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 = 3𝑛2 − 3𝑛 + 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:
𝐿 = (2 × 10−7 ) 𝑙𝑛(
𝐺𝑀𝐷
𝐺𝑀𝑅
) → 𝐻/𝑚
 𝐺𝑀𝐷 = 59 𝑖𝑛𝑐ℎ𝑒𝑠
𝐿 𝑇 = (4 × 10−7 ) 𝑙𝑛(
35×10−3
4×10−3
= [𝑙𝑛 (
4.735 = 𝑙𝑛(
𝐺𝑀𝑅 = (
59
𝑒 4.735
𝐺𝑀𝑅
)
)] (20, 000)
𝐺𝑀𝑅
59 𝑖𝑛𝑐ℎ𝑒𝑠
𝐺𝑀𝑅
59 𝑖𝑛𝑐ℎ𝑒𝑠
𝐺𝑀𝐷
)
)
𝐺𝑀𝑅 = (0.743 𝑖𝑛𝑐ℎ𝑒𝑠)(
 𝑮𝑴𝑹 = 𝟏. 𝟖𝟗 𝒄𝒎
2.54 𝑐𝑚
1 𝑖𝑛𝑐ℎ
)
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 =
6
√(52 )(52 )(82 )
 𝐺𝑀𝐷 = 5.84 𝑚
𝑟=(
0.25 𝑖𝑛
2
)(
2.54 𝑐𝑚
1 𝑖𝑛
)
 𝑟 = 0.3175 𝑐𝑚
𝐺𝑀𝑅 = 0.7788 𝑟
𝐺𝑀𝑅 = 0.7788(0.3175 𝑐𝑚)
 𝐺𝑀𝑅 = 0.247 𝑐𝑚
𝐿 = (2 × 10−7 ) 𝑙𝑛(
𝐿 = (2 × 10−7 ) 𝑙𝑛(
 𝐿 = 1.55
𝜇𝐻
𝑚
𝐺𝑀𝐷
)
𝐺𝑀𝑅
5.84 𝑚
0.247
100
)
/∅
𝑋𝐿 = 2𝜋𝑓𝐿
𝑋𝐿 = 2𝜋(60)(1.55 × 10−6 )(1, 000)
 𝑿𝑳 = 𝟎. 𝟓𝟖𝟔
Ω
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 (𝐷13 = 2𝐷12 = 2𝐷23 ). 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:
𝑉12
𝑉23
𝑉13
3
𝐺𝑀𝐷1 = √12(12)(12)
 𝐺𝑀𝐷1 = 12 𝑓𝑡
𝐺𝑀𝐷2 = 3√𝐷12 𝐷23 𝐷13
3
𝐺𝑀𝐷2 = √(𝐷12 )2 2𝐷12
3
𝐺𝑀𝐷2 = √2𝐷12 3
 𝐺𝑀𝐷2 = √2 𝐷12
𝐿1 = 𝐿2
(2 × 10−7 ) 𝑙𝑛(
𝐺𝑀𝐷1
𝐺𝑀𝑅1
=
𝐺𝑀𝐷1
𝐺𝑀𝑅1
) = (2 × 10−7 ) 𝑙𝑛(
𝐺𝑀𝐷2
𝐺𝑀𝑅2
)
𝐺𝑀𝐷2
𝐺𝑀𝑅2
12 = (3√2)(𝐷12 )
 𝑫𝟏𝟐 = 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:
