EE6501 Power system Analysis Department of EEE 2016-2017

UNIT – I INTRODUCTION
PART – A
1. What is the advantage of per unit method over percentage method?
The per unit method has an advantage over the percent method because the product of two quantities
expressed in per unit is expressed in per unit itself, but the product of two quantities expressed in
percent must be divided by 100 to obtain result in percent.
2. What is the need of base values?
The components or various sections of power system may operate of different voltage and power
levels. It will be convenient for analysis of power system if the voltage, power, current and impedance
ratings of components of power system are expressed with reference to a common value called base
value. Hence for analysis purpose, a base value is chosen for voltage, power, current and impedance
ratings of the components are expressed as a percent of per unit of the base value.
3. Why the value of voltage and 3-phase KVA are directly used for per unit calculation in 3-phase
system?
The per-unit value of a line-to-neutral (V LN) voltage on the line-to-neutral voltage base value (V b,L-N) is
equal to the per unit value of the line-to-line voltage(V LL) at the same point on the line-to-line voltage
(VbLL) if the system is (VLL) at the same point on the line-to-line voltage base(V bLL)if the system is

balanced. .The per unit value of a 3-phase KVA on the 3-phase (KVA) base is identical
to the per unit value of KVA per phase on the KVA per phase base.i.e.,
.Therefore in 3-phase systems the line value of voltage and 3-
phase KVA are directly used for unit calculations.
4. What is single line diagram? (Nov 2015)
A single line diagram is diagrammatic representation of power system in which the components are
represented by their symbols and the inter connection between them are shown by a single straight line
(even though the system is 3-phase system). The ratings and the impedances of the components are
also marked on the single line diagram.
5. What are the components of power system? (May 2012)
The components of power system are generators, power transformers, motors, transmission lines,
substation transformers, distribution transformers and loads.
6. Define per unit value. (Nov 2015)
The per unit value of any quantity is defined as the ratio of the actual value of the quantity to the base
value expressed as a decimal. The base value is an arbitrary chosen value of the quantity.

7. Write the equation for converting the p.u. impedance expressed in one base to another?

8. What are the advantages (needs) of per unit computation? (Nov 2014)
i) Manufactures usually specify the impedance of a device or machine in per unit on the base of the
name plate rating .ii) The p.u.Values of widely different rating machines lie within a narrow range,
even though the ohmic values has a very large range.iii) The p.u. Impedance of circuit element
connected by transformers expressed on a power base will be same is if it is referred to either side of a
transformer.
iv) The p.u. impedance of a 3ϕ transformer is independent of the type of winding connection(Y or Δ)
9. How the loads are represented in reactance or impedance diagram?
The resistive and reactive loads can be represented by any one of the following representation.
i) Constant power representation, Load power

ii) Constant current representation, Load Current

iii) Constant impedance representation. Load impedance

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10. A generator rated at 30MVA, 11KV has a reactance of 20% calculate its p.u reactance for a
base of 50 MVA and 10KV.

2
=0.2 x (11/10) x (50/30) = 0.403pu

11. The base KV and base MVA of a 3-phase transmission line is 33KV and 10 MVA respectively
calculate the base current and base impedance?

Base current,

Base impedance, 108.9Ω

12. What is impedance diagram?
The impedance diagram is the equivalent circuit of power system in which the various components of
power system are represented by their approximate or simplified equivalent circuits. The impedance
diagram is used for load flow studies.
13. What is reactance diagram?
The reactance diagram is the simplified equivalent circuit of power system in which the various
components are represented by their reactance. The reactance diagram can be obtained from impedance
diagram if all the resistive components are neglected. The reactance diagram is used for fault
calculations.
14. What are the approximations made in reactance diagram?
i) The neutral reactance are neglected ii) Shunt branches in the equivalent circuits of transformers are
neglected iii) The resistance are neglected. iv) All static loads and induction motors are neglected. v) the
capacitance of the transmission lines are neglected
15. Give equations for transforming base KV on LV side to HV side of transformer.

16. What is bus?

The meeting point of various components in a power system is called as bus. The bus is a conductor made
of copper or aluminum having negligible resistance. The buses are considered as points of constant
voltage in a power system.
17. What are the disadvantages of per unit system?
The disadvantages of per unit system are some equations that hold in the unscaled case are modified when
scaled into per unit factors such as√3 and 3 are removed or added in this method. Equivalent circuits of
the components are modified making them somewhat more abstract. Sometimes these shifts that are
clearly present in the unscaled circuit vanish in per unit circuit.
18. What is “Tap changing” transformer? State its types.
The transformer is wounded with tapping on either primary or secondary winding to adjust the voltage.
The device used to give the constant output voltage is called tap changing transformer. The types arei)ON
load automatic tap changer ii)OFF load tap changer.
19. What is off nominal transformation ratio?
When the voltage (or) turns ration of a transformer is not used to decide the ratio of base kV, its voltage
(or) turns ration is called off-nominal ratio. Usually the voltage ratio of regulation transformer will be
off-nominal ratio.
20.Write the four ways of adding an impedance to an existing system so as to modify ZBus matrix.
1. Adding a branch of impedance Z b from a new bus p to the reference bus. 2. Adding a branch of
impedance Zb from a new bus p to an existing bus. 3. Adding a branch of impedance Z b from an existing
bus q to the reference bus. 4. Adding a branch of impedance Zb between two existing buses p and q.
21.What are the methods available for forming bus impedance matrix?
(i) Form the bus impedance matrix and then take its inverse to get bus impedance matrix.

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(ii) Directly form the bus impedance matrix from the reactance diagram. This method utilizes the
techniques of modifications of existing bus impedance matrix due to addition of new bus.

22. What are the representation of loads? (May 2014)

i) Constant power representation ii) Constant current representation iii) Constant impedance
representation
23. What is the purpose of providing third winding (tertiary) in a transformer?
i) Third winding may be used for interconnecting three transmission line at different voltages.
ii) It is sometimes used for other purposes such as connecting shunt capacitors (or) suppression of third
harmonics voltages.iii) To get supply power for substation internal purposes. iv) Tertiary winding can
serve the purpose of measuring voltage of an HV testing transformer.
24. What are the advantages of per unit system? (May 2011)
a) calculations are simple. b) It will be convenient for analysis of power system if the voltage, power,
current and impedance ratings of components of power system are expressed with reference to a common
value called base value
25. Draw a simple per-phase model for a cylindrical rotor synchronous machine. (May 2011)

26. If the reactance in ohms is 15, find the p.u value for a base of 15KVA and 10KV? (May 2012)

27. Draw the equivalent circuit of a three winding transformer. (Nov 2012)(May 2013)

28. What is meant by percentage reactance? (May 2013)

Percentage reactance of a transformer (or in general, a circuit) is the percentage of phase voltage drop
when full load current flows through it, i.e %X= (IX/V)*100.
29. What are the functions of Modern power system? (Nov 2013)
The modern power system is a network of electric components which is used to supply (generating
station), transmit (transmission system) and distribute (distribution system) the electrical power.
30. Name the diagonal and off diagonal elements of bus impedance matrix. (Nov 2013)
The diagonal elements are called as driving point impedances and off diagonal elements are called as
transfer impedances.
31. Draw the impedance diagram for the given single line representation of the power system.

(May 2014)
Impedance Diagram
32. What are the main divisions of Power System? (Nov 2014)

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The main divisions of power systems are: i) Generation ii) Transmission iii) Distribution.

PART – B
1. Explain modern power system in detail and draw basic components of power system.
(Nov 2014)
2. The three phase power and line-line ratings of the electric power system are given
below:G1:60MVA,20KV,X=9%;T1:50MVA,20/200KV,X=10%;T2:50MVA,200/20KV,X=10%;
M: 43.2MVA, 18KV, X=8%; Line: 200KV, Z=120+j200 ohm. Draw an impedance diagram
showing all impedances in per-unit on a 100-MVA base. Choose 20KV as the voltage base for
generator.

Solution

The per unit impedance of

Generator G: j0.15p.u,
Transformer T1:j0.2 p.u,
Transformer T2:j0.2 p.u,
Motor M:j0.167 p.u
For transmission line:Zp.u: 0.3+0.5j p.u

A 20MVA, 11KV three phase synchronous generator has a sub transient reactance of 10%. It is
connected through three identical single phase -Y connected transformer 5000KVA 11/127.02KV
with a reactance of 15% to a high voltage transmission line having a total series reactance of
80ohm. At the end of HT transmission line, three identical single phase star/star connected
transformer of 5000KVA, 127.02/12.702KV with a reactance of 20%. The load is drawing 15MVA,
at 12.5 KV and 0.9p.f lagging. Draw single line diagrams of the network choose a common base of
15MVA and 12.5KV and determine the reactance diagram.

