++=
orooo 9025)87.3613.53()5(5AB =+=
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o13.5354j3*A ==
For example, given A=3+j4=553.13o andB=-4-j3=5-143.13o
o13.14353j4*B =+=
Definition: The conjugateof a complex numberA=a+jb=||||A||||A is definedas
AAjba*A ==a
Imag
Real
jb A
-jb A*
Conjugate of a Complex Number
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Division
Division of complex numbers can be done using therectangular-coordinate or polar form.
Given A = a+jb =||||A||||A and B = c+jd =||||B||||B, thenin the rectangular-coordinate form, we get
jdc
jba
B
A
++
=jdc
jdc
22 dc bdjbcjadac + ++=or
2222 dc
adbcj
dc
bdac
B
A
+
+++
=
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Given A = a+jb =||||A||||A and B = c+jd =||||B||||B, then
B
A
B
A
B
A
=
)(B
A
B
ABA =
The rule is divide magnitude and subtract angles.We get
For example, given A=3+j4=553.13o andB=4-j3=5-36.87o
1j90187.365
13.535
B
A oo
o
==
=
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Phasor Transformation
Define a transformation from the time domain tothe complex frequency domain such that
)tcos(F)t(f m +=
=2
F)j(F m
For example, given f1(t)=311 cos (377t+60o) volts
and F2(j)=1020o Amps
V60220)j(F o1 =
A)20tcos(14.14)t(f o2 +=
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From Ohms law, we get tcosRIRiv mRR ==
Consider a resistor. Let thecurrent be described by
tcosIi mR =
R
+ -vR
iR
omR 0
2
I)j(I = omR 0
2
RI)j(V =
Transformation gives
and
=
R)j(I
)j(V
R
RDividing, we get
The Resistor
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Consider an inductor. Let thecurrent be described by
tcosIi mL =
LiL
+ -vL
From vL= , we get tsinLIv mL =dt
diL L
Transformation gives
omL 0
2
I)j(I = omL 90
2
LI)j(V
=and
==
Lj90L
)j(I
)j(V o
L
LDividing, we get
The Inductor
=LImcos(t+90o)
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iC
vC+ -
CConsider a capacitor. Let thecurrent be described by
tcosIi mC =
From vC= , we get tsinC
Iv mC
= dtiC1
C
=
=
Cj
1
90C
1
)j(I
)j(Vo
C
CDividing, we get
Transformation gives
and
om
C 02
I
)j(I =om
C 90C2
I
)j(V =
The Capacitor
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Note:
(2) For an inductor, ZL = jL = jXL in
Impedance
)j(I
)j(VZ
=
Definition: The ratio of transformed voltage totransformed current is defined as impedance.
(1) For a resistor, Z(1) For a resistor, ZRR = R in
(3) For a capacitor, ZC = 1/jC = -jXC in
(4) XL and XC are the reactance of L and C,respectively.
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Admittance
(3) For a capacitor, YC = jC = jBC in -1
)j(V
)j(I
Z
1Y
==
Definition: The ratio of transformed current totransformed voltage is defined as admittance.
(1) For a resistor, Y(1) For a resistor, YRR = 1/R in -1
(2) For an inductor, YL = 1/jL =-jBL in -1
Note:
(4) BL and BC are the susceptance of L andC, respectively.
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2. All the methods of analysis developed forresistive networks (e.g. Mesh Analysis, NodalAnalysis, Superposition, Thevenins and NortonsTheorems) apply to the transformed network.
Summary
1. The equation describing any impedance isalgebraic; i.e. no integrals, no derivatives.
)j(IZ)j(V = (Ohms Law)
3. The phasor transformation was defined for acosine function. The magnitude is based on theRMS value. Other phasor transformations exist.
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Solution 1: Use network reduction to get the inputimpedance.
Transformed network
+= 12j6Z1 )j(I1
1Z
V(j)+
-2Z CZ
)j(I2
)j(I3 += 5j5Z2= 10jZC
10j5j5
)5j5(10j
ZZ
ZZZ
C2
C2eq +
+=
+=
=
= 105j5
50j50
+=+= 12j16ZZZ eq1in
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Apply current division to get I2(j).
)87.36071.7(5j5
10j)j(I
ZZ
Z)j(I o1
C2
C2
=
+=
=
+==
87.3620042.141
12j16042.141
Z)j(V)j(I
oo
in
1
Solve for current I1(j).
A87.36071.7 =
45071.7
)87.36071.7)(9010(o
oo
=
A87.810.10 o=
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)j(I)j(I)j(I 213 =
Use KCL to get I3(j).
oo 87.810.1087.3607.7 =
)9.9j41.1()24.4j66.5( =
A13.5307.766.5j24.4 o=+=
Inverse transform I1(j), I2(j), and I3(j).
