Power Systems

Test Paper-1Power System Test Paper

Source Book: GATE Multiple Choice Questions Electrical Engineering (Vol-1 and Vol-2)

Author: RK Kanodia & Ashish Murolia Edition: 1st

ISBN: 9788192276212, 9788192276229

Publisher : Nodia and Company

Visit us at: www.nodia.co.in

Q. No. 1 - 10 Carry One Mark Each

MCQ 1.1 The per-unit impedance of an alternator corresponding to base values 13.2 kV and 30 MVA is 0.2 p.u. The pu value of the impedance for base values 13.8 kV and 50 MVA in pu will be(A) 0.131 (B) 0.226

(C) 0.305 (D) 0.364

SOL 1.1

Zpu new Z MVAMVA

kVkV 2

pu oldbase new

base old

base old

base new= c bm l

0.2 .. 0.30530

5013 813 2 pu

2= =b bl l

Hence (C) is correct option.

MCQ 1.2 When the load on a transmission line is a equal to the surge impedance loading(A) the receiving end voltage is less than the sending end voltage.

(B) the sending end voltage is less than the receiving end voltage.

(C) the receiving end voltage is equal to the sending end voltage.

(D) none of these.

SOL 1.2 When the load on a transmission line is equal to surge impedance loading(SIL), the receiving end voltage is equal to the sending end voltage.Hence (C) is correct option.

MCQ 1.3 For a given transmission line the expression for voltage regulation is given by

100%VV V

R

S R#

−. Hence

(A) this must be a ‘short’ line

(B) this may either be a ‘medium line’ or a ‘short line’

(C) this expression is true for any line

Power System Test Paper Test Paper-1

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(D) this may either be a ‘medium line’ or a ‘long line’

SOL 1.3 %regulation 100VV V

R

R R

FL

NL FL#= −

In case of short transmission line the receiving end voltage under no load is the same as sending end voltage under full load conditionV VR RFL = , V VR SNL =

%regulation 100VV V

R

S R#=

−

Hence (A) is correct option.

MCQ 1.4 “Expanded ACSR” are conductors composed of(A) larger diameter individual strands for a given cross section of the aluminium

strands.

(B) larger diameter of the central steel strands for a given overall diameter of the conductor.

(C) larger diameter of the aluminium strands only for a given overall diameter of conductor.

(D) A filler between the inner steel and the outer aluminium strands to increase the overall diameter of the conductor.

SOL 1.4 Expanded ACSR contains a filler such as a paper which separates the inner steel strands from the outer aluminum strands. The paper provides a larger diameter or less corona for a given conductivity and tensile strength.Hence (D) is correct option.

MCQ 1.5 Consider the following statements in respect of load flow studies in power systems :1. Bus admittance matrix is a sparse matrix

2. Gauss-Seidel method is preferred over Newton-Raphson method for load flow studies

3. One of the buses is taken as slack bus in load flow studies

Which of the statements given above are correct ?(A) 1, 2 and 3 (B) 1 and 2

(C) 1 and 3 (D) 2 and 3

SOL 1.5 1. BUS admittance matrix is a sparse matrix.

2. GS method is easier but it is less accurate and has a slow convergence rate compare to NR method .So, GS method is not preferred over NR method.

3. One of the buses is taken as slack bus in power flow studies.



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MCQ 1.6 Which portion of the power system is least prone to faults?(A) Alternators (B) Switchgear

(C) Transformers (D) Overhead lines.

SOL 1.6 In a power system alternators are least prone to faults.Hence (A) is correct option.

MCQ 1.7 A three-phase transformer having zero-sequence impedance of Z0 has the zero sequence network as shown in the figure. The connections of its windings are

(A) star-star

(B) delta-delta

(C) star-delta

(D)delta-star with neutral grounded

SOL 1.7 Since a T circuit provides no return path for zero-sequence current, no zero sequence current can flow into a T-T bank, but it can circulate within the T winding.Hence (B) is correct option.

MCQ 1.8 Tick the correct statement:(A) The negative and zero sequence voltages are maximum at the fault location

and decrease towards neutral.

(B) The negative and zero sequence voltages are minimum at the fault point and increase towards neutral.

(C) The negative sequence is maximum and zero-sequence is minimum at the fault point and decrease and increase respectively towards the neutral.

