A.

Year / Chapter Semester Subject Topic

2013 / 5 6 Power system Unsymmetrical faults
analysis and
stability
5.1 Introduction:
The concept of faults has been already introduced in chapter 2 which was
dedicated to the treatment of symmetrical faults. In this chapter, we shall deal
with unsymmetrical faults. The unsymmetrical faults are basically categorized into
two types, namely,
1)Shunt type of faults and
2)Series type of faults.
Shunt type of fault involves short circuit between conductors or between the
conductors and ground. They are characterized by an increase in current and fall in
voltage and frequency in the faulted phase. Shunt type of faults are in turn
classified as:
1)Single line to ground (LG) fault
2)Line to line (LL) fault
3)Double line to ground (LLG) fault.
When one or two lines in a thee phase system get opened while other lines
or line remain intact, such faults are called as series type of faults. They are
characterized by increase in voltage and frequency and fall in current in the faulted
phase series type of faults cab be grouped as:
1)One conductor open fault
2)Two conductor open fault
We will individually consider each of these faults in this chapter. Before that,
let us look into the typical relative frequencies of different kinds of faults in a
power system in order of decreasing severity.
Symmetrical faults (3L) -5%
Double line to ground (LLG) faults -10%
Double line (LL) faults -15%
Single line to ground (LG) fault- 70%
It can be observed that three phase faults (3L) has the maximum severity,
though its occurrence is infrequent. Hence the rapturing capacity of circuit breakers
are calculated on the basis of a three phase symmetrical fault. However, for relay
setting, single phase switching and performing the system stability studies, the
analysis of unsymmetrical faults are very important. Since any unsymmetrical fault
causes unbalanced currents to flow in the system, the method of symmetrical
components is very useful in an analysis to determine the currents and voltages in
all parts of the system after the occurrence of the fault. Also the sequence
networks of the system will come quite handy in this process. First, we shall
discuss fault at the terminals of an unloaded synchronous generator. Then, we
shall consider faults on a power system by applying Thevenin's theorem, which

S J P N Trust's Author TCP04

Hirasugar Institute of Technology, Nidasoshi-591236 Pramod M V 1.1
Tq: Hukkeri, Dt: Belgaum, Karnataka, India, Web:www.hsit.ac.in Page No. EEE
Phone:+91-8333-278887, Fax:278886, Mail:principal@hsit.ac.in
1 FEB 2013
 A.Year / Chapter Semester Subject Topic
2013 / 5 6 Power system Unsymmetrical faults
analysis and
stability
allows us to find the current in the fault by replacing the entire system by a single
generator (voltage source) and series impedance.
5.2 Fault calculations of synchronous generator.
Consider a balanced three phase synchronous generator (alternator), which
is subjected to some unsymmetrical fault F at the terminal, as shown in fig 5.1.

The fault may be unsymmetrical one. But to the left of the fault point F, the
system (alternator) is completely symmetrical. Hence, in such a system, currents
of a given sequence produce voltage drops of the same sequence only. The
sequence impedances are uncoupled. Since the generates balanced voltages only
(positive sequence voltages only), the following equations are applicable to a
synchronous generator, even during an unsymmetrical fault.
Va1=Ea-Ia1. Z1 ..............5.1
Va2=-Ia2. Z2 .................5.2
Va0=-Ia0.Z0 .................5.3
These equations cab be called the system equations. For any fault at the
terminals of synchronous generator, the quantities that are to be determined are
the three sequence currents (Ia1, Ia2, Ia0) and the three sequence terminal voltages
(Va1, Va2, Va0). Out of the six unknowns, only three quantities are linearly
independent. Hence to determine these three linearly independent quantities, three
terminal conditions are to be specified for any type of fault at the terminals of the
generator.
Before proceeding to the analysis of faults at the terminals of an unloaded
generator, it is good enough to remember that the single phase representation of
the positive sequence network of a synchronous generator consists of positive
sequence generated emf Ea1 in series with positive sequence impedance Z1(fig
4.5b). The negative sequence network consists of negative sequence impedance Z2
with no negative sequence generated voltage (fig 4.6b). The zero sequence
network consists of zero sequence impedance Z0 with no zero sequence generated
voltage (fig 4.7b).

S J P N Trust's Author TCP04

Hirasugar Institute of Technology, Nidasoshi-591236 Pramod M V 1.1
Tq: Hukkeri, Dt: Belgaum, Karnataka, India, Web:www.hsit.ac.in Page No. EEE
Phone:+91-8333-278887, Fax:278886, Mail:principal@hsit.ac.in
2 FEB 2013
 A.Year / Chapter Semester Subject Topic
2013 / 5 6 Power system Unsymmetrical faults
analysis and
stability
5.2.1 Single line to ground (LG) fault on an unloaded generator:
The circuit diagram for an LG fault on an unloaded star connected generator
with its neutral grounded through a reactance is shown in fig 5.2. Here it is
assumed that phase a is shorted to ground directly. The condition at the fault are
represented by the following terminal conditions.
Terminal conditions:
Va=0 ..................5.4
Ib=0 ....................5.5
Ic=0 ...................5.6
These three terminal conditions in terms of line currents and phase voltage
are to be transformed to conditions in terms of symmetrical components.

Symmetrical components relations:

Since two conditions are available regarding the line currents, it is convenient
to transform them to conditions in terms of symmetrical components.
Ia0=(1/3)(Ia+Ib+Ic)=(1/3)(Ia+0+0)=(1/3).Ia
Ia1=(1/3)(Ia+a.Ib+a2.Ic)=(1/3)(Ia+0+0)=(1/3).Ia
Ia2=(1/3)(Ia+a2.Ib+a.Ic)=(1/3)(Ia+0+0)=(1/3).Ia
so Ia1=Ia2=Ia0=(1/3).Ia ................................5.7
The terminal conditions Va=0 gives,
Va0+Va1+Va2=0 .............................5.8
As per eq. 5.7, all sequence currents are equal and as per eq. 5.8, the sum
of sequence voltage equals zero. Therefore, these equations suggest a series
connection of sequence networks through a short circuit as shown in fig 5.3.
Interconnection of sequence networks:

S J P N Trust's Author TCP04

Hirasugar Institute of Technology, Nidasoshi-591236 Pramod M V 1.1
Tq: Hukkeri, Dt: Belgaum, Karnataka, India, Web:www.hsit.ac.in Page No. EEE
Phone:+91-8333-278887, Fax:278886, Mail:principal@hsit.ac.in
3 FEB 2013
 A.Year / Chapter Semester Subject Topic
2013 / 5 6 Power system Unsymmetrical faults
analysis and
stability
Sequence quantities:
The following relations can be directly obtained from fig 5.3
Ia1=Ia2=Ia0=Ea / (Z1+Z2+Z0) ...................................5.9
Va1=Ea1 – Ia1.Z1= Ea – (Ea/(Z1+Z2+Z0)).Z1
=Ea((Z2+Z0) / (Z1+Z2+Z0)) .................5.10
Va2=-Ia2.Z2= -(Ea.Z2 /(Z1+Z2+Z0)) .....................5.11
Va0= -Ia0.Z0 = -(Ea.Z0 /(Z1+Z2+Z0)) .....................5.12
Fault current;
The fault current If in this case is equal to the current in phase a i.e Ia. Hence
the fault current is given as,
If=Ia=3.Ia0 .............in view of eq. 5.7
=3(Ea /(Z1+Z2+Z0)) .....................5.13
In case the neutral of the generator is not grounded, then
Z0=Zg0+3Zn=Zg0+∞=∞. Therefore, the fault current in such a condition is,
If=3(Ea /(Z1+Z2+∞)=0 .........................5.14
Thus, it can be inferred that fault current in the system is zero if the neutral is not
grounded in the case of an LG fault.

5.2.2 Line to line (L-L) fault on an unloaded generator:

The circuit diagram for an LL fault on an unloaded star connected generator
with its neutral grounded through a reactance is as shown in fig 5.4. Here it is
assumed that phase b and phase c are shorted.

Terminal conditions:
The condition at the fault are expressed by the following terminal conditions:
Ia=0 ......................5.15

S J P N Trust's Author TCP04

Hirasugar Institute of Technology, Nidasoshi-591236 Pramod M V 1.1
Tq: Hukkeri, Dt: Belgaum, Karnataka, India, Web:www.hsit.ac.in Page No. EEE
Phone:+91-8333-278887, Fax:278886, Mail:principal@hsit.ac.in
4 FEB 2013
 A.Year / Chapter Semester Subject Topic
2013 / 5 6 Power system Unsymmetrical faults
analysis and
stability
Ib+Ic=0 i.e. Ic= -Ib ..........................5.16
Vb=Vc ...........................5.17
Symmetrical components relations:
Since there are two conditions regarding current, analysing them first, we get
Ia0=(1/3)(Ia+Ib+Ic)
=(1/3)(0+Ib-Ib)
=0 ..............in view of eq. 5.16
Ia1=(1/3)(Ia+a.Ib+a2.Ic)
=(1/3)(0+a.Ib-a2.Ib)
=(1/3)(a-a2).Ib
Ia2=(1/3)(Ia+a2.Ib+a.Ic)
=(1/3)(0+a2.Ib-a.Ib)
=(1/3)(a2-a).Ib
So, we have,
Ia0=0 .....................................5.18
Ia2= -Ia1 .......................................5.19
Regarding sequence terminal voltages,
Va1=(1/3)(Va+a.Vb+a2.Vc)
=(1/3)(Va+(a+a2)Vb) ....................in view of eq. 5.17
=(1/3)(Va-Vb) .................because a+a2= -1
Va2=(1/3)(Va+a2.Vb+a.Vc)
=(1/3)(Va+(a2+a)Va) ....................in view of eq.5.17
=(1/3)(Va-Vb)
Since Ia0=0; the zero sequence terminal voltage
Va0= -Ia0.Z0= -0.Z0 =0
so, we have
Va0=0 ........................5.20
Va1=Va2 ................................5.21
Equations 5.19 and 5.21 suggest parallel connection of positive and negative
sequence networks. Since Ia0=Va0=0, the zero sequence networks is connected
separately and shorted on itself as shown in the following diagrams.
Interconnection of sequence networks:

S J P N Trust's Author TCP04

Hirasugar Institute of Technology, Nidasoshi-591236 Pramod M V 1.1
Tq: Hukkeri, Dt: Belgaum, Karnataka, India, Web:www.hsit.ac.in Page No. EEE
Phone:+91-8333-278887, Fax:278886, Mail:principal@hsit.ac.in
5 FEB 2013
 A.Year / Chapter Semester Subject Topic
2013 / 5 6 Power system Unsymmetrical faults
analysis and
stability

Sequence quantities:
The following relations can be directly obtained from fig 5.5
Ia1= -Ia2 =Ea / (Z1+Z2) .....................................5.22
Ia0=Va0=0 ........................................5.23
Va1=Va2=Ea – Ia1.Z1 = Ea (Z2/Z1+Z2)) ......................5.24
Fault current:
The fault current in this case is,
If=Ib (or Ic)
=Ia0+a2.Ia1+a.Ia2
=0+(a2-a)Ia1
= -j√3Ia1
or |If|=√3 Ia1 = √3(Ea / (Z1+Z2)) .....................5.25
In case the neutral is not grounded, then Z0=Zg0+3Zn=Zgo+∞=∞. But since
the expression for fault current is independent of the value of Z 0, the presence or
absence of a grounded neutral at the generator does not affect the fault current.

5.2.3 Double line to ground (LLG) fault on an unloaded generator:

The circuit diagram for an LLG fault on an unloaded star connected alternator
having grounded neutral is shown in fig 5.6. We assume that the fault takes place
in phases b and c.

S J P N Trust's Author TCP04

Hirasugar Institute of Technology, Nidasoshi-591236 Pramod M V 1.1
Tq: Hukkeri, Dt: Belgaum, Karnataka, India, Web:www.hsit.ac.in Page No. EEE
Phone:+91-8333-278887, Fax:278886, Mail:principal@hsit.ac.in
6 FEB 2013
 A.Year / Chapter Semester Subject Topic
2013 / 5 6 Power system Unsymmetrical faults
analysis and
stability
Terminal conditions:
The conditions at the fault are expressed by the following equations:
Vb=0 ...............................5.26
Vc=0 .................................5.27
Ia=0 ................................5.28
Symmetrical components relations:
Since there are the condition regarding the voltages analysing (transforming
to symmetrical components) then first we get.
Va0=(1/3)(Va+Vb+Vc)
=(1/3)(Va+0+0)
=(1/3).Va
Va1=(1/3)(Va+a.Vb+a2.Vc)
=(1/3)(Va+0+0)
=(1/3).Va
Va2=(1/3)(Va+a2.Vb+a.Vc)
=(1/3)(Va+0+0)
=(1/3).Va
so, Va0=Va1=Va2 ...........................................5.29
coming to sequence currents, we have,
Ia=0
i.e Ia0+Ia1+Ia2 = 0 ....................................................5.30
Equations 5.29 and 5.30 indicates that the sequence networks should be
connected in parallel as shown in fig. 5.7.
Interconnection of sequence networks:

Sequence quantities:
The following relations can be directly obtained from the fig 5.7
Va1=Va2=Va0=Ea=Ia1.Z1 .....................................5.31
Ia1=Ea / [Z1+{Z2Z0/(Z2+Z0)}] ......................... ...........5.32
S J P N Trust's Author TCP04
Hirasugar Institute of Technology, Nidasoshi-591236 Pramod M V 1.1
Tq: Hukkeri, Dt: Belgaum, Karnataka, India, Web:www.hsit.ac.in Page No. EEE
Phone:+91-8333-278887, Fax:278886, Mail:principal@hsit.ac.in
7 FEB 2013
 A.Year / Chapter Semester Subject Topic
2013 / 5 6 Power system Unsymmetrical faults
analysis and
stability
Ia2= -Ia1 .[Z0/(Z2+Z0)] ............................5.33
Ia0= -Ia1.[Z2/(Z2+Z0)] ............................5.34
Equations 5.33 and 5.34 are direct consequences of current division formula.
Fault Current:
The fault current If in this case is given by,
If=Ib+Ic
=(Ia0+a2.Ia1+a.Ia2) + (Ia0+a.Ia1+a2.Ia2)
=2Ia0+(a+a2)Ia1+(a+a2)Ia2
=2Ia0 – Ia1 – Ia2 because (a+a2)= -1
=2Ia0 -(Ia1+Ia2)
It can be observed from fig 5.7 that (Ia1+Ia2)= -Ia0.
Substituting this in the expression for fault current, we get
If=2Ia0 -(-Ia0)
=3Ia0 ....................................5.35
=-3.Ia1.[Z2/(Z2+Z0)] --------------in view of eq. 5.34
If the neutral grounding is absent, then Zn=∞.
Therefore, Z0=Zg0+3Zn=Zg0+∞=∞.
Hence,
If = -3Ia1.[Z2/(Z2+∞)]
Therefore, If=0 ....................5.36
Thus, it can be inferred that fault current in the system is Zero, if the neutral
is not grounded in the case of LLG fault.

5.3 Fault through impedances:

All the faults discussed in the preceding section consisted of direct short
circuits between line and between one or two line to ground. In these cases, the
impedance between the fault points is considered as zero. There may be situation
in which the fault path includes an impedance between the faulted points. In these
situation the analysis is carried similar to that of the previous section, except that
the fault impedance is concluded at appropriate points in the circuits obtained by
connecting sequence networks. Hence the theory is not elaborated in much detail.
5.3.1 Single line to ground (LG) fault on an unloaded generator through a fault
impedance:
The circuit diagram for an LG fault on an unloaded generator through a fault
impedance Zf is shown in fig 5.8.

S J P N Trust's Author TCP04

Hirasugar Institute of Technology, Nidasoshi-591236 Pramod M V 1.1
Tq: Hukkeri, Dt: Belgaum, Karnataka, India, Web:www.hsit.ac.in Page No. EEE
Phone:+91-8333-278887, Fax:278886, Mail:principal@hsit.ac.in
8 FEB 2013
 A.Year / Chapter Semester Subject Topic
2013 / 5 6 Power system Unsymmetrical faults
analysis and
stability

Terminal conditions:
Va=Ia. Zf ..................................5.37
Ib=0 .....................................5.38
Ic=0 ....................................5.39

Symmetrical components relations:

The following relations can be obtained from the terminal conditions
Ia0=(1/3)(Ia+Ib+Ic)
=(1/3)(Ia+0+0)
=(1/3).Ia
Ia1=(1/3)(Ia+a.Ib+a2.Ic)
=(1/3)(Ia+0+0)
=(1/3).Ia
Ia2=(1/3)(Ia+a2.Ib+a.Ic)
=(1/3)(Ia+0+0)
=(1/3).Ia
So Ia1=Ia2=Ia0=(1/3).Ia ..........................................5.40
The terminal condition Va= Ia.Zf gives
Va0+Va1+Va2=Ia.Zf=3.Ia0.Zf ..................................5.41
As per equations 5.40 and 5.41 all sequence currents are equal and the sum of
sequence voltages equals 2.Ia0.Zf. Therefore, these equations suggest a series
connection of sequence networks through an impedance 3.Zf as shown in fig 5.9.

Interconnection of sequence networks:

S J P N Trust's Author TCP04
Hirasugar Institute of Technology, Nidasoshi-591236 Pramod M V 1.1
Tq: Hukkeri, Dt: Belgaum, Karnataka, India, Web:www.hsit.ac.in Page No. EEE
Phone:+91-8333-278887, Fax:278886, Mail:principal@hsit.ac.in
9 FEB 2013
 A.Year / Chapter Semester Subject Topic
2013 / 5 6 Power system Unsymmetrical faults
analysis and
stability

Sequence quantities:
The following equations can be directly obtained from the fig 5.9
Ia0=Ia1=Ia2=Ea / (Z1+Z2+Z0+3Zf) ............................5.42
Va1=Ea-Ia1.Z1=Ea.(Z2+Z0+3Zf) /(Z1+Z2+Z0+3Zf) ...................5.43
Va2= -Ia2.Z2 = -EaZ2 / (Z1+Z2+Z0+3Zf) ............................5.44
Va0= -Ia0Z0= -EaZ0 / (Z1+Z2+Z0+3Zf) ............................5.45
Fault current:
The fault current in this case is given as,
If=Ia=3.Ia0=3 [Ea / (Z1+Z2+Z0+3Zf)] ..................................5.46
From the above expression, it can be observed that the fault current is reduced by
the fault impedance. Even in this case, if the neutral is left ungrounded, Zn=∞
i.e Z0=∞ and hence If=0.

5.3.2 Line to line (LL) fault on an unloaded generator through a fault impedance:
The circuit diagram for an LL fault on an unloaded generator through a fault
impedance Zf is shown in fig 5.10.

Terminal conditions:
S J P N Trust's Author TCP04
Hirasugar Institute of Technology, Nidasoshi-591236 Pramod M V 1.1
Tq: Hukkeri, Dt: Belgaum, Karnataka, India, Web:www.hsit.ac.in Page No. EEE
Phone:+91-8333-278887, Fax:278886, Mail:principal@hsit.ac.in
10 FEB 2013
 A.Year / Chapter Semester Subject Topic
2013 / 5 6 Power system Unsymmetrical faults
analysis and
stability
Ia=0 .................5.47
Ib+Ic=0 ; Ic=-Ib ...............5.48
Vb=Vc+Ib.Zf ........................5.49

Symmetrical components relations:

The following relations can be obtained from the terminal conditions
Ia0=(1/3)(Ia+Ib+Ic)
=(1/3)(Ia+Ib-Ib)
=0
Ia1=(1/3)(Ia+a.Ib+a2.Ic)
=(1/3)(0+a.Ib-a2.Ic)
=(1/3)(a-a2)Ib
Ia2=(1/3)(Ia+a2.Ib+a.Ic)
=(1/3)(0+a2 .Ib-a.Ic)
=(1/3)(a2-a)Ib
so, Ia0=0 .....................................5.50
Ia1= -Ia2 ....................................5.51
Next,
Va1=(1/3)(Va+a.Vb+a2.Vc)
Va2=(1/3)(Va+a2.Vb+a.Vc)
Therefore,
Va1-Va2= (1/3) [(a-a2)Vb+(a2-a)Vc]
=(1/3) [(a-a2)(Vb-Vc)]
=(1/3) (a-a2)(Ib.Zf) ------------in view of eq. 5.49
= Ia1.Zf
Thus, Va1=Va2+Ia1.Zf ................................................5.52
Since, Ia0=0, Va0= -Ia0.Z0=0 ................................5.53
Equations 5.51 and 5.52 suggest parallel connection of positive and negative
sequence networks through a series impedance Zf as shown in fig 5.11. Since
Ia0=Va0=0, the zero sequence network is connected separately through a short as
shown in fig 5.12.

Interconnection sequence networks:

S J P N Trust's Author TCP04

Hirasugar Institute of Technology, Nidasoshi-591236 Pramod M V 1.1
Tq: Hukkeri, Dt: Belgaum, Karnataka, India, Web:www.hsit.ac.in Page No. EEE
Phone:+91-8333-278887, Fax:278886, Mail:principal@hsit.ac.in
11 FEB 2013
 A.Year / Chapter Semester Subject Topic
2013 / 5 6 Power system Unsymmetrical faults
analysis and
stability

Sequence quantities:
The following equations can be directly obtained from the diagrams shown above.
Ia1= -Ia2 =Ea / (Z1+Z2+Zf) ............................5.54
Ia0=Va0=0 ................................5.55
Va1=Ea – Ia1Z1
=Ea.(Z2+Zf) /(Z1+Z2+Zf) ...................5.56
Va2= -Ia2.Z2
= - Ea.Z2 /(Z1+Z2+Zf) ...................5.57

Fault current:
In this case the fault current is equal to the current in phase-b (or phase c). Hence
If=Ib=Ia0+a2.Ia1+a.Ia2
=0+a2.Ia1 -a.Ia1
=(a2 – a)Ia1
= - j√3Ia1
or |If|=√3.Ia1
=√3.Ea / (Z1+Z2+Zf) ............................5.58

Since Z0 does not appear in the above equation, the presence or absence of a
grounded neutral does not affect the fault current.

5.3.3 Double line to ground fault (LLG) on an unloaded generator through a fault
impedance:
The circuit diagram for the aforesaid case is shown in fig 5.13

S J P N Trust's Author TCP04

Hirasugar Institute of Technology, Nidasoshi-591236 Pramod M V 1.1
Tq: Hukkeri, Dt: Belgaum, Karnataka, India, Web:www.hsit.ac.in Page No. EEE
Phone:+91-8333-278887, Fax:278886, Mail:principal@hsit.ac.in
12 FEB 2013
 A.Year / Chapter Semester Subject Topic
2013 / 5 6 Power system Unsymmetrical faults
analysis and
stability

Terminal conditions:
Ia=0 .........................................................5.59
Vb=(Ib+Ic)Zf .........................................5.60
Vc=(Ib+Ic)Zf ...........................................5.61
Symmetrical component relations:
Consider,
Va1=(1/3)(Va+a.Vb+a2.Vc)
=(1/3)[(Va+(a+a2)Vb]
=(1/3)(Va-Vb)
Va2=(1/3)(Va+a2.Vb+a.Vc)
=(1/3)[(Va+(a2+a)Vb]
=(1/3)(Va-Vb)
Va0=(1/3)(Va+Vb+Vc)
=(1/3)(Va+2.Vb)
Thus,
Va1=Va2 .................................5.62
Va0 -Va2 =(1/3).3Vb
=Vb
=(Ib+Ic)Zf ..................from eq. 5.60
=3 .Ia0. Zf --------This will be proved to the expression for fault current.
Thus, Va0=Va2+3.Ia0.Zf ................................5.63
The condition Ia=0 gives Ia0+Ia1+Ia2=0 ........................5.64

S J P N Trust's Author TCP04

Hirasugar Institute of Technology, Nidasoshi-591236 Pramod M V 1.1
Tq: Hukkeri, Dt: Belgaum, Karnataka, India, Web:www.hsit.ac.in Page No. EEE
Phone:+91-8333-278887, Fax:278886, Mail:principal@hsit.ac.in
13 FEB 2013
 A.Year / Chapter Semester Subject Topic
2013 / 5 6 Power system Unsymmetrical faults
analysis and
stability
Equations 5.62 , 5.63 and 5.64 suggest the connection of sequence network as
shown in fig 5.14.
Interconnection of sequence networks:

Ia1 = Ea / [Z1+{Z2.(3Zf+Z0)/(Z2+3Zf+Z0)}] ............................5.65

Ia2 =-Ia1. (Z0+3Zf)/(Z0+Z2+3Zf) ............................................5.66
Ia0 = -Ia1.Z2/(Z0+Z2+3Zf) ;using current division eq.)...........................5.67
Equations 5.66 and 5.67 are obtained from the current division fromula.
Fault current:
In this case, the fault current is given as,
If=Ib+Ic
=(Ia0+a2.Ia1+a.Ia2) + (Ia0+a.Ia1+a2.Ia2)
=2Ia0+(a+a2)Ia1+(a+a2)Ia2
=2Ia0 – Ia1 – Ia2 because (a+a2)= -1
=2Ia0 -(Ia1+Ia2)
It can be observed from fig 5.64 that (Ia1+Ia2)= -Ia0.
Substituting this in the expression for fault current, we get
If=2Ia0 -(-Ia0)
=3Ia0 ....................................5.68
=-3.Ia1.[Z2/(Z0+Z2+3Zf)] --------------in view of eq. 5.69
In absence of the neutral grounding, Zn=∞. That is, Z0=∞ and Hence, If=0.
Note:
1)Instead of remembering the various results for a particular fault, it is often easier
to visualize the connection of sequence networks to represent the fault and then
proceed suitably.
2)The fault current in the case of faults involving ground(LG,LLG) is given as,
If=3.|Ia0|
3)In case of LL fault, the fault current is given as If=√3.|Ia1|
S J P N Trust's Author TCP04
Hirasugar Institute of Technology, Nidasoshi-591236 Pramod M V 1.1
Tq: Hukkeri, Dt: Belgaum, Karnataka, India, Web:www.hsit.ac.in Page No. EEE
Phone:+91-8333-278887, Fax:278886, Mail:principal@hsit.ac.in
14 FEB 2013
 A.Year / Chapter Semester Subject Topic
2013 / 5 6 Power system Unsymmetrical faults
analysis and
stability
Example 5.1:
A three phase generator with an open circuit voltage of 400V is subjected to an LG
fault through a fault impedance of j2Ω. Determine the fault current if Z1=j4Ω,
Z2=j2Ω and Z0=j1Ω. Repeat the problem for LL and LLG fault.
Solution:
i)LG fault:
The interconnection of sequence networks for an LG fault is shown in fig. 5.15

In this case,
Ia1=ia2=Ia0=Ea / (Z1+Z2+Z0+3Zf)
=(400∠0° /√3) / j(4+2+1+6)
=-j17.765 A
Fault current=If=3.|Ia0|
=3(17.765)
=53.295 A
ii)LL fault:
The interconnection of sequence networks to represent an LL fault is shown if fig
5.16.

Here, Ia1=Ea / (Z1+Z2+Zf)

=(400∠0° /√3) / j(4+2+2)
= -j28.87 A
S J P N Trust's Author TCP04
Hirasugar Institute of Technology, Nidasoshi-591236 Pramod M V 1.1
Tq: Hukkeri, Dt: Belgaum, Karnataka, India, Web:www.hsit.ac.in Page No. EEE
Phone:+91-8333-278887, Fax:278886, Mail:principal@hsit.ac.in
15 FEB 2013
 A.Year / Chapter Semester Subject Topic
2013 / 5 6 Power system Unsymmetrical faults
analysis and
stability
Therefore, fault current, If=√3.|Ia1|
= √3(28.87)
=50A.
Iii) LLG fault:
The sequence networks are interconnected as shown in fig 5.17

Here, Ia1 = Ea / [Z1+{Z2.(Z0+3Zf)/(Z2+Z0+3Zf)}]

= (400∠0° /√3) / [4+{2(1+6)/(2+1+6)}]
= -j41.57 A
Therefore, using current division equation,
Ia0 = -Ia1.Z2/(Z2+Z0+3Zf)
= -j41.57 (2/ ((2+1+6))
=j9.24 A
Hence, the fault current is,
If=3.|Ia0|
=3(9.24)
=27.72 A

-------------END-------------

S J P N Trust's Author TCP04

Hirasugar Institute of Technology, Nidasoshi-591236 Pramod M V 1.1
Tq: Hukkeri, Dt: Belgaum, Karnataka, India, Web:www.hsit.ac.in Page No. EEE
Phone:+91-8333-278887, Fax:278886, Mail:principal@hsit.ac.in
16 FEB 2013
 A.Year / Chapter Semester Subject Topic
2013 / 6 6 Power system Unsymmetrical faults,Series type of faults
analysis and
stability
5.4 Unsymmetrical faults on power system:
The unsymmetrical faults on the power system are analyzed using Thevenin's
theorem. The Thevenin's equivalent of positive, negative and zero sequence
networks are obtained with respect to the fault point.
The prefault voltage at the fault point is the Thevenin's voltage of positive
sequence network. The negative and zero sequence components of prefault voltage
at the fault point is absent.
Let,
Z1= Thevenin's impedance of positive sequence network.
Z2= Thevenin's impedance of negative sequence network.
Z0= Thevenin's impedance of zero sequence network.
VTH=prefault voltage at the fault point.
=Thevenin's impedance of positive sequence network.
Thevenin's equivalent of positive, negative and zero sequence networks of
the power system with respect to the fault point will be as shown in fig 5.29, 5.30
and 5.31 respectively.

Using Using Kirchoff's law to the circuits shown below, we get

Va1=VTH-Ia1.Z1 ...........................5.70
Va2= -Ia2.Z2 ............................5.71
Va0= -Ia0.Z0 ...........................5.72
These equations are similar to that of a synchronous generator. They are
useful in the analysis of unsymmetrical faults on the power system. We shall now
consider the various types of unsymmetrical faults on a general power system.

5.4.1 Single line to ground (LG) fault:

fig 5.32 shows an LG fault at F in a power system through a fault impedance
Zf. The phases are so labeled that the fault occurs on phase a.

S J P N Trust's Author TCP04

Hirasugar Institute of Technology, Nidasoshi-591236 Pramod M V 1.1
Tq: Hukkeri, Dt: Belgaum, Karnataka, India, Web:www.hsit.ac.in Page No. EEE
Phone:+91-8333-278887, Fax:278886, Mail:principal@hsit.ac.in
1 FEB 2013
 A.Year / Chapter Semester Subject Topic
2013 / 6 6 Power system Unsymmetrical faults,Series type of faults
analysis and
stability

Terminal conditions:
Va=Ia.Zf ...............................5.73
Ib=0 .................................5.74
Ic=0 ................................5.75
Symmetrical component relations:
The following relations can be obtained from the terminal conditions.
Ia0=(1/3)(Ia+Ib+Ic)=(1/3)(Ia+0+0)=(1/3).Ia
Ia1=(1/3)(Ia+a.Ib+a2.Ic)=(1/3)(Ia+0+0)=(1/3).Ia
Ia2=(1/3)(Ia+a2.Ib+a.Ic)=(1/3)(Ia+0+0)=(1/3).Ia
so Ia1=Ia2=Ia0=(1/3).Ia ................................5.76

The terminal condition Va=Ia.Zf gives

Va0+Va1+Va2=Ia.Zf=3Ia0.Zf .....................5.77

Equations 5.76 and 5.77 suggest a series connection of sequence networks

through a impedance 3.Zf as shown in fig 5.33

Interconnection of sequence networks:

Fault current:
The fault current in this case in given as,

S J P N Trust's Author TCP04

Hirasugar Institute of Technology, Nidasoshi-591236 Pramod M V 1.1
Tq: Hukkeri, Dt: Belgaum, Karnataka, India, Web:www.hsit.ac.in Page No. EEE
Phone:+91-8333-278887, Fax:278886, Mail:principal@hsit.ac.in
2 FEB 2013
 A.Year / Chapter Semester Subject Topic
2013 / 6 6 Power system Unsymmetrical faults,Series type of faults
analysis and
stability
If=Ia=3Ia0= 3VTH / (Z1+Z2+Z0+3Zf) ..............................5.78
Note:
In the absence of fault impedance, replace Zf by zero in the above calculations.

5.4.2 Line to line (LL) fault:

Fig 5.34 shows a LL fault at F in a power system on phase b and c through a
fault impedance Zf

Terminal conditions:
Ia=0 .................5.79
Ib+Ic=0 ; Ic=-Ib ...............5.80
Vb=Vc+Ib.Zf ........................5.81
Symmetrical components relations:
The following relations can be obtained from the terminal conditions
Ia0=(1/3)(Ia+Ib+Ic)
=(1/3)(Ia+Ib-Ib)
=0
Ia1=(1/3)(Ia+a.Ib+a2.Ic)
=(1/3)(0+a.Ib-a2.Ic)
=(1/3)(a-a2)Ib
Ia2=(1/3)(Ia+a2.Ib+a.Ic)
=(1/3)(0+a2 .Ib-a.Ic)
=(1/3)(a2-a)Ib
so, Ia0=0 .....................................5.82
Ia1= -Ia2 ....................................5.83
Next,

S J P N Trust's Author TCP04

Hirasugar Institute of Technology, Nidasoshi-591236 Pramod M V 1.1
Tq: Hukkeri, Dt: Belgaum, Karnataka, India, Web:www.hsit.ac.in Page No. EEE
Phone:+91-8333-278887, Fax:278886, Mail:principal@hsit.ac.in
3 FEB 2013
 A.Year / Chapter Semester Subject Topic
2013 / 6 6 Power system Unsymmetrical faults,Series type of faults
analysis and
stability
Va1=(1/3)(Va+a.Vb+a2.Vc)
Va2=(1/3)(Va+a2.Vb+a.Vc)
Therefore,
Va1-Va2= (1/3) [(a-a2)Vb+(a2-a)Vc]
=(1/3) [(a-a2)(Vb-Vc)]
=(1/3) (a-a2)(Ib.Zf)
= Ia1.Zf
Thus, Va1=Va2+Ia1.Zf ................................................5.84
Since, Ia0=0, Va0= -Ia0.Z0=0 ................................5.85
Equations 5.83 and 5.85 suggest parallel connection of positive and negative
sequence networks through a series impedance Zf as shown in fig 5.35. Since
Ia0=Va0=0, the zero sequence network is connected separately and a shorted as
shown in fig 5.36.
Interconnection sequence networks:

Fault current:
If=Ib=Ia0+a2.Ia1+a.Ia2
=0+a2.Ia1 -a.Ia1
=(a2 – a)Ia1
= - j√3Ia1
or |If|=√3.Ia1
=√3.VTH / (Z1+Z2+Zf) ............................5.86

Note:

In the absence of fault impedance, replace Zf by zero in the above calculations.

5.4.3 Double line to ground fault (LLG):

S J P N Trust's Author TCP04

Hirasugar Institute of Technology, Nidasoshi-591236 Pramod M V 1.1
Tq: Hukkeri, Dt: Belgaum, Karnataka, India, Web:www.hsit.ac.in Page No. EEE
Phone:+91-8333-278887, Fax:278886, Mail:principal@hsit.ac.in
4 FEB 2013
 A.Year / Chapter Semester Subject Topic
2013 / 6 6 Power system Unsymmetrical faults,Series type of faults
analysis and
stability
Fig 5.37 shows an LLG fault at F in a power system. The fault may in general
have an impedance Zf as shown.

Terminal conditions:
Ia=0 .........................................................5.87
Vb=(Ib+Ic)Zf .........................................5.88
Vc=(Ib+Ic)Zf ...........................................5.89
Symmetrical component relations:
Consider,
Va1=(1/3)(Va+a.Vb+a2.Vc)
=(1/3)[(Va+(a+a2)Vb]
=(1/3)(Va-Vb)
Va2=(1/3)(Va+a2.Vb+a.Vc)
=(1/3)[(Va+(a2+a)Vb]
=(1/3)(Va-Vb)
Va0=(1/3)(Va+Vb+Vc)
=(1/3)(Va+2.Vb)
Thus,
Va1=Va2 .................................5.90
Va0 -Va2 =(1/3).3Vb
=Vb
=(Ib+Ic)Zf
=3 .Ia0. Zf --------This will be proved to the expression for fault current.
Thus, Va0=Va2+3.Ia0.Zf ................................5.91
The condition Ia=0 gives Ia0+Ia1+Ia2=0 ........................5.92
S J P N Trust's Author TCP04
Hirasugar Institute of Technology, Nidasoshi-591236 Pramod M V 1.1
Tq: Hukkeri, Dt: Belgaum, Karnataka, India, Web:www.hsit.ac.in Page No. EEE
Phone:+91-8333-278887, Fax:278886, Mail:principal@hsit.ac.in
5 FEB 2013
 A.Year / Chapter Semester Subject Topic
2013 / 6 6 Power system Unsymmetrical faults,Series type of faults
analysis and
stability
From equations 5.90 , 5.91 and 5.92 we can draw connection of sequence
networks as shown in fig 5.38
Interconnection of sequence networks:

Fault current:
In this case, the fault current is given as,
If=Ib+Ic
=(Ia0+a2.Ia1+a.Ia2) + (Ia0+a.Ia1+a2.Ia2)
=2Ia0+(a+a2)Ia1+(a+a2)Ia2
=2Ia0 – Ia1 – Ia2 because (a+a2)= -1
=2Ia0 -(Ia1+Ia2)
It can be observed that (Ia1+Ia2)= -Ia0.
Substituting this in the expression for fault current, we get
If=2Ia0 -(-Ia0)
=3Ia0
=-3.Ia1.[Z2/(Z0+Z2+3Zf)] .................................5.93
Note:
In the absence of fault impedance, Zf is replaced by zero in the calculations.

The steps briefed below are used in solving the following problems:
1)The positive, negative and zero sequence networks for the given system are
drawn.
2)The Thevenin's equivalent circuit of each of the networks with respect to the
fault point is calculated.
3)These networks are interconnected suitably to simulate the particular type of
fault condition.

S J P N Trust's Author TCP04

Hirasugar Institute of Technology, Nidasoshi-591236 Pramod M V 1.1
Tq: Hukkeri, Dt: Belgaum, Karnataka, India, Web:www.hsit.ac.in Page No. EEE
Phone:+91-8333-278887, Fax:278886, Mail:principal@hsit.ac.in
6 FEB 2013
 A.Year / Chapter Semester Subject Topic
2013 / 6 6 Power system Unsymmetrical faults,Series type of faults
analysis and
stability
Example 3.8:
A synchronous motor is receiving 10MW of power at 0.8pf lag at 6kV. An LG fault
takes place at the middle point of the transmission line as shown in fig 5.39. Find
the fault current. The ratings of the generator motor and transformer are as under.
Generator: 20MVA, 11kV, X1=0.2p.u; X2=0.1p.u; X0=0.1p.u.
Transformer T1: 18MVA, 11.5Y-34.5Y kV, X=0.1p.u.
Transmission line: X1=X2=5Ω; X0=10Ω.
Transformer T2: 15MVA, 6.9Y-34.5Y kV, X=0.1p.u.
Motor : 15MVA, 6.9kV, X1=0.2p.u; X2=X0=0.1p.u.

Solution:
base values:
Let we choose,
base MVA=20
base kV on the generator=11
we calculate,
base kV on the transmission line=11(34.5/11.5)=33
base kV on the motor= 33(6.9/34.5)=6.6

Sequence reactances of generator:

X1 = X1,p.u, old × ( (MVA)B, new / (MVA)B, old ) × ( (kV)2B, old / (kV)2B, new )
= j0.2 × (20 / 20 ) × (112 / 112)
= j 0.2 p.u
X2 = X2,p.u, old × ( (MVA)B, new / (MVA)B, old ) × ( (kV)2B, old / (kV)2B, new )
= j0.1 × (20 / 20 ) × (112 / 112)
= j 0.1 p.u
X0 = X0,p.u, old × ( (MVA)B, new / (MVA)B, old ) × ( (kV)2B, old / (kV)2B, new )
= j0.1 × (20 / 20 ) × (112 / 112)
= j 0.1 p.u

Sequence reactances of transformer T1: (calculated primary side of it)

X1 =X2 =X0 = Xp.u, old × ( (MVA)B, new / (MVA)B, old ) × ( (kV)2B, old / (kV)2B, new )
= j0.1 × (20 / 18 ) × (11.52 / 112)
= j 0.12 p.u
Sequence reactances of transmission line:
X1TL=X2TL= XTL( Ω ) ×(MVA)B / (kV)2B
= 5 × 20 / 332
= 0.092p.u
X0TL= 10 × 20 / 302
=0.184p.u
S J P N Trust's Author TCP04
Hirasugar Institute of Technology, Nidasoshi-591236 Pramod M V 1.1
Tq: Hukkeri, Dt: Belgaum, Karnataka, India, Web:www.hsit.ac.in Page No. EEE
Phone:+91-8333-278887, Fax:278886, Mail:principal@hsit.ac.in
7 FEB 2013
 A.Year / Chapter Semester Subject Topic
2013 / 6 6 Power system Unsymmetrical faults,Series type of faults
analysis and
stability
Sequence reactances of transformer T2: (calculated secondary side of it)
X1 =X2 =X0 = Xp.u, old × ( (MVA)B, new / (MVA)B, old ) × ( (kV)2B, old / (kV)2B, new )
= j0.1 × (20 / 15 ) × (6.92 / 6.62)
= j 0.146 p.u
Sequence reactances of motor:
X1 = X1,p.u, old × ( (MVA)B, new / (MVA)B, old ) × ( (kV)2B, old / (kV)2B, new )
= j0.2 × (20 / 15 ) × (6.92 / 6.92)
= j 0.29 p.u
X2 =X0 = X2,p.u, old × ( (MVA)B, new / (MVA)B, old ) × ( (kV)2B, old / (kV)2B, new )
= j0.1 × (20 / 15 ) × (6.92 / 6.92)
= j 0.145 p.u

Positive sequence Network (PSN):

Using the calculated values of positive sequence impedances, the PSN is drawn as
shown in fig 5.40.

To find the voltage at the fault point (VTH):

The current drawn by the motor Im= ((10×106) / (√3×6×103×0.8)) ∠-cos-10.8
=1202.8∠-36.87° A
The base current in the motor (Im)B= (1000×20) / (√3×6.6)
=1749.55 A
Therefore,
Im in p.u =Im/(Im)B= (1202.8/ 1749.55) ∠-36.87°
= 0.687∠-36.87° p.u
Vm in p.u =6/6.6=0.909∠0° p.u
VTH=Vm+Im. Zfm
=0.909+((0.687∠-36.87°)(0.182∠90°)),where Zfm=j0.046+j0.146=0.182∠90°
=0.909+0.132∠53.13°
=0.909+0.0792+j0.106
=0.9882+j0.106
=0.994∠6.1°

To find the Thevenin's impedance Z1TH

The Thevenin's impedance as seen from point F is,
Z1TH=j[(0.2+0.12+0.046)||(0.046+0.146+0.29)]
=j(0.366||0.482)
=j0.208 p.u

S J P N Trust's Author TCP04

Hirasugar Institute of Technology, Nidasoshi-591236 Pramod M V 1.1
Tq: Hukkeri, Dt: Belgaum, Karnataka, India, Web:www.hsit.ac.in Page No. EEE
Phone:+91-8333-278887, Fax:278886, Mail:principal@hsit.ac.in
8 FEB 2013
 A.Year / Chapter Semester Subject Topic
2013 / 6 6 Power system Unsymmetrical faults,Series type of faults
analysis and
stability

Hence the equivalent PSN of the system is as shown below:

Negative sequence Network (NSN):

The Thevenin's equivalent impedance with respect to the fault point is:
Z2TH=j[(0.1+0.12+0.092)||(0.046+0.146+0.145)]
=j(0.266||0.337)
=j0.149 p.u

Zero sequence network (ZSN):

The Thevenin's zero sequence impedance is,

Z0TH=j[(0.1+0.12+0.092)||(0.092+0.146+0.145)]
=j(0.312||0.383)
=j0.172 p.u

Interconnection of sequence networks:

The sequence networks are connected as shown in fig 5.44 to represent LG fault.

S J P N Trust's Author TCP04

Hirasugar Institute of Technology, Nidasoshi-591236 Pramod M V 1.1
Tq: Hukkeri, Dt: Belgaum, Karnataka, India, Web:www.hsit.ac.in Page No. EEE
Phone:+91-8333-278887, Fax:278886, Mail:principal@hsit.ac.in
9 FEB 2013
 A.Year / Chapter Semester Subject Topic
2013 / 6 6 Power system Unsymmetrical faults,Series type of faults
analysis and
stability

Here,
Ia1=Ia2=Ia0= (0.994∠6.1°) / j(0.208+0.149+0.172)
=1.88∠-83.9° p.u

Hence fault current:

|If|=3.|Ia0|
=3(1.88)
=5.64 p.u
Fault current in amperes is:
=|If|p.u × (ITL)B
=5.64×((1000×20) / (√3×33))
=1973.49 A

5.5 Series type of faults:

We have so far discussed the various shunt type of faults that occur in a
power system. But unsymmetrical faults in the form of open conductors (series
type) also do take place in power system. It is required to determine the sequence
components of line currents and the voltages across the broken ends of the
conductors.
Fig 5.56 shows a system wherein an open conductor fault takes place.

The ends of the system on the sides of the fault are identified as F, F', while
the conductor ends are denoted by aa', bb' and cc'. The voltage across the
S J P N Trust's Author TCP04
Hirasugar Institute of Technology, Nidasoshi-591236 Pramod M V 1.1
Tq: Hukkeri, Dt: Belgaum, Karnataka, India, Web:www.hsit.ac.in Page No. EEE
Phone:+91-8333-278887, Fax:278886, Mail:principal@hsit.ac.in
10 FEB 2013
 A.Year / Chapter Semester Subject Topic
2013 / 6 6 Power system Unsymmetrical faults,Series type of faults
analysis and
stability
conductors are denoted by Vaa', Vbb' and Vcc'. The symmetrical components of these
voltages are (Vaa')1, (Vaa')2,(Vaa')0. The sequence networks as seen from the two
ends FF' of the system are schematically shown in fig 5.57.
These are suitably interconnected depending on the type of fault (one or two
conductors open).

i) One conductor open fault:

Let us assume that the conductor 'a' of a system gets opened as shown in fig
5.58.

Terminal conditions:
As seen from the two ends F and F'. The terminal conditions that are applicable
are:
Ia=0 .............................5.94
Vbb'=0 .............................5.95
Vcc'=0 ..............................5.96

Symmetrical components relations:

(Vaa')1=(1/3)(Vaa'+a.Vbb'+a2.Vcc')
=(1/3)(Vaa'+0+0)
=(1/3)(Vaa')

S J P N Trust's Author TCP04

Hirasugar Institute of Technology, Nidasoshi-591236 Pramod M V 1.1
Tq: Hukkeri, Dt: Belgaum, Karnataka, India, Web:www.hsit.ac.in Page No. EEE
Phone:+91-8333-278887, Fax:278886, Mail:principal@hsit.ac.in
11 FEB 2013
 A.Year / Chapter Semester Subject Topic
2013 / 6 6 Power system Unsymmetrical faults,Series type of faults
analysis and
stability
(Vaa')2=(1/3)(Vaa'+a2.Vbb'+a.Vcc')
=(1/3)(Vaa'+0+0)
=(1/3)(Vaa')
(Vaa')0=(1/3)(Vaa'+Vbb'+Vcc')
=(1/3)(Vaa'+0+0)
=(1/3)(Vaa')
Thus,
(Vaa')1=(Vaa')2=(Vaa')0=(1/3)Vaa' ..........................................5.97
The condition Ia=0 gives the result
Ia0+Ia1+Ia0=0 ...........................................5.98
Condition 5.97 and 5.98 in terms of symmetrical components are similar to a
double line to ground fault. These suggest that the three sequence networks
should be connected in parallel to represent the fault, as shown in fig 5.59.

ii)Two conductors open fault:

Let us assume that the two conductors b and c get open at the points F, F' as
shown in fig 5.60.

Terminal conditions:
As seen from the points F and F', the terminal conditions that are applicable to this
fault are:
Ib=0 ......................5.99
Ic=0 ......................5.100
Vaa'=0 .......................5.101

S J P N Trust's Author TCP04

Hirasugar Institute of Technology, Nidasoshi-591236 Pramod M V 1.1
Tq: Hukkeri, Dt: Belgaum, Karnataka, India, Web:www.hsit.ac.in Page No. EEE
Phone:+91-8333-278887, Fax:278886, Mail:principal@hsit.ac.in
12 FEB 2013
 A.Year / Chapter Semester Subject Topic
2013 / 6 6 Power system Unsymmetrical faults,Series type of faults
analysis and
stability
Symmetrical components relations:
consider,
Ia0=(1/3)(Ia+Ib+Ic)=(1/3)(Ia+0+0)=(1/3).Ia
Ia1=(1/3)(Ia+a.Ib+a2.Ic)=(1/3)(Ia+0+0)=(1/3).Ia
Ia2=(1/3)(Ia+a2.Ib+a.Ic)=(1/3)(Ia+0+0)=(1/3).Ia
so Ia1=Ia2=Ia0=(1/3).Ia ................................5.102
The terminal conditions Vaa'=0 gives the result,
(Vaa')0+(Vaa')1+(Vaa')2=0 ..........................................5.103
These conditions are similar to those of line to ground fault and suggest that
the three sequence networks be connected in series and shorted as shown in
fig5.61.

-----------END------------

S J P N Trust's Author TCP04

Hirasugar Institute of Technology, Nidasoshi-591236 Pramod M V 1.1
Tq: Hukkeri, Dt: Belgaum, Karnataka, India, Web:www.hsit.ac.in Page No. EEE
Phone:+91-8333-278887, Fax:278886, Mail:principal@hsit.ac.in
13 FEB 2013