UNIT IV

FAULT ANALYSIS –
UNBALANCED
FAULTS
Symmetrical components – Sequence
impedances – Sequence circuits of
synchronous machine, transformer and
transmission line – Sequence networks
– Analysis of unsymmetrical faults:
single-line to-ground, line-to-line and
double-line-to-ground using Thevenin’s
theorem and Z-Bus- computation of
post fault currents in symmetrical
component and phasor domains.
 Introduction to Symmetrical
Components
So far, we have considered both normal and
abnormal (short circuit) operating conditions under
completely balanced system.
Under such operations the system impedances in
each phase are identical and the three-phase
voltages and currents throughout the system are
completely balanced, i.e. they have equal
magnitudes in each phase and are progressively
displaced in time phase by 120◦ (phase a leads/lags
phase b by 120◦ and phase b lags/leads phase c by
120◦)
 In a balanced system, analysis can proceed on a
single-phase basis. The knowledge of voltage and
current in one phase is sufficient to completely
determine voltages and currents in other two
phases.
Real and reactive powers are simply three times
the corresponding per phase values.
Unbalanced system operation can result in an
balanced system due to unsymmetrical fault, e.g.
Single Line-to-ground fault or line-to-line fault.
System operation may also become unbalanced
when loads are unbalanced as in the presence of
large single-phase loads.
 Analysis under unbalanced conditions has to be
carried out on a three-phase basis.
Alternatively, a more convenient method of
analysing unbalanced operation is through
symmetrical components, “where the three-phase
voltages (and currents) which may be unbalanced
are transformed into three sets of balanced voltages
(and currents) are symmetrical components”.
Fortunately, in such a transformation the
impedance presented by various power system
elements (Synchronous generator, transformers and
lines) to symmetrical components are decoupled
from each other resulting in independent system
networks for each components (balanced set).
 Symmetrical Component
Transformation
A set of three balanced voltages (phasors) Va, Vb, Vc is
characterised by equal magnitude and inter-phase
difference of 120◦.
The set is said to have a phase sequence of abc
(positive sequence) if Vb lags Va by 120 ◦ and Vc lags Vb by
120◦. The three phasors can then be expressed in terms
of the reference phasor Va as,

Va = Va, Vb = α2Va, Vc = αVa

Where the complex number operator α is defined as

α = ej120 ◦ = cos 120◦ + jsin 120◦ = - 0.5 + j0.866
 It has the following properties:
 2  e j 240  e  j120    
 2  
( )  
 3  (1)
 1 
1     2  
 0 
If the phase sequence is acb (negative sequence), then
Va = Va, Vb = αVa, Vc = α2Va
Thus a set of balanced phasors is fully characterised by
its reference phasors (say Va) and its phase sequence
(positive or negative)
Suffix 1 is commonly used to indicate positive
sequence. A set of (balanced ) positive sequence phasor
is written as
Va1, Vb1 = α2Va1, Vc1 = αVa1 (2)
 Similarly, suffix 2 is used to indicate negative sequence.
A set of (balanced) negative sequence phasors is written
as
Va2, Vb2 = αVa2, Vc2 = α2Va2 (3)
A set of three voltage (phasors) equal in magnitude
and having the same phase is said to have zero
sequence. Thus a set of zero sequence phasors is
written as
Va0, Vb0 = Va0, Vc0 = Va0 (4)
 Consider now a set of three voltage (phasors) Va, Vb, Vc
which in general may be unbalanced. According to
Fortesque’s theorem the three phasors can be expressed
as the sum of positive, negative and zero sequence
phasors. Thus
Va = Va1 + Va2 + Va0 (5)
Vb = Vb1 + Vb2 + Vb0 (6)
Vc = Vc1 + Vc2 + Vc0 (7)

The three phasor sequences (positive, negative and

zero) are called the symmetrical components of the
original phasor set Va, Vb, Vc. The addition of
symmetrical components as per Eqs. (5) to (7) to
generate Va, Vb, Vc is indicated by the phasor diagram of
Fig 1.
 Fig 1 – Graphical addition of the symmetrical components to
obtain the set of phasors Va, Vb, Vc (Unbalanced in general)

Let us now express Eqs. (5) to (7) in in terms of reference

phasors Va1, Va2, and Va0. Thus
Va = Va1 + Va2 + Va0 (8)
Vb = α2Va1 + αVa2 + Va0 (9)
Vc = αVa1 + α2Va2 + Va0 (10)
These equations can be expressed in the matrix form