(D) The negative sequence and zero-sequence currents do not exist at the fault location.

SOL 1.8 Hence (A) is correct option.

MCQ 1.9 For the system shown in figure the per unit reactance values are marked in the figure. The transfer reactance would be

(A) 0.12 pu (B) 0.5 pu

(C) 0.2 pu (D) 0.3 pu


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SOL 1.9 The equivalent circuit of the power system is shown in fig.

The transfer reactance X between the generator and the finite bus is given below. X 0.2 0.3 0.5 .p uX X"

d i= + = + =Hence (B) is correct option.

MCQ 1.10 Three generators are feeding a load of 100 MW. The details of the generators are

Rating(MW)

Efficiency(%)

Regulation (Pu.)( on 100 MVA base)

Generator-1 100 20 0.02

Generator-2 100 30 0.04

Generator-3 100 40 0.03

In the event of increased load power demand, which of the following will happen ?(A) All the generator will share equal power(B) Generator-3 will share more power compared to Generator-1(C) Generator-1 will share more power compared to Generator-2(D) Generator-2 will share more power compared to Generator-3

SOL 1.10 Given that three generators are feeding a load of 100 MW. For increased load power demand, Generator having better regulation share More power, so Generator -1 will share More power than Generator -2.Hence (C) is correct option.

Q. No. 11- 21 Carry Two Mark Each

MCQ 1.11 The power system shown in figure has the following specification:

Generator G1 : 20 MVA, 6.6 kV, 0.10 puXG1 =Generator G2 : 25 MVA, 11 kV, 0.20 puXG2 =Transformer T1 : 25 MVA, 6.6/132 kV, 0.08 puXT1 =


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Transformer T2 : 30 MVA, 11/132 kV, 0.10 puXT2 =Transmission line : Line-to-line voltage = 132 kV, Impedance ZLine ( )j30 120 Ω= +If all the voltage bases are same as rated and base MVA is 50 MVA, then per unit impedance diagram is

SOL 1.11 We know that

Zpu(new) Z VV

MVAMVA2

pu(old)base(new)

base(old)

base(old)

base(new)= c em o

XG1 on system base values

X 1,G pu (0.1)( . ) ( )( . ) ( )6 6 206 6 50

2

2

#

##=

base MVA 50 MVA

base votage 6.6 kV=

=

0.25 pu=

XG2 on system base values

X ,G2 pu (0.2)( ) ( )( ) ( )11 2511 50

2

2

#

##=

base MVA 50 MVA

base votage 11 kV

==

0.4 pu=XT1 on system base values


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X 1,T pu (0.08)( ) ( )( ) ( )132 25132 50

2

2

#

##=

base MVA 50 MVA

base votage 132 kV

==

0.16 pu=XT2 on system base values

X 2,T pu (0.10)( ) ( )( ) ( )132 30132 50

2

2

#

##=

base MVA 50 MVA

base votage 132 kV

==

0.167 pu=For transmission line

Zpu [ ]Z in ohmskV

MVA2

base

base=^ h

base MVA 50 MVA

base votage 132 kV

==

( )

( )j132

30 120 502#= +

Zpu (0.086 0.344)j= +Thus, the pu diagram of the given power system on a system base of 50 MVA is shown as


MCQ 1.12 The impedance diagram of a power system is shown in figure. The bus admittance matrix YBUS is


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(A)

...

...

.

..

...

Y j

8 52 55 00

2 58 755 00

5 05 022 5

12 5

00

12 512 5

SBUS =

−−

−−

R

T

SSSSSS

V

X

WWWWWW

(B)

.

.

.

.

.

.

.

.

.

...

Y j

1 60 40 20

0 41 40 20

0 20 21 20 8

00

0 80 8

SBUS =

R

T

SSSSSS

V

X

WWWWWW

(C)

.

.

.

...

....

.

.

Y j

8 52 55 00

2 58 755 00

5 05 0

22 512 5

00

12 512 5

SBUS = −−

R

T

SSSSSS

V

X

WWWWWW

(D)

...

..

.

.

..

...

Y j

1 60 40 20

0 41 4

0 20

0 20 21 2

0 8

00

0 80 8

SBUS =

−−

−−

R

T

SSSSSS

V

X

WWWWWW

SOL 1.12 The admittance diagram for the system is shown below:

YBUS

YYYY

YYYY

YYYY

YYYY

11

21

31

41

12

22

32

42

13

23

33

43

14

24

34

44

=

R

T

SSSSSS

V

X

WWWWWW

...