A)36.87-(10tcos10)t(i o1 =A)81.87-(10tcos14.14)t(i o2 =
A)53.13(10tcos10)t(i o3 +=
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Solution 2: Use meshanalysis.
)j(I1
1Z
V(j)+
-2Z CZ)j(I3
+= 12j6Z1+= 5j5Z2= 10jZC
)]j(I)j(I[Z)j(IZ)j(V 31211 +=mesh 1:
mesh 2: )](jI-)(j[IZ)j(IZ0 1323C +=
Substitution gives
)]j(I)j(I)[5j5()j(I)12j6(2.141 311 +++=
)]j(I)j(I)[5j5()j(I10j0 133 ++=
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)j(I)5j5()j(I)17j11(2.141 31 ++=
Simplifying the equations, we get
(1)
)j(I)5j5()j(I)5j5(0 31 ++= (2)
)j(I1j)j(I901 11o ==
From (2), we get
)j(I45071.7
45071.7)j(I
5j5
5j5)j(I 1o
o
13
=+
=
Substitute in (1)
)j(jI)5j5()j(I)17j11(2.141 11 ++=
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Solve for I1(j). We get
)j(I)12j16(2.141 1 +=
o1 87.3620
2.141
12j16
2.141)j(I
=
+=
or
A87.36071.7 o=
A1.53071.7 o=
Solve for I3(j). We get
)(jI)901()j(jI)j(I 1o
13 ==
)87.36071.7)(901( oo =
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Finally, I2(j) can be found using KCL.
)j(I)j(I)j(I 312 =
A87.810.1090.9j41.1 o==
oo 53.137.071-87.36071.7 =
)66.5j24.4()24.4j66.5( +=
Inverse transform I1(j), I2(j), and I3(j).
A)36.87-(10tcos10)t(i o1 =
A)81.87-(10tcos14.14)t(i o2 =
A)53.13(10tcos10)t(i o3 +=
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Xi10
5 H5.0
vs
+
-
.01F is
Example: Given
vs=100cos10t voltsis=10cos(10t+30
o)amps. Find iX.
Transform the network
V071.70)j(V oS =
A30071.7)j(I oS =
==
= 10j)01.0)(10(j
1
Cj
1ZC
=== 5j)5.0)(10(jLZL
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Transformed network 1Z
Vs(j)+
-2Z 3Z Is(j)
Ix(j)Z1=5+j5
Z2=10
Z3=-j10
)j(VX +
Solution 1: Nodal AnalysisREF
3
X
2
X
1
SXS
Z
)j(V
Z
)j(V
Z
)j(V)j(V)j(I
+
+
=
Substitution gives
5j5
71.70)j(V
10j
1
10
1
5j5
130071.7 X
o
+
++
+=
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Evaluate the coefficient of VX(j)
2.0
1.0j1.050
5j51.0j1.0
5j5
5j5
5j5
1
=
++
=++
+
o
o
oo
451045071.7
071.70
5j5
071.70=
=+
Evaluate the constant term
Substitution giveso
Xo 4510)j(V2.030071.7 =
]451030071.7[)j(V oo2.0
1X +=
or
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Solve for Ix(j).
A1583.610
)j(V)j(I oXX =
=
V1533.687.17j0.66 o==
Simplifying, we get
]07.7j07.754.3j12.6[5)(jVx ++=
Thus, using inverse transformation, we get
A)15-(10tcos66.9)t(i oX =
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Solution 2: Superposition
=++
= 5j5
1j1
1j1
1j1
10j
Get the input impedance.
10j10
)10j(10
ZZ
ZZZ
32
32eq
=
+=
=+= 105j5ZZ 1in
Thus,
Consider the voltagesource alone.
1Z
Vs(j)
+
- 2Z 3Z
Ix1(j)Is1(j)
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A0071.710
071.70Z
)j(V)j(I oo
in
s1s ===
The source current is
A54.3j54.3455 o ==
Using current division, we get
)j(IZZ
Z)j(I 1s
32
31X +
=
)0071.7(4514.14
9010 oo
o
=
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)j(V10j
1
10
1
5j5
1)j(I XS
++
+=
From KCL, we get
)(VX +
REF
Consider the current
source alone.
