1. Why do faults occur in a power system?

The faults occur in a power system due to insulation failure of equipment or flashover
of lines initiated by a lightning stroke or through accidental faulty operation. When a
line, which has been made safe for maintenance by clamping all the three phases to earth,
is accidentally made alive or when due to slow fault clearance, an earth fault spreads
across to the other two phases.
2. What are all the assumptions to be made to simplify the short circuit study?
a. Representing each machine by constant voltage source behind proper reactances.
b. Prefault load currents are neglected
c. Transformer taps are assumed to be nominal.
d. Shunt elements in the transformer model that accounting for magnetizing current and
core losses are neglected.
e. Shunt capacitance of the transmission line is ignored.
f. Series resistance of transmission lines is neglected.
3. Define short circuit capacity.
The short circuit capacity at a bus is defined as the product of the magnitudes of the
prefault bus voltages and post fault current.
Short Circuit MVA(3φ)=| | *| | *
4. What is bolted or solid fault?
A fault represents a structural network change equivalent with that caused by the addition
of impedance at the place of fault. If the fault impedance is zero, then the fault is referred
as bolted fault or solid fault.
5. For a fault at a given location, rank the various faults in the order of severity.
Types of faults Relative frequency of occurrence
Three phase fault 5%
Double line to ground fault 10%
Line to line fault 15%
Single line to ground fault 70%
6. Express the unbalanced voltage in terms of symmetrical components.

[ ] =[ ][ ]

7. Draw the zero-sequence network of Υ/Δ transformer with neutral ungrounded.

8. Why the neutral grounding impedance Zn appears as 3Zn in zero sequence
equivalent circuit?
= = = + + =3
= - ( +3 )
Therefore becomes 3 for the zero sequence network.
9. What is sequence networks?
The single phase equivalent circuit of a power system consists of impedances to
current of any one sequence only is called sequence network
10. Write the features of zero sequence current?
 Zero sequence current flows to maintain three phase symmetry during unbalanced
conditions.
 Zero current shows there is a path for ground.

11. a. A synchronous generator and motor each rated 30MVA, 13.2kV and both have
subtransient reactance of 20% and the line reactance of 12% on a base of machine
ratings. The motor is drawing 25MW at 0.85p.f leading. The terminal voltage is 12kV
when a three phase short circuit fault occurs at motor terminals. Find the subtransient
current in generator, motor and fault (13)
Solution:

Terminal voltage=12kV

Prefault voltage = = = 0.909〈

Motor is drawing 25MW at 0.85p.f leading

Load in p.u= = = 0.98 p.u

Prefault current, = = 1.07〈

 = j0.123 p.u

Fault current = = -j7.39 p.u

Base current= = 1.312kA

√

Actual fault current= 7.395*1.312 = 11.64kA

= -j2.842 p.u = -j4.5479 p.u
= = 0.909 – j2.278 p.u = 2.452〈 p.u
= = 0.909 – j5.11 p.u =5.19〈 p.u
= 1.5746k*2.452 =3.86kA
= 1.5746k*5.19 = 8.172kA

11. Describe the bus impedance matrix method based fault current calculation with neat
algorithm and flow chart. (13)
Step-I: Draw the prefault per phase network(Positive sequence network)
• Each Machine represented by a constant voltage source behind proper reactance.
• Transmission line reactance are represented in p.u on a common base MVA
• Assume 3-φ fault at bus q with Zf.

Prefault Bus Voltages

It is obtained from power flow solution

Initial bus voltages Vbus0 =

[ ]
Step 2: Obtain Zbus matrix using bus building algorithm
Assume one node as reference node and s.c all the voltage sources

Step 3:
The prefault currents are negligible

Fault current= Zqq =diagonal element

Step 4: Obtain the Thevenin’s networkby inserting the Thevenin’s voltage source Vq0 in series with
Zf and calculate change in bus voltage.

The current entering in

every bus is zero except at
fault bus.
I1=I2=……….=IN=0 except Iq
Iq =-If

Ibus= =

[ ] [ ]
Ibus(F)=Ybus.ΔVbus solving, ΔVbus =Zbus .I bus(F)
Change in bus voltage
 = = =

[ ] [ ] [ ]

Step 5:
Post fault bus voltages are obtained by superposition of the prefault bus voltages and the change in the
bus voltages.
Vbus(f)=Vbus0+ ΔVbus

= +

[ ]
[ ] [ ] [ ]