Va   1 1 1 Va1  (11)

Vb    2  1 Va 2
    
Vc     1 Va 0 
2

Or
Vp = AVs (12)
where Va 
Vp = Vb  = vector of original phasors
 
Vc 

Va1 
Vs = Va 2 = vector of symmetrical components
 
Va 0 
 1 1 1
A =  2  1 (13)
   2 1
We can write Eq. 12 as
Vs = A-1Vp (14)
Computing A-1 and utilizing relations (1), we get
1   2 
1 
A = 1  2  
-1 (15)
3
1 1 1 

In expanded form we can write Eq. 14 as
1
Va1 = (Va  Vb   2Vc) (16)
3
1
Va2 = (Va   2Vb  Vc) (17)
3
1
Va0 = (Va  Vb  Vc ) (18)
3
 Equations (16) to (18) give the necessary relations for
obtaining symmetrical components of the original phasors,
while Ens. (5) to (7) give the relationships for obtaining
original phasors from the symmetrical components.
The symmetrical component transformations though
given above in terms of voltages hold for any set of
phasors and therefore automatically apply for a set of
currents. Thus
Ip = AIs (19)
And
Is = A-1Ip (20)

 Ia   Ia1 
 Ib   Ia 2 
Where Ip =   and Is =  
 Ic   Ia 0 
 In expanded form the relations (19) and (20) can be
expressed as follows:
(i) Construction of current phasors from their
symmetrical components:
Ia = Ia1 + Ia2 + Ia0 (21)
Ib = α2Ia1 + αIa2 + Ia0 (22)
Ic = αIa1 + α2Ia2 + Ia0 (23)
(ii) Obtaining symmetrical components of current
phasors:
1
Ia1 = ( Ia  Ib   Ic )
2
(24)
3
1
Ia2 = 3 ( Ia   2
Ib  Ic ) (25)
1
Ia0 = 3 ( Ia  Ib  Ic) (26)
 Certain observations can be made regarding a three-phase
system with neutral return in Fig 2

Fig 2
The sum of the three line voltages will always be zero.
Therefore, the zero sequence component of line voltages is
always zero, i.e.
1
Vabc  (Vab  Vbc  Vca )  0 (27)
3
 On the other hand, the sum of phase voltages (line to
neutral) may not be zero, so that their zero sequence
components Va0 may exist.
Since the sum of the three line currents equals the current
in the neutral wire, we have
1 1
Iao  ( Ia  Ib  Ic )  In (28)
3 3
( ∵Ia + Ib + Ic = In )
i.e. The current in the neutral is three times the zero
sequence line current. If the neutral connection is removed,
1
Iao  In  0 (29)
3
i.e. In the absence of a neutral connection the zero
sequence line current is always zero.
Power Invariance
We shall now show that the symmetrical component
transformation is power invariant, which means that the
sum of powers of the three symmetric components equals
the three-phase power.
Total complex power in a three-phase circuit is given by

S VpT I *p Va I a*  Vb I b*  Vc I c* (30)

Or
(31)
S  AVs  AI s 
T *

 VsT AT A* I sT
Now
1  2    1 1 1 1 0 0
 
AT A*  1   2     2 1  30 1 0  3U (32)
1 1 1   2  1 0 0 1
 

S  3VsTUI s*  3VsT I s*
 3Va1 I a*1  3Va 2 I a*2  3Va 0 I a*0 (33)
= sum of symmetrical component powers
 EX 1: The line to ground voltages on high voltage side of
a step up transformer are 100 kV, 33 kV and 38 kV on
phases a, b and c respectively. The voltage of phase ‘a’
leads that of phase ‘b’ by 100◦ and lags that of phase ‘c’ by
176.5◦. Determine the symmetrical components of
voltages.
Sol:
Va = 100∠0◦
Vb = 33∠-100 ◦
Vc = 38 ∠175.5 ◦

1
Va0 = [Va  Vb  Vc]
3
= 0.333[100+j0 – 5.7303-j32.4986 -37.9291+j2.3198]
 = 18.78 - j10.05595
= 21.3045 ∠-28.1757 ◦

1
Va1 = [Va  Vb   Vc ]
2

3
= 0.333[100 ∠0 + 1∠120 ◦(33 ∠-100 ◦) +
1∠240 ◦(38 ∠176.5◦)
= 52.66 + j14.3255
= 52.646 ∠15.7897◦

