SYMMETRICAL COMPONENTS

SYMMETRICAL COMPONENTS
SUMIT K RATHORE
EE DEPT, GCET
Instruction Objective of lesson
 Upunitl now we have studied faults that are symmetrical
(all the three lines are shorted) but this kind of fault rare in
system, and unsymmetrical faults are major and to study
them or analyze them we need a simple technique this is
called symmetrical component analysis discovered by C L
Fortescue.
SYNTHESIS OF UNSYMMETRICAL PHASOR FROM THEIR
SYMMETRICAL COMPONENT
 CL Fortescue suggested that n related unbalanced
phasors can be resolved into n system of balanced
phasors called the symmetrical components of the
original phasors.
 The n phasors of each set of components are equal in
magnitude(length) and the angle between adjacent
phasors of the set are equal.
 This method is applicable to any unbalanced systems
we will confine attention to three phase system.
SYNTHESIS OF UNSYMMETRICAL PHASOR FROM THEIR
SYMMETRICAL COMPONENT
 According to CL Fortescue unbalanced phasors of three phase system can be
resolved into three balanced systems of phasors. The balanced sets of
components are :
 1. positive sequence component consisting of three phasors equal in
magnitude, displaced from each other by 120 in phase and having same
phase sequence as the original phasor.
 2. negative sequence components are consisting of three phasors equal in
magnitude, displaced from each other by 120 in phase and having phase
sequence opposite to that of original phasor.
 3. zero sequence components consisting of three phasors equal in magnitude
and with zero phase displacement from each other.
SYNTHESIS OF UNSYMMETRICAL PHASOR FROM THEIR
SYMMETRICAL COMPONENT
SYNTHESIS OF UNSYMMETRICAL PHASOR FROM THEIR
SYMMETRICAL COMPONENT
 It is customary when solving problem by symmetrical components to
designate the three phases of system as a, b and c of the voltages and currents
of sequence of positive sequence components of unbalanced phasors is abc,
negative sequence as acb.
 If original voltage sequence is Va, Vb, and Vc then positive sequence sets of
symmetrical components are designated by subscript 1, that becomes Va1,
Vb1, Vc1.
 Similarly negative sequence components are Va2, Vb2, Vc2 and the zero
sequence components are Va0, Vb0 and Vc0.
 And phasor representing currents are represented by I followed by subscripts
SYNTHESIS OF UNSYMMETRICAL PHASOR FROM THEIR
SYMMETRICAL COMPONENT
 Since each of the original unbalanced phasors is the sum of its components,
the original phasors expressed in terms of their components are
 Synthesis of set of three unbalanced phasors from the three sets of
symmetrical components is show in figure (next slide)
SYNTHESIS OF UNSYMMETRICAL PHASOR FROM THEIR
SYMMETRICAL COMPONENT
 The advantages of analysis by symmetrical components will become
apparent as we apply the method to study unsymmetrical faults on otherwise
symmetrical systems.
 Method consist of finding symmetrical components of current at fault
 Then value of current and voltages at various points in the system can be
found.
 The method is simple and leads to accurate prediction of system behavior
OPERATORS
 Because of phase displacement in phase components of sysmmetrical
components of voltage and currents in three phase system it is convenient to
have short hand method to represent the rotation of phasor through 120
degree.
 Result of multiplication of two complex number is the product of their
magnitudes and sum of their angles.
 If complex number expressing a phasor is multiplied by a complex number
of unit magnitude and angle theta the resulting complex number represent a
phasor equal to the original phasor displaced by the angle theta.
 The complex number of unit magnitude and associated angle theta is an
operator that rotates the phasor on which it operates through the angle theta.
OPERATORS
 Operator j which causes the rotation through 90 and operator -1 which causes
rotation through 180.
 Two successive applications of operator j causes rotation through 90+90
which leads up to the conclusion that j*j causes rotation through 180 and
thus we recognize to the conclusion that j2 =-1
 the letter a is commonly used to designate the operator that cause a rotation
of 120 in counter clock wise direction.
 Such an operator is a complex number of unit magnitude and with angle of
120 and defined by