Is(j)
1Z
2Z 3Z
Ix2(j)
Substitution gives
)j(V2.030071.7 Xo =
V3036.35)j(V oX =or
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A77.1j06.33054.3 o +==
Applying, superposition, we get
)j(I)j(I)j(I 2X1Xx +=
A1583.677.1j6.6 o==
Solving for the current, we get
10
3036.35
Z
)j(V)j(I
o
2
X2x
=
=
A77.1j06.354.3j54.3 ++=
A)15-(10tcos66.9)t(i oX =Thus,
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1Z
Vs(j)+
-3Z Is(j)
I1(j)Vth(j)
+
-
Solution 3:Thevenins Theorem
For mesh 1,we get
)](jI)(j[IZ)j(IZ)j(V s1311s ++=
)](jI)(j[I10j)j(I)5j5()j(Vs11s
++=
Substitution gives
5j5
)j(I10j)j(V)j(I ss1
+=
Solve for I1(j). We get
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5j5
24.61j36.35
+
=
Simplifying, we get
5j5
)3007.7(10j071.70)j(I
oo
1
+=
o
o
o
1051045071.7
6071.70=
=
The Thevenin voltage is
)]j(I)j(I[10j)j(V s1th +=
]3007.710510[10j oo +=
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V156.13636.35j94.131 o==
=
+=
+= 10
5j5
)10j)(5j5(
ZZ
ZZZ
31
31th
Simplifying, we get
)54.3j12.666.9j59.2(10j)j(Vth +++=
Find the Theveninimpedance 1
Z
3Z
a
babth ZZ =
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A1583.620
156.136 oo
=
=
V156.136)j(V oth =
= 10Zth
The Thevenin equivalent networkZ
th
Vth(j)+
-10)j(IX
Finally, we put back the 10resistor and solve for thecurrent.
10Z
)j(V)j(I
th
thX +
=
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Power Equations
Consider a voltage source, a current source or anetwork of passive elements (R, L and/or C). Let
i(t)=Im cos (t+ I) and v(t)= Vm cos (t+V).
VoltageSource
i(t)v(t)
+
-
CurrentSource
i(t) v(t)
+
-
PassiveNetwork
i(t) v(t)
+
-
Note: The current flows from positive to negativeterminal for the passive network.
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)tcos()tcos(IVv(t)i(t)p IVmm ++==
The instantaneous power supplied by the voltageor current source or delivered to the passivenetwork is
Trigonometric Identities:
(1) =+ sinsincoscos)cos(
(3) +=+ sincoscossin)sin(
(4) )2cos1(cos212 +=
(2) += sinsincoscos)cos(
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From (1) and (2), we get
)]cos()[cos(coscos21 ++=
The instantaneous power can be expressed as
)]cos()t2[cos(IVp IVIVmm21 +++=
)]()2t2cos[(IV IVImm21 ++=
)cos(IV IVmm21 +
Simplify using trigonometric identity (1). We get
)cos()t2[cos(IVp IVImm21 ++++====
)]cos()sin()t2sin( IVIVI ++++++++
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Collecting common terms, we get
)]t(2cos1)[cos(IVp IIVmm21 ++=
)t(2sin)sin(IV IIVmm21 +
Using the RMS values of the voltage and current,we get
)]t(2cos1)[cos(VIp IIV ++=
)t(2sin)sin(VI IIV +
Note: The instantaneous power consists of aconstant term plus two sinusoidal components.
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From Ohms law, we get
)tcos(RIRiv ImRR +==
Consider a resistor. Let thecurrent be described by
)tcos(Ii ImR +=
R
+ -v
R
iR
The instantaneous power delivered to (dissipatedby) the resistor is
)]t(2cos1[RI)t(cosRIp I2m2
1I
22mR ++=+=
The Resistor
or )]t(2cos1[RIp I2
R ++=
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The Inductor
Consider an inductor. Let thecurrent be described by
)tcos(Ii ImL +=
LiL
+ -vL
From vL= , we get )tsin(LIv ImL +=dt
diL L
The instantaneous power delivered to the inductoris
)tcos()tsin(LIp II2mL ++=
)t(2sinLI I2m2
1 +=or )t(2sinXIp IL
2L +=
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The Capacitor
iC
vC+ -
CConsider a capacitor. Let thecurrent be described by
)tcos(Ii ImC +=
From vC= , we get )tsin(C
Iv I
mC += dtiC
1C
The instantaneous power delivered to the capacitoris
)tcos()tsin(C
Ip II
2m
C ++=
or )t(2sinXIp IC2
C +=
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Real or Active Power
Definition: Real or active power is defined as theaverage value of the instantaneous power. It is thepower that is converted to useful work or heat.
Recall the instantaneous power supplied by asource or delivered to a passive network.
)]t(2cos1)[cos(VIp IIV ++=
)t(2sin)sin(VIIIV
+
)cos(VIP IV = in Watts
Since the average of any sinusoid is zero, the realor active power is
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Recall the instantaneous power delivered to aresistor, inductor or capacitor.