...

.

..

...

j

8 52 55 00

2 58 755 00

5 05 022 5

12 5

00

12 512 5

S=

−−

−−

R

T

SSSSSS

V

X

WWWWWW

Where Y11 y y y10 12 13= + + ; Y y y y y22 20 12 23 24= + + + Y33 y y y y30 13 23 34= + + + ; Y y y y44 40 24 34= + +


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Y12 Y y21 12= =− ; Y13 Y y31 13= =− Y23 Y y32 23= =−and Y34 Y y43 34= =− ; Y Y y24 42 24= =− Y14 Y y14 14= =−Hence (A) is correct option.

MCQ 1.13 For a single phase power system shown in figure, transformer T1 and T2 are identical and their specification are given as followingTransformer ratio 2 kV/11 kV=The resistance on LV side 0.04 Ω= and HV side 1.3 Ω=Reactance on LV 0.125 Ω= and HV side 4.5 Ω=

The efficiency of the transmission is(A) 80% (B) 96.3%

(C) 46.5% (D) 74.4%

SOL 1.13 The transmission line equivalent impedance when referred to LV side will be

Z j10 112 30 11

22 2

# #= +b bl l

( . . )j0 33 0 99= +Impedance of each transformer referred to low voltage side

ZT1 0.04 1.3 0.125 4.5Z j j112

112

T

2 2

2 # #= = + + +b bl l

( . . )j0 083 0 273= +The equivalent circuit for the total system

The line current I 125 amps2000250 1000#= =

The line loss PLoss (125) 0.496 7.7 kWI R2 2#= = =

The output Pout 250 0.8 200 kW#= =

Efficiency η 100P PP

out Loss

out#= +


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. . %200 7 7200 100 96 3#= + =

Hence (B) is correct option.

MCQ 1.14 Two overhead lines ‘P’ and ‘Q’ are connected in parallel to supply a load of 10 MW at 0.8 pf lagging. The resistance and reactance of line ‘P’ are 3 Ω and 4 Ω, respectively and of the line ‘Q’ are 4 Ω and 3 Ω respectively. The power supplied by line ‘P’ is(A) 6.30 MW (B) 4.46 MW

(C) 6.73 MW (D) 5.88 MW

SOL 1.14 Impedance of line P ZP .j3 4 5 53 13c Ω= + =

Impedance of line Q ZQ .j4 3 5 36 86c Ω= + =

Z ZP Q+ ( ) ( )j j3 4 4 3= + + + j7 7 7 2 45c Ω= + =

Total load supplied in kVA

S . . 12500 .0 810000 36 86 36 86c c= − = −

Load supplied by line P,

SP SZ ZZ

P Q

Q= +

12500 . .36 867 2 455 36 86

cc

c#= −

6313.457 2

12500 5 45 kVA# c= − = at pf 0.707 (lag)

Power supplied by line P P . 0.7 . kW6313 45 07 4463 6#= =Hence (B) is correct option.

MCQ 1.15 The network shown in the given figure has impedances in p.u. as indicated. The diagonal element Y22 of the bus admittance matrix YBUS of the network is


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(A) 19.8j− (B) 20.0j+

(C) 0.2j+ (D) 19.95j−

SOL 1.15 Y22 ?= I1 ( )V Y V V Y1 11 1 2 12= + − 0.05 10( ) 9.95 10V j V V j V j V1 1 2 1 2= − − =− + I2 ( ) ( )V V Y V V Y2 1 21 2 3 23= − + − . .j V j V j V10 9 9 0 11 2 3= − − Y22 Y Y Y11 23 2= + + 9.95 9.9 0.1j j j=− − − 19.95j=−Hence (D) is correct option.

MCQ 1.16 Two generators G1 and G2 are connected to a transformer T as shown in figure. The specification of components are as following

Generator G1 : 15 MVA, 11 kV, G 0.10 puX 1 =m

Generator G2 : 10 MVA, 11 kV, 0.10 puX G2 =m

Transformer T : 15 MVA, 11/66 kV, 0.06X puT = .