=1ej2л/3= -0.5+j0.866
OPERATORS
 If operator is applied twice in succession the phasor is rotated through 240
and three succession
 a2= 1/_240o=-0.5-j0.866
 And
 a3=1/_360o=1/_0o=1
 An important difference must be noted between uses of operators j and a. the
operator j is unit magnitude at 90 and –j means that complex number j is
changed by an angle of 180 to give unit magnitude at 270 thus -j

j=1/_90 and –j=1/_270=1/_-90
 Hence it is said sometimes that +j indicate rotation through 90 and –j indicate
rotation through -90 statement is correct but a similar statement does not
apply to operator a since
 A=1/_120
OPERATORS
 -a=1/_120 x 1/-180 = 1/_ -60
-a2
a
-1 -a3
1 a3
a2
-a
SYMMETRICAL COMPONENT OF UNSYMMETRICAL
PHASORS
 Lets see how to solve unsymmetrical phasors into their symmetrical
components.
 Synthesis of three unsymmetrical phasors form three sets of
symmetrical phasors done by use of equation as below
 Now lets examine how to resolve three unsymmetrical phasors into
their symmetrical components.
SYMMETRICAL COMPONENT OF UNSYMMETRICAL
PHASORS
 We have seen that Number of unknown quantity can be reduced by
expressing each component of Vb and Vc as the product of some
function of the operator α and component a of Va.
SYMMETRICAL COMPONENT OF UNSYMMETRICAL
PHASORS
 Substituting values yield the
 Or in matrix form
SYMMETRICAL COMPONENT OF UNSYMMETRICAL
PHASORS
 For convenience we let
 Then as
 Rearrange the equation we ll obtain
SYMMETRICAL COMPONENT OF UNSYMMETRICAL
PHASORS
 Which shows us how to resolve three unsymmetrical
phasors into their symmetrical components. So separate
equation in ordinary form
 Similarly if we require to find out components Vbo. Vb1,
Vb2, Vco. Vc1, and Vc2, can be found out.
SYMMETRICAL COMPONENT OF UNSYMMETRICAL
PHASORS
 This shows that no zero sequence component exist if the
sum of the unbalanced phasors is zero. Since the sum of the
line to line voltage phasors in a three phase system is
always zero, zero sequence components are never present in
the line voltages, regardless of the amount of unbalance.
 The sum of three line to neutral voltage phasor is not
necessarily zero and voltages to neutral may contains zero
sequence component.
SYMMETRICAL COMPONENT OF UNSYMMETRICAL
PHASORS
 Preceding equation could have been written for any set of related phasors
and we might have written them for currents instead of for voltages. They
may be solved either analytically or graphically . Because some of the
preceding equations are so fundamental they are summarized for currents
SYMMETRICAL COMPONENT OF UNSYMMETRICAL
PHASORS
 In three phase system sum of line currents is equal to the currents In in the
return path through the neutral
Ia+Ib+Ic=In
 If currents are equal then In=3Iao
 In the absence of path through the neutral of three phase system In is zero
and the line currents contain no zero sequence components
 A delta connected load provides no path for neutral current and line
currents flowing to a delta connected load can contain no zero sequence
components.
POWER IN TERMS OF SYMMETRICAL COMPONENT
 If the symmetrical components of current and voltage are
known the power expended in three phase circuit can be
computed directly from the components.
POWER IN TERMS OF SYMMETRICAL COMPONENT
Which shows complex power can be computed from symmetrical components of voltage and
currents of unbalanced three phase circuit.
PHASE SHIFT OF SYMMETRICAL COMPONENT IN STARDETLA TRANSFORMER BANK
 Before we learn phase shift let us examine standard method of marking
transformer terminals.
 Primary and secondary is wound on core as in figure.
 High voltage winding marked by H1, H2 and low tension winding is
marked as X1, X2.
 Current flowing from H1 to H2 produces flux in common core in the same
directions current flowing from X1 to X2.
 Transformer theory shows that current must flow out at terminal X1 when
current is flowing into terminal H1 with magnetizing current neglected for
the magneto motive forces produced by the currents
in two coils must cancel each other all times.
 The flux common to both the coil is due
to magnetizing current alone.
PHASE SHIFT OF SYMMETRICAL COMPONENT IN STARDETLA TRANSFORMER BANK
 For that we have to mark the terminal by dot that indicate
direction of current flow from the termianls whether the
current Ip and Is are in phase or out of phase by 180 degree.
 The direction of arrow and dot marked on winding terminal
H1 and X1 are positive at the same time with respect to H2