)]t(2cos1[RIp I2
R ++=
)t(2sinXIp IL2
L +=
)t(2sinXIp IC2
C +=
Since the average of any sinusoid is zero, the realor active power delivered to R, L and C are
RIP 2R = in Watts
0PL =
0PC =
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Consider a series LC circuit. Letthe current be described by
i=Im cos (t+I). The voltagesvL and vC can be shown to be
vC
+ -
CLi
+ -vL
ivL
vC
Reactive Power
)tsin(C
Iv I
mC +=
)tsin(LIv ImL +=
90 180-90t, deg
270 360
I
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The energy stored in the magnetic and electricfields are
)t(cosLiW I22
L21
L +=
Plots of the energy are shown below.
i
WLWC
90 180-90t, deg
270 360
I
)t(sinCvW I22
C21
C +=
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Complex Power
)(VI)I)(V(*IVS IVIV ===vv
)sin(jVI)cos(VI IVIV +=
Note: The complex power S is a complex numberwhose real and imaginary components are the realand reactive power, respectively.
Definition: The product of the phasor voltage andthe conjugate of the phasor current is defined asthe complex power S. Let and .III =
v
VVV =v
jQPS +=or
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Recall the transformation from the time domainto the complex frequency domain defined by
)tcos(F)t(f m +=
= F)j(F
Note: Unlike complex numbers, a phasor quantityis a complex representation of a sinusoidal function.
The quantity F(j) is referred to as a phasor. Forsimplicity, we make a change in notation:
== F)j(FFv
Phasor Notation
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Transform the network.
volts0220V o=v
== 154.1j)00306.0(377jZ 1L== 846.4j)012854.0(377jZ 2L
Example: In the circuit shown, v(t) = 311 cos377tvolts. Find the power and reactive power deliveredto the load.
v(t)
+
-
i(t)
0.5 3.06 mH
7.5
12.854 mHLoad
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+
-
0.5 j1.154
7.5
j4.846Vv I
v
+
-
xVv
TransformedNetwork
Get the totalimpedance.
846.4j5.7154.1j5.0Zeq +++=
=+= 87.360.106j8 o
Find the current
A87.362287.3610
0220
Z
VI o
o
o
eq
=
==
v
v
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Find the voltage across the load
)846.4j5.7(IVX +=vv
oo
32.87(8.929)87.3622( = volts0.445.196 o=
Find the complex power delivered to the load.
)36.87(22)445.196(IVjQP oo*XLL ==+vv
o87.328.4321 =
346,2j630,3 +=
Thus, PL=3,630 Watts and QL=2,346 Vars.
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Example: Givenv(t)=100 cos 10t volts,find all Ps and Qs. v(t)
+
-
i(t)
8
0.6H
Transform the network
V071.70V o=v
=+= o87.36106j8Z
+
-
8
j6Vv I
v
A87.36071.787.3610
071.70
Z
VI o
o
o
=
==
v
v
Solve for the current
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Power and Reactive Power delivered to R and L
watts400)8(071.7RIP 22R ===
arsv300)6(071.7XIQ 2L2
L ===
Power and reactive power supplied by the source
oIVs 87.36cos)071.7(71.70)cos(VIP ==
watts400=
Vars30087.36sin)071.7(71.70Q os ==
300j40087.36500 o +==
)87.36071.7)(071.70(*IVS oo ==vvor
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Example: Given
v=200 cos 10tVolts. Find allreal power andreactive power.
i1+
-
R1=6 L1=1.2H
R2=5
L2=0.5H0.01F
i2 i3
v
The transformed network
V042.141V o=v
= 12jZ1L
= 5jZ2L
= 10jZC
Vv 1I
v
+
-
6 j12
5
j5-j10
2Iv
3Iv
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In a previous example, we found
A87.36071.7I o1 =v
A87.8110I o2 =v
A13.53071.7I o3 =v
watts500)5(10RIP 22222R ===
Average power dissipated by the resistors
watts300)6(071.7RIP 212
11R ===
Reactive Power delivered to the capacitor
vars500)10(071.7XIQ 2C2
3C ===
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Reactive Power delivered to the inductors
vars600)12(071.7XIQ 21L2
11L ===
arsv500)5(10XIQ 22L2
22L ===
Power and reactive power supplied by the source
ooIV 87.36)87.36(0 ===
o1S 87.36cos)071.7(42.141cosVIP ==
watts800=
vars600sinVIQ 1S ==
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We can also use the complex power formula
*
1SS IVjQPvv
=+
600j80087.361000 o +==)87.3607.7)(042.141( oo =
Thus, PS=800 watts and QS=600 vars.