Solution

The per unit reactance of

Generator G: j0.2323 p.u
Transformer T1:j0.464 p.u,
Transformer T2:j0.6195 p.u,
Load :0.9+0.436j p.u
Load Current: 1∟-36.87º p.u
For transmission line:Zp.u: 0.0768j p.u

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3. Write short notes on the following: i) per-phase analysis of a generator ii) per-phase analysis of
3-windig transformer.
Per-phase analysis of a generator

per-phase analysis of 3-windig transformer

4. With the help of single line diagram, explain the basic components of a power system.(May
2011)

5. i)Write detailed notes about the per phase model of a three phase transformer.(May 2011)
ii) Draw an impedance diagram for the electric power system shown in figure, showing all the
impedances in per unit on a 100 MVA base .Choose 20 KV as the voltage base for generator.
The 3ϕpower and line rating are given below.

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G1:90MVA,20KV,X=9%.; Tr1:80MVA,20/220KV,X=16%
Tr2:80MVA, 200/20KV,X=20%; G2:90MVA,18KV,X=9%; Line:200KV,X=120Ω,
Load:200 KV, S=48MW+j64MVAR.

Solution

The per unit impedance of

Generator G1: j0.1 p.u
Generator G2: j0.081 p.u
Transformer T1:j0.2 p.u,
Transformer T2:j0.25 p.u,
Load :0.75+j p.u
For transmission line:Zp.u: 0.3j p.u

6. i. What are the advantages of per unit computations.(May 2012)

ii. Draw the reactance diagram for the power system shown in figure. Neglect resistance and use a
base of 100 MVA, 220 KV in 50Ω line. The ratings of the generator, motor and transformer are
given as Generator: 40 MVA, 25 KV, X” =20%; Synchronous motor: 50 MVA, 11KV, X”=30%
Y- Y Transformer: 40 MVA, 33/220KV, X=15% ;
Y - ∆ Transformer: 30MVA, 11/220KV, (∆/Y), X=15%

Solution

The per unit reactance of

Generator G: j0.378p.u

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Transformer T1:j0.375 p.u,

Transformer T2:j0.5 p.u,
Motor M :j0.6p.u
For transmission line:Zp.u: 0.103j p.u

8. Find the bus impedance matrix for the 4 – bus system shown in figure. Consider bus – 4 as the
reference bus. (May 2012)

Solution:

ZBus =

9. (i) The one-line diagram of a power system is shown in figure. The three-phase power and line
ratings are given below. (13 Marks) (Nov 2012)
G: 80 MVA 22KV X=9% Tr1: 50 MVA 22/220 KV X=10%
Tr1: 40 MVA 220/22 KV X=6.0% Tr3, Tr4: 40 MVA 22/110 KV X=6.4%
Line 1: 200 KV X=121Ω Line 2: 110 KV X= 42.35Ω
M: 68.85 MVA 20 KV X=22.5% Load: 10 MVAR, 4KV Δ-Connected
Capacitor. Draw an impedance diagram showing all impedance in per –unit on a 100
MVA base. Choose 22KV as the voltage base for generator.

Solution

The per unit impedance of

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Generator G: j0.1125 p.u

Transformer T1:j0.2 p.u,
Transformer T2:j0.15 p.u,
Transformer T3:j0.16 p.u,
Transformer T4:j0.16 p.u,
Line 1 :0.25j p.u Line 2:0.35j p.u
Motor M:0.27jp.u

(ii) State the applications of bus admittance matrix (3 Marks)

Represents the admittance relationships between nodes, which then determine the voltages, currents and power
flows in the system.
10. Form the bus impedance matrix for the network shown by building algorithms. (Nov2012) (May
2013)

Solution:

ZBus =

11. For the system shown in figure obtain the impedance diagram. Take a base of 100 MVA and
210 KV in the transmission line. (May 2013)

Solution

The per unit impedance of

Generator G: j p.u
Transformer T:j0.4 p.u,
Line :0.206j p.u Load:1.03 p.u

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12. Why is per unit system used in power system analysis? And list its advantages. (May 2013)

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13. A 90 MVA 11 KV 3 phase generator has a reactance of 25%.The generator supplies two motors
through transformer and transmission line as shown in figure. The transformer T 1 is a 3-phase
transformer,100 MVA, 10/132 KV, 6% reactance. The transformer T 2 is composed of 3 single
phase units each rated, 300 MVA; 66/10 KV, with 5% reactance. The connection of T 1& T2 are
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shown .The motors are rated at 50 MVA and 400 MVA both 10 KV and 20% reactance.
Taking the generator rating as base draw reactance diagram and indicate the reactance in per
unit. The reactance of line is 100 Ω. (Nov 2013)

The per unit reactance of

Generator G: j 0.25 p.u
Transformer T1:j0.044 p.u,
Line :0.426j p.u
Transformer T2:j0.0309p.u
Motor M1:j0.068p.u
Motor M2:j0.055p.u

14. (i) Determine YBusfor the 3-bus system shown in figure. The line series impedance as follows.
Line (bus to bus) Impedance(pu)
1-2 0.06 + j0.18
1-3 0.03 + j0.09
2-3 0.08 + j0.24
Neglect the Shunt capacitance of the lines

Solution:

YBus =

(ii) What are impedance and reactance diagram? Explain. (Nov 2013)
The impedance diagram is the equivalent circuit of power system in which the various components of
power system are represented by their approximate or simplified equivalent circuits. The impedance
diagram is used for load flow studies.
The reactance diagram is the simplified equivalent circuit of power system in which the various
components are represented by their reactance. The reactance diagram can be obtained from
impedance diagram if all the resistive components are neglected. The reactance diagram is used for
fault calculations.

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15. The Single line diagram of a power system is shown in figure along with components data .
Determine the new per unit values and draw the reactance diagram. Assume 25 MVA and 20
KV as new base on generator G1. (May, 2014)

Solution

The per unit impedance of

Generator G1: j0.079 p.u
Generator G2: j0.226p.u
Generator G3: j0.20 p.u
Transformer T1:j0.10 p.u,
Transformer T2:j0.20 p.u,
Transformer T3:j0.10 p.u,
Line 1 :0.0625j p.u Line 2:0.093j p.u

16. Describe the ZBus building algorithms in detailed by using a three bus system. (May 2014)

17.(i)Describe about the representation of loads. (Nov 2014)

i) Constant power representation
ii) Constant current representation
iii) Constant impedance representation
(ii) Draw the per unit equivalent circuit of single- phase transformer?

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Z(pu)=Zp(pu) + Zs(pu)

18. Obtain the per unit Impedence diagram of the Power system of fig shown below: (Nov 2014)

Fig one line diagram representation of a simple power system.

Generator No:- 1:30 MVA, 10.5 kv, X’’= 1.6 ohms
Generator No:- 2: 15 MVA, 6.6 kv, X’’= 1.2 ohms
Generator No:- 3:25 MVA, 6.6 kv, X’’= 0.56 ohms
Transformer T1(3 phase):- 15 MVA , 33/11 Kv , X= 15.2 ohms per phase on high tension side .
Transformer T2(3 phase):- 15 MVA , 33/6.2 Kv , X= 16 ohms per phase on high tension side
Transmission line : 20.5 ohms per phase
Load A, 15MW,11KV,0.9 lagging power factor
Load B, 40MW,6.6KV,0.85 lagging power factor.
Solution:
G1 Xpu =j0.441pu
T1 Xpu=j0.3265pu
TLine Xpu=j0.619pu
T2 Xpu=j0.4837pu
G2 Xpu= j0.3122pu
G3 Xpu= j0.2123 pu
Load A=0.192+j0.32pu
Load B=0.215+j0.354 pu
19. Using the method of building algorithm find the bus impedance matrix for the network shown
in figure. (May 2015).