and X2 if dot is on upper side in both winding that is called
in phase
 If on low tension side dot is on bottom of winding that
indicate both current are out of phase by 180degree.
PHASE SHIFT OF SYMMETRICAL COMPONENT IN STARDETLA TRANSFORMER BANK
 For three phase transformer high tension terminals are
marked H1, H2 and H3 & for low tension terminals are
marked X1, X2 and X3.
 For the figure YY or ∆ ∆ transformers the markings are
such that voltages to neutral from terminals H1, H2 and H3
are in phase with the voltages to neutral from terminals X1,
X2 and X3 respectively .
PHASE SHIFT OF SYMMETRICAL COMPONENT IN STARDETLA TRANSFORMER BANK
 High tension terminals H1, H2, H3 are connected to
termoinals A,B,C respectively and phase sequence is ABC
similarly for low tension winding X1,X2,X3.
 Now carefully see winding AN of Y connected side which
is linked magnetically with phase winding bc on ∆
connected side.
 Location of dots on windings shows that VAN is in phase
with Vbc
PHASE SHIFT OF SYMMETRICAL COMPONENT IN STARDETLA TRANSFORMER BANK
 American standard for designating terminals H1 and X1 on
Y- ∆ transformer requires that positive sequence voltage
drop from H1 to neutral lead the positive sequence voltage
drop from X1 to neutral by 30 degree regardless of whether
Y or ∆ winding is on high tension side.
 Similarly voltage from H2 to neutral leads the voltage from
X2 to neutral by 30 and voltage from H3 to neutral leads
the voltage from X3 to neutral by 30 degree.
 VA1 leads Vb1 by 30degree which enables us to determine
that terminal to which phase b is connected should be
labeled X1.
PHASE SHIFT OF SYMMETRICAL COMPONENT IN STARDETLA TRANSFORMER BANK
Va2 lags VA2 by 90
PHASE SHIFT OF SYMMETRICAL COMPONENT IN STARDETLA TRANSFORMER BANK
 Figure(b) shows that normal designation of winding
terminals by which we normally designate the transformer
winding but figure (a) shows nomenclature is most
convenient for computations in former case we need to
exchange a for b, b for c and c for a.
PHASE SHIFT OF SYMMETRICAL COMPONENT IN STARDETLA TRANSFORMER BANK
 Inspection of positive and negative sequence phasor
diagram shows that Va1 leads VA1 by 90 amd Va2 lags
VA2 by 90.
 The diagram shows VA1 and VA2 in phase which is not
necessarliy ture but phase shift between VA1 and VA2 does
not alter the 90 relation between VA1 and Va1 or between
VA2 and Va2.
PHASE SHIFT OF SYMMETRICAL COMPONENT IN STARDETLA TRANSFORMER BANK
 Direction specified for IA is away from the dot in
transformer winding and direction of Ibc is also away from
dot in its winding these current are 180degree out of phase.
 Therefore the phase relation between Y ∆ current is in
figure shows that Ia1 leads IA1 by 90 and Ia2 lags IA2 by
90.
 Summarizing relations between the symmetrical
components of line currents on two sides of transformer
gives Va1=+jVA1
Va2=-jVA2
Ia1=+jIA1
Ia2=-jIA2
PHASE SHIFT OF SYMMETRICAL COMPONENT IN STARDETLA TRANSFORMER BANK
UNSYMMETRICAL SERIES IMPEDANCE
 System becomes unbalanced only upon the occurrence of
unsymmetrical fault. Let see the system which having
unequal impedance.(We should reach to conclusion that is
important in analysis by symmetrical components)
 Figure shows unequal impedances Za, Zb, Zc.
 If we assume no mutual coupling or (mutual inductance )
the voltage drop across the part of the system given by
matrix equation
UNSYMMETRICAL SERIES IMPEDANCE
And in terms of symmetrical components of voltage and
current
Pre multiplying both side by equation A-1 gives the result
If all impedances are equal
UNSYMMETRICAL SERIES IMPEDANCE
 Thus we conclude that symmetrical components of
unbalanced currents flowing in balanced series impedances
produce voltage drop of like sequence only condition is
that no coupling between phases.
 If coupling is exist so square matrix will have some off
diagonal elements .
 In transmission line system assumption of transposition
yields equal series impedances.
UNSYMMETRICAL SERIES IMPEDANCE
 Thus component currents of any one sequence produce
voltage drop of like sequence only in transmission line that
is positive sequence current produces positive sequence
voltage drops only, negative sequence produces negative
sequence voltage drops and zero sequence currents
produces zero sequence voltage drop only.
 We could study variation of these equation for special cases
such as single phase loads where Zb=Zc=0.
SEQUENCE IMPEDANCE AND SEQUENCE NETWORK