= - = - …… = - …… = -

𝑓
In General,
𝑉𝑖 = 𝑉𝑖 -𝑍𝑖𝑞 𝐼𝑓

Bus Voltages during the fault,

= - i q = i=q

If the short circuit is solid or bolted Fault,

Fault current If =

=0

= - ; i q

Step 6: Post Fault Line Currents. Post fault line current =

 Start

Read the line data, bus data, fault bus, sub-transient reactance of each machine

Assume prefault load currents, shunt elements in transformer, transformer taps, shunt
capacitance, series capacitance, series resistance of lines lines are neglected

Draw the prefault per phase network obtain Zbus matrix using bus building
algorithm

Obtain prefault bus voltages from power flow

𝑉
𝑉
equationVbus0 =
𝑉𝑞

[𝑉𝑁 ]

Draw the thevenin’s equivalent circuit and obtain If =

𝑉𝑞
𝑍𝑞𝑞 𝑍𝑓

Compute change in bus voltages using network equation

𝛥𝑉 𝑍 𝑍 𝑍𝑞 𝑍𝑁
𝛥𝑉 𝑍 𝑍 𝑍𝑞 𝑧𝑁

𝛥𝑉𝑞 = 𝑍 𝑍𝑞 𝑍𝑞𝑛 = 𝐼𝑓
𝑞 𝑍𝑞𝑞

[𝛥𝑉𝑁 ] [𝑍𝑁 𝑍𝑁 𝑍𝑁𝑄 𝑍𝑁𝑁 ] [ ]

Compute change in bus voltages using network equation,

𝑓
𝑉 = 𝑉 -𝑍 𝑞 𝐼𝑓
𝑓
𝑉𝑞 = 𝑉𝑞 -𝑍𝑞𝑞 𝐼𝑓
𝑓
𝑉𝑁 = 𝑉𝑁 -𝑍𝑁𝑞 𝐼𝑓

Post fault line currents

𝑓 𝑓
𝑓 𝑉 𝑉𝑗
𝐼𝑖𝑗 = 𝑍 𝑖
𝑖𝑗𝑠𝑒𝑟𝑖𝑒𝑠

Print If, Post fault voltages, post fault line currents, etc.,
12.a) a. A synchronous generator and motor each rating 30MVA, 11kV having 20%
subtransient reactance are connected through transformer and a line as shown in fig.
The transformers are rated 30MVA, 11/66kV and 66/11kV with leakage reactance of
10% each. The line has a reactance of 10% on a base of 30MVA, 66kV. The motor is
drawing 20MW at 0.8p.f leading and a terminal voltage of 10.65kV when a symmetrical
three phase fault occurs at the terminals of motor. Find the subtransient current in the
motor and Generator. (13)

Terminal voltage = 10.65kV

Prefault voltage = = = 0.9636〈

Motor is drawing 20 MW at 0.8p.f leading

Load in p.u= = = 0.833 p.u

Prefault current, = = 0.865〈

= j0.1428 p.u

Fault current = = -j6.745 p.u

Base current = = 1.5746kA

√

Actual fault current = 6.745*1.5746 = 10.62KA

= - j1.9272 p.u = - j4.818 p.u

= 0.692 – j1.408 p.u

= 0.692 – j5.337 p.u

12. b. Using building algorithm method, determine ZBus for the network shown in
figure, where the impedances labeled are shown in per unit. (13)

Solution:

J0.1 and j0.1 are connected in series

Z=j0.2

Step1: add an element j0.25 between ref node and new node 1.

=[ ]

Step2: Add an element j0.2 between existing node (1) and Node (2)

[ ]

Step3: Add an element j0.25 between existing node (2) and ref. node

=[ ]

Fictitious node (3) can be eliminated by using,

= - n=3, i=1,2,3; j=1,2,3

 [ ]

Step 4: Add an element j0.2 between the existing nodes (1) and (2).

=[ ]

Fictitious node (3) can be eliminated by using,

= - n=3, i=1,2,3; j=1,2,3

[ ]

13.a. Generator G1 and G2 are identical and rated as 11kV, 30MVA and have a
transient reactance of 0.3 p.u. at own MVA base. The transformers T1 and T2 are also
identical and are rated 11/66kV, 10MVA and have a reactance of 0.075p.u to their own
MVA base. The line is 60km long, each conductor has a reactance of 0.92Ω/km. the
three phase fault is assume at point F, which is 25km from G1. Find the short circuit
current. (13)

Solution:

Choose common base MVA as 30MVA,The reactance of various elements are,

Generator 1:

= *( )*( )

=j0.3 ( ) ( ) =j0.3 p.u

Transformer 1:

= *( )*( )

