Board Exam
Problems on
POWER
SYSTEM
1. The fact that the outer layer of the conductor carries more current as compared to
core is known as
.
A. corona
B. permeability
C. fault
D. skin effect
2. How many strands are there for a three-layer stranded conductor?
A. 19
B. 37
C. 54
D. 27
Solution:
¿ of strands = x
x=3 n2−3 n+1
n=number of layers
@n=1

x=1
@n=2

x=7
@n=3

x=19
3. An ACSR conductor having seven steel strands surrounded by 25 aluminum conductor
is specified as
.
A. 7/25
B. 25/7
C. 14/50
D. 50/14
REE – May 2008
4. A single phase, 20 km line has a total inductance of 35 mH. The distance between the
two conductors is 59 inches. Find the GMR of the conductor.
A. 1.75 cm
B. 1.89 cm
C. 1.65 cm
D. 1.99 cm
Solution:
−7
GMD
L= ( 2 ×1 0 ) ln (
)→ H / m
GMR

GMD =59 inches
GMD
−7
)
LT = ( 4 × 10 ) ln (
GMR
[ ( )]
59
35 × 10−3
= ln
(20, 000)
−3
GMR
4 ×1 0
59inches
GMR
(¿)
4.735=ln ¿
59 inches
GMR=( 4.735 )
e
2.54 cm
)
GMR=(0.743 inches)(
1 inch

GMR=1.89 cm
REE – April 2011
5. A three-phase, 60 Hz, transmission line has its conductors arranged in a triangular
configuration so that the two distances between conductors are 5m and the third is
8m. The conductors have an outside diameter of 0.25 inch. Find the inductive
reactance in ohm per km per phase of the transmission line.
A. 0.567
B. 0.283
C. 0.586
D. 0.293
Solution:
5m
5m
8m
GMD=√ (5 )(5 )(8 )
GMD =5.84 m

0.25∈ ¿
2
1∈¿
2.54 cm
¿
¿¿
r =¿

r=0.3175 cm
GMR=0.7788 r
6
2
2
2
GMR=0.7788 (0.3175 cm )

GMR=0.247 cm
−7
GMD
L= ( 2 ×1 0 ) ln(
)
GMR
5.84 m
−7
)
L= ( 2 ×1 0 ) ln (
0.247
100

L=1.55
μH
/∅
m
X L=2 πfL
−6
X L=2 π (60 )(1.55 × 10 )(1, 000)
Ω

/∅
X L=0.586
km
REE – April 2007
6. A three-phase transposed distribution line is designed with equilateral spacing of 12
ft. It is decided to build the line with horizontal spacing (D 13=2 D 12 =2 D 23 ) . The
conductors are transposed. What should be the spacing between adjacent conductors
in order to obtain the same inductance as in the original design?
A. 9.5246
B. 9.5056
C. 9.6437
D. 3.6190
Solution:
V 12
V 23
V 13
GMD 1=√ 12 (12)(12 )
GMD 1=12 ft

3
GMD 2=√3 D 12 D 23 D 13
D
(¿¿ 12)2 2 D 12
3
GMD 2= √ ¿
GMD 2=√ 2 D 123
3

GMD 2=√ 2 D 12
L1=L2
G MD1
G MD 2
)=( 2× 10−7 ) ln(
)
G MR1
GMR2
G MD1 G MD 2
=
G MR1 GMR2
( 2 ×1 0−7) ln(
12=( 3 √ 2)(D 12 )

D 12=9.524 ft
REE – April 2007
7. A three-phase circuit, 60 Hz, 230 kV transposed transmission line is composed of two
1,272 MCM 54 / 19 conductors per phase with horizontal configuration. The bundle
conductors are 50 cm in distance while phase spacing between the centers of the
bundle is 10 meters. If the GMR of the ACSR conductors is 0.0466 ft., find the inductive
reactance in ohm per km per phase of the transmission line.
A. 0.3775
B. 0.3398
C. 0.3628
D. 0.3964
Solution:
1m
GMR=0.0466 ft ×
3.281 ft

GMR=0.0142 m
GMD=√ (1 0 )(20 )

GMD =12.6 m
42
0.01 ¿
¿
50
100
¿
¿
¿
GMR=√4 ¿

GMR=0.0843 m
GMD
L= ( 2 ×1 0−7 ) ln(
)
GMR
12.6
−7
)
L= ( 2 ×1 0 ) ln (
0.0843
μH

L=1
per ∅
m
X L=2 πfL
3
2
X L=2 π (60 )(1 × 10−6)(1, 000)
Ω
X L=0.377
per ∅

km
8. A double circuit line consists of 300, 000 mil 26/6 ACSR Ostrich conductor arranged
vertically with distances between phases of 15 ft, 15 ft and 30 ft. The horizontal
distance between the circuits is 23 ft. Determine the inductive reactance of the line in
milliohms per mile per phase if the GMR of each conductor is 0.0229 ft.
A. 0.28
B. 0.46
C. 0.56
D. 0.39
Solution:
a
a’
15 ft
b
c
b’
23 ft
c’
GMR=0.0229 ft
√
GMD ab =√ d ab d a b d a b d a b = ( 15) ( 15 ) ( √ 15 + 23
GMD ab =20.295 ft