𝐺𝑀𝑅 = 0.0466 𝑓𝑡 ×
1𝑚
3.281 𝑓𝑡
 𝐺𝑀𝑅 = 0.0142 𝑚
3
𝐺𝑀𝐷 = √(102 )(20)
 𝐺𝑀𝐷 = 12.6 𝑚
4
𝐺𝑀𝑅 = √(0.0142)2 (
50 2
)
100
 𝐺𝑀𝑅 = 0.0843 𝑚
𝐿 = (2 × 10−7 ) 𝑙𝑛(
𝐿 = (2 × 10−7 ) 𝑙𝑛(
 𝐿=1
𝜇𝐻
𝑚
𝐺𝑀𝐷
)
𝐺𝑀𝑅
12.6
0.0843
)
𝑝𝑒𝑟 ∅
𝑋𝐿 = 2𝜋𝑓𝐿
𝑋𝐿 = 2𝜋(60)(1 × 10−6 )(1, 000)
 𝑿𝑳 = 𝟎. 𝟑𝟕𝟕
Ω
𝒌𝒎
𝒑𝒆𝒓 ∅
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
b’
c
23 ft
c’
𝐺𝑀𝑅 = 0.0229 𝑓𝑡
𝐺𝑀𝐷𝑎𝑏 = 4√𝑑𝑎𝑏 𝑑𝑎′𝑏 𝑑𝑎𝑏′ 𝑑𝑎′𝑏′ =
4
√(15)(15)(√152 + 232 )(√152 + 232 )
 𝐺𝑀𝐷𝑎𝑏 = 20.295 𝑓𝑡
𝐺𝑀𝐷𝑏𝑐 = 4√𝑑𝑏𝑐 𝑑𝑏′𝑐 𝑑𝑏𝑐′ 𝑑𝑏𝑐
𝐺𝑀𝐷𝑎𝑏 = 𝐺𝑀𝐷𝑏𝑐
 𝐺𝑀𝐷𝑏𝑐 = 20.295 𝑓𝑡
𝐺𝑀𝐷𝑐𝑎 = 4√𝑑𝑐𝑎 𝑑𝑐′𝑎 𝑑𝑐𝑎′ 𝑑𝑐′𝑎′ = 4√(30)(30)(23)(23)
 𝐺𝑀𝐷𝑐𝑎 = 26.268 𝑓𝑡
∴ 𝐺𝑀𝐷 = 3√𝐺𝑀𝐷𝑎𝑏 𝐺𝑀𝐷𝑏𝑐 𝐺𝑀𝐷𝑐𝑎 = 3√20.295 (20.295 )(26.268)
 𝐺𝑀𝐷 = 22.117 𝑓𝑡
4
4
𝐺𝑀𝑅𝑎 = √𝐺𝑀𝑅 2 (𝑑𝑎′𝑎 ) = √0.02292 (√302 + 232 )
 𝐺𝑀𝑅𝑎 = 0.93 𝑓𝑡
4
4
𝐺𝑀𝑅𝑏 = √𝐺𝑀𝑅 2 (𝑑𝑏′𝑏 ) = √0.02292 (23)
 𝐺𝑀𝑅𝑏 = 0.726 𝑓𝑡
𝐺𝑀𝑅𝑐 = 𝐺𝑀𝑅𝑎
 𝐺𝑀𝑅𝑐 = 0.93 𝑓𝑡
∴ 𝐺𝑀𝑅 = 3√𝐺𝑀𝑅𝑎 𝐺𝑀𝑅𝑏 𝐺𝑀𝑅𝑐 = 3√(0.93)(0726)(0.93)
 𝐺𝑀𝑅 = 0.856 𝑓𝑡
𝐿 = (2 × 10−7 ) 𝑙𝑛(
𝐿 = (2 × 10−7 ) 𝑙𝑛(
𝐺𝑀𝐷
)
𝐺𝑀𝑅
22.117
0.856
)
 𝐿 = 0.65
𝜇𝐻
𝑚
𝑝𝑒𝑟 ∅
𝑋𝐿 = 2𝜋𝑓𝐿
𝑋𝐿 = 2𝜋(60)(0.65 × 10−6 )(
 𝑿𝑳 = 𝟎. 𝟑𝟗𝟒
𝒎Ω
𝒎
1.6𝑚𝑖𝑙𝑒
1𝑚
)
𝒑𝒆𝒓 ∅
9. The conductors are bundled primarily to
A. increase reactance
C. reduce reactance
.