In the method of symmetrical components, to calculate the effect of a fault on a power
system, the sequence networks are developed corresponding to the fault condition. These
networks are then interconnected depending on the type of fault. The resulting network is then
analyzed to find the fault current and other parameters.
The positive-sequence network is obtained by determining all the positive-sequence voltages
and positive-sequence impedances of individual elements, and connecting them according to
the SLD. All the generated emfs are positive-sequence voltages. Hence all the per unit
reactance/impedance diagrams obtained in the earlier chapters are positive-sequence
networks. The negative-sequence generated emfs are not present. Hence, the negativesequence network for a power system is obtained by omitting all the generated emfs and
replacing all impedances by negative-sequence impedances from the positive-sequence
networks.
Since all the neutral points of a symmetrical three-phase system are at the same potential
when balanced currents are flowing, the neutral of a symmetrical three-phase system is the
logical reference point. It is therefore taken as the reference bus for the positive- and
negative-sequence networks. Impedances connected between the neutral of the machine and
ground is not a part of either the positive- or negative- sequence networks because neither
positive- nor negative-sequence currents can flow in such impedances.
SEQUENCE IMPEDANCE AND SEQUENCE NETWORK
 Impedance of symmetrical component depends on type of power
system equipment i.e generator, transformer, or transmission line.
 Zero sequence impedance of over head line depends on presence of
ground wires, tower footing resistance and grounding.
 Line capacitance of over head lines is ignored in SC calculations.
 While estimating sequence impedances of power system
components is one problem constructing zero, positive and negative
sequence impedance network is the first step for unsymmetrical
fault current calculations
SEQUENCE IMPEDANCE AND SEQUENCE NETWORK
 The sequence network are constructed as viewed from the fault
point which can be defined as point at which unbalance occurs in a
system i.e fault point or unbalance load.
 The voltages for sequence network are taken as line to neutral
voltages
 Only active network contains voltage source is positive sequence
networks
 Sequence networks have per phase impedance values
 Normally sequence networks are constructed on the basis of per unit
values on common MVAbase and 100 MVA is in common use.
SEQUENCE IMPEDANCE AND SEQUENCE NETWORK
 The analysis of an unsymmetrical fault on a symmetrical system consist in
finding the symmetrical component of unbalanced currents that are flowing
 Since the component current of one phase sequence cause voltage drop of
like sequence only and are independent of current of other sequence.
 The single phase equivalent circuit composed of impedance to current of
any one sequence only is called the sequence network for that particular
sequence.
 The sequence network includes any generated emfs of like sequence
 To calculate the effect of fault by method of symmetrical components it is
essential to determine the sequence impedances and to combine them to
form the sequence network
SEQUENCE NETWORK OF UNLOADED GENERATOR
 An unloaded generator, grounded through a reactor.
 When fault occurs at terminal of generator current Ia, Ib
and Ic flows in a lines.
 If fault involves ground the current flowing into the neutral
of generator is designated by In
 One or two of the line currents may be zero but the current
can resolved into their symmetrical components regardless
of how unbalanced they may be.
SEQUENCE NETWORK OF UNLOADED GENERATOR
 Drawing sequence networks is very simple. The generated
voltages are of positive sequence only(and as reference)
since generator is designed to supply balanced three phase
voltages.
 Therefore positive sequence network is composed of an emf
in series with the positive sequence impedance of the
generator.
 The negative and zero sequence network contains no emf
but include the impedances of the generator to negative and
zero sequence respectively ,
 Current are flowing through impedance of their own
sequence only.
SEQUENCE NETWORK OF UNLOADED GENERATOR
 The generated emf in the positive sequence network is the no load terminal
voltage to neutral which is also equal to voltages behind transient and sub
transient, transient or synchronous reactance since the generator is not
loaded.
The reactance in the positive sequence
network is sub transient, transient or
synchronous reactance depending on
whether sub transient, transient or steady
state conditions are being studied.
SEQUENCE NETWORK OF UNLOADED GENERATOR
 The reference bus for the positive and negative sequence
networks is the neutral of the generator. So far as positive
negative sequence components are concerned the neutral of
generator is at ground potential since only zero sequence
current flows in the impedance between neutral and ground.
 There reference bus for zero sequence network is the
ground at generator.
SEQUENCE NETWORK OF UNLOADED GENERATOR
 The current flowing in the impedance Zn between neutral
and ground is 3Iao.
 Voltage drop of zero sequence from point a to ground it 3IaoZn - IaoZgo where Zgo is zero sequence impedance per
phase of the generator.
 The zero sequence network which is single phase circuit
assumed to carry only the zero sequence current of one
phase must therefore have an impedance of 3Zn+Zgo
 The total zero sequence impedance through which Iao flows
is Zo=3Zn+Zgo
SEQUENCE IMPDEANCES OF CIRCUIT ELEMENTS