4
4
'
'
'
'
2
2
)( √1 52 +232 )
4
GMD bc =√ d bc d b ' c d bc ' d bc
GMD ab =GMD bc

GMD bc =20.295 ft
GMD ca=√4 d ca d c' a d ca ' d c ' a ' =√4 (30 )(30 )(23 )(23 )

GMD ca=26.268 ft
∴GMD= √3 GMD ab GMD bc GMD ca=√3 20.295 (20.295 )( 26.268)

GMD=22.117 ft
√
√
4
2
2
2
2
GMRa= GMR ( d a a ) = 0.0229 ( √ 30 + 23 )

4
'
GMRa=0.93 ft
2
GMRb= √GMR2 ( db ' b )=√ 0.0229 (23 )
GMRb=0.726 ft

GMRc =GMRa
4

4
GMRc =0.93 ft
∴GMR=√ GMR a GMR b GMRc =√(0.93 )(0726 )(0.93)
GMR=0.856 ft

GMD
−7
)
L= ( 2 ×1 0 ) ln(
GMR
22.117
−7
L= ( 2 ×1 0 ) ln(
)
0.856
μH

L=0.65
per ∅
m
X L=2 πfL
1.6 mile
−6
)
X L=2 π (60 )(0.65 × 10 )(
1m
mΩ

X L=0.394
per ∅
m
3
9. The condu ctors are bundled primarily to
A. increase reactance
C. reduce reactance
REE – September 2010
3
.
B. reduce ratio interference
D. reduce resistance
10. A three-phase 60 Hz line has flat horizontal spacing. The conductors have an outside
diameter of 3.28 cm with 12 m between conductors. Determine the capacitive
reactance to neutral of the line in ohms if its length is 125 miles.
A. 2, 023 Ω
B. 1, 619 Ω
C. 3, 238 Ω
D. 2, 605 Ω
Solution:
GMD=√3 d ab d bc d ca =√3 (12 )(12)(24 )

GMD =15.12 m
d 3.28
GMR= =
2
2

GMR=1.64 cm
2 π εo
2 π (8.85 ×1 0−12 )
=
Cn =
15.12
GMD
ln(
)
)
ln(
0.164
GMR
Cn =8.15 pF / m

1
1
=
X C=
ωC 2 π (8.15 × 10−12)
3.281 ft 1 mile
)
X C =325.49 MΩ−m (
)(
5280 ft
1m

X C =202.198 kΩ−mi
202, 198Ω−mi
X CT =
125 mile
X CT =1, 617 ohms

REE – September 2010
11. A string efficiency of 100% means that
A. one of the insulator discs is shorted
B. the potential across each disc is zero
C. potential across each disc is the same
D. potential across the large disc is very large
.
REE – May 2009
12. A three-phase, 60 Hz overhead line has a horizontal configuration. The conductors
have an outside diameter of 3.28 cm. With 12 m distance between conductors,
determine the capacitive reactance to neutral.
A. 0.2631 × 106 Ω−mi ¿ neutral
B. 0.1864 ×10 6 Ω−mi ¿ neutral
C. 0.1946 ×106 Ω−mi ¿ neutral
D. 0.2024 ×106 Ω−mi ¿ neutral
Solution:
GMD=√3 12 ( 12)( 24 )
GMD =15.12 m
2 π εo
2 π (8.85 ×1 0−12 )
=
GMD
15.12
)
)
ln(
ln(
0.164
GMR
Cn =8.15 pF / m

1
1
X C=
=
ωC 2 π (8.15 × 10−12)
3.281 ft 1 mile
)(
)
X C =325.49 MΩ−m (
1m
5280 ft

X C =0.2022× 10 6 Ω−mi ¿ neutral
Cn =
13. Calculate the capacitive reactance in kΩ-km of a bundled 60 Hz, three-phase line
having three conductors per bundle with 45 cm between conductors of the bundle. The
outside diameter is 0.175 in and the spacing between bundle centers is 10, 10 and 20
m.
A. 243
B. 765
C. 486
D. 382
Solution:
10 m
45 cm
GMD=√ (10 )(10 )(20 )

GMD =12.6 m
0.175∈ ¿
2
1∈¿
2.54 cm
¿
¿¿
r=¿

r=0.22225 cm
25
0.222 ¿
¿
5
4¿
¿
¿
GMR=√9 ¿

GMR=7.6635 cm
3
2 π εo
2 π (8.85 ×1 0−12 )
=
GMD
12.6
)
)
ln(
ln (
0.076635
GMR
C n =10.9 pF / m