20. Draw the reactance diagram for the power system shown in figure. Neglect resistance and use a
base of 50MVA and 13.8KV on generator G1. (Nov 2015)
G1: 20 MVA, 13.8 kv, X’’= 20%
G2: 30 MVA, 18.0 kv, X’’= 20%
G3:30 MVA, 20 kv, X’’= 20%
Transformer T1:- 25 MVA , 220/13.8 Kv , X= 10%
Transformer T2:- 3 single phase unit each rated 10 MVA , 127/18 Kv , X= 10%

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Transformer T2:- 35 MVA , 220/22 Kv , X= 10%

Determine the new values of per unit reactance of G1, T1, Transmission line 1, Transmission line 2,
G2, G3, T2 and T3.
21. Form Ybus of the test system shown in figure using singular transformation method. The
impedance data is given in table. Take (1) as reference node. (Nov 2015)

Element Self Mutual

No. Bus Impedance Bus Impedance
code Code
1 1-2 0.5 1-2 0.1
2 1-3 0.6
3 3-4 0.4
4 2-4 0.3
UNIT – II POWER FLOW ANALYSIS
PART – A
1.What is power flow study or load flow study? (Nov 2014)
The study of various methods of solution to power system network is referred to as load flow study. The
solution provides the voltages at various buses, power flowing in various lines and line-losses.
2.What is the need for load flow study?(Nov 2015)
The load flow study of a power system is essential to decide the best operation of existing system and for
planning the future expansion of the system. It is also essential for designing a new power system.
3.What are the different types of buses in a power system?
The buses of a power system can be classified into three types based on the quantities being specified for
the buses. The different types of buses are,(i) Load bus or PQ bus (ii) Generator bus or voltage controlled
bus or PV bus(iii) Slack bus (or) swing bus (or) reference bus
4. When the generator bus is treated as load bus? (Nov 2013) (May 2014) (Nov 2015)
When the generator bus is treated as load bus, the reactive power of the bus is equated to the limit it has
violated, and the previous iteration value of bus voltage is used for calculating current iteration value.
5.What are the advantages of G-S method?
i) Calculations are simple so the programming task is less ii) the memory requirement is less iii) Useful
for small systems
6.What are the disadvantages of G-S method?
i) Requires large number of iterations to reach convergence. ii) Not suitable for large systems iii)
Convergence time increases with size of the system
7.What are the advantages of N-R method?
i) The N-R method is faster, more reliable and the results are accurate ii) Requires less number of
iterations for convergence.iii) The number of iterations is independent of the size of the system.
vi)Suitable for large size system.
8.What are the disadvantages of N-R method?
i) Programming is more complex ii) The memory requirement is more iii) Computational time per

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iteration is higher due to large number of calculations per iteration.

9.How the disadvantages of N-R method are overcome?
The disadvantages of large memory requirement can be overcome by decoupling the weak coupling
between P- δ and Q-V (i.e using de coupled load flow algorithm). The disadvantage of large
computational time per iteration can be reduced by simplifying the decoupled load flow equations. The
simplifications are made based on the practical operating conditions of a power system.
10. How are the diagonal elements of Ybus known as?
The diagonal elements of Ybus are known as the short circuited driving point admittance or self-admittance
of the buses.
11. State the major steps involved in load flow studies?
The major steps involved in load flow studies are i) Mathematical modeling of the power system; this
would be a set of non-linear algebraic equations. ii) Solution of the non-linear equations through an
iterative technique.
12. Why acceleration factor is used in the G-S method?
To increase the rate of convergence of the iterative process, acceleration factor is used.
13. What are the approximations made in FDLF method?
i) Real power at a bus does not change appreciably for a small change in the voltage magnitude
ii) Reactive power at a bus does not change appreciably for a small change in bus voltage phase angle.
14. What is the need of load flow solution?
The load flow solution is essential for designing a new power system and for planning extension as well
as operation of the existing one for increased power demand.
15. What is load bus?
A load bus is one at which the active power and reactive power are specified. In this bus its voltage can
be allowed to vary within permissible values. i.e ±5%. Also bus voltages phase angle is not very
important for the load.
16. How the convergence of N-R method is speeded up?
The convergence of N-R method is speeded up using fast decoupled load flow (FDLF) method. In FDLF,
the weak coupling between P-V and Q- δ are decoupled and the equations are further simplified equations
are further simplified using the practical operating conditions of the power system.
17. What are the advantages of decoupled method over N-R method?
i) This method is simple and computationally efficient than the N-R method.
ii) It requires less memory compared to N-R method.
18. What is the need for voltage control in a power system?
The various components of a power system (or equipments connected to power system) are designed to
work satisfactorily at rated voltages. If the equipments are not operated at rated voltages then the
performance of the equipments will be poor and the life of the equipments will reduce. Hence the
voltages at various points in a power system should be maintained at rated value (specified value)
19. How the reactive power of a generator is controlled?
The reactive power of a generator is controlled by varying the magnitude and phase of induced emf,
which in turn varied by varying excitation. For an increase in reactive power the magnitude of induced
emf is increased and its phase angle is decreased. For a reduction in reactive power the magnitude of
induced emf is decreased and its phase angle is increased.
20. What is Slack or swing bus?(May 2011)
A bus is called swing bus when the magnitude and phase of bus voltage are specified for it. The swing
bus is the reference bus for load flow solution and it is required for accounting line losses. Usually one of
the generator bus is selected as the swing bus.
21. What is Jacobian matrix? How the elements of Jacobian matrix are determined?(May 2011)
The matrix formed the first order derivatives of load flow equations is called Jacobian matrix (J).The
elements of Jacobian matrix will change in every iteration. In each iteration the elements of this matrix
are obtained by partial differentiating the load flow equations with respect to an unknown variable and
then calculating the first derivatives using the solution of previous iteration.
22. What are the information that are obtained from a power flow study? (May 2012)
Bus voltages, Line / transformer power flows, and transmission power losses.
23. Compare Gauss-seidal and Newton Raphson methods of load flow solutions. (May 2012) (Nov
2015)

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S. N Gauss seidal Newton Raphson

1. Reliable More reliable
2. Require large number of iterations to Faster. Require less number if iteration to
reach convergence. It has linear reach convergence It has quadratic
convergence characteristics convergence characteristics.
3. Programming task is less Programming is more complex.
4. Suitable for small size system and not Suitable for large size system.
suitable for large system. Number Number of iterations does not depend on size
iterations increases with increase in size. of the system.
5. Memory required is less Memory required is more.
24. Why power flow analysis is made?(Nov2012)
Power flow analysis is performed to calculate the magnitude and phase angle of voltage at the buses and
also the active power and reactive volt amperes flow for the given terminal or bus conditions. The
variables associated with each bus or node are i) magnitude of voltage (v) ii) phase angle of voltage (δ)
iii) active power (P) iv) reactive volt amperes (Q).
25. What is acceleration factor?(Nov2012) (May 2013)
The acceleration factor is a numerical multiplier which is used to increase which is used to increase the
rate of convergence in an iterative process. The previous value at the bus is multiplied by the acceleration
factor to obtain a correction to be added to previous values.
26. What is the need of slack bus? (May 2013) (May 2014)
The slack bus is needed to account for transmission line losses. In a power system the total power
generated will be equal to sum of power consumed by loads and losses. In a power system only the
generated power and load power are specified for buses. The slack bus is assumed to generate the power
required for losses. Since the losses are unknown the real and reactive power are not specified for slack
bus .They are estimated through the solution of load flow equations.
27. Why do YBus used in load flow study instead of ZBus? (Nov 2013)
Ybus is sparsity matrix ie. Number of non-zero elements is less compared to zero elements. Hence
formation of Ybus needs less memory.
28. Define voltage controlled bus (Nov 2014)
These are the buses where generators are connected. Therefore the power generation in such buses is
controlled through a prime mover while the terminal voltage is controlled through the generator
excitation. Keeping the input power constant through turbine-governor control and keeping the bus
voltage constant using automatic voltage regulator, we can specify constant PGi and | Vi |for these buses.
This is why such buses are also referred to as P-V buses. It is to be noted that the reactive power supplied
by the generator QGi depends on the system configuration and cannot be specified in advance.
Furthermore we have to find the unknown angle δi of the bus voltage.
29. Why is Bus impedance matrix preferred for fault analysis? (May 2015)
Bus impedance matrix is preferred for the fault analysis because fault analysis required full matrix for
calculating line flows in all the lines. The admittance matrix is sparse in nature, so it is not preferred for
fault analysis.

PART – B
1. State the load flow problem and derive load flow equation.
The study of various methods of solution to power system network is referred to as load flow study.
The solution provides the voltages at various buses, power flowing in various lines and line-losses.
The load flow study of a power system is essential to decide the best operation of existing system and
for planning the future expansion of the system. It is also essential for designing a new power system.
The buses of a power system can be classified into three types based on the quantities being specified
for the buses. The different types of buses are,(i) Load bus or PQ bus (ii)Generator bus or voltage
controlled bus or PV bus(iii)Slack bus (or) swing bus (or) reference bus
2. (a) What are the practical application of the power flow analysis ?
(b) Derive the mathematical model of phase shifting transformer to be used in a power flow
analysis.

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Power flow analysis is performed to calculate the magnitude and phase angle of voltage at the buses
and also the active power and reactive volt amperes flow for the given terminal or bus conditions. The
variables associated with each bus or node arei) magnitude of voltage (v) ii) phase angle of voltage (δ)
iii) active power (P)iv) reactive volt amperes (Q).
The load flow solution is essential for designing a new power system and for planning extension
as well as operation of the existing one for increased power demand.
(b) Derive the mathematical model of phase shifting transformer to be used in a power flow analysis.