1
Va2 = [Va   2Vb  Vc]
3
= 0.333[100 ∠0 + 1∠240 ◦(33 ∠-100 ◦) +
1∠120 ◦(38 ∠176.5◦)
 = 30.55 – j4.265
= 30.846 ∠-7.947◦

Ex 2: The line currents in amperes in phases a, b and c

respectively are 500 + j150, 100 – j600 and -300 + j600
referred to the same reference vector. Find the
symmetrical components of currents.
Sol:
Ia = 500 + j150
Ib = 100 – j600
Ic = -300 + j600

1
Ia0 = [ Ia  Ib  Ic ]
3
Ia0 = 100 + j50 Amps

1
Ia1 = [ Ia   Ib   2
Ic ]
3
= 546.4 + j165.46 Amps

1
Ia2 = [ Ia   2
Ib  Ic ]
3
= 48.8 – j21.82 Amps

α = -0.5 + j0.866
α2 = -0.5 - j0.866
Sequence Impedance of Transmission Lines
Figure 3 shows the circuit of a fully transposed line
carrying unbalanced currents. The return path for In is
sufficiently away for the mutual effect to be ignored.
Let,
Xs = self reactance of each line
Xm = mutual reactance of any line pair

Fig 3
The following KVL equations can be written as

Va  V ' a  jXsIa  jXmIb  jXmIc

(34)
Vb  V b  jXmIa  jXsIb  jXmIc
'

Vc  V ' c  jXmIa  jXmIb  jXsIc

or in matrix form


V 
 a 
V ' a 
  Xs Xm Xm   Ia 

j  Xm Xm   Ib 
  (35)
V  V 'b  
 b 
 

 Xs
   
V
 c 


' 
V c 
   Xm Xm Xs   Ic 


or Vp – V’p = ZIp (36)

or AVs – AV’s = ZAIs (37)
or (Vs – V’s) = A-1 ZAIs (38)
 Now
1   2   jXs jXm jXm   1 1 1
1 
A ZA  1 
1 2
   jXm jXs jXm   2  1 (39)
3
1 1 1   jXm jXm jXs     2 1


 Xs  Xm 
 j  Xs  Xm 

 Xs  2 Xm 

Thus Eq. (38) can be written as



V 
1  V ' 1   Xs  Xm 0 0  I 1 
 
 j  0 0   I 2
   ' 


V 2

V 2 Xs  Xm (40)
   


V 0


V ' 
0  0 0 Xs  2 Xm   I 0 
 
 Z 1 0 0   I 1  (41)
 j  0 Z 2 0   I 2
 0 0 Z 0  I 0 
where,
Z1 = j(Xs-Xm) = positive sequence impedance (42)
Z2 = j(Xs-Xm) = negative sequence impedance (43)
Z0 = j(Xs + 2Xm) = zero sequence impedance (44)
we conclude that fully transposed transmission has:
i) equal positive and negative sequence impedances.
ii) zero sequence impedance much larger than the positive
or negative sequence impedance (it is approximately
2.5 times).
 It is further observed that the sequence circuit Eqn. 41 in
decoupled form, i.e. There is no mutual sequence
inductances. Eqn. 41 can be represented in network form
as Fig. 4

Fig. 4
The decoupling between sequence networks of a fully
transposed transmission holds also in 3-phase synchronous
machines and 3-phase transformers. This fact leads to
considerable simplifications in the use of symmetrical
components method in unsymmetrical fault analysis.
 Sequence Impedance and Sequence
Network Of Synchronous Machine
Fig. 5 depicts an unloaded synchronous machine
(generator or motor) grounded through a reactor
(impedance Zn). Ea, Eb and Ec are the induced emfs of the
three phases. When a fault takes place at machines
terminals, currents Ia, Ib and Ic flow in the lines. Whenever
the fault involves ground currents In = Ia + Ib + Ic flows to
neutral from ground via Zn.

Fig. 5
 Unbalanced line currents can be resolved into their
symmetrical components Ia1, Ia2 and Ia0.
Positive sequence Impedance and network:
Since a synchronous machine is designed with
symmetrical windings, it induces emfs of positive sequence
only, i.e. no negative or zero sequence voltages are
induced in it.