1 km
1
1
(
)
=
X C=
−12
ωC 377 ( 10.9 ×10 ) 1, 000 m
Cn =

X C =243.35 kΩ−km
REE – September 2008
14. A single circuit 745 kV, 60 Hz three-phase transposed transmission line is composed of
four ACSR 1, 272, 000 cmils, 54/19. Pheasant conductors per phase with horizontal
configuration. The phase spacing between the line center of the bundles is 14 m and
the bundle spacing is 45 cm. The conductors have a diameter of 1.382 inch and a GMR
of 0.5592 inch. Determine the capacitance per phase per km of the transmission line in
microfarad per kilometer.
A. 0.01424
B. 0.01266
C. 0.01583
D. 0.01234
Solution:
14 m
45 cm
d=1.382∈¿
GMR=0.5592∈¿
GMD=√3 (14 )(14 )(28)

GMD =17.64 m
1.382∈ ¿
2
1∈¿
2.54 cm
¿
¿¿
r=¿

r=1.76 cm
√
16
4
8
GMR= ( 1.76 ) ( 45 ) ( 45 √ 2) (

4
1m
)
100 cm
GMR=0.21823 m
2 π εo
2 π (8.85 ×1 0−12 )
=
Cn =
17.64
GMD
ln (
)
)
ln(
0.21823
GMR

Cn =0.01267
μF
per ∅
km
REE – September 2001
15. A 250 km transmission line has the following parameters: resistance per kilometer
0.05 ohm, capacitive reactance per kilometer 625, 000 ohms, and inductive reactance
per kilometer 0.2 ohm. What is the series impedance?
A. 37.5+ j 150 Ω
B. 12.5− j2, 450 Ω
C. 12.5+ j 50 Ω
D. 75.5− j2, 450 Ω
Solution:
Ω
)(250 km)
Rt =(0.05
km

Rt =12.5 ohms
Ω
)(250 km)
X L=(0.2
km
X L=50 ohms

Z =R t + jX L

Z =12.5+ j 50 ohms
REE – April 2004
16. A 30-mile, three-phase transmission line is to deliver 20, 000 kW at 69 kV at 85%
power factor. The line resistance is 0.324 ohm per mile and inductive reactance is
0.737 ohm per mile. What is the line loss?
A. 1, 050 kW
B. 376.7 kW
C. 997 kW
D. 1, 130.3 kW
Solution:
Ω
)(30 mile)
RT =(0.324
mile
RT =9.72 ohms

Ω
)(30 mile)
X ¿ =( 0.737
mile

X ¿ =22.11ohms
P=S p . f
( 20, 000 ) ( 103 )
S=
0.85
S=23.529 MVA

S= √ 3 V L I L
6
23.529 (10 )
I L=
√ 3(69, 000)

I L =196.876 Amp .
2
PLoss=3(I L )( RT )
PLoss=3(196.8762 )(9.72)
PLoss=1,130.25 Watts

REE – September 2009
17. A 69 kV, three-phase transmission line is 50 km long. The resistance per phase is
0.3006 ohm per mile and the inductance per phase is 0.451 ohm per mile. Neglect the
shunt capacitance. Using the short-line model, what is the voltage regulation when the
line is supplying a three-phase load of 20 MVA at 0.80 lagging power factor at 69 kV?
A. 6.7 %
B. 7.2 %
C. 7.6 %
D. 6.3 %
Solution:
S ∠cos−1 p . f
IT =
√ 3(V L )
1 06 0∠ cos−1 0.8
¿
20 ¿
I T =¿
I T =167.35 ∠−36.7 ° A

(
)
Ω
1 mile
)(50 km)
(
mile 1.61 km
RT =9.3354 ohms
RT = 0.3006

(
)
Ω
1 mile
)(50 km)
(
1.61
km
mile

X T =14.01 ohms
Z =R T + jX T
Z =9.3354+ j14.01

V s ∅=V R ∅+ ZI R
69,000
V s ∅=
+(9.3354 + j 14.01 )( 167.35 ∠−36.7 ° )
√3
V s ∅=42,501 ∠1.28 °Volts

V −V RL V S ∅−V R ∅
=
%VR= SL
V R∅
V RL
69, 000
V R ∅=
√3
X T = 0.451

V R ∅=39, 837. 17 Volts
%VR=

( 42,504 ) −(39,837. 17 )
39,837. 17
%VR=6.7 %
18. Shunt capacitance is neglected in
A. short
B. medium length
transmission lines.
C. long
D. all of these
REE – September 2007
19. A 69 kV, three-phase short transmission line is 16 km long. The line has a per phase
series impedance of 0.125+ j0.4375 ohm per km . Determine the transmission
efficiency when the line delivers 70 MVA, 0.80 lagging power factor at 64 kV.
A. 98.75 %
B. 96.36 %
C. 94.67 %
D. 95.90 %
Solution:
6
Pout=S p . f =(70 )(10 )(0.8 )

Pout=56 MW
Ω
)(16 km)
Z T =( 0.125+ j 0.4375
km
Z T =2+ j 6.97 ohms

IT =
S ∠cos−1 p . f
√ 3(V L )
(70 )(106 )∠ cos−1 0.8
√ 3(69,000)
I T =631.48 ∠−36.87 ° Amp .