Note: Real and reactive power must always bebalanced. That is,
watts800PPP 2R1RS =+=
vars600QQQQ C2L1LS =++=
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From a previous example, we found
V153.68V oX =v
A1583.6I oX =v
and
Example: Given
V071.70V oS =v
A30071.7I oS =v
Find all P and Q.
+= 5j5Z1= 10Z2= 10jZ3
XVv
+
SIv
SVv
REF
1Iv
XIv
CIv
1Z
+
-
2Z 3Z
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We can also find and .1Iv
CIv
( )
11
XS1
Z
68.17j97.6571.70
Z
VVI
=
=
vv
v
o
o
45071.7
753.18
5j5
68.17j737.4
=++
=
A3059.2 o=
o
o
C
XC
9010
153.68
Z
VI
==
v
v
A7583.6 o=
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Power and reactive power delivered to Z1
watts5.33)5(59.2RIP 21Z2
11Z ===
vars5.33XIQ 1Z2
11Z ==
watts5.466)10(83.6RIP 222
X2R ===
Power dissipated by the resistor R2
vars5.466XIQ C2
CC ==
Reactive power delivered to the capacitor
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SVv
Power and reactive power supplied by
)3059.2)(71.70(IVjQP o*
1SVV ==+vv
watts5.158PV = vars5.91QV =
5.91j5.15830183 o ==
)30071.7)(153.68(IVjQP oo*
SXII ==+vv
Power and reactive power supplied by SIv
o45-95.824 = 5.341j5.341 =
watts5.341PI = vars5.341QI =
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vars100,3)pf(costanPQ 11
11 ==
V0220V o1 =v
Example: Findthe power andpower factorof generator 2.Assume
Gen. 1
1Vv
1Iv
LIv
+
-2V
v
+
-
0.3+j0.4 0.2+j0.2
LVv
+
-
2Iv
Gen. 2Load
P1=5 kW PL=10 kW
pf1=0.85 lag pfL=0.8 lag
For generator 1,watts000,5P1 =
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From the complex power formula, we get
1
11*
1V
jQPI vv +
=
08.14j73.22220
100,3j000,5+=
+=
08.14j73.22I1 =v
A79.3174.26 o=
Thus,
From KVL, we get the voltage at the load
11L I)4.0j3.0(VVvvv
+=
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Substitution gives
)31.79-)(26.7413.535.0(VV oo1L =vv
o1 34.2137.13V =v
)86.4j45.12(220 +=
86.4j55.207 =
V34.16.207V oL =v
vars7,5000.8)tan(cos000,10Q -1L ==
At the load,
watts000,10PL =
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o
o
34.16.207
87.36500,12
=
A24.37j31.47 =
o34.16.207
500,7j000,10
+
=
L
LL*
LVjQPI v
v
+=
From the complex power formula, we get
A21.3821.60I oL =v
Thus,
Department of Electrical and Electronics Engineering
1L2 IIIvvv
=
From KCL, we get the current supplied by Gen 2.
)08.14j73.22()24.37j31.47( =
16.23j58.24 =
A3.4377.33I o2 =v
Thus,
From KVL, we get the voltage of Generator 2
L22 VI)2.0j2.0(Vvvv
++=
LV)16.23j58.24)(2.0j2.0(v
++=
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Simplifying, we get
86.4j55.20728.0j55.9V2 ++=v
V21.115.21758.4j1.217 o==
Applying the complex power formula,
*
2222 IVjQPvv
=+
)3.4377.33)(21.115.217( oo =
915,4j443,51.42334,7o
+==
lag74.0)P
Qcos(tanpf
2
212 ==
The power is 5,443 watts while the power factor is
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Example: A small industrial shop has the followingconnected load:
Load L1: Induction motor 2 kW, 0.85 pf lagLoad L2: Electric Heater 3 kW, 1.0 pfLoad L3: Lighting Load 500 W, 0.9 pf lagLoad L4: Outlets 1 kW, 0.95 pf lag
The voltage across the load is 220 V RMS. Find thecurrent through each load and the total currentsupplied to the shop.