B. reduce ratio interference
D. reduce resistance
REE – September 2010
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√𝑑𝑎𝑏 𝑑𝑏𝑐 𝑑𝑐𝑎 = 3√(12)(12)(24)
 𝐺𝑀𝐷 = 15.12 𝑚
𝐺𝑀𝑅 =
𝑑
2
=
3.28
2
 𝐺𝑀𝑅 = 1.64 𝑐𝑚
2𝜋𝜀𝑜
𝐶𝑛 =
𝑙𝑛(
𝐺𝑀𝐷
)
𝐺𝑀𝑅
=
2𝜋(8.85×10−12 )
15.12
)
0.164
𝑙𝑛(
 𝐶𝑛 = 8.15 𝑝𝐹/𝑚
𝑋𝐶 =
1
𝜔𝐶
=
1
2𝜋(8.15×10−12 )
3.281 𝑓𝑡
𝑋𝐶 = 325.49 𝑀Ω − 𝑚 (
1𝑚
)(
1 𝑚𝑖𝑙𝑒
5280 𝑓𝑡
)
 𝑋𝐶 = 202.198 𝑘Ω − 𝑚𝑖
𝑋𝐶𝑇 =
202,198 Ω−𝑚𝑖
125 𝑚𝑖𝑙𝑒
 𝑿𝑪𝑻 = 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 𝑡𝑜 𝑛𝑒𝑢𝑡𝑟𝑎𝑙
B. 0.1864 × 106 Ω − 𝑚𝑖 𝑡𝑜 𝑛𝑒𝑢𝑡𝑟𝑎𝑙
C. 0.1946 × 106 Ω − 𝑚𝑖 𝑡𝑜 𝑛𝑒𝑢𝑡𝑟𝑎𝑙
D. 𝟎. 𝟐𝟎𝟐𝟒 × 𝟏𝟎𝟔 Ω − 𝒎𝒊 𝒕𝒐 𝒏𝒆𝒖𝒕𝒓𝒂𝒍
Solution:
3
𝐺𝑀𝐷 = √12(12)(24)
𝐺𝑀𝐷 = 15.12 𝑚
𝐶𝑛 =
2𝜋𝜀𝑜
𝑙𝑛(
𝐺𝑀𝐷
)
𝐺𝑀𝑅
=
2𝜋(8.85×10−12 )
15.12
)
0.164
𝑙𝑛(
 𝐶𝑛 = 8.15 𝑝𝐹/𝑚
1
𝑋𝐶 =
=
𝜔𝐶
1
2𝜋(8.15×10−12 )
3.281 𝑓𝑡
𝑋𝐶 = 325.49 𝑀Ω − 𝑚 (
1𝑚
6
)(
1 𝑚𝑖𝑙𝑒
5280 𝑓𝑡
)
 𝑋𝐶 = 0.2022 × 10 Ω − 𝑚𝑖 𝑡𝑜 𝑛𝑒𝑢𝑡𝑟𝑎𝑙
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 = 3√(10)(10)(20)
 𝐺𝑀𝐷 = 12.6 𝑚
𝑟=(
0.175 𝑖𝑛

2
)(
2.54 𝑐𝑚
1 𝑖𝑛
)
𝑟 = 0.22225 𝑐𝑚
9
𝐺𝑀𝑅 = √(0.22225)3 (45)6
 𝐺𝑀𝑅 = 7.6635 𝑐𝑚
𝐶𝑛 =
2𝜋𝜀𝑜
𝑙𝑛(
𝐺𝑀𝐷
)
𝐺𝑀𝑅
=
2𝜋(8.85×10−12 )
𝑙𝑛(
12.6
)
0.076635
 𝐶𝑛 = 10.9 𝑝𝐹/𝑚
1
𝑋𝐶 =
𝜔𝐶
=
1
(
1 𝑘𝑚
377(10.9×10−12 ) 1,000 𝑚
)
 𝑿𝑪 = 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 𝑖𝑛
𝐺𝑀𝑅 = 0.5592 𝑖𝑛
𝐺𝑀𝐷 = 3√(14)(14)(28)
 𝐺𝑀𝐷 = 17.64 𝑚
𝑟=(
1.382 𝑖𝑛
2
)(
2.54 𝑐𝑚
1 𝑖𝑛
)
 𝑟 = 1.76 𝑐𝑚
16
4
𝐺𝑀𝑅 = √(1.76)4 (45)8 (45√2) (