3. The following is the system data for a load flow solution: (Nov 2015)
Bus code Admittance
1-2 2.0-j8.0
1-3 1.0-j3.0
2-3 0.6-j2.0
2-4 1.0-j4.0
3-4 2.0-j8.0
The schedule of active and reactive power is
Bus code P Q V Remarks
1 - - 1.05+j0.0 Slack
2 0.5 0.2 1.0+j0.0 PQ
3 0.4 0.3 1.0+j0.0 PQ
4 0.3 0.1 1.0+j0.0 PQ
Determine the voltage at the end of first iteration using G-S method. Take acceleration factor= 1.4.
Solution:
Form the Y-BUS

YBus=

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V14acc=1.96-0.09055
4. Compare N-R and FDLF methods of load flow analysis.

5. With neat flow chart explain the computational procedure for load flow solution using Fast
decoupled method when the system contains all the types of buses.(May 2011)

6. Figure shows a five bus system. Each line has an impedance of (0.05+ j0.15) pu. The line shunt
admittance may be neglected. The bus power and voltage specifications are given in table.
(May 2012)
Bus PL QL PG QG V Bus Specification
1 1.0 0.5 - - 1.02 Slack bus
2 0 0 2 - 1.02 PV bus
3 0.5 0.2 0 0 - PQ Bus
4 0.5 0.2 0 0 - PQ bus
5 0.5 0.2 0 0 - PQ bus
(i) Form Ybus (ii) Find Q2, δ2, V3, V4, and V5 after first iteration using Gauss seidal method.
Assume Q2min=0.2pu, Q2max= 0.6pu.

Solution:
Form YBus Matrix:

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YBUS=

Q21=0.2448pu

V21=1.02∟5.11º p.u
V31=.98∟0.76º p.u
V41=0.963∟-1.53º p.u
V21=0.9836∟-0.04º p.u
7. What is Jacobian Matrix? How the elements of Jacobian matrix are computed? (May 2012)
(Nov2012)

7. Write the step by step procedure for load flow analysis by Newton Raphson method.
(May 2012) (Nov 2014) (May 2014)

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9. Compare G-S and N-R methods of load flow analysis.(Nov 2012)

S. N Gauss seidal Newton Raphson
1. Reliable More reliable
2. Require large number of iterations to Faster. Require less number if iteration to
reach convergence. It has linear reach convergence It has quadratic
convergence characteristics convergence characteristics.
3. Programming task is less Programming is more complex.
4. Suitable for small size system and not Suitable for large size system.
suitable for large system. Number Number of iterations does not depend on size
iterations increases with increase in size. of the system.
5. Memory required is less Memory required is more.

10. The figure given below shows a power system. (Nov 2012)
Bus 1: Slack bus ESpecified=1.05∟0º ;Bus 2: PV bus |E|Specified= 1.2 p.u PG= 3 p.u; Bus 3: PQ bus
PL= 4 p.u QL=2 p.u .Carry out one iteration ofload flow solution by Gauss-Seidal method. Take
Q limits of generator 2 as Takeα = 1.

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Solution:

YBUS=

Q2=-0.25p.u, P2=3
V2’=0.9836+0.18j
V3’=0.7737-0.1134j
11. Consider the power system with the following data:
Bus No. Type Generation Load Voltage
P Q P Q Magnitude Angle
1 Slack - - - - 1.0 00
2 P-V 5.0 - 0 0 1.05 -
3 P-Q 0 0 3.0 0.5 - -

YBus =

Obtain the power flow solution (one iteration) for the given system . The line admittance are in
per unit on a 100 MVA base. Use fast decoupled load flow method.
Solution:

Q2=0.6

V2’=0.8838+0.5667j

V3’=0.8794-0.0916j
12. A three bus power system is shown in figure. The relevant per unit line admittance on 100 MVA
base are indicated on the diagram and bus data are given in table. FormYBusand determine the
voltage at bus 2 and bus 3 after first iteration using GS method. Take the acceleration factor α =
1.6 (Nov, 2013)

Bus No. Type Generation Load Voltage

P Q P Q Magnitude Angle
1 Slack - - - - 1.02 00

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2 P-Q 25 15 50 25 - -
3 P-Q 0 0 60 30 - -
Solution:

YBus =

V21=1.038+0.0625j
V31=0.61+0.568j
13. (i) Give the classification of various types of buses in a power system for load flow
studies. (Nov 2014) (Nov 2014)
The buses of a power system can be classified into three types based on the quantities being
specified for the buses. The different types of buses are,(i) Load bus or PQ bus (ii)Generator bus
or voltage controlled bus or PV bus(iii)Slack bus (or) swing bus (or) reference bus
(ii) Give the advantages and limitations of Newton Raphson method.
(iii)What is meant by decoupled load flow method? (Nov 2013)
More reliable
Faster. Require less number if iteration to reach convergence It has quadratic convergence
characteristics.
Programming is more complex.
Suitable for large size system.
Number of iterations does not depend on size of the system.

Memory required is more.

14. Formulate the power flow equation for ‘n’ bus system. (May 2014)
The complex power injected by the source into the ith bus of a power system is
Si=Pi+JQi
The current injected at the ith node is given by

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15. Describe the step by step procedure for load flow solution from Gauss siedal method, if PV and
PQ buses are present along with slack bus. (May 2011) (May 2013), (May 2014), (Nov 2015)

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16. Fig. shown below a three bus power system Bus 1: Slack bus VSpecified=1.05∟0º ;Bus 2: PV bus |
V|Specified= 1.0 p.u ,PG= 3 p.u; Bus 3: PQ bus PL= 4 p.u QL=2 p.u .Carry out one iteration of load
flow solutions by Gauss Seidel method. Neglect limits on reactive power generation? (Nov 2014)

Solution

Y bus= -j5.38 j2.5 j3.33

j2.5 -j7.5 j5
j3.33 j5 -j8.33

V2=1∟-2.6 o
V3=0.9874∟-5.22 o
17. In the power system network shown in figure, bus 1 is slack bus with V 1= 1.0+j 0.0 pu and bus 2
is a load bus with S2= 280MW+j60 MVAr. The line impedance on a base of 100 MVA is Z=
0.02+j0.04 pu. Using Gauss Seidal method, determine V 2. Use an initial estimate of V2(0) =
1.0+j0.0 and perform four iterations. Also find S1 and the real, reactive power loss in the line,
assuming that the bus voltages have converged. (May 2015)

UNIT – III FAULT ANALYSIS - BALANCED FAULT

PART – A
1. What is Short Circuit MVA and how it is calculated? (May 2015)

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The short circuit capacity or the short circuit MVA at a bus is defined as the product of the magnitudes of
the rated bus voltage and the fault current. S.C MVA capacity of the circuit breaker =  3 x pre fault
voltage in KV x S.C current in KA.
2. What are the types of faults?
SERIES FAULT: a) One open conductor fault b) Two open conductor fault
SHUNT FAULT: (a) Symmetrical or balanced fault (i) Three phase Fault(LLLG)(b) Unsymmetrical or
unbalanced fault ( i) Line to line fault(LL)(ii) Line to ground fault (LG)(iii) Double line to ground fault.
(LLG).
3. What are the factors to be considered for selecting the C.B.?
The factors to be considered in selecting a circuit breaker for a protection scheme are: Normal operating
voltage, Momentary, interrupting current. Speed of the breaker and S.C interrupting MVA.
4. What you mean by symmetrical faults? (November 2014)
The fault is called symmetrical fault if the fault current is equal in all the phases and the phase difference
between any two phases is equal.
5. What you mean by doubling effect?
The first peak of the resultant current will become twice the peak value of the final steady current. This
effect is called as doubling effect.
6. What you mean by transient and sub transient reactance?
Xd’ (transient reactance) is the ratio of no load e.m.f and the transient symmetrical r.m.s current.
Xd’’ (sub transient reactance) is the ratio of no load e.m.f and the sub transientsymmetricalr.m.s current.
7. What is the application of transient reactance?
The transient and sub transient reactance helps in calculating the interrupting and maximum momentary
s.c currents.
8. Give the various assumptions made for fault analysis.
The assumptions made in analysis of faults are:i) Each synchronous machine model is represented by an
e.m.f behind a series reactance ii) In the transformer models the shunt that account for core loss and
magnetizing components are neglected.iii) In the transmission line models the shunt capacitances are
neglected. iv)All series resistances in generators, transformers, lines are neglected. v) In the normal
operating conditions the pre fault voltage may be considered as 1.0 p.u.vi) Load impedances are
neglected; hence the pre fault system may be treated as unloaded. vii) As the pre fault currents are much
smaller than the post fault currents the pre fault currents can be neglected.
9. Name any methods of reducing short circuit current.
By providing neutral reactances and by introducing a large value of shunt reactances between buses.
10.What are the reactances used in the analysis of symmetrical faults on the synchronous machines
as its equivalaent reactances.
i) Subtransient reactance Xd” ii) Transient Reactance Xd’ iii) Synchronous reactance Xd
11. What is synchronous reactance?
It is the ratio of induced emf and the steady state r.m.s current. X d =Eg / I
It is the sum of leakage reactance and the armature reaction reactances. It is given byX d = Xl + Xa,Xd =
Synchronous reactance. Xl = Leakage reactance Xa = Armature reaction reactance.
12.What are the causes of fault in power system. (May 2015)
A fault may occur on a power system due to a number of reasons. Some of the causes are(i) Insulation
failure of the system(ii) Falling of a tree along a line(iii) Wind and ice loading on the transmission
lines(iv) Vehicles colliding with supporting structures(v) Overloading of underground cables(vi) Birds
shorting the lines.
13. Name the main differences in representation of power system for load flow and short circuits
studies
S.N Load flow studies Short circuit studies
1 The resistances and reactances are The resistances are neglected
considered
2 To solve load flow analysis, the bus To solve load flow analysis, the bus
admittance matrix is used impedance matrix is used
3 It is used to determine the exact voltages and Pre fault voltages are assumed to be 1 p.u
currents and the pre fault current can be neglected
14. Find the fault current in figure, if the pre fault voltage at the fault point is 0.97p.u?