IT =
2
PLOSS =3(I T )( R )
8
631.4 ¿
¿
PLOSS =3 ¿
PLOSS =2.39 MW

P out
ɳ=
Pout + Ploss
56 MW
ɳ=
56 MW +2.39 MW

ɳ=95.91%
REE – April 2001
20. The capacitive reactance of a 40 km, 34.5 kV line is 90, 000 ohms-kilometer. What is
the total capacitive reactance of the line?
C. 6, 750 Ω
D. 3.6 ×106 Ω
A. 2, 250 Ω
B. 1.08 ×107 Ω
Solution:
l=40 km
X C =90, 000Ω−km
X
X CT = C
l
90, 000 Ω− km
X CT =
40 km

X CT =2, 250 Ω
REE – September 2006
21. A three-phase short transmission line having per phase impedance of 2+ j 4 ohms
has an equal line to line receiving end and sending end voltages of 115 kV, while
supplying a load of 0.8 p.f leading. Find the power supplied by the line?...line to the
load?
A. 872.8 MW
B. 860.2 MW
C. 846.4 MW
D. 822.4 MW
Solution:
115, 000
V ∅=
√3

V ∅=66, 395.28Volts

V R ∅=66,395.28 ∠ 0°
V S ∅=V R + IZ
66,395.28 ∠ 0 °=66, 395.28+(I R )(2+ j 4)
−1
−1
θ L=cos p . f =cos (0.8)

θ L=36.87 °

I R=¿ I R /∠36.87 ° Amp .

I R=¿ I R /(1 ∠36.87 °) Amp .

I R=¿ I R /(0.8+ j 0.6) Amp .
¿ I R /¿
66,395.28 ∠ 0 °=66, 395.28+(2 + j 4 )(0.8+ j 0.6)¿
°
66,395.28 ∠ 0 ¿
¿
¿
4.4 /I R ¿
¿
66,395.28−0.8 /I R /¿2+¿
¿

¿ I R /¿ 5, 311.62 Amp .
PR =√ 3V L I L p . f =√3 ( 115, 000 )( 5, 311.62) (0.8)
PR =846.4 MW

REE – May 2009
22. What is the maximum power that can be transmitted over a three-phase short
transmission line having a per phase impedance of 0.3+ j0.4 Ω if the receiving end
voltage is 6, 351 volts per phase and the voltage regulation is not to exceed 5 percent?
A. 103.5 MW
B. 114.4 MW
C. 108.9 MW
D. 105.6 MW
Solution:
V −6, 351
V −V R ∅
0.05= s ∅
% V . R= s ∅
VR∅
6, 351
V s ∅=6, 668.55Volts

S
(¿¿ T )( PT )
3(V R ∅)
PRmax =
( p . f ) =¿
Z
6,351
¿
¿
¿2
¿
3¿
3 V R ∅ V s∅ Z
PRmax =(
−Rline )=¿
)(
2
VR∅
Z
PRmax =108.86 MW

23. When the load at the receiving end of a long transmission line is removed or the line is
lightly loaded, the sending end voltage is less than the receiving end voltage. This
phenomenon is called
.
A. Ferranti Effect B. Proximity Effect C. Kelvin Effect
D. Skin Effect
24. Which of the following in not one of the classes of arresters?
A. transmission class
B. station class
C. distribution class
D. intermediate class
REE – October 1996
25. A 60 Hz, three phase transmission line delivers 20 MVA to a load at 66 kV and 80 %
power factor lagging. The total series impedance of each line is 15+ j 75 ohms . If
nominal “pi” circuit is used, what would be the transmission efficiency if the
admittance is j 6 ×10−4 mhos ?
A. 90.8 %
B. 91.7 %
C. 93.5 %
D. 92.6 %
Solution:
P=S p . f
6
Pout =(20 )( 10 )(0.8 )

Pout=16 MW
66, 000
V R ∅=
√3
V R ∅=38.11 kV
By KCL:
I Z= I R + I Y
6
−1
S ∠ cos−1 p . f (20 )(1 0 )∠ cos 0.8
=
I R=
√ 3(V L)
√ 3(66, 000)

I R=174.95 ∠−36.87° Amp .
V
V Y
IY = R ∅ = R ∅
ZC
2
(38,110 )( j6 × 10−4 )
2

I Y =11.43 ∠90 ° Amp .
I Z =174.95 ∠−36.87 ° +11.43∠ 90 °
I Z =168.34 ∠−33.76 ° Amp .