When a three phase fault occurs on the high voltage side of the transformer, what are the values of subtransient current in generator G1 and G2 respectively ?(A) 5 3.33kA, kA (B) 3.94 2.62kA, kA

(C) 2.27 1.51kA, kA (D) 0.47 0.31kA, kA

SOL 1.16 Choose a base 15 MVA X ( )G new1

m 0.10 puj=

X G (new)2m 0.10 0.15 puj j10

15#= =

X ( )T new 0.06 puj=


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Fault current

If . . 8.33 pujV

j j0 12 0 1210= = =−

Sub transient current in G1

I G1m

( . . ). ( 8.33) 5.0 pu

jj j j

0 1 0 150 15

#= + − =−

Sub transient current in G2

GI 2m

( . . ). ( 8.33) 3.33 pu

jj j j

0 1 0 150 10

#= + − =−

Base current

IB 11

15 787.3 Amp3 kV

MVA#

= =

I G1m 5 787.3 3.936 kA#= =

GI 2m 3.33 787.3 2.621 kA#= =

Hence (B) is correct option.

MCQ 1.17 An 11.8 kV busbar is fed from three synchronous generators as shown in the figure. The generator specifications are as followingGenerator G1: 20 MVA, 0.08 puX =l

Generator G2: 60 MVA, 0.1puX =l

Generator G3: 20 MVA, 0.09 puX =l


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The voltage base is taken as 11.8 kV and the VA base as 60 MVA. If a three-phase symmetrical fault occurs on the busbars then the fault current is(A) 1.07 kA (B) 90.75 kA

(C) 40.60 kA (D) 52.40 kA

SOL 1.17 The transient reactance of the generators are

XG1 0.08 0.242060 pu#= =

XG2 0. 0.1 6060 1 pu#= =

XG3 0.0 0.29 2060 7 pu#= =

There values are shown in the equivalent circuit in the figure below.

As the generator e.m.f.s are assumed to be equal, one source may be used which is also shown in figure.

The equivalent reactance is

Xeq / . / . / .0.056 pu

1 0 24 1 0 27 1 0 11= + + =

Therefore fault MVA

. 1071MVA0 05660= =


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and fault current

52402 A3 1180

1071 106

#

#= =

Hence (D) is correct option.

MCQ 1.18 The following figure shows a synchronous generator whose neutral is grounded through a reactance Xn . The generator has balanced emfs and sequence reactances

,X X1 2 and X0. The value of neutral grounding reactance for which the LG fault current is less than the three phase fault current is

(A) X X X31<n 1 0−^ h (B) X X X> 1 0n +^ h

(C) X X X31>n 1 0−^ h (D) X X X2>n 1 0+

SOL 1.18 Connection of sequence networks for a solid LG fault is shown below from which we can write the fault current as

ILG

a X X XE

2 33

n

a

1 0= + +


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Similarly for a solid three-phase fault

Ia3φ

XE

XE

33a a

1 1= =

We have for LG fault current to be less than three-phase fault current, by comparing above two expressions

X X XE

2 33

n

a

1 0+ + XE

33

<a

1

or X X X2 3 n1 0+ + X3> 1

or

Xn ( )X X31> 1 0−


MCQ 1.19 A 3-phase fault occurs at the middle point F on the transmission line as shown in figure. The transfer reactance appearing between the generator and the infinite bus is


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(A) 0.9j pu (B) 0.j 575 pu

(C) 0.j 62 pu (D) 0.j 65 pu

SOL 1.19 The equivalent circuit for the case of the fault at middle point F on line-2 is shown in fig.

We have to equivalent reactance between the generator and the infinite bus In the above circuit, the star network consisting of generator reactance 0.2 pu, Line-1 reactance 0.3 pu and faulted line half reactance 0.15 pu is converted into equivalent delta network as shown in figure below

Transfer reactance X12

. . .. .j j j

j j0 2 0 3 0 150 2 0 3#= + +

0.9 puj=


MCQ 1.20 Consider the protection system shown in the figure below. The circuit breakers numbered from 1 to 7 are of identical type. A single line to ground fault with zero fault impedance occurs at the midpoint of the line (at point F), but circuit breaker 4 fails to operate (‘‘Stuck breaker’’). If the relays are coordinated correctly, a valid sequence of circuit breaker operation is


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(A) 1, 2, 6, 7, 3, 5 (B) 1, 2, 5, 5, 7, 3

(C) 5, 6, 7, 3, 1, 2 (D) 5, 1, 2, 3, 6, 7

SOL 1.20 Due to the fault ‘F’ at the mid point and the failure of circuit-breaker ‘4’ the sequence of circuit-breaker operation will be5, 6, 7, 3, 1, 2 (as given in options)(due to the fault in the particular zone, relay of that particular zone must operate first to break the circuit, then the back-up protection applied if any failure occurs.)Hence (C) is correct option.