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Fault Current
15. What is the reason for transients during short circuits?
The fault or short circuits are associated with sudden change in currents. Most of the components of the
power system have inductive property which opposes any sudden change in currents and so the faults
(short circuit) are associated with transients.
16. What is the significance of transient reactance in short circuit studies?
The transient reactance is used to estimate the transient value of fault current. Most of the circuit breakers
open their contacts only during this period. Therefore, for a circuit breaker used for fault clearing, its
interrupting short – circuit rating should be less than the transient fault current.
17. What is the significance of sub - transient reactance in short circuit studies?
The sub - transient reactance is used to estimate the initial value of fault current immediately on the
occurrence of the fault. The maximum momentary short circuit current rating of the circuit breaker used
for protection or fault clearing should be less than this fault clearing value.
18. How to conduct fault analysis of a power system network?
By using equivalent circuit representation and by using bus impedance matrix
19. What is meant by fault calculations?
The fault condition of a power system can be dived into sub transient, transient and steadystate periods.
The currents in the various parts of the system and in the fault are different in these periods. The
estimation of these currents for various types of faults at variouslocations in the system are commonly
referred as fault calculations.
20. Mention the objectives of short circuit studies or fault analysis. (May 2011)(Nov 2012),
(Nov2014)
The short circuit studies are essential in order to design or develop the protective schemes for various
parts of the system. The protective scheme consists of current and voltage sensing devices, protective
relays and circuit breakers. The selection or proper choice of these mainly depends on various currents
that may flow in the fault conditions.
21. Write down the balanced and unbalanced faults occurring in a power system. (May 2011)
BALANCED FAULT: 3 phase short circuit fault
UNBALANCED FAULT: Single line to ground fault, line to line fault and double- line to ground fault.
22. Distinguish symmetrical and unsymmetrical fault. (Nov 2012) (May 2013)
The fault is called Symmetrical fault if the fault current is equal in all the phases.eg. 3ϕ short circuit fault.
The fault is called unsymmetrical fault if the fault current is not equal in all the three phases. eg.i) single
line to ground fault ii) line to line fault iii) double line to ground fault iv) open conductor fault
23. What is meant by fault level? (May 2013)
It relates to the amount of current that can be expected to flow out of a bus in to a 3 phase fault.
Fault level in MVA at bus .
24. Give the frequency of various faults occurrence in ascending order(Nov 2013) (May 2014)
Types of Faults Relative Frequency of Occurrence of Faults
3 phase fault 5%
Double Line to Ground Fault 10%
Line to Line Fault 15%
Single Line to Ground Fault 70%
25. Define bolted fault. (May 2014)

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A fault represents a structural network change equivalent with that caused by the addition of impedance at
the place of the fault. If the fault impedance is zero, then the fault is referred as bolted or solid fault.

26. For a system, the bus impedance matrix was found to be Z= .

The impedances are in pu. A three phase symmetrical fault occurs at bus 3 through a fault impedance
of Zf = j0.19 pu. Find out the post fault voltage at bus 2 assuming zero prefault current. (May 2015).

Solution:

The post fault voltage,

27. What is direct axis reactance? (Dec 2015)

The direct axis is defined as the direction along the rotor that the field winding current causes magnetic
flux to flow. Xd - direct axis reactance.
PART – B
1.A Synchronous generator rated 500 KVA, 400, 0.1 p.u, sub transient reactance is supplying a
passive load of 400KW at 0.8 lag p.f. Calculate the initial symmetrical RMS current for a 3ϕ fault
at the generator terminals.
2.Two generating stations having S.C capacities of 1500MVA & 1000MVA respectively and
operating at 11 KV are linked by a interconnected cable having s reactance of 0.6/phase.
Determine S.C capacity of each station.
3.Two synchronous motors are connected to the bus of a large system through a short transmission
line as shown. The ratings of the various components are: Motor each: 1MVA, 440V, 0.1p.u
reactance. Line: 0.05  reactance. Large system S.C MVA at 440V bus is 8.0. When two motors
are in operation at 440V, calculate the S.C current (symmetrical) fed into a 3 phase fault at the
motors.
4. A small generating station has a bus bar divided into three sections. Each section is connected to
a tie- bar with reactors each rated at 5MVA, 0.1p.u reactance. A generator of 8 MVA rating and
0.15 p.u reactance is connected to each section of the bus bar. Determine the S.C capacity of the
breaker if a 3 phase fault takes place on one of the sections of the bus bar.
5. An alternator and a synchronous motor each rated for 50 MVA, 13.2 KV having sub transient of
20% are connected through a transmission link of reactance 10% on the base of machine ratings.
The motor acts as a load of 30 MW at 0.8 p.f lead and terminal voltage 12.5 KV when a 3 phase
fault takes place at the motor terminals. Determine the sub transient current in the alternator,
the motor and the fault.
6. A Station operating at 33 KV is divided into sections A & B. Section A consists of three
generators 15 MVA each having a reactance of 15% and section B is fed from the grid through a
75 MVA transformer of 8% reactance. The ckt breakers have each a rupturing capacity of 750
MVA. Determine the reactance of the reactor to prevent the breakers being over loaded if a
symmetrical S.C occurs on an outgoing feeder connected to A
7. The per unit impedance matrix of a four bus power system shown in figure below,

ZBus=

Calculate the fault current for a solid three symmetrical fault at bus 4. Also calculate the post
fault bus voltages and line currents.
8. Explain symmetrical fault analysis using Z-bus matrix with neat flow chart. (May 2011)(Nov
2012)(May 2013)
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9. A 3ϕ 5MVA 6.6 KV alternator with a reactance of 8% is connected to a feeder of series

impedance 0.12+j0.48 Ω/Km.The transformer is rated at 3 MVA 6.6kV/33 KV and has a
reactance of 5%.Determine the fault current supplied by the generator operating under no load
with a voltage of 6.9 KV, when a 3ϕ symmetrical fault occurs at a point 15 Km along the feeder..
(May 2012).
10. The bus impedance matrix of 4-bus system with values in p.u is given by,

ZBus=

In this system generator are connected to buses 1 and 2 and their sub transient reactances
included when finding ZBus. If pre-fault current is neglected, find sub transient current in p.u in
the fault for a 3-ph fault voltage as 1 p.u. If the sub transient reactance of generator in Bus 2 is
0.2 p.u., find the sub transient fault current supplied by generator. (May 2012)
11.For the radial network shown, a 3φ -fault occurs at F. Determine the fault current and the line
voltage at 11 KV bus under fault condition. (Nov 2012) (Nov 2014)