IY =
2
Plosses=3 (I Z )( R )
2
Plosses=3 (168.34 )(15)
Plosses=1.275 MW

P out
16
ɳ=
=
Pout + Ploss 16 + 1.275
ɳ=92.62%

26. A single circuit, 60 Hz, three-phase transmission line is 300 miles long and has the
Ω
μF
mH
, C=0.014
. What is the surge
following parameters: R=0.30 , L=2.10
mi
mi
mi
impedance loading of the line if the receiving end voltage is 132 kV?
A. 45 MW
B. 54 MW
C. 15 MW
D. 38 MW
Solution:
2
SIL=
SIL=

V
L
C
√
√
132,000 2
(2.1 × 10−3 )
(0.014 ×1 0−6 )
SIL =44.99 MW
REE – April 1997
27. In transmission lines, the most effective protection against lightning strikes is one of
the following. Which one is this?
A. lightning rods
B. lightning arresters
C. Peterson coils
D. Overhead wires
REE – September 2001
28. A combination of switch and fuse
A. fuse cut-out B. relay
C. safety switch
D. circuit breaker
REE – September 2007
29. A three-phase, 115 kV, 60 Hz, transmission line has a per phase series impedance of
Z =0.05+ j 0.45 ohm / km
and
a
per
phase
shunt
admittance
of
−6
Y = j3.4 × 10 siemens / km . The line is 120 km long. Using the nominal “pi” line
model. Determine the transmission line D constant.
B. 0.968 + j 0.001199
A. 0.999 + j 0.001248
D. 0.001212
C. 0.989 + j 0.001224
Solution:
YZ
D=1+
2
−6
( j 3.4 × 10 )(120 )(0.05+ j0.45 )(120 )
D=1+
2

D=0.989+ j 0.001224
REE – April 2011
30. A 132 kV, 60 Hz, three-phase transmission line delivers a load of 50 MW at 0.8 power
factor lagging at the receiving end. The generalized constants of the transmission line
are:
A=D =0.95 ∠1.4 °
C=0.0015 ∠90 °
Find the sending end line voltage.
A. 160.9 kV
B. 161.9 kV
Solution:
V S ∅= A V R ∅ + B I R
132, 000
V R ∅=
√3

V R ∅=76.21 kV
S
∠ co s−1 p . f
p. f
I R=
√ 3(V L )
B=96 ∠ 78 °
C. 162.9 kV
D. 163.9 kV
50(1 06 )
∠ co s−1 0.8
0.8
I R=
√ 3(132, 000)
I R=273.37∠−36.87 ° Amp

V S ∅= (0.95 ∠ 1.4 ° )(76.21 ) ( 103 ) +(96 ∠ 78° )( 273.37 ∠−36.87 ° )

V S ∅=94.09 kV
V SL= √3(94.09)
V SL =162.9 kV

31. For a line terminated by its characteristic impedance, the reflected wave is equal to
A. zero
C. half of the incident wave
.
B. equal to incident wave
D. twice of the incident wave
32. A three-core sheathed cable is being tested at 15 kV. The capacitance measured
between any two conductors is 0.2 μF /km . Determine the charging current per km
of the cable at 60 Hz.
A. 0.65 A
B. 1.3 A
C. 1.8 A
D. 2.6 A
Solution:
(C N )(C N ) C N2 C N
C C =C N ∈series withC N
CC =
=
=
C N +C N
2C N 2
C N =2 C C =2(0.2)
C N =0.4 μF /km

15, 000
V∅
√3
=
I charging=
Xc
1
−6
2 π ( 60)( 0.4 )(1 0 )
I charging=1.3 Amp .

REE – October 1994
33. A lead sheath cable for underground service has a copper conductor
(diameter=0.35 inch) surrounded by 0.2 inch wall of rubber insulation. Assuming a
dielectric constant of 4.3 for rubber, calculate the capacitance of the cable.
A. 1.01 μF /mile
D.
0.252
B. 0.504 μF /mile C. 0.76 μF /mile
μF /mile
Solution:
2 π εo ε r
C=
ro
ln( )
ri
0.35
r i=
=0.175∈¿
2
r i=0.175∈¿

r o =r i +t=0.175+0.2

r o =0.375∈¿
2 π (8.85 ×1 0−12 )(4.3)
0.375
)
ln(
0.175
−10 F

C=3.14 ×1 0
m
5280 ft
F 1m
)(
)
CT =3.14 × 10−10 (
m 3.281 ft 1 mile
CT =0.504 μF /mile

C=
REE – October 2000
34. What arrester rating shall be used to protect an 11 kV ungrounded system?
A. 18 kV
B. 16 kV
C. 12 kV
D. 9 kV
Solution:
arrester rating=(regulation factor )(voltage rise factor )(V ∅)
regulation factor for ungrounded :1.05
11, 000
A . R=( 1.05)( √ 3)(
)
√3

A . R=11.55kV
REE – April 2002
35. The distribution system is 34.5 kV, grounded. Which arrester shall be installed to
protect a distribution transformer on the system?
A. 27 kV
B. 30 kV
C. 34.5 kV
D. 38 kV
Solution:
arrester rating=(regulation factor )(V ∅)
regulation factor for grounded :1.35
34, 500
A . R=(1.35)(
)
√3