=j0.075 ( ) ( ) =j0.225p.u
Transmission Line:

Xp.u= *MVAb
 *30 = j0.1584 p.u

Transformer 2:

=j0.075 ( ) ( ) =j0.225p.u

Generator 2:

=j0.3 ( ) ( ) =j0.3p.u

〈
Fault current = = = –j2.8023 p.u =2.8023〈

Base current= = 262.43 A

√

Actual fault current= 735.413Amps

13.b. It is Proposed to conduct fault analysis on two alternative configurations of the 4-

bus system. A Symmetrical fault occurs at bus 4 for the system shown in figure.
Determine the fault current using Thevenin equivalent circuit. (13)

G1,G2 : 100 MVA, 20kV, X+=15%

Transformer : 100MVA, 20/345kV, Xleakage=8%
L1,L2 : X+=15% on a base of 100MVA
(13)

Solution:

Reactance diagram
Thevenin’s Reactance:

Thevenin’s Equivalent circuit:

〈
Fault current = = = –j9.33 p.u =9.33〈

Base current= = 2.88kA

√

Actual fault current= 26.93kAmps

14.a) a. A salient pole generator without dampers is rated 25 MVA, 13.2 kV and has a
direct axis subtransient reactance of 0.25 p.u. The negative and zero sequence
reactances are 0.35 and 0.1 p.u. respectively. The neutral of the generator is solidly
grounded. Determine the subtransient current in the generator and the line to line
voltage for subtransient conditions when a single line to ground fault occurs at the
terminals of an unloaded generator. (13)
Solution:
= j0.25p.u = j0.35p.u = j0.1p.u
Prefault voltage = 1〈
Symmetrical components of fault current,
= = = = =-j1.4286 p.u

Fault current in p.u,

= =3 =-j4.2857p.u

Base current= = 1093.46 Amps

√

Fault current = 4686.28Amps

Subtransient Current,

[ ] =[ ][ ]

Ia= Ia0+ Ia++ Ia- = 0 –j4.2858p.u Ic = Ia0+ aIa++a2 Ia = 0 P.u Ib= Ia0+ a2Ia++a Ia- = 0p.u
Symmetrical component of voltages,
=- -0.1429 p.u
= - = 0.6429 p.u
= - = -0.5 p.u

[ ] =[ ][ ]

Va= Va0+ Va++ Va- = 0

Vb= Va0+ a2Va++a Va- -0.2144 – j0.9898
Vc= Va0+ aVa++a2 Va- =-0.2144 + j0.9898
Line - Line Voltages,
Vab = Va - Vb = 0.2144 + j0.9898
Vbc = Vb - Vc = -j1.9796 p.u
Vca = Vc - Va = - 0.2144 – j0.9898
14. b. Derive the expression for fault current in line to line fault on unloaded generator.
Draw an equivalent network showing the interconnection of networks to simulate
double line to ground fault. (13)

=- =0 - =
Symmetrical components of currents are,

[ ] = [ ][ ]

[ ] = [ ][ ]

= [ ]=0

Ia+= [ ]

= [ ]

Ia+ = - and =0
From sequence networks of generator, the symmetrical voltages are,

[ ]=[ ]-[ ][ ]

=- =- =0

= -

=- =

The phase currents are given by,

 [ ] =[ ][ ] =[ ][ ]

= -a = ( – a)
=a - = ( )
=- =0
The voltages through the zero sequence network must be zero since there are no zero
sequence sources and because = 0, current is not being injected into that network due to
the fault. Hence LL fault calculation do not involve zero sequence network.
Phase voltages are ,

[ ] =[ ][ ]

=0 = +
= +
= +

From the condition,

- =
+ - - =
( - a) - ( – a) =
– a[ ]= = ( – a)
[ ]=
- - =
 =

The phase currents are given by,

[ ] =[ ][ ] =[ ][ ]

=- = ( – a) =-j1.732
√
= =

Direct Short Circuit or Bolted Fault

Fault impedance =0

The conditions of the fault at bus,

=- =0 =
Ia+ = -
√
=

15.a. Determine the fault current in p.u., current in phase domain for a double line to
ground fault occurs between phases b and c at bus 4. For the system given below.