+
220VtI
v
1Iv
2Iv
3Iv
4Iv
L4L3L2L1
-
Department of Electrical and Electronics Engineering
For load L1, P1=2,000 watts, pf1=0.85 lag
vars1,2400.85)(costan000,2Q -11 ==
A79.317.1063.5j09.9 o==
For load L2, P2=3,000 watts. Since pf2=1, thenQ2=0. Thus
A064.13220
0j000,3I o2 =
=
v
220
240,1j000,2
V
jQPI
*
1
111
=
= v
v
From the complex power formula, we get
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For load L3, P3=500 watts, pf3=0.90 lag
vars2420.9)(costan500Q -13 ==
A84.2552.21.1j27.2 o==220
242j500I3 =v
For load L4, P4=1,000 watts, pf4=0.95 lag
vars3290.95)tan(cos000,1Q -14 ==
220
329j1000I4
=
v
A19.1878.449.1j54.4 o==
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From KCL, the total current is
4321t IIIIIvvvvv
+++=
A55.1566.3022.8j54.29 o==
)0j64.13()63.5j09.9( +=
)49.1j54.4()1.1j27.2( ++
or
watts500,6PPPPP 4321t =+++=
vars811,1QQQQQ 4321t =+++=
A22.8j54.29220
811,1j500,6It =
=
v
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Phasor Diagrams
v(t)
+
-
i(t)
8
0.6H
vR +
-
+ -
vL
Consider the circuit shown.Let v=100 cos(10t+) V.
Phasor diagrams show graphically how KVL andKCL equations are satisfied in a given circuit.
The transformed network
V71.70V =v
= 8ZR= 6jZL
+
-
+
-
+ -
Vv
RVv
LVv
ZL
ZR
Iv
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)8)](87.36(071.7[ZIV oRR ==vv
V)87.36(57.56 o=
V)13.53(43.42 o+=
)906)](87.36(071.7[ZIV ooLL ==vv
oT 87.3610
71.70
6j8
71.70
Z
VI
=+
==
v
v
Solve for the phasor current and voltages. We get
A)87.36(07.7 o=
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Assume =0o. We get
V071.70V o=v
V36.87-56.57V oR =v
V53.1342.43V oL =v
A87.3607.7I o=v
The phasor diagramis shown.
Note:LR
VVVvvv
+=
Iv
is in phase with RVv
LVv
Iv
lags by 90o
RVv
V
v
LVv
Iv 36.87o
53.13o
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RVv
Vv
LVv
Iv
36
.87o
53
.13o
Assume =60o. We get
V6071.70V o=v
V3.13256.57V oR =v
V3.131142.43V oL =v
A13.2307.7I o=v
The phasor diagramis shown.
Note: The entire phasordiagram was rotated byan angle of 60o.
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Assume =120o. We get
V12071.70V o=v
V3.13856.57V oR =v
V3.131742.43V oL =v
A13.8307.7I o=v
The phasor diagram is shown.
Note: The entire phasordiagram was rotated byanother 60o.
RVv
Vv
LVv
Iv3
6.8
7o
53.1
3o
Note: The magnitude and phase displacementbetween the phasors is unchanged. The phasorsare rotating in the counterclockwise direction.
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Power and reactive power supplied by the source
SSSS cosIVP =
W40087.36cos)071.7(71.70 o ==
vars30087.36sin)071.7(71.70Q oS ==
Power dissipated by the resistor
watts400)8()071.7(RIP 22R ===
vars300)6()071.7(XIQ 22L ===
Reactive power delivered to the inductor
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The total impedance seen by the source
C21T Z//ZZZ +=
10j5j5
)5j5(10j12j6
++
++=
=+= 87.362012j16 o
Example:
V042.141V o=v
+= 5j5Z2
+= 12j6Z1 1Vv
2Vv
Vv
1Iv
+
-
6 j12
5
j5-j10
2Iv
3I
v
+ -
+
-
Note: Refer to a previous problem.
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Source current
A87.3607.787.3620
042.141
Z
VI o
o
o
T
1 =
==
v
v
Apply current division to get
A87.810.10I o2 =v
A13.5307.7I o3 =v
)12j6)(87.3607.7(ZIV o111 +==vv
V56.2687.94 o=
Solve for the voltage1
Vv
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)5j5)(87.8110(ZIV o222 +==vv
Solve for the voltage2V
v
V87.3671.70 o=
Phasor Diagram
Vv
1Vv
2Vv
21 VVVvvv
+=
Note:
1Iv
2Iv
3Iv
321 IIIvvv
+=
Department of Electrical and Electronics Engineering
)87.36071.7)(042.141(oo
=600j80087.361000 o +==
Power supplied by the source
*
1SS IVjQPvv
=+
or
watts80087.36cos)071.7(42.141P oS ==
vars60087.36sin)071.7(42.141Q oS ==
vars500)10()07.7(XIQ 2C2
CC ===
Power delivered to ZC
Department of Electrical and Electronics Engineering
We can also use the complex-power formula
*
1111 IVjQPvv
=+
)87.36071.7)(56.2687.94( oo =
600j30043.6382.670 o +==
Power delivered to Z1
watts300)6()071.7(RIP 212
11 ===
vars600)12()071.7(XIQ 212
11 ===
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Power delivered to Z2
watts500)5()10(RIP 222
22 ===
vars500)5()10(XIQ 222
22 ===
or
*
2222 IVjQPvv
=+
)87.8110)(87.3671.70( oo =
500j500451.707 o +==
Note: It can be shown that power balance issatisfied.