A . R=26.89 kV
36. A transmission line delivers 1200 MW at 500 kV and 85 % power factor. The series
ohms
and the shunt impedance to neutral is
impedance of the line 0.42+ j 0.65
mi
ohms
− j 0.12× 106
. What is the velocity of propagation of the line?
mi
A. 560 ×106 km / hr
B. 890 ×106 km / hr
C. 750 ×106 km / hr
D. 510 ×106 km / hr
Solution:
δ → propagation constant

δ=α + jβ = √ZY

α → attenuation constant →meter /mile

β → phase constant → rad /mile
√
δ= √ ZY =
δ=
√√
Z se
Z sh
0.42 + j 0.65
j(0.12 ×1 06 )
δ= 6.45 ×1 06 ∠ 163.48 °
√ ¿ Z /∠θ=(¿ Z /∠ θ )
2
A
¿ A /¿ ∠ ARG
2
¿
Z
/∠
θ=
¿
√
√



δ=7.18 ×10−4 + j 2.44 ×10−3
−4
α =7.18 ×10 meter /mile
−3
β=2.44 × 10 rad /mile
ω
β
2 π (60 )(rad / sec) 1.61 km 3, 600 sec
)
v=
)(
(
1 hr
1 mile
−3 rad
2.44 ×10
mile

v =895 ×1 06 kph
v=
37. Which of the following protects a cable against mechanical injury during handling
and laying?
A. serving
B. armouring
C. sheath
D. bedding
38. The transfer bus scheme has the following characteristics except one. Which one is
this?
A. It allows the disconnection of circuit breaker for maintenance without
interrupting the service.
B. It is more flexible.
C. It allows better continuity of service as compared with the single bus
D. It is more costly than that of the single bus system.
39. An overhead transmission line has a span of 300 m. it is supported by two towers of
heights 20.5 m and 30.5 m, respectively. If the required clearance between the
conductor and the ground midway between towers is 19 m and the weight of a
conductor is 0.80 kg/m, what is the tension in the conductor?
A. 1524 kg
B. 1458 kg
C. 1232 kg
D. 1385 kg
Solution:
x
Σ M A =W X
−T ( y)
2
()
2
Wx
=Ty
2
Wx2
y=
2T
2
Wx
d 1= 1
2T
hT
x 1=l−

2 ωl
C=h1− d1 + d2

and
d 2=
W (l−x 1 )
2T
and
x 2=l+
2
19=20.5 −
Wx 1 W (l−x 1)
+
2T
2T
X1
150−¿
¿
−1.5 (2T )
=−X 12 +¿
0.8
−3.75 T =−X 12 +(1 502−300 X 1 + X 12 )
−3.75 T =22,500−300 X 1
X 1=150−
(10)(T )
2(0.8 )( 150 )
10(T )
3.75 T +22,500
=150−
300
2(0.8 )(150 )

T =1384.62
hT
2 ωl
REE – September 2010
40. A system operates at 220 kVA and 11 kV. Using these quantities as base values, what is
the base impedance of the system?
A. 240
B. 220
C. 505
D. 550
Solution:
S b=220 kVA
V b=11 kV
Zb=
Vb
Sb
2
2
11, 000
220, 000
Z b =550

Zb=
41. The conductor of a transmission line has a diameter of 19.6 mm and weighs 0.865
kg/m its ultimate strength is 8070 kg. If the permissible slant sag with a horizontal
wind pressure of 3.91 g/c m2 is 6.28 m, calculate the maximum span between two
consecutive level supports considering a factor of safety of 3.
A. 171 m
B. 296 m
C. 342 m
D. 592 m
Solution:
U .S
F . S=
A.S
8, 070
3=
A.S→T
8,070
T=
3

T =2, 690
W w=Pd
100 cm 2
1m
¿ (19.66 mm)(
)
1m
1, 000 mm
3.91 g
1 kg
)¿
W w =(
)(
c m 2 1, 000 g
W w=0.766 kg/ m

2
2
W T =√ W w +W C
W T =√ 0.7662 + 0.8652

W T =1.155 kg /m
y=
WT X
2T
2
6.28=
1.155 X 2
2(2, 690)
√
6.28(2)(2, 690)
1.155

X =171m
max span=(x)(2)=2 (171 m)

max span =342 m
X=
42. A string of three insulators is used to suspend one conductor of a 33 kV, three-phase
overhead line. The air capacitance between each cap/pin junction and the tower is
1/10 of the capacitance of each unit. Determine the string efficiency.
A. 86.8 %
B. 88.6 %
C. 68.8 %
D. 78.6 %
Solution:
I1 '
I2 '
Ce
I1
I2
V1
a
V2
b
C e=0.1C
I3
V3
KCL @a :
c
I 2 = I 1 + I 1'
V2 V1 V1
;ωC V 2=ωC V 1+ ωC e V 1
= +
X c X c X ce
C V 2=C V 1+(0.1 C)(V 1)
V 2=1.1 V 1