MCQ 1.21 The incremental cost curves in Rs/MWhr for two generators supplying a common load of 700 MW are shown in the figures. The maximum and minimum generation limits are also indicated. The optimum generation schedule is :

(A) Generator A : 400 MW, Generator B : 300 MW

(B) Generator A : 350 MW, Generator B : 350 MW

(C) Generator A : 450 MW, Generator B : 250 MW

(D) Generator A : 425 MW, Generator B : 275 MW

SOL 1.21 Given incremental cost curve


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P PA B+ 700= MWFor optimum generator ?PA = , ?PB =a From curve, maximum incremental cost for generator A 600= at 450 MWand maximum incremental cost for generator B 800= at 400 MWminimum incremental cost for generator B 650= at 150 MW

a Maximim incremental cost of generation A is less than the minimum incremental constant of generator B. So generator A operate at its maximum load 450= MW for optimum generation. PA 450= MW PB (700 450)= − 250= MWHence (C) is coerrect option.

Common data for Question 22 to Q. 23A power system consisting two generating plants G1 and G2 has minimum cost dispatch with 150 MWPG1 = and 275 MWPG2 = and the loss coefficients are B11 0.10 10 MW2 1

#= − −

B12 0.01 10 MW2 1#=− − −

B22 0.13 10 MW2 1#= − −

MCQ 1.22 The penalty factor of plant G1 is(A) 1.3245 (B) 4.08

(C) 2.452 (D) 1.825

SOL 1.22 For a system having two generation plants, transmission loss is given by PL P B P P B P B21

211 1 2 12 2

222= + +


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PPL

122 P B P B2 2 01 11 2 12= + +

. ( . )2 150 0 1 10 2 275 0 01 102 2# # # # #= + −− −

. . .0 3 0 055 0 245= − =

Penalty factor for plant G1

L1 . .

PP1

11 0 245

1 1 3245L

122

=−

= − =


MCQ 1.23 To increase the total load on the system by 1 MW will cost an additional Rs 200 per hour. The additional cost in Rs per hour to increase the output of plant G1 by 1 MW(A) 151.0 Rs/MWh (B) 81.56 Rs/MWh

(C) 49.0 Rs/MWh (D) 109.6 Rs/MWh

SOL 1.23 It is given that incremental cost of the system λ Rs per MWh200=Let the incremental cost of plant G1 is /dC dP1 1 then

L dPdC

11

1 l=

. dPdC1 3245

1

1 200=

dPdC

1

1 . 151 /Rs MWh1 3245200= =


Common data for Question 24 to Q. 25A 25 MVA, 11 kV star-connected synchronous generator has its neutral point solidly grounded. The generator is operating at no load at rated voltage. Its sequence reactances are 0.20X X pu1 2= = and 0.08 puX0 = .

MCQ 1.24 What is the symmetrical sub-transient line current for single line-to-ground fault ?(A) 2.5 puj− (B) 2.08 puj−

(C) 6.25 puj− (D) 0.16 puj−

SOL 1.24 Single line-to-ground fault current

If X X XE3 a

1 2 0= + + 1E pua =


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( . . . )

6.25 puj

j0 2 0 2 0 08

3= + + =−


MCQ 1.25 The symmetrical sub-transient line current for double line faults is(A) 4.33 pu− (B) 2.5 pu

(C) .0 25 pu− (D) 1.30 pu

SOL 1.25 Fault current for line-to-line fault is

If Z Zj E3 a

1 2= +

−

. 4.33 pujj

0 43 1#= − =−


Answer Sheet

1. (C) 6. (A) 11. (A) 16. (B) 21. (C)2. (C) 7. (B) 12. (A) 17. (D) 22. (A)3. (A) 8. (A) 13. (B) 18. (C) 23. (A)4. (D) 9. (B) 14. (B) 19. (A) 24. (C)5. (C) 10. (C) 15. (D) 20. (C) 25. (A)

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