 𝐺𝑀𝑅 = 0.21823 𝑚
𝐶𝑛 =
2𝜋𝜀𝑜
𝐺𝑀𝐷
𝑙𝑛(
)
𝐺𝑀𝑅
=
2𝜋(8.85×10−12 )
𝑙𝑛(
 𝐶𝑛 = 0.01267
17.64
)
0.21823
𝜇𝐹
𝑘𝑚
𝑝𝑒𝑟 ∅
1𝑚
100 𝑐𝑚
)
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 + 𝑗150 Ω
B. 12.5 − 𝑗2, 450 Ω
C.𝟏𝟐. 𝟓 + 𝒋𝟓𝟎 Ω
D. 75.5 − 𝑗2, 450 Ω
Solution:
Ω
𝑅𝑡 = (0.05
𝑘𝑚
)(250 𝑘𝑚)
 𝑅𝑡 = 12.5 𝑜ℎ𝑚𝑠
𝑋𝐿 = (0.2
Ω
𝑘𝑚
)(250 𝑘𝑚)
 𝑋𝐿 = 50 𝑜ℎ𝑚𝑠
𝑍 = 𝑅𝑡 + 𝑗𝑋𝐿
 𝒁 = 𝟏𝟐. 𝟓 + 𝒋𝟓𝟎 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:
𝑅𝑇 = (0.324
Ω
𝑚𝑖𝑙𝑒
)(30 𝑚𝑖𝑙𝑒)
 𝑅𝑇 = 9.72 𝑜ℎ𝑚𝑠
𝑋𝐿𝑇 = (0.737
Ω
𝑚𝑖𝑙𝑒
)(30 𝑚𝑖𝑙𝑒)
 𝑋𝐿𝑇 = 22.11 𝑜ℎ𝑚𝑠
𝑃 = 𝑆 𝑝. 𝑓
𝑆=
(20,000)(103 )
0.85
 𝑆 = 23.529 𝑀𝑉𝐴
𝑆 = √3𝑉𝐿 𝐼𝐿
𝐼𝐿 =
23.529(106 )
√3(69,000)
 𝐼𝐿 = 196.876 𝐴𝑚𝑝.
𝑃𝐿𝑜𝑠𝑠 = 3(𝐼𝐿 2 )(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:
𝐼𝑇 =
𝐼𝑇 =
𝑆∠𝑐𝑜𝑠 −1 𝑝.𝑓
√3(𝑉𝐿 )
20(106 0∠𝑐𝑜𝑠 −1 0.8
√3(69,000)
 𝐼𝑇 = 167.35∠ − 36.7° 𝐴
𝑅𝑇 = (0.3006
Ω
𝑚𝑖𝑙𝑒
)(
1 𝑚𝑖𝑙𝑒
1.61 𝑘𝑚
)(50 𝑘𝑚)
 𝑅𝑇 = 9.3354 𝑜ℎ𝑚𝑠
𝑋𝑇 = (0.451
Ω
𝑚𝑖𝑙𝑒
)(
1 𝑚𝑖𝑙𝑒
1.61 𝑘𝑚
)(50 𝑘𝑚)
 𝑋𝑇 = 14.01 𝑜ℎ𝑚𝑠
𝑍 = 𝑅𝑇 + 𝑗𝑋𝑇
 𝑍 = 9.3354 + 𝑗14.01
𝑉𝑠∅ = 𝑉𝑅∅ + 𝑍𝐼𝑅
𝑉𝑠∅ =
69,000
√3
+ (9.3354 + 𝑗14.01)(167.35∠ − 36.7° )
 𝑉𝑠∅ = 42, 501∠1.28° 𝑉𝑜𝑙𝑡𝑠
%𝑉𝑅 =
𝑉𝑅∅ =
𝑉𝑆𝐿 −𝑉𝑅𝐿
𝑉𝑅𝐿
69,000
=
𝑉𝑆∅ −𝑉𝑅∅
𝑉𝑅∅
√3
 𝑉𝑅∅ = 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:
𝑃𝑜𝑢𝑡 = 𝑆 𝑝. 𝑓 = (70)(106 )(0.8)
 𝑃𝑜𝑢𝑡 = 56 𝑀𝑊
𝑍𝑇 = (0.125 + 𝑗0.4375
Ω
𝑘𝑚
)(16 𝑘𝑚)
 𝑍𝑇 = 2 + 𝑗6.97 𝑜ℎ𝑚𝑠
𝐼𝑇 =
𝐼𝑇 =
𝑆∠𝑐𝑜𝑠 −1 𝑝.𝑓
√3(𝑉𝐿 )
(70)(106 )∠𝑐𝑜𝑠 −1 0.8
√3(69,000)
 𝐼𝑇 = 631.48∠ − 36.87° 𝐴𝑚𝑝.