KCL @b :
I 3 = I 2 +I 2 '
C V 3=C V 2+(0.1 C)(V 1 + V 2)
V 3=1.1 V 2+ 0.1V 1
1
1. ¿
¿
V 3=¿
V 3=1.31 V 1

V +V + V
ɳ= 1 2 3
3 (V 3)
V 1 + 1.1V 1 + 1.31 V 1
ɳ=
3(1.31 V 1)

ɳ=86.77 %
43. A 500 kV line has a total corona loss of 280 kW. When energized at 230 kV, the corona
loss is 42 kW. What will be the corona loss if used to transmit power at a voltage of
385 kV?
A. 117 kW
B. 153 kW
C. 207 kW
D. 183 kW
Solution:
−V 0
V∅¿
¿
CPLα ¿
@case I
−V 0
V ∅1 ¿
¿
CPL=k ¿
@case II
−V 0
V ∅2 ¿
¿
CPL=k ¿
−V 0
V ∅1 ¿
¿
¿2
¿
−V 0
V ∅2 ¿
¿
¿2
¿
−V 0
500
¿
√3
¿
¿2
¿
−V 0
200
¿
√3
¿
k¿
CPL 1
=¿
CPL 2

V 0=34.25 kW
−V 0
V ∅3 ¿
¿
¿2
¿
−V 0
V ∅1 ¿
¿
k¿
CPL 3
=¿
CPL 1
5
385
−34.2 ¿
√3
¿
¿
¿
5
500
−34.2 ¿
√3
¿
¿
¿
¿
CPL3=280 ¿
CPL3=152.93 kW

REE – April 2001
44. The percent impedance of a line is 6 % at 34.5 kV and 100 MVA base. What is the
ohmic impedance?
A. 2.32
B. 3
C. 0.72
D. 1.2
Solution:
2
V base
Z base =
Sbase
2
34,500
Z base =
6
100(1 0 )
Z base =11.9 ohms

Z actual =Z p . u Z base
Z actual =( 0.06 )(11.9)
Z actual =0.72

REE – September 2009
45. A 20 kVA, 480 V, single phase generator supplies power to a load through a
transmission line. The load impedance is 2+ j 5 Ω and the transmission line
impedance is 1 + j 3 Ω . If a 1:2 step-up transformer is placed at the output of
generator and a 10:1 step-down transformer is placed at the load of the transmission
line, what is the per unit line current assuming transformers are ideal?
A. 0.0862 p.u.
B. 0.0953 p.u.
C. 0.0851 p.u .
D. 0.0970 p.u.
Solution:
10
a2 =
1

a2=10
2
Z p . u=a (2+ j 5)
2
Z p . u=10 (2+ j5)

Z p . u =200+ j500
480 1
=
V II 2

V II =960 Volts
V II
960
I L=
=
Z p .u Z base (200+ j500)(1 + j3)

I L =1.77 ∠−68.22 °
S
20, 000
I base = base =
V base
480
I base=20.83 Amp .
1.77 ∠−68.22 °
I p . u=
20.83
I p . u =0.085 p .u .

REE – April 2002
46. At a certain point in an electric network, the available fault MVA is 400. A 15 MVA,
34.5 kV, 2.5 % impedance, wye-grounded transformer is installed at that location.
Determine the short circuit MVA at the secondary side of the transformer.
A. 600
B. 625
C. 240
D. 500
Solution:
S base
X p . u=
S sc
15
X p . u=
400

X p . u =0.0375 p .u .
X p . u2 =X p .u 1 + X p . uT
X p . u2 =0.0375+0.025

X p . u2 =0.0625 p . u .
S sc 2=

15 (10 6)
0.0625
S sc 2=240 MVA
REE – April 2001
47. At a 34.5 kV substation, the available fault current is 10 p.u. What is the available fault
MVA if the base is 50 MVA?
A. 50 MVA
B. 100 MVA
C. 250 MVA
D. 500 MVA
Solution:
1
I F p . u .=
X1p.u
1
I F p . u .=
10 p .u .