Fig 6
 Fig. 6a shows the three-phase positive sequence network
model of a synchronous machine.
Zn does not appear in the model as In = 0 for positive
sequence currents.
Since it is a balanced network it can be represented by
the single-phase network model of Fig. 6b for purposes of
analysis.
The reference bus for a positive sequence network is at
neutral potential. Further, since no current flows from
ground to neutral, the neutral is at ground potential.
With reference to Fig. 6b, the positive sequence voltage
of terminal a with respect to the reference bus is given by
Va1 = Ea – Z1Ia1 (45)
 Negative Sequence impedance and Network:
Synchronous machines has zero negative sequence induced
voltages.
With the flow of negative sequence currents in the stator a
rotating field is created which rotates in the opposite
direction to that of the positive sequence field and therefore ,
at double synchronous speed with respect to rotor. Currents
at double the stator frequency are therefore induced in rotor
field and damper winding.

Fig 7
 Negative sequence network models of a synchronous
machine, on a three-phase basis are shown in Fig. 7. The
reference bus is at neutral potential which is the same as
ground potential.
From Fig. 7b the negative sequence voltage of terminal a
with respect to reference bus is
Va2 = - Z2Ia2 (46)

Zero Sequence Impedance and network:

No zero sequence voltages are induced in an
synchronous machine.
The flow of zero sequence currents creates three mmfs
which are in time phase but are distributed in space phase
by 1200.
 Zero sequence network models on a three and single
phase basis are shown in Fig. 8. In Fig. 8a, the current
flowing in the impedance Zn between neutral and ground
is In = 3Ia0. The zero sequence voltage of terminal a with
respect to ground, the reference bus is therefore
Va0 = – 3ZnIa0 – Z0gIa0 = – (3Zn + Z0g)Ia0 (47)

Fig 8
where Z0g is the zero sequence impedance per phase of
the machine.
 Since the single-phase zero sequence network of Fig. 8b
carries only per phase zero sequence current, its total zero
sequence impedance must be
Z0 = 3Zn + Z0g (48)
In order for it to have the same voltage from a to
reference bus. The reference bus here is, of course at
ground potential.
From Fig. 8b zero sequence voltage of a point a with
respect to the reference bus is
Va0 = - Z0Ia0 (49)
Sequence Impedance and Networks
of Transformers
Positive sequence series impedance of a transformer
equals its leakage impedance. Since a transformer is a
static device, the leakage impedance does not change with
alternation of phase sequence of balanced applied
voltages. The transformer negative sequence impedance
is also therefore equal to its leakage reactance. Thus, for a
transformer
Z1 = Z2 = Zleakage (50)
Assuming such transformer connections that zero
sequence currents can flow on both sides, a transformer
offers a zero sequence impedance which may differ slightly
from the corresponding positive and negative sequence
values.
 Zero Sequence Networks of Transformers:
Before considering the zero sequence networks of
various types of transformer connections, three important
observations are made:
i) When magnetizing current is neglected, transformer
primary would carry current only if there is current flow on
secondary side.
ii) Zero sequence currents can flow in the legs of a star
connection only if the star point is grounded which

Fig 9
 provides the necessary return path for zero sequence
currents. This fact is illustrated in Fig. 9a and 9b.
iii) No zero sequence currents can flow in the lines
connected to a delta connection as no return path is
available for these currents. Zero sequence currents can,
however, flow in the legs of a delta-such currents are
caused by the presence of zero sequence voltage in the
delta connection. This fact is illustrated by Fig. 10.

Fig 10
 Various conditions can be taken into account by use of
equivalent circuit shown in Fig. 11. Z0 is the zero sequence
impedance of the transformer.

Fig. 11
The general rules to find the correct connections are:
1. Close switch ‘a’ if the primary winding is star
connected with neutral grounded
 2. Close switch ‘c’ if the primary winding is delta
connected. Keep both ‘a’ and ‘c’ open if the primary
winding is star connected with neutral isolated (not
grounded)
3. Close switch ‘b’ if the secondary winding is star
connected with neutral grounded.
4. Close switch ‘d’ if the secondary winding is delta
connected. Keep ‘b’ and ‘d’ open if the secondary winding
is star connected with neutral isolated (not grounded)
 Fig. 12 shows some typical transformer connections and
the zero sequence networks.

Fig . 12
 a) connection: The grounding of both the
neutral provides a path for zero sequence current to flow
in the primary and secondary. Therefore zero sequence
network shows transformer leakage impedance per phase
connected.

b) connection: The zero sequence currents

can flow in the primary because the neutral in grounded.
But no zero sequence current can leave delta connected
winding. Therefore the secondary sides is isolated.

c) connection: Since the neutral is not

grounded, zero sequence currents cannot flow in
transformer winding. Hence the network has an open
circuit.
 d) connection: The zero sequence currents
can circulate in delta windings but cannot leave the delta
terminals.