𝑃𝐿𝑂𝑆𝑆 = 3(𝐼𝑇 2 )(𝑅)
𝑃𝐿𝑂𝑆𝑆 = 3(631.48)2 (2)
 𝑃𝐿𝑂𝑆𝑆 = 2.39 𝑀𝑊
ɳ=
ɳ=
𝑃𝑜𝑢𝑡
𝑃𝑜𝑢𝑡 +𝑃𝑙𝑜𝑠𝑠
56 𝑀𝑊
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?
A. 2, 250 Ω
B. 1.08 × 107 Ω
C. 6, 750 Ω D. 3.6 × 106 Ω
Solution:
𝑙 = 40 𝑘𝑚
𝑋𝐶 = 90, 000 Ω − 𝑘𝑚
𝑋𝐶𝑇 =
𝑋𝐶𝑇 =
𝑋𝐶
𝑙
90,000 Ω−𝑘𝑚
40 𝑘𝑚
 𝑿𝑪𝑻 = 𝟐, 𝟐𝟓𝟎 Ω
REE – September 2006
21. A three-phase short transmission line having per phase impedance of 2 + 𝑗4 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:
𝑉∅ =
115,000
√3
 𝑉∅ = 66, 395.28 𝑉𝑜𝑙𝑡𝑠
 𝑉𝑅∅ = 66, 395.28 ∠0°
𝑉𝑆∅ = 𝑉𝑅 + 𝐼𝑍
66, 395.28∠0° = 66, 395.28 + (𝐼𝑅 )(2 + 𝑗4)
𝜃𝐿 = 𝑐𝑜𝑠 −1 𝑝. 𝑓 = 𝑐𝑜𝑠 −1 (0.8)
 𝜃𝐿 = 36.87°
 𝐼𝑅 =/𝐼𝑅 /∠36.87° 𝐴𝑚𝑝.
 𝐼𝑅 =/𝐼𝑅 /(1∠36.87°)𝐴𝑚𝑝.
 𝐼𝑅 =/𝐼𝑅 /(0.8 + 𝑗0.6) 𝐴𝑚𝑝.
66, 395.28∠0° = 66, 395.28 + (2 + 𝑗4)(0.8 + 𝑗0.6)(/𝐼𝑅 /)
(66, 395.28∠0°)2 = (66, 395.28 − 0.8/𝐼𝑅 /)2 + (4.4/𝐼𝑅 /)2
 /𝐼𝑅 /= 5, 311.62 𝐴𝑚𝑝.
𝑃𝑅 = √3𝑉𝐿 𝐼𝐿 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. 𝑅 =
𝑉𝑠∅ −𝑉𝑅∅
0.05 =
𝑉𝑅∅
𝑉𝑠∅ −6,351
6,351
 𝑉𝑠∅ = 6, 668.55 𝑉𝑜𝑙𝑡𝑠
𝑃𝑅𝑚𝑎𝑥 =
3(𝑉𝑅∅ )
𝑃𝑅𝑚𝑎𝑥 = (
𝑍
3𝑉𝑅∅
𝑍2
(𝑝. 𝑓) = (𝑆𝑇 )(𝑃𝑇 )
)(
𝑉𝑠∅ 𝑍
𝑉𝑅∅
− 𝑅𝑙𝑖𝑛𝑒 ) =
 𝑷𝑹𝒎𝒂𝒙 = 𝟏𝟎𝟖. 𝟖𝟔 𝑴𝑾
3(6,351)2 (6,668.55)(0.5)
0.52
[
6351
− 0.3)]
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 + 𝑗75 𝑜ℎ𝑚𝑠. If
nominal “pi” circuit is used, what would be the transmission efficiency if the
admittance is 𝑗6 × 10−4 𝑚ℎ𝑜𝑠?