12. A synchronous generator and motor are rated 30 MVA, 13.2 KV and both have sub transient
reactance of 20 %.The line connecting them has reactance of 10% on the base of machine ratings.
The motor is drawing 20,000 KW at 0.8 p.f leading and terminal voltage of 12.8 KV when a
symmetrical 3 phase fault occurs at the motor terminals. Find the sub transient current in the
generator, motor and fault by using internal voltages of machines.( May 2013) (Nov 2015)
13. A 11 KV, 100 MVA alternators having a sub –transient reactance of 0.25 p.u is supplying a 50
MVA motor having a sub –transient reactance of 0.2 p.u through a transmission line. The line
reactance is 0.05 pu on a base of 100 MVA. Motor is drawing 40 MW at 0.8 power factor leading
with a terminal voltage of 10.95 KV when a 3-phase fault occurs at the generator terminals.
Calculate the total current in the generator and motor under fault conditions. (Nov 2013) (May
2011).
14. The figure shows a generating station feeding a 132 KV system. Determine the total fault current ,
fault level and fault current supplied by each alternator for a 3ϕ fault at the receiving end bus .
The line is 200 Km long. Take a base of 100 MVA, 11 KV for LV side and 132 KV for HT side.
(Nov 2013)

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15. A generator is connected through a five cycle circuit breaker to a transformer is rated 100 MVA,
18 KV with reactances Xd”=20%,Xd’=25% and Xd= 110%. It is operated on no-load and at rated
voltage.When a 3-phase fault occurs between the breaker and the transformer, find,
(i) Short circuit current in circuit breaker; (ii) The initial symmetrical rms current in the circuit
breaker (iii) The maximum possible dc component of the short circuit current in the breaker;
(iv)The current to be interrupted by the breaker;(v)The interrupting MVA(May 2014)
16. With the help of a detailed algorithm, Explain how a symmetrical fault can be analysed using
ZBus (May 2014)
17. For the three bus network Fig. shown below, obtain Z bus by building algorithm (Nov 2014)

18.Two generators are connected in parallel to the low voltage side of a 3ϕ delta star transformer as
shown in figure. Generator 1 is rated 60,000KVA, 11KV. Generator 2 is rated 30,000 KVA, 11Kv.
Each generator has a subtransient reactance of X d”= 25%. The transformer is rated 90,000 KVA at
11kv-∆/66kv- Ү with a reactance of 10%. Before a fault occurred, the voltage is unloaded and there
is no circulating current between the generators. Find the subtransient current in each generator
when a three phase fault occurs on the hv side of the transformer. (May 2015)

19. Generator G1 and G2 are identical and rated 11Kv, 20 MVA and have a transient reactance of
0.25pu at own MVA base. The transformer T1 and T2 are also identical and are rated 11/66KV,
5MVA and have a reactance of 0.06pu tp their own MVA base. A 50km long transmission line is
connected between the two generators. Calculate three phase fault current, when fault occurs at
middle of the line as shown in figure. (Nov 2015)

UNIT – IV FAULT ANALYSIS –UNBALANCED FAULT

PART – A
1. Name the faults involving ground.
The faults involving ground are: single line to ground fault ii) double line to ground fault iii) Three phase
fault
2. Define positive sequence impedance.
The negative sequence impedance of equipment is the impedance offered by the equipment to the flow of
positive sequence currents.
3. In what type of fault the +ve sequence component of current is equal in magnitude but opposite
in phase to negative sequence components of current?
Line to line fault.
4. In which fault the negative and zero sequence currents are absent?
In three phase fault the negative and zero sequence currents are absent.
5. What are the boundary condition in line-to-line fault?
Ia=0; Ia+Ic=0; Vb=Vc
6. Write down the boundary condition in double line to ground fault?
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Ia=0; Vb=0; Vc=0

7. Give the boundary condition for the 3-phase fault.
Ia + Ib = Ic=0; Va=Vb=Vc=0
8. Name the fault in which positive, -ve and zero sequence component currents are equal.
(May 2012)
In single line to ground fault the +ve, -ve and zero sequence component currents are equal
9. Name the various unsymmetrical faults in a power system.
i) single line to ground fault ii) line to line fault iii) double line to ground fault iv) open conductor fault
10.Write a short notes on Zero sequence network.
While drawing the zero sequence network of a given power system, the following points may be kept in
view.The zero sequence currents will flow only if there is a return path i.e.path from neutral to ground or
to another point in the circuit.In the case of a system with no return path for zero sequence currents, these
currents cannot exist.
11.Write a short notes on negative sequence network.
The negative sequence network can be readiley obtained from positive sequence network with the
following modifications:i) Omit the emfs of 3 – phase generators and motors in the positive sequence
network. It is because these devices have only positive sequence generated voltages.ii) Change, if
neccesary, the impedances between the generators neutral and ground pass no negative sequence current
and hence are not included in the negative seqeunce network.iii) For static devices such as transmission
lines and transformers, the negative sequence impedances have the same value as the corresponding
positive sequence impedances.
12.Write a short notes on positive sequence network.
While drawing the positive sequence network of a given power system, the following points may be
kept in view: i) Each generator in the system is represented by the generated voltage in series with
appropriate reactance and resistance. ii) Current limiting impedances between the generators neutral and
ground pass no positive sequence current and hence are not included in the positive sequence network. Iii)
All resistance and magnetizing currents for each transformer are neglected as a matter of simplicity.For
transmission lines, the shunt capacitances and resistances are generally neglected.
13.How will you express positive, negative and zero – sequence impedances of Y – connected loads?
Positive seqence impedance .Negative sequence impedance
Zero sequence impedance Where, Zs = self impedance of Y – connected load, Z n= load
neutral impedance Zm = Mutual impedance.
14. Which is the most frequently occurring fault?
Single line to ground fault is the most frequently occurring fault
15. Define unsymmetrical fault.
The fault is called unsymmetrical fault if the fault current isnot same in all the three phases.
16. Which is the most severe fault in power system?
Three phase fault is the most severe and rarely occurring fault in the power system.
17. What is sequence network? (May 2011) (Nov 2014)
The network which is used to represent the positive, negative and zero sequence components of
unbalanced system is called as sequence network
18. What are the symmetrical components of a three phase system? (May 2011)(Nov 2012) (Nov
2014) (Nov 2015)
1) Positive sequence 2) negative sequence 3) Zero sequence
19. What is meant by a Fault? (May 2012)
A fault in a circuit is any failure which interferes with the normal flow of current .The faults are
associated with abnormal change in current, voltage and frequency of the power system. The faults may
cause damage to the equipment if it is allowed to persist for a long time.
20. List the various symmetrical and unsymmetrical faults in a power system.(May 2012)
Symmetrical fault: 3 phase short circuit fault.
Unsymmetrical fault: i) single line to ground fault ii) line to line fault iii) double line to ground fault
iv) open conductor fault
21. Define negative sequence impedance? (May 2013)
The negative sequence impedance of an equipment is the impedance offered by the equipment to the flow
of negative sequence current.
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22. Draw the sequence network connections corresponding to L-L fault at bus. (May 2013)

23. What are the observations made from the analysis of various faults? (Nov 2013)
i) To check the MVA ratings of the existing circuit breakers, when new generation are added into a
system; ii) To select the rating for fuses, circuit breaker and switch gear in addition to setting up of
protective relays; iii) To determine the magnitudes of currents flowing throughout the power system at
various time intervals after a fault occurs.
24. Write the boundary conditions for single line to ground fault. (Nov 2013)
The boundary conditions are Va = 0; Ib=Ic=0
25. What are the features of zero sequence current? (May 2014)
As zero sequence currents in three phases are equal and of same phase, three systems operate like single
phase as regards zero sequence currents. Zero sequence currents flow only if return path is available
through which circuit is completed.
26. Write the symmetrical component current of phase ‘a’ in terms of 3ϕ currents. (May 2014).