Ex 3: Fig 13 shows a power system network. Draw the

zero sequence network for this system.

Fig. 13
The system data is:
Generator G1 50 MVA 11 kV, X0 = 0.8 pu
Transformer T1 50 MVA 11/220 kV, X0 0.1 pu
 Generator G2 30 MVA 11 kV, X0 = 0.7 pu
Transformer T2 30 MVA 11/220 kV, X0 = 0.09 pu
Zero sequence reactance of line is 555.6 ohms.
Sol:
Base MVA = 50 MVA
base kV = 11 kV for LT side and 220 kV for HT side of the
transformers
220  220
Base impedance of line =  968 ohms
50
j555.6
pu impedance for line =  j0.574
968
j0.09  50
pu impedance of T2 =  j0.15
30
 j0.07  50
pu impedance of G2 =  j0.117
30
11  11
Base impedance for generator =  2.42 ohms
50
j3
Pu impedance of neutral reactor =  j1.24
2.42

The zero sequence network is shown in Fig. 14. The

neutral impedance appears as three times its actual value
in the circuit.

Fig.14
 EX 4: A 25 MVA, 11 kV, three-phase generator has a
subtransient reactance of 20%. The generator supplies
two motors over a transmission line with transformers at
both ends as shown in the one-line diagram of Fig. 15. The
motors have rated inputs of 15 and 7.5 MVA, both 10 kV
with 25% subtransient reactance. The three-phase
transformers are both rated 30 MVA, 10.8/121 kV,
connection ∆-Y with leakage reactance of 10% each. The
series reactance of the line is 100 ohms. Draw the positive
and negative sequence networks of the system with
reactances marked in per unit.

Fig .15
 Assume that the negative sequence reactance of each
machine is equal to its subtransient reactance. Omit
resistances. Select generator rating as base in the
generator circuit.
Sol: A base of 25 MVA, 11 kV in the generator circuit
requires a 25 MVA base in all other circuits and the
following voltage bases.
121
Transmission line voltages base = 11 123.2 kV
10.8
10.8
Motor voltage base = 123.2  11 kV
121
The reactances of transformers, lines and motors are
converted to pu values on appropriate base as follows :
2
25  10.8 
Transformer reactance = 0.1     0.0805 pu
30  11 
 100  25
Line reactance = 2
 0.164 pu
(123.2)
2
25  10 
Reactance of motor 1 = 0.25      0.345 pu
15  11 
2
Reactance of motor 2 = 0.25  25   10   0.69 pu
7.5  11 

The required positive sequence network is presented in

Fig. 16

Fig. 16
 Since all the negative sequence reactance of the system
are equal to the positive sequence reactances, the
negative sequence network is identical to the positive
sequence network but for the omission of voltage sources.
The negative sequence network is shown in Fig. 17

Fig. 17
 Ex 5: Draw the zero sequence network for the system
described in Ex 4. Assume zero sequence reactances for
the generator and motor of 0.06 pu. Current limiting
reactors of 2.5 ohms each are connected in the neutral of
the generator and motor No.2. The zero sequence
reactance of the transmission line is 300 ohms.
Sol: the zero sequence reactance of the transformer is
equal to its positive sequence reactance. Hence,
Transformer zero sequence reactance = 0.0805 pu
Generator zero sequence reactance = 0.06 pu
2
25  10 
Zero sequence reactance of motor 1 = 0.06    
15  11 
= 0.082 pu
 2
25  10 
Zero sequence reactance of motor 2 = 0.06   
7.5  11 
= 0.164 pu
2.5  25
Reactance of current limiting reactors = 2
 0.516pu
(11)
Reactance of current limiting reactor included in zero
sequence network = 3  0.516 = 1.548 pu
300  25
Zero sequence reactance of transmission line =
(123.2)2
= 0.494 pu.