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Let volts, the reference phasoro
R 0120V =v
A0620
0120
20
VI o
oR
S =
==
v
v
Then
Example: A voltmeter reads these voltages for the
network shown below:RMSV220VS =
RMSV120VR =
RMSV150VC =CV
v
SIv
+
-
+
-SV
v
RVv
+ -
20
R C
a) Find R and C.
b) Find P and Q supplied by the source.
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Note: (1) Since the network is capacitive, iS mustlead vS.
CRS VVVvvv
+=(2) From KVL,
Phasor DiagramRV
v
SIv
SVv
CV
v
Apply the cosine law
+= cos(220)(120)2-120220150 222
o25.40=763.0cos = orWe get
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o12.31=86.0cos =We get or
Check: 14.142j92.47120VV CR +=+vv
14.142j92.167 =
So V25.40220
v
==
Apply the cosine law again
volts0120V oR =v
volts25.40220V oS =v
volts37.71150V oC =v
Thus
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038.0j013.0 += CjR
1+=
o
o
o
C
S 37.7104.0
37.71150
06
V
I=
=v
v
The admittance for the parallel RC branch
o25.401320 =
853j1008 =
We get R=78.26 and C=0.04-1.
*
SSSS IVjQPvv
=+ )06)(25.40220( oo =
Power supplied by the source
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Balanced Three Phase Voltages
Three sinusoidal voltages whose amplitudes areequal and whose phase angles are displaced by120o are three-phase balanced.
)tcos(V(t)v ma +=
The RMS value of the voltages is
mm V71.02
VV =
120)tcos(V(t)v mc ++=
120)tcos(V(t)v mb +=
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Balanced Three Phase Voltages
Note: The synchronousgenerator is a three phasemachine that is designed togenerate balanced three-phase voltages.
Transforming to phasors, we get
Va = V Vb = V -120Vc = V +120
aVr
bVr
cVr
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Balanced Three-Phase Currents
are three-phase balanced.
The currents )t(cosIi ma +=
)120tcos(Ii mb +=
)120tcos(Ii mc ++=
In phasor form, we get
= IIar
ob 120II =
r
oc 120II +=
r
aIr
bIr
cIr
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Balanced Three-Phase SystemA balanced three-phase system consists of :
1. Balanced three-phase sinusoidal sources;
3. The connecting wires have equal impedances.
2. Balanced three-phase loads; and
a) Equal impedances per phase or
A balanced three-phase load has:
Note: The load may be connected in wye or delta.
b) Equal P and Q per phase
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ExampleExample: Given Va=200/0
o volts, Vb=200/-120o
volts and Vc=200/120o volts. Find the phasor
currents Ia, Ib and Ic. Also, find P and Q supplied toZa, Zb and Zc.
n n'
aVr
bVr
cV
r
cIr
aIr
bIr
Za
Za=Zb=Zc=7+j5
Zb Zc
2Ir
1Ir
ZF
ZF
ZF=1+j1
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Example
Mesh equations using loop currents I1 and I2.Va Vb = 2(Zf+ ZL)I1 (Zf+ ZL)I2Vb Vc = -(Zf+ ZL)I1 + 2(Zf+ ZL)I2
Substitution gives300 + j173.2 = (16 + j12)I1 (8 + j6)I2
- j346.4 = -(8 + j6)I1 + (16 + j12)I2
Solving simultaneously we getI1 = 20-36.87AI2 = 20-96.87 A
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Example
Solving for currents Ia, Ib and Ic, we getIa = I1 = 20-36.87AIb = I2 I1 = 20-156.87AIc = -I2 = 2083.13A
Note: Currents Ia, Ib and Ic are balanced.
Power and Reactive Power supplied to load impedancesZa, Zb and Zc.
Pa = Pb = Pc = (20)2(7) = 2800 Watts
Qa = Qb = Qc = (20)2(5) = 2000 Var
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Comments1. The sum of 3 balanced phasors is zero.2. If a neutral wire is connected between n and n, no
currents will flow through the neutral wire.
3. The nodes n and n are at the same potential.4. We can analyze the circuit using single-phase
analysis.
n n'neutral line
aVr
Za=7+j5
ZF=1+j1
aIr
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CommentsUsing KVL, we get
Va = (Zf+ Za)Ia + Vnn
Since Vnn = 0, we get forphase a
Ia =Zf + Za
Va= 20-36.87A
8 + j6
2000=
Pa = (20)2(7) = 2800 Watts
Qa = (20)2(5) = 2000 Watts
Forphases b and c, we getIb = Ia-120= 20-156.87 AIc = Ia120 = 2083.13 A
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Line-to-line and Phase Voltages
Consider a 3-phase wye-connected generator or a3-phase wye-connected load.