27. Derive the expression for neutral grounding reactance such that the single line to ground fault
current is less than the three phase fault current. (May 2015).
In case of bolted L-G fault

Whereas in LLL fault,

28. State the reason why, the negative sequence impedance of a transmission line is taken as equal
to positive sequence impedance of the line. (May 2015).
A transmission line is a passive and bilateral device. By passive, we mean there are no voltage or
current sources present in the equivalent model of a transmission line. Bilateral means the line behaves
the same way regardless of the direction of the current. Note that although a single transmission line is
bilateral. Because of a transmission line’s passive and bilateral properties, the phase sequence of the
applied voltage makes no difference, as a-b-c (positive-sequence) voltages produce the same voltage
drops as a-c-b (negative-sequence) voltages. This means that the positive- and negative-sequence
impedances of a transmission line are identical, provided that the line is transposed. Transposition means
physically exchanging the position of each phase conductor along the length of the line such that
conductor #1 occupies: position #1 for 1/3 of the line length, position #2 for 1/3 of the line length, and
position #3 for 1/3 of the line length. Conductors #2 and #3 are rotated similarly.
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29. What is sequence Operator? (Nov 2015)

In balanced problem, to find the relationship between phase voltagesand phase currents, we use sequence
operator. An operator which will turn a phasor through 120 degree in three phase problems. This operator
is called as ' a ' operator.
PART – B
1. A generator of negligible resistance having 1.0 per voltage behind transient reactance is
subjected to different types of faults
Type of fault Resulting fault current in p.u
3-phase 3.33
L-L 2.23
L-G 3.01
Calculate the per unit value of 3 sequence reactance’s.
2. A 50Hz 80MVA, 11kV generator has positive, negative and zero sequence impedances of j0.4,
j0.3 and j0.1p.u respectively. The generator is connected to a busbar A through a transformer
X1=X2=X0=j0.4p.u.on 100MVA base and rated voltage. Determine the ohmic resistance and
rating of earthing resistor such that for LG fault on busbar B, the fault current of the generator
does not exceed full load current. A reactor of reactance 0.08p.u on 100 MVA base is connected
between bus bars A and B.
3. Develop the expressions for analyzing double line to ground fault in a large power system using
Zbus matrix.
4. A 50Hz, 13.2 KV, 15MVA alternator has X1=X2=20% and X0=8% and the neutral is grounded
through a reactor of 0.5ohm. Determine the initial symmetrical rms current in the ground
reactor when a double line to ground fault occurs at the generator terminals at a time when the
generator voltage was 12KV.
5. Derive the necessary equations for calculating the fault current and bus voltages for a single line
to ground fault.
6. A 3-phase, 10 MVA, 11KV, generator with solidity earthed neutral point supplies a feeder. The
relevant impedances of the generator and feeder in ohm are as below:
Generator Feeder
(a) +ve sequence j1.2 j1.0
(b )-ve sequence j0.8 j1.0
(c) zero sequence j0.4 j3.0
If the line to line fault occurs at the far end of the feeder, calculate the fault current.
7. A salient pole generator is rated 20 MVA, 13.8 kV and has X1=0.25p.u X2=0.35p.u and
X0=0.1p.u. The neutral of the generator is solidly grounded. Compute fault current in the
generator and line to line to ground fault at its terminals. Neglect initial load on the generator.
The reactance of a generator are X’’=X2=0.15p.u and X0=0.05p.u. The generator ratings are 10
MVA, 6KV. The generator is star connected with neutral point grounded through a reactor of
0.5ohm reactance. Compute fault current in amps when a single line to ground fault occurs at
the generator terminals.
8. Two 25 MVA, 11KV synchronous generators are connected to a common bus bar which supplies
a feeder. The star point one of the generators is grounded through a resistance of 1 ohm and
that of the other generator is isolate. A line to ground fault occurs at the far end of the feeder.
Determine the fault current.
9. Develop the expressions for analyzing single line to ground fault in a large power system using
Zbus matrix.
11.What are the assumptions made in short circuit studies? Deduce and show the sequence network
for a line to line fault at the terminals of a unloaded generator. (May 2011)
12. Two 11KV, 20MVA .Three phase star connected generators operate in parallel as shown in
figure. The positive, negative and zero sequence reactance are j0.18,j0.15,j0.10 pu. The star
point of one of the generator is isolated and that of the other is earthed through 2.0 ohms
resistor. A single line to ground fault occurs at the terminals of one of the generators. Estimate

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i) Fault current ii)current in the grounding resistor and iii)the voltage across the grounding
resistor. (May 2011)

13. Derive the necessary equation to determine the fault current for a single line to ground fault.
Draw a diagram showing the interconnections of sequence networks. (May 2012)
14. A11 kV, 30MVA alternator has Z1=Z2=-j0.2 pu and Z0=-j0.05 pu. A line to ground fault occurs
on the generator terminals. Determine the fault current and line to line voltages during faulted
conditions. Assume that the generator neutral is solidly grounded and the generator is operating
at no load and at the rated voltage during the occurrence of the fault. (May 2012)
15. A 50 MVA 11 KV alternators was subjected to different types of faults. The faults are 3ϕ fault
1870A, Line to Line Fault 2590A, Single line to ground fault 4130 A. The alternator neutral is
solidly grounded. Find the per unit values of the three sequence reactance of an alternator. (May
2012).
16. Draw the sequence network connection for a double line to ground fault at any point in a power
system and from that obtain an expression for the fault current. (Nov 2012)
17. Derive an expression for the total power in a three phase system in terms of sequence
components of voltages and currents. (ii) Discuss in detail about the sequence impedances of
transmission lines. (Nov 2012) (Nov 2015)
18. Discuss in detail about the sequence impedances and networks of synchronous machines,
transmission lines, transformers and load. (May 2013)
19. A single line diagram of a power network is shown in the figure. (May 2013)

The system data is given in the tables below:

Element Positive sequence Negative sequence Zero sequence
reactance reactance reactance
Generator G 0.1 0.12 0.05
Motor M1 0.05 0.06 0.025
Motor M2 0.05 0.06 0.025
Transformer 0.07 0.07 0.07
Tr1
Transformer 0.08 0.08 0.08
Tr2
Line 0.10 0.10 0.10
Generator grounding reactance is 0.5 pu. Draw sequence networks and calculate the fault for a
line to line fault on phase b and c at point P. Assume 1.0 pu pre fault voltage throughout.
20. The figure shows the power system network .Draw zero sequence network for this system. The
system data is as under.
Generator G1:50 MVA ,11 KV, X0= 0.08 pu Transformer T1 :50 MVA,11/220 KV, X0= 0.1pu
Generator G2 :30 MVA ,11 KV, X0= 0.07pu Transformer T2 ;30 MVA ,11/220 KV ,X0= 0.09 pu
Zero sequene reactance of line is 555.6 Ω. Choose base MVA 50 and base voltage 11 KV for LT
side and 220 KV for HT side. (Nov 2013)

21. A 25 MVA , 13.2 KV alternator with solidly grounded neutral has a sub transient reactance of
0.25 p.u. The negative and zero sequence reactance are 0.35 and 0.01 p.u. respectively. If a

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double line-to-ground fault occurs at the terminals of the alternator, determine the fault
current and line to line voltage at the fault. (May 2014)
22. Obtain the expression for fault current for a line to line fault taken place through an impedance
Zb in a power system.(Nov 2013)(May 2014)
23. Explain about the concepts of symmetrical component. (Nov 2014)
24. A single line to ground fault occurs on Bus 1 of the system of the fig. shown below.Find
i) Current in the fault
ii) SC current in phase A of generator
iii) Voltage of the healthy phases of the bus1 using Z bus method

Given: Rating of the each machine 1200KVA, 600V,with X=X2=10%,Xo=5% each three phase
transformer is rated 1200 KVA , 600V-∆/3000V-Y with leakage reactance of 5% the reactance of
the transmission line are X1 = X2 =20% and Xo=40% on the base of 1200 KVA ,3300V,the
reactance of the neutral reactors are 5% on the KVA and voltage base of the machine. (Nov
2014)
25. Calculate the sub- transient current in each phase for a dead short circuit on one phase
to ground at bus ‘q’ for the system shown in figure.

26. A 30 MVA, 11Kv, 3ϕ synchronous generator has a direct subtransient reactance of 0.25 pu. The
negative and zero sequence reactance are 0.35 and 0.1pu respectively. The neutral of the
generator is solidly grounded. Determine the subtransient current in the generator and the line
to line voltages for subtransient conditions when a single line to ground fault occurs at the
generator terminals with the generstor operating unloaded at rated voltages. (Nov 2015)

UNIT–V STABILITYANALYSIS
PART – A
1. Define Dynamic stability of a power system.
Dynamic stability is the stability given to an inherently unstable system by automatic control devices and
this dynamic stability is concerned with small disturbances lasting for times of the order of 10 to 30
seconds.
2. Define the inertia constants M & H.
Angular momentum (M) about a fixed axis is defined as the product of moment of inertia about that axis
and the associated angular velocity. M = I.  watt/rad/Sec2.Inertia constant (H) is the K.E in Mega joules
to the three phase MVA rating of the machine.
3. Define load angle of a generator.
Load angle:- This is the angle between the generated e.m.f or the supply voltage (E ) and the terminal
voltage. This angle is also called as torque or power angle of the machine.
4. State equal area criterion of stability.
The system is stable if the area under accelerating power (Pa) -  curve reduces to zero at some value of.
In other words positive area under Pa - curve must be equal to the negative area and hence the name
equal area criterion of stability.
5. What are limitations of equal area criterion?

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The limitations of equal area criterion are: i) one drawback of equal area criterion approach is that critical
clearing time cannot be calculated even though the critical clearing angle is known. Hence numerical
methods such as Runge-kutta method, point by point or Euler’s method are employed.
ii) It’s a more simplified approach.
6. If two machines with inertia’s H 1, H2 are swinging together, what will be the inertia of the
equivalent machine?