Fig. 18
 Ex 6: Fig. 19 shows a power system network. Draw
positive sequence network, negative sequence network
and zero sequence network. The system data is under:
Equipment MVA Voltage X1 X2 X0
rating rating
Generator G1 100 11 kV 0.25 0.25 0.05
Generator G2 100 11 kV 0.2 0.2 0.05
Transformer T1 100 11/220 kV 0.06 0.06 0.06
Transformer T2 100 11/220 kV 0.07 0.07 0.07
Line 1 100 220 kV 0.1 0.1 0.3
Linn 2 100 220 kV 0.1 0.1 0.3

Fig. 19
 Unsymmetrical Fault analysis
The various types of unsymmetrical faults that occur in
power systems are:
Shunt Type Faults
i) Single line-to-ground (LG) fault
ii)Line-to-line (LL) fault
iii)Double line-to-ground (LLG) fault
Series Type Faults
i) Open conductor (one or two conductors open) fault
Though, three-phase faults being most sever, there are
situations when LG fault can cause greater fault current
when the fault location is close to large generating units.
Symmetrical Component Analysis of
Unsymmetrical Faults
Consider a general power system network as shown in
Fig. 20. It is assumed that a shunt type of fault occurs at
point F in the system, as a result of which currents Ia, Ib, Ic
flow out of the system, and Va, Vb, Vc are voltages of lines
a, b, c with respect to ground.

Fig. 20
 Let us assume that the system is operating at no load
before occurrence of the fault. Therefore, the positive
sequence voltages of all synchronous machines will be
identical and will equal the prefault voltage at F. Let this
voltage be labelled as Ea.
As seen from F, the power system will present positive,
negative and zero sequence networks, which are
schematically represented by Figs. 21a, b and c.
The reference bus is indicated by a thick line and the
point F is identified on each sequence network. Sequence
voltages at F and sequence currents flowing out of the
networks at F are also shown on the sequence networks.
Fig. 22a, b and c respectively give the Thevenin
equivalents of the three sequence networks.
Fig. 21 Sequence networks Fig. 22 Thevenin equivalents
of Sequence networks
 Recognizing that voltage Ea is present only in the positive
sequence network and that there is no coupling between
sequence networks, the sequence voltages at F can be
expressed in terms of sequence currents and Thevenin
sequence impedances as
Va1   Ea  Z 1 0 0   Ia1 
Va 2   0    0 Z 2 0   Ia 2 (51)
      
Va 0   0   0 0 Z 0  Ia 0 

Depending upon the type of fault, the sequence currents

and voltages are constrained, leading to a particular
connection of sequence networks.
Single Line-to- Ground (LG) Fault
Fig. 23 shows a line-to-ground
fault at F in a power system
through a fault impedance Zf.
The phases are so labelled
that the fault occurs on Fig. 23
phase a. At the fault F, the currents out of the power
system and the line-to-ground voltages are constrained as
follows:
Ib = 0 (52)
Ic = 0 (53)
Va = ZfIa (54)
 The symmetrical components of the fault currents are

 Ia1  1   2   Ia 
 Ia 2   1 1  2    0 
  3  
 Ia 0  1 1 1   0 
 
From which it is easy to see that
Ia1 = Ia2 = Ia0 = Ia /3 (55)
Expressing Eq. 54 in terms of symmetrical components,
we have
Va1 + Va2 + Va0 = ZfIa = 3ZfIa1 (56)
As per Eqs. (55) and (56) all sequence currents are equal
and the sum of sequence voltages equals 3ZfIa1. Therefore,
these equations suggest a series connections of sequence
networks through an impedance 3Zf as shown in Fig. 24a
and b.
Fig. 24
 In terms of the Thevenin equivalent of sequence
networks, we can write from Fig. 24b
Ea
Ia1  (57)
(Z1  Z2  Z0)  3Z f

Fault current Ia is then given by

3Ea
Ia  3Ia1 
(Z1  Z2  Z0)  3Z f (58)

The above results can also be obtained directly from Eqs.

(55) and (56) by using Va1, Va2 and Va0 from Eq (51). Thus
(Ea-Ia1Z1) + (-Ia2Z2) + (-Ia0Z0) = 3ZfIa1
Or
[(Z1+Z2+Z0)+3Zf]Ia1 = Ea
 Ea
Ia1 
(Z1  Z2  Z0)  3Z f (59)

The voltage of line b to ground under fault condition is

Vb = α2Va1 + αVa2 + Va0
 Ia   Ia   Ia 
 α 2  Ea  Z1   α  Z2     Z0 
 3  3  3

Substituting for Ia from Eq. 58 and reorganising, we get

3α2Z f  Z2(α2  α)  Z0(α2  1)

Vb  Ea (60)
(Z1  Z2  Z0)  3Z f

The expression for Vc can be similarly obtained.

Line-To-Line (LL) Fault
Fig. 25 shows a line-to-line fault at F in a power system on
phase b and c through a fault impedance Zf. The phases
can always be relabelled, such that the fault is on phases b
and c.