Van, Vbn and Vcn are line-to-neutral (or phase) voltagesVab, Vbc and Vca are line-to-line (or line) voltages
nnb
a
c
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Line-to-line and Phase Voltages
From KVL, Vab = Van + Vnb = Van VbnVbc = Vbn + Vnc = Vbn VcnVca = Vcn + Vna = Vcn Van
If Van, Vbn and Vcnare balanced3-phase voltages,we get the phasordiagram shown.
Vca
VcnVab
Van
Vbn
Vbc
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Comments1. The line-to-line voltages Vab, Vbc and Vca are also
balanced 3-phase voltages;
2. The magnitude of the line-to-line voltage issquare root of three times the magnitude of theline-to-neutral voltage; and
3. Vab leads Van by 30, Vbc leads Vbn by 30and Vcaleads Vcn by 30.
oanab 30V3V =
rr
obnbc 30V3V =
rr
o
cnca 30V3V =rr
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-Y Conversion for GeneratorsGiven a balanced 3-phase delta connectedgenerator, what is its equivalent wye ?
a a
Note: The line-to-line voltages must be the same.
c
b
c
b
Vca Vab
Vbc
Vcn
Van
Vbn
If Vab = VL, then Van = (1/3) VL-30
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-Y Conversion for Loads
Note: For equivalence, Zab, Zbc and Zca must bethe same for both networks.
If the impedance of the delta load is specified,convert the impedance to wye.
Y Z3
1Z =
a
c b
YZ
YZYZ
c
a
b
Z Z
Z
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Power EquationsFor a single-phase systemPp = VpIp cos Qp = VpIp sin
VAp = VpIp
For a three-phase systemP3 = 3Pp = 3VpIp cos = 3VLIp cos Q3 = 3Qp = 3VpIp sin = 3VLIp sin
VA3 = 3VAp = 3VpIp = 3VLIpwhere
Vp = magnitude of voltage per phaseIp = magnitude of current per phase = angle of Vp minus angle of Ip
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ExampleA balanced 3-phase load draws a total power of 75KW at 0.85 pf lag from a 440-volt line-to-linesupply. Find the current drawn by the load?
A115.8IP =
cosIV3P PL3 =
cos V3
PI
L
3
P =
From
we get
)(440)(0.853
75,000=
or Department of Electrical and Electronics Engineering
Example
For load 1, P1=75 kw
For load 2, P2=60 kw
kvars48.46)85.0tan(cosPQ 111 ==
kvars06.29)9.0tan(cosPQ 122 == Get P and Q drawn by combined load
kw135PPP 21T =+=kvars54.75QQQ 21T =+=
Another 3-phase load rated 60 KW at 0.9 pf lag isconnected in parallel with the load in the previousexample. Find the total current drawn.
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Example
A203(440)3
154,700IP ==
The total volt-ampere is
kva7.154QPVA2
T
2
TT =+=
Since
PL3 IV3VA =
we get
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Single-Phase Analysis of a BalancedThree-Phase System
A balanced three-phase system can bereplaced by a single-phase equivalent circuitprovided:
1. All generators are connected in wye; and2. All loads are connected in wye.
Note: To get the single-phase equivalent,draw the neutral line and isolate one phase.
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ExampleFind the phasor current Is, I1 and I2 and the total power andreactive power supplied by the three-phase voltage source.
Source Load 1 Load 2
Van = 120 /0oV PT = 4.5 kW Z = 8 + j6
at 0.92 pf lag
c
b
a
IsI1 Z
ZZ
I2
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ExampleSingle-Phase Equivalent Circuit
Van = 1200 V P1=1,500 W Z=8+j60.92 pf lag
For load 1, P1 = 1,500 wattsQ1 = P1tan(cos
-1 0.92) = 639 vars
A1.236.13 o=
Is I1 I2Van
I1 =1500 j369
120= 12.5 j5.3 A= 13.6-23.1 A
Z
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ExampleFor load 2,
For the source,Is = I1 + I2 = 22.1 j12.5 A = 25.4 -29.5Ps + jQs = VanIs* = 120 (25.4 -29.5)
= 2652 + j1503For the three phase source
arsv509,4)503,1(3Q3 ==
watts956,7)652,2(3P3 ==
I2 =VanZ
120 0120 0=
= 12 -36.87 = 9.6 j7.2 A