H1 and H2 is the Inertia constant of M1 and M2; G1 and G2 is the capacity of M1 and M2.
Hs is the equivalent inertia of M1 and M2 ; Gs is the equivalent capacity of M1 and M2.
7. On what basis do you conclude that the given synchronous machine has lost stability?
Following a sudden disturbance on a power system rotor speeds, rotor angular differences and power
transfer undergo fast changes whose magnitude is dependent on the severity of the disturbance. If these
disturbances leads to growing oscillations in the power system even after some period of time say more
than 30 seconds then system said are in asynchronous state and it has lost synchronism.
8. On what a factor does the critical clearing angle depends.
The critical clearing angle depends upon the clearing time, which depends upon auto closing/reclosing
and opening of circuit breakers.
9. Define steady state stability limit. (Nov 2014)
It is the maximum power that can be transferred without the system becoming unstable when the system
is subjected to small disturbances.
10. Mention methods of improving the steady state stability limit.
Pmax = ( E.V / X ). The steady state stability limit can be increased by i) Reducing the X, in case of
transmission lines by using double circuit lines. ii) Use of series capacitors to get better voltage. iii)
Higher excitation systems and quick excitation system are employed.
11. A 50Hz, 4 pole turbo alternator rated at 20 MVA, 13.2 KV has as inertia constant H = 4 KW –
sec/ KVA. Find the K.E stored in the rotor at synchronous speed.
F = 50Hz. P = 4, G = 20 MVA, H = 4 KW – Sec/ KVA. Stored K.E = 4 x 20 = 80MJ.
12. Mention the methods used for the solution of swing equation.
Methods used for solution of swing equation are: Point by point method, Modified Euler’s method and
Runge-kutta method.
13. Give methods used for improving the transient stability.
The following methods are employed to increase the transient stability limit of the power system-
(i)Increase of system voltages,(ii) use of AVR.(iii)Use of High speed excitation systems.(iv)Reduction in
transfer reactance.(v)Useof high speed reclosing breakers.
14. Define the term synchronizing power coefficient of a synchronous machine?
The rate(dp/d), ie, the differential power increase obtained per differential load angle increase is called
the synchronizing power coefficient or electrical stiffness of a synchronous machine.
15. What are the applications of equal area criterion?
(i) Switching operation. (ii) Fault and subsequent circuit isolation. (iii) Fault, circuit isolation and
reclosing
16. What are the classifications of angle stability?
Small signal stability (steady state) and transient stability (large signal).Small signal is further classified
as Oscillatory and Non oscillatory stability.Oscillatory includes Inter area mode, control mode and
Torsional mode
17. Define critical clearing angle and time? (May 2011)(May 2012)(Nov 2012) (Nov 2014) (May
2015)
Critical clearing angle ‘c ’corresponds to critical clearing time tc,in which the fault in the line is cleared
by the circuit breaker above which the system goes out of synchronism.
18. Write swing equation (May 2011)
2 2
Pm-Pe= Md δ/dt . Pm- Input Mechanical power: Pe – output electrical power; M- Angular momentum
19. Define transient stability and stability limit. (May 2012)
The maximum power that can be transferred through the system during a very large disturbance without
loss of synchronism is called transient stability limit.
20. Distinguish between steady state and transient state stability. (Nov 2012)
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Steady state stability is basically concerned with the ability of the system to restore back to its stable state
upon a small disturbance whereas the transient stability is concerned with large disturbances.
21. What is meant by power angle curve? (May 2013) (Nov 2015)
The graphical plot of real power versus power/torque angle is called as power angle curve.
Pe = Pm sin . Pm = E1E2 / X.
22. Define Infinite bus in power system. (Nov 2012)(May 2013)
The capacity of a system comprising of many machines is so large, that its voltage & frequency may
be taken as constant. The connection or disconnection of a single machine does not change the |V| and
frequency. Such a constant voltage and frequency system is called as Infinite bus.
23. Differentiate between voltage stability and rotor angle stability. (Nov 2013)
Voltage stability is the ability of a power system to maintain steady acceptable voltage at all buses in the
system under normal operating conditions and after being subjected to a disturbance.
Rotor angle stability is the ability of interconnected synchronous machines of a power system to remain
in synchronism.
24. Define swing curve? What is the use of this curve? (Nov 2013)
A graph of  versus time in seconds is called swing curve. The stability of the machine is calculated by
using swing curve. This curve is obtained by solving the swing equation of the machine. The critical
angle and critical clearing time is calculated by using swing curve.
25. Define dynamic stability (May 2014)
The dynamic stability study is concerned with the study of nature of oscillations and its decay for small
disturbances.
26. Find the frequency of oscillation for a synchronizing co-efficient of 0.6, inertia constant H= 4
and system frequency of 50 Hz. (May 2014)

Frequency of oscillation = ;

Frequency of oscillation =

27. A four pole , 60 Hz synchronous generator has a rating of 200 MVA, 0.8 power factor lagging.
The moment of inertia of the motor is 45,100kg-m2. Determine M and H. (Nov 2015)
Solution.
Ns= = 1800rpm,
ns= Ns/60 = 30 rps, ωs = 2π ns= 188.4
Kinetic Energy= =

Inertia Constant H =

M=
28. How is the power system stability classified? (May 2015)

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PART B
1. A 50Hz generator is supplying 60% of Pmax to an infinite bus through a reactive network. A
fault occurs which increases the reactance’s of the network between the generator internal
voltage and infinite bus by 400%. When the fault is cleared, compute the max value of critical
clearing angle by applying equal area criteria.
2. State the bad effects of instability. Distinguish between steady state and transient state stabilities.
3. Two power stations A & B are located close together. Station A has 4 identical generator sets
each rated 100MVA and having an inertia constant of 9 MJ / MVA, whereas station B has three
sets each rated 200MVA , 4 MJ/ MVA. Calculate the inertia constant of single equivalent
machines on a base of 100 MVA.
4. Explain the Euler’s method of solving the stability problem.
5. A motor is receiving 30% of the power that it is capable of receiving from an infinite bus. If the
load on the motor is doubled. Calculate the max value of  during the swinging of the rotor
around its new equilibrium position. (Nov 2015).
6. Describe the Runge-Kutta method of solution of swing equation for multi machine systems.
(May 2011)
7. Derive the Power angle equation for a i) SMIB system and also draw the power angle curve.
(May 2012) What are the techniques available to improve steady state stability? (May 2015)
ii) A generator having Xd=0.7pu delivers power at power factor of 0.8 Lagging. Determine
Pe,Qe, E and δ .
8. Using Equal area criterion derive an expression for critical clearing angle for a system having a
generator feeding a large system through a double circuit line. (May 2012) (Nov 2014)
9. A 3 phase generator delivers1.0 pu power to an infinite bus through a transmission network
when fault occurs. The maximum power which can be transferred during pre-fault, during
fault and post fault condition are 1.75pu, 0.4pu and 1.25 pu. Find the critical clearing angle.
(May 2012)
10. (a) State and explain the equal area criterion. (b) Indicate how you will apply equal area
criterion.(i) To find the max additional load that can be suddenly added. (ii) In a two circuit
transmission system sudden loss of one circuit. (Nov 2012) (May 2013)
11. Derive the swing equation of a synchronous machine swinging against an infinite bus. Clearly
state the assumptions made in deducing the swing equation. State the usefulness of this
equation. State the reasons for its nonlinearity and extend the derivation for two parallel
connected coherent and incoherent machines.(May 2014) (May 2013) (Nov 2014) (Nov 2015).
12. (i)Distinguish between steady state, transient and dynamic stability. (Nov 2013)
13. (i) Explain the methods of improving power system stability. (Nov 2013)
(ii) Explain the terms critical clearing angle and critical clearing time in connection with the
transient stability of a power system. (May 2011)
14. Describe the algorithm for modified Euler method of finding solution for power system stability
problem studies.(Nov 2012) (May 2014).
15. A 60 Hz synchronous generator has a transient reactance of 0.2 per unit and an inertia constant
of 5.66 MJ/MVA. The generator is connected to an infinite bus through a transformer and a
double circuit transmission line, as shown in the figure. Resistances are neglected and
reactances are expressed on a common MVA base and are marked on the diagram. The
generator is delivering a real power of 0.77 per unit to bus bar 1. Voltage magnitude at bus 1 is
1.1pu. The infinite bus voltage V= 1.06 . Determine the generator excitation voltage and
obtain the swing equation.

16. The moment of inertia of a 4 pole, 100 MVA, 11Kv, 3ϕ, 0.8 power factor, 50 Hz turbo alternator
is 10000kg-m2. Calculate H and M. (Nov 2015).

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