Fig . 25
 The currents and voltages at the fault can be expressed
as
Ia  0 

IP  Ib  ; V  V  I Z f
 b c b (61)
Ic   Ib 
The symmetrical components of the fault currents are
Ia1  1 α α 2  0 
Ia2   1 1 α 2 
α  Ib 
  3
Ia0  1 1 1   Ib 


From which we get

Ia2 = -Ia1 (62)
Iao = 0 (63)
 The symmetrical components of voltages of F under fault
are
Va1  1 α α 2  Va 
Va2   1 1 α 2 
α  Vb

  3  (64)
Va0  1 1 1  Vb  Z fIb 

Writing the first two equations, we have
3Va1 = Va + (α + α2)Vb – α2ZfIb
3Va2 = Va + (α + α2)Vb – αZfIb
From which we get
3(Va1 –Va2) = (α - α2) ZfIb = j3ZfIb (65)
Now Ib = α2Ia1 + αIa2 + Ia0
Ib = (α2 - α) Ia1 (since Ia2 = - Ia1; Ia0 = 0)
= -j3Ia1 (66)
 Substituting Ib from Eq. (66) in Eq. (65), we get
3(Va1 –Va2) = j3Zf (-j3Ia1)
Va1 –Va2 = ZfIa1 (67)
Equations (62) and (67) suggest parallel connection of
positive and negative sequence networks through a series
impedance Zf as shown in Fig. 26a and b. Since Ia0 = 0 as
per Eq. (61), the zero sequence network is unconnected.

Fig 26
 In terms of the Thevenin equivalents, we get from Fig. 26b

Ea
Ia1  (68)
Z1  Z2  Z f
From Eq. (66), we get

 j 3Ea
Ib   Ic  (69)
Z1  Z2  Z f
Knowing Ia1, we can calculate Va1 and Va2 from which
voltages at the fault can be found
Double Line-to-Ground (LLG) Fault
Fig. 27 shows a double line-to-ground fault at F in a
power system. The fault may in general have an impedance
Zf as shown.

3Ia0

Fig. 27
The current and voltage (to ground) conditions at the fault
are expressed as

Ia  0 
  (70)
Ia1  Ia2  Ia0  0
Vb = Vc = Zf(Ib+Ic) = 3ZfIa0 (71)

The reason is:

1
Ia 0  ( Ia  Ib  Ic)
3
 Ia  0
Ib  Ic  3Ia 0
The symmetrical components of voltages are given by

Va1  1 α α 2  Va 
Va2   1 1 α 2  
α  Vb 
  3 (72)
Va0  1 1 1  Vb 

From which it follows that
1

Va1  Va2  Va  (α  α 2 )Vb
3
 (73)

1
Va0  Va  2Vb  (74)
3
From Eqs. (73) and (74),

Va0  Va1 
1
3
 
2  α  α 2 Vb  Vb  3Z fIa0 (α+ α2+1=0)
 Or
Va0 = Va1 + 3ZfIa0 (75)
From Eqs (70), (73) and (75), we can draw the connection
of sequence networks as shown in Fig. 28a and b. This can
be verified using mesh and nodal equations for these
figures

Fig. 28
 In terms of the Thevenin equivalents, we can write from
Fig. 28b
Ea
Ia1 
Z1  Z2 || (Z0  3Z f )

Ea
Ia1  (76)
Z1  Z2(Z0  3Z f )/(Z2  Z0  3Z f )
 Ex 7: A 30 MVA, 11 kV generator has Z1= Z2 = j0.2 pu,
Z0 = j0.05 pu. (a) A line to ground fault occurs on the
generator terminals. Find the fault current and line to line
voltages during fault conditions. Assume that the
generator neutral is solidly grounded and that the
generator is operating at no load and at rated voltage at
the occurrence of fault. (b) Find the line current for a 3-
phase fault.
Sol:
Base MVA = 30 MVA
Base voltage = 11 kV

30  106
Base current =  1574.6 A
3  11  10 3

Zf = 0 (solidly grounded)
 (a) Since the fault is at generator terminals no network
reduction is necessary. The values of Z1, Z2 and Z0 of the
generator can be directly used to get the results,
E 10
Ia1  Ia2  Ia0  
Z1  Z2  Z0 j0.2  j0.2  j0.05
= -j2.222 pu
Fault current = Ia = 3Ia1 = -j6.666 pu
= -j6.666  1574.6 i.e. 10496.3∠-90ﾟA