A. 90.8 %
B. 91.7 %
C. 93.5 %
D. 92.6 %
Solution:
𝑃 = 𝑆 𝑝. 𝑓
𝑃𝑜𝑢𝑡 = (20)(106 )(0.8)
 𝑃𝑜𝑢𝑡 = 16 𝑀𝑊
𝑉𝑅∅ =
66,000
√3
𝑉𝑅∅ = 38.11 𝑘𝑉
By KCL:
𝐼𝑍 = 𝐼𝑅 + 𝐼𝑌
𝐼𝑅 =
𝑆∠𝑐𝑜𝑠 −1 𝑝.𝑓
√3(𝑉𝐿 )
=
(20)(106 )∠𝑐𝑜𝑠 −1 0.8
√3(66,000)
 𝐼𝑅 = 174.95∠ − 36.87° 𝐴𝑚𝑝.
𝐼𝑌 =
𝐼𝑌 =
𝑉𝑅∅
𝑍𝐶
=
𝑉𝑅∅ 𝑌
2
(38,110)(𝑗6×10−4 )
2
 𝐼𝑌 = 11.43∠90° 𝐴𝑚𝑝.
𝐼𝑍 = 174.95∠ − 36.87° + 11.43∠90°
 𝐼𝑍 = 168.34∠ − 33.76° 𝐴𝑚𝑝.
𝑃𝑙𝑜𝑠𝑠𝑒𝑠 = 3(𝐼𝑍 2 )(𝑅)
𝑃𝑙𝑜𝑠𝑠𝑒𝑠 = 3(168.342 )(15)
 𝑃𝑙𝑜𝑠𝑠𝑒𝑠 = 1.275 𝑀𝑊
ɳ=
𝑃𝑜𝑢𝑡
𝑃𝑜𝑢𝑡 +𝑃𝑙𝑜𝑠𝑠
=
16
16+1.275
 ɳ = 𝟗𝟐. 𝟔𝟐 %
26. A single circuit, 60 Hz, three-phase transmission line is 300 miles long and has the
Ω
𝑚𝐻
following parameters: 𝑅 = 0.30 𝑚𝑖 , 𝐿 = 2.10 𝑚𝑖 , 𝐶 = 0.014
𝜇𝐹
𝑚𝑖
. What is the surge
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:
𝑆𝐼𝐿 =
𝑉2
√
𝐿
𝐶
132,0002
𝑆𝐼𝐿 =
√
(2.1×10−3 )
(0.014×10−6 )
 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 + 𝑗0.45 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.
A. 0.999 + 𝑗0.001248
B. 0.968 + 𝑗0.001199
C. 𝟎. 𝟗𝟖𝟗 + 𝒋𝟎. 𝟎𝟎𝟏𝟐𝟐𝟒
D. 0.001212
Solution:
𝐷 =1+
𝐷 =1+
𝑌𝑍
2
(𝑗3.4×10−6 )(120)(0.05+𝑗0.45)(120)
2
 𝑫 = 𝟎. 𝟗𝟖𝟗 + 𝒋𝟎. 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°
𝐵 = 96∠78°
𝐶 = 0.0015∠90°
Find the sending end line voltage.
A. 160.9 kV
B. 161.9 kV
C. 162.9 kV
D. 163.9 kV
Solution:
𝑉𝑆∅ = 𝐴𝑉𝑅∅ + 𝐵𝐼𝑅
𝑉𝑅∅ =
132,000
√3
 𝑉𝑅∅ = 76.21 𝑘𝑉
𝐼𝑅 =
𝐼𝑅 =
𝑆
∠𝑐𝑜𝑠 −1 𝑝.𝑓
𝑝.𝑓
√3(𝑉𝐿 )
50(106 )
∠𝑐𝑜𝑠 −1 0.8
0.8
√3(132,000)
 𝐼𝑅 = 273.37∠ − 36.87° 𝐴𝑚𝑝
𝑉𝑆∅ = (0.95∠1.4°)(76.21)(103 ) + (96∠78°)(273.37∠ − 36.87°)
 𝑉𝑆∅ = 94.09 𝑘𝑉
𝑉𝑆𝐿 = √3(94.09)
 𝑽𝑺𝑳 = 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:
𝐶𝐶 = 𝐶𝑁 𝑖𝑛 𝑠𝑒𝑟𝑖𝑒𝑠 𝑤𝑖𝑡ℎ 𝐶𝑁
𝐶𝑁 = 2𝐶𝐶 = 2(0.2)
 𝐶𝑁 = 0.4 𝜇𝐹/𝑘𝑚
𝐼𝑐ℎ𝑎𝑟𝑔𝑖𝑛𝑔 =

𝑉∅
𝑋𝑐
=
15,000
√3
1
2𝜋(60)(0.4)(10−6 )
𝑰𝒄𝒉𝒂𝒓𝒈𝒊𝒏𝒈 = 𝟏. 𝟑 𝑨𝒎𝒑.
𝐶𝐶 =
(𝐶𝑁 )(𝐶𝑁 )
𝐶𝑁 +𝐶𝑁
=
𝐶𝑁 2
2𝐶𝑁
=
𝐶𝑁
2
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
B. 0.504 𝝁𝑭/𝒎𝒊𝒍𝒆 C. 0.76 𝜇𝐹/𝑚𝑖𝑙𝑒
D. 0.252 𝜇𝐹/𝑚𝑖𝑙𝑒
Solution:
𝐶=
2𝜋𝜀𝑜 𝜀𝑟
𝑟
ln( 𝑜 )
𝑟𝑖
𝑟𝑖 =
0.35
2
= 0.175 𝑖𝑛
 𝑟𝑖 = 0.175 𝑖𝑛
𝑟𝑜 = 𝑟𝑖 + 𝑡 = 0.175 + 0.2
 𝑟𝑜 = 0.375 𝑖𝑛
𝐶=
2𝜋(8.85×10−12 )(4.3)
𝑙𝑛(
0.375
)
0.175
 𝐶 = 3.14 × 10−10
𝐶𝑇 = 3.14 ×
𝐹
𝑚
1𝑚
5280 𝑓𝑡
−10 𝐹
10
(
)(
)
𝑚 3.281 𝑓𝑡
1 𝑚𝑖𝑙𝑒
 𝑪𝑻 = 𝟎. 𝟓𝟎𝟒 𝝁𝑭/𝒎𝒊𝒍𝒆
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
𝐴. 𝑅 = (1.05)(√3)(
11,000
√3
)
 𝑨. 𝑹 = 𝟏𝟏. 𝟓𝟓 𝒌𝑽
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
𝐴. 𝑅 = (1.35)(

34,500
√3
)
𝑨. 𝑹 = 𝟐𝟔. 𝟖𝟗 𝒌𝑽
36. A transmission line delivers 1200 MW at 500 kV and 85 % power factor. The series
impedance of the line 0.42 + 𝑗0.65
−𝑗0.12 × 106
𝑜ℎ𝑚𝑠
𝑚𝑖
6
𝑜ℎ𝑚𝑠
𝑚𝑖
and the shunt impedance to neutral is
. What is the velocity of propagation of the line?
A. 560 × 10 𝑘𝑚/ℎ𝑟
C. 750 × 106 𝑘𝑚/ℎ𝑟
Solution:
B. 𝟖𝟗𝟎 × 𝟏𝟎𝟔 𝒌𝒎/𝒉𝒓
D. 510 × 106 𝑘𝑚/ℎ𝑟
 𝛿 → 𝑝𝑟𝑜𝑝𝑎𝑔𝑎𝑡𝑖𝑜𝑛 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝛿 = 𝛼 + 𝑗𝛽 = √𝑍𝑌
 𝛼 → 𝑎𝑡𝑡𝑒𝑛𝑢𝑎𝑡𝑖𝑜𝑛 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 → 𝑚𝑒𝑡