For obtaining the sequence networks, the component voltages/ currents and the component
impedances of all the elements of the network are to be determined. The usual elements of a power
system are: passive loads, rotating machines (generators/ motors), transmission lines and
transformers. The positive- and negative-sequence impedances of linear, symmetrical, static circuits
are identical.
The sequence impedances of rotating machines will generally differ from one another. This is due to
the different conditions that exist when the sequence currents flow. The flux due to negativesequence currents rotates at double the speed of rotor while that due to positive-sequence currents is
stationary with respect to the rotor. The resultant flux due to zero-sequence currents is ideally zero
as these flux components add up to zero, and hence the zero-sequence reactance is only due to the
leakage flux. Thus, the zero-sequence impedance of these machines is smaller than positive- and
negative-sequence impedances.
The positive- and negative-sequence impedances of a transmission line are identical, while the zerosequence impedance differs from these. The positive- and negative-sequence impedances are
identical as the transposed transmission lines are balanced linear circuits. The zero-sequence
impedance is higher due to magnetic field set up by the zero-sequence currents is very different
from that of the positive- or negative-sequence currents. The zero-sequence reactance is generally 2
to 3.5 times greater than the positive- sequence reactance. It is customary to take all the sequence
impedances of a transformer to be identical, although the zero-sequence impedance slightly differs
with respect to the other two.
POSITIVE AND NEGATIVE SEQUENCE NETWORK
 Obtaining the values of sequence impedances of power system is to enable
us to construct the sequence network for the complete system.
 The network of particular sequence shows all the paths for the flow of
current of that sequence in system.
 Three phase synchronous generator and motors have internal voltages of
positive sequence only since they are designed to generate balance
voltages.
 Since the positive and negative sequence impedances are same in static
symmetrical system.
 conversion of positive sequence network to negative sequence network is
accomplished by changing if necessary only the impedances that represents
rotating machinery and by omitting the emfs
 Electromagnetic forces are omitted on assumption of balanced generated
voltages and the absence of negative sequence voltages induces from
outside sources
POSITIVE AND NEGATIVE SEQUENCE NETWORK
 Since all the neutral points of symmetrical three phase
system are at the same potential when balanced three phase
curents are flowing, all the neutral points must be at same
potential for either positive or negative sequence currents
 Therefore neutral of symmetrical three phase system is the
logical reference potential for specifying positive and
negative sequence voltage drop and is the reference bus for
positive and negative sequence network.
 Impedance connected between neutral of machine and
ground is not a part of either the positive or negative
sequence network because neither positive nor negative
sequence current can flow in impedance so connected.
ZERO SEQUENCE NETWORK
 Zero sequence currents are same in magnitude and phase at
any point in all the phases of system
 Therefore zero sequence current will flow only if return
path exists through which a completed circuit is provided.
 The reference for zero sequence voltages is potential of the
ground at the point in the system at which any particular
voltage is specified.
 Since zero sequence current may be flowing in ground, the
ground is not necessarily at same potential at all points and
the reference bus of zero sequence network does not
represent a ground of uniform potential.
ZERO SEQUENCE NETWORK
 The zero-sequence components are the same both in magnitude and in
phase. Thus, it is equivalent to a single-phase system and hence, zerosequence currents will flow only if a return path exists. The reference point
for this network is the ground.
If a circuit is Y-connected, with no connection from the neutral to ground
or to another neutral point in the circuit, no zero-sequence currents can
flow, and hence the impedance to zero-sequence current is infinite. This is
represented by an open circuit between the neutral of the Y-connected
circuit and the reference bus,
ZERO SEQUENCE NETWORK
 if the neutral of the Y-connected circuit is grounded through zero
impedance, a zero-impedance path (short circuit) is connected between the
neutral point and the reference bus.

If an impedance Zn is connected between the neutral and the ground of a
Y-connected circuit, an impedance of 3Zn must be connected between the
neutral and the reference bus (because, all the three zero-sequence currents
(3Ia0) flows through this impedance to cause a voltage drop of 3Ia0 Z0 )
ZERO SEQUENCE NETWORK
 A ∆-connected circuit can provide no return path; its impedance
to zero-sequence line currents is therefore infinite. Thus, the
zero-sequence network is open at the ∆ -connected circuit, as
shown in the figure Zero-sequence equivalent networks of ∆ connected load. However zero-sequence currents can circulate
inside the ∆ -connected circuit.
ZERO SEQUENCE NETWORK TRANSFORMER
ZERO SEQUENCE NETWORK
 The zero-sequence equivalent circuits of three-phase
transformers deserve special attention. The different possible
combinations of the primary and the secondary windings in Y
and ∆ alter the zero-sequence network. The possible
connections of two-winding transformers and their equivalent
zero-sequence networks are shown in the above figure. The
networks are drawn remembering that there will be no primary
current when there is no secondary current, neglecting the noload component. The arrows on the connection diagram show
the possible paths for the zero-sequence current. Absence of
an arrow indicates that the connection is such that zerosequence currents cannot flow. The letters P and Q identify the
corresponding points on the connection diagram and
equivalent circuit.
ZERO SEQUENCE NETWORK

Case 1: Y-Y Bank with one neutral grounded: If either one of the neutrals of a Y-Y bank is
ungrounded, zero-sequence current cannot flow in either winding ( as the absence of a path through
one winding prevents current in the other). An open circuit exists for zero-sequence current
between two parts of the system connected by the transformer bank.
Case 2: Y-Y Bank with both neutral grounded: In this case, a path through transformer exists for the
zero-sequence current. Hence zero-sequence current can flow in both sides of the transformer
provided there is closed path for it to flow. Hence the points on the two sides of the transformer are
connected by the zero-sequence impedance of the transformer.
Case 3: Y- ∆ Bank with grounded Y: In this case, there is path for zero-sequence current to ground
through the Y as the corresponding induced current can circulate in the ∆. The equivalent circuit
must provide for a path from lines on the Y side through zero-sequence impedance of the
transformer to the reference bus. However, an open circuit must exist between line and the
reference bus on the ∆ side. If there is an impedance Zn between neutral and ground, then the zerosequence impedance must include 3Zn along with zero-sequence impedance of the transformer.
Case 4: Y- ∆ Bank with ungrounded Y: In this case, there is no path for zero-sequence current. The
zero-sequence impedance is infinite and is shown by an open circuit.

Case 5: ∆ -∆ Bank: In this case, there is no return path for zero-sequence current. The zerosequence current cannot flow in lines although it can circulate in the ∆ windings.
ZERO SEQUENCE NETWORK
Conclusion



The sequence networks are three separate networks which are the singlephase equivalent of the corresponding symmetrical sequence systems.
These networks can be drawn as follows: For the given condition (steady
state, transient, or subtransient), draw the reactance diagram (selecting
proper base values and converting all the per unit values to the selected
base, if necessary). This will correspond to the positive-sequence network.
Determine the per unit negative-sequence impedances of all elements (if
the values of negative sequence is not given to any element, it can
approximately be taken as equal to the positive-sequence impedance).
Draw the negative-sequence network by replacing all emf sources by short
circuit and all impedances by corresponding negative-sequence
impedances in the positive-sequence network.
Determine the per unit zero-sequence impedances of all the elements and
draw the zero-sequence network corresponding to the grounding conditions
of different elements.
Conclusion
 Unbalanced voltage and currents can be resolved into their symmetrical
components.
 Problem are solved by treating each set of components separately and
super imposing the results
 In balanced network having no coupling between phases the currents of
one phase sequence induces voltage drop of like sequence only
 Knowledge of positive sequence is necessary for load flow studies on
power system
 Fault calculation and stability studies that involves unsymmetrical faults
that needed negative and zero sequence components also.
 Synthesis of zero sequence network requires particular care because the
zero sequence network may differ from others considerably
Reference


Elements of power system analysis by Willian D Stevenson Jr. International student edition
Power system analysis by JB Gupta


http://elearning.vtu.ac.in/P2/EE61/Ch03/html/0019.htm
http://nptel.iitm.ac.in/courses/Webcourse-contents/IIT-KANPUR/power-system/chapter_8/8_5.html
THANK YOU

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