Sequence voltages at fault point are

Va1 = E - Ia1Z1 = 1 ∠0ﾟ- (-j2.222)(j0.2) = 0.5556 pu

Va2 = - Ia2Z2 = - (-j2.222)(j0.2) = -0.4444 pu
Va0 = - Ia0Z0 = - (-j2.222)(j0.05) = -0.1111 pu
 Line to neutral voltages at fault point are

Va   1 1 1  0.5556  0 
Vb  α2 α 1  0.4444    0.1667  j0.866
  
Vc   α α2 1  0.1111   0.1667  j0.866
Line to line voltages at fault point are
Vab = Va - Vb = 0.1667 + j0.866 = 0.882∠79ﾟpu
Vbc = Vb - Vc = - j1.732 = 1.732∠270ﾟpu
Vca = Vc - Vc = -0.1667 + j0.866 = 0.882∠100.9ﾟpu
Since the generated line to neutral voltage has been
taken as 1 pu in the calculation of sequence currents, the
above voltages are in pu of line to neutral voltage.
Therefore, the actual values of line to line post fault
voltages are
 Vab = (11/3)  0.882 ∠79.1ﾟ= 5.6 ∠79.1ﾟkV
Vbc = (11/3)  1.732 ∠270ﾟ= 11 ∠270ﾟkV
Vca = (11/3)  0.882 ∠100.9ﾟ= 5.6 ∠100.9ﾟkV

Before the fault each line voltage was equal to 11 kV. It is

seen that a fault causes the post fault line voltages to be
different from pre-fault voltages.

10
(b) For a 3-phase fault, line current =  5  90 pu
j0.2

Actual value of line current = 5 1574.6 ∠-90ﾟ

= 7873 ∠-90ﾟ
 Ex 8: A line to line fault occurs on the terminals of the
generator of Example 7. Find the line current and line to
neutral voltages under fault conditions.
Sol: E = 1∠0ﾟpu, Z1 = Z2 = j0.2 pu, Zf = 0
10
Ia1    j2.5pu
j(0.2  0.2)
Ia2 = -Ia1 = j2.5 pu
The line currents are
Ia = 0
Ib = -j3 Ia1 = 4.33 ∠180ﾟpu
Ic = -Ib = 4.33 ∠0ﾟpu
Base current = 1574.6 A
Ia = 0
Ib = 1547.6  4.33 ∠180ﾟ= 6818 ∠ 180ﾟA
 Ic = 1547.6  4.33 ∠0ﾟ= 6818 ∠ 0ﾟA
Va1 = Va2 = 1- (-j2.5)(j0.2) = 0.5 pu
Va0 = 0
The line to neutral voltages at fault point are
Va   1 1 1 0.5  1 
Vb   α2 α 1 0.5   0.5 α2+ α+1 = 0
α2+ α = -1
      
Vc   α α 1  0   0.5
2

Since the line to neutral voltages has been taken as 1 pu

in the calculation of sequence currents, the actual values
of line to neutral voltages are
Va = 11/3 = 6.35 kV
Vb = (-0.5) 11/3 = -3.175 kV
Vc = (-0.5) 11/3 = -3.175 kV
 Ex 9: A double line to ground fault occurs on the
terminals of generator of example 7. Determine the line
currents, fault current and line to neutral voltages under
fault conditions.
Sol: Fig 28b shows the sequence networks connected for
a double line to ground fault when Zf =0
10
Ia1    j4.167 pu
(j0.2)(j0.05)
j0.2 
j0.25
j0.05
Ia2   ( j4.167)  j0.8334 pu
j0.05  j0.2
j0.2
Ia0   ( j4.167)  j3.3336 pu
j0.05  j0.2
The line currents are

Ia   1 1 1  j4.167  0 
Ib   α2 α 1  j0.8334   6.61130.9
  
Ic   α α2 1  j3.3336  6.6149.1 

The actual values of line currents are

Ia = 0
Ib = 6.61  1574.6 ∠ 130.9 = 10408.1 ∠ 130.9 A
Ic = 6.61  1574.6 ∠ 49.1 = 10408.1 ∠ 130.9 A
Fault current to ground = 3Ia0 = 10∠ 90 pu or
= 15746 ∠90 A
The sequence voltages are
Va1 = Va2 = Va0 = 1- (-j4.167)(j0.2) = 0.1666
The line to neutral voltages are

Va   1 1 1 0.1666 0.4998

Vb   α2 α 1 0.1666   0 
  
Vc   α α2 1 0.1666  0 

Actual values of line to neutral voltages are

Va = (0.4998)11/3 = 3.174 kV
Vb = Vc = 0