I F p . u .=0.1 p . u
S base
S sc =
X p .u
50,000, 000
S sc =
0.1

S sc =500 MVA
48. The causes of nearly all high voltage flashovers in transmission lines are due to one of
the following. Which one is the following?
A. high humidity
B. dust and dirt
C. corona
D.
lightning
discharges
REE – May 2010
49. Consider
a
system
with
sequence
impedance
of
+¿= j 0.2577 p . u
.,
Z¿
−¿= j 0.2085 p . u
., and Z (0)= j 0.14 p .u , determine the fault current at phase A
Z¿
for a single line to ground fault at A.
B. – j3 p .u
C. – j11 p .u
A. – j6.4 p . u
D.
– j5 p .u
Solution:
3
I FC=
( j0.2577)( j0.2085)( j0.14 )

I FC =− j5 p .u
REE – April 2011
Z 1= j 0.2577 p . u .,
50. Consider a system with sequence impedances of
Z 2= j 0.2085 p . u ., and Z 0 = j 0.14 p .u ., determine the fault current I a1 for a
single line to ground fault at A.
A. – j2.13 p .u .
B. – j1 p .u .
C. – j3.67 p .u .
D. – j1.6 5 p .u .
Solution:
I a1 =I a 2 =I a 0
I a = I a 1+ I a 2 + I a 0

I a=3 I a1
3
( j 0.2577 ) +( j 0.2085 ) +( j 0.14 )
I a=− j4.9

I
I a1 = a
3
− j 4.9
I a1 =
3
I a1 =− j 1.65 p . u

I a=
REE – September 2002
51. At a certain point in a 69 kV transmission line, the positive sequence impedance is
j 0.15 p . u ., and the zero sequence impedance is j 0.55 p . u . Calculate the fault
current if a line to line fault occurs. The base is 50 MVA.
A. 3, 511 A
B. 1, 890 A
C. 420 A
D. 2, 414 A
Solution:
Sbase
I FC=
(V base )( X 1 p . u + X 2 p .u )
50,000, 000
I FC=
(69, 000)(− j0.15+ j0.15 )
I FC=2, 415 Amp .

52. At a certain point in a system, the thevenin’s equivalent impedance of the network is
0.2 p.u. at a 100 MVA base. A 115/34.5 kV, 10 MVA transformer with 5 % impedance is
tapped at this point. If a three-phase fault occurs at the secondary, find the fault
current at the primary.
A. 2390
B. 718
C. 1380
D. 1240
Solution:
10 MV A
)
100 MVA

X T p .u = j0.02
X 1 p .u = j 0.02+ j 0.05
X 1 p .u = j 0.07

1
1
=
I F p . u=
X 1 p .u j 0.07
I F p . u=14.29 p . u
SB
SB
I B p=
=
√ 3V B p √3 (115, 000)
I B p=50.20 Amp .

I FC pri=I F p .u + I B p =( 14.29 ) (50.2)
I FC pri=717.36 Amp.

X T p .u =( j 0.2)(
REE – April 2001
53. The transformer used to serve a customer is rated 5 MVA, 13.8/0.48 kV, 5 %
impedance. The cable connecting the breaker to the transformer has an impedance of
0.032 ohm per phase. What is the fault current if a three-phase fault occurs at the
breaker?
A. 8, 000 A
B. 5, 000 A
C. 6, 000 A
D. 1, 200 A
Solution:
V2
480 2
Z base =
=
Sbase 5, 000,000
Z base =0.046 p . u

0.032
X p . u=
0.046
X p . u =0.7 p .u
X p . uT =0.7 +0.05
X p . uT =0.75 p . u

SB
SB
I FC=
=
√3 V B X p .uT √ 3(480)( 0.75 )
I FC =8018.75 Amp .

54. NPC supplies energy to Clark Development Corporation at 69 kV from a 50 MVA
transformer whose impedance is 4 %. The short circuit MVA at the primary of the
transformer is 600 MVA. To limit the fault current, a three-phase limiting reactor is
connected on the secondary of the transformer. What is the reactance of the reactor
required to limit the short circuit MVA at the secondary to 200 MVA.
A. 0.127 p.u.
B. 0.167 p.u.
C. 0.173 p.u.
D. 0.139 p.u.
Solution:
SB
SB
50 MVA
=
=
S SC S SC pri 600 MVA

X 1 p .u =0.083 p . u
S
X 1 prs= B
S SC
50 MVa
X 1 prs=
200 MVA
X 1 prs=0.25 p . u

X R p .u =X 1 prs− X 1 p .u − X p .uT
X R p .u =0.25−0.083−0.04
X 1 p .u =

X R p .u =0.127 p . u
REE – September 2009
55. A synchronous generator and motor are rated 30 MVA, 13.2 kV, and both have
subtransient reactances of 20%. The line connecting them has a reactance of 12% on
the base of the machine ratings. The motor is drawing 20, 000 kW at 80% power
factor leading and a terminal voltage of 12.8 kV when a symmetrical three-phase fault
occurs at the motor terminals. Find the three-phase fault current at the motor
terminals.
A. 10, 600 A
B. 10, 150 A
C. 10, 350 A
D. 10, 800 A
Solution:
j 0.1
j 0.
+
-
j 0.
+
-
N
12, 800
13, 200

V TH =0.97
V TH
I F p . u=
Z TH
V TH =
0.97
j 0.123

I F p . u =− j7.89 p .u
I F = I F p . u + I base
30, 000, 000
√ 3(13,200)
I F =( 7.89) +¿
I F =10,352. 94 Amp .

I F p . u=