G1,G2 : 100 MVA, 11kV, X+= X-=15%, X0=5% Xn=6%

Transformer : 100MVA, 20/220kV, Xleakage=8%
L1, L2 : X+= X-=15%, X0=10% on a base of 100MVA. (13)
Step1: Positive sequence thevenin equivalent viewed from bus 4:

Step2: Negative sequence thevenin equivalent viewed from bus 4:

Step3: Zero sequence thevenin equivalent viewed from bus 4:

Step4:Sequence Network
Step5:
Prefault voltage = 1〈
= = -j5.5322

=- * = j3.77 p.u

=- * = j1.7621 p.u
Fault current = 3 = j5.2863 p.u
Current in phase domain:
Ia= Ia0+ Ia++ Ia- = 0 p.u
Ib= Ia0+ a2Ia++a Ia- = -8.0557 + j2.6431 p.u
Ic = Ia0+ aIa++a2 Ia- = 8.0557 + j2.6431p.u
IN = Ia+ Ib+ IC = j5.286 p.u

15.b The currents flowing in the lines towards a balanced load connected in Δ are
=100〈 . =141.4〈 , and =100〈 . Find the symmetrical components of the
given line currents and draw phasor diagram of positive and negative sequence
currents. (13)
Solution:
Symmetrical components of currents,

[ ] =[ ] [ ]

= [ ] = 0.007〈

Ia+= [ ] = 111.528〈

= [ ] =29.88〈

Ib0 = Ia0 = 0.007〈

Ib+ = a2 Ia+ = 111.528〈
Ib-= a Ia- = 29.88〈
IC0 = Ia0 = 0.007〈
Ic+= a Ia+ = 111.528〈
Ic-= a2 Ia- = 29.88〈
16. a (i) Derive the necessary equation to determine the fault current for a single line to
ground fault. Draw a diagram showing the inter-connection of sequence networks.
(8)
Single line to ground fault:

A single line ground fault occurs on phase ‘a’ connected to ground through impedance Zf

Initially the generator is at no load.

= = =0 =
Symmetrical components of currents are,

[ ] = [ ][ ]

Put = =0

[ ] = [ ][ ]

= = =

= = = =

From sequence networks of generator, the symmetrical voltages are,

[ ]=[ ]-[ ][ ]

=- =-

= -

=- = -
Phase voltages are ,

[ ] =[ ][ ]

Va= Va0+Va++ Va =

- + - +(- )= =

= =3 =

16. a. ii) A 30MVA, 11kV generator has Z1=Z2=j0.2p.u. Z0=j0.05p.u. A line to ground
fault occurs on the generator terminals. Find the fault current 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.
Solution:
= j0.2p.u = j0.2p.u = j0.05p.u
Prefault voltage = 1〈
Symmetrical components of fault current,
= = = = =-j2.22 p.u

Fault current in p.u,

= =3 =-j6.66p.u

Base current= = 1.57kAmps

√

Fault current = 10.46kAmps

Subtransient Current,
 [ ] =[ ][ ]

Ia= Ia0+ Ia++ Ia- = -j2.22p.u Ic = Ia0+ aIa++a2 Ia = 0 p.u Ib= Ia0+ a2Ia++a Ia- = 0p.u
Symmetrical component of voltages,
=- = 0.111 p.u
= - = 0.556 p.u
= - = 0.444 p.u

[ ] =[ ][ ]

Va= Va0+ Va++ Va- = 0

Vb= Va0+ a2Va++a Va- = -0.389 – j0.096
Vc= Va0+ aVa++a2 Va- = -0.389 + j0.096
Line - Line Voltages,
Vab = Va - Vb = 0.389+j0.096 p.u
Vbc = Vb - Vc = j0.192 p.u
Vca = Vc - Va = - 0.389+j0.096 p.u

16.b. A 25MVA, 13.2kV alternator with solidly grounded neutral has a sub transient
reactance of 0.25p.u. The negative and zero sequence reactances are 0.35 and 0.1p.u
respectively. If a double line-to-ground fault occurs at the terminals at the terminals of
the alternator, determine the fault current and line-to line voltages at the fault.
(15)

Solution:
Prefault voltage = 1〈

= -j3.0508
 =- * = j0.678p.u

=- * = j2.373 p.u
Fault current =3 = j7.119 p.u
Base current = = 1093.46kAmps
√
Fault current = 7.784kAmps
Symmetrical component of voltages:

=- = 0.2373p.u
= - = 0.2373 p.u
= - = 0.2373 p.u

[ ] =[ ][ ]

Va= Va0+ Va++ Va- = 0.7119 p.u

Vb= Va0+ a2Va++a Va- = 0 p.u
Vc= Va0+ aVa++a2 Va- = 0p.u
Line - Line Voltages,
Vab = Va - Vb = 0.7119 p.u
Vbc = Vb - Vc = 0 p.u
Vca = Vc - Va = - 0.7119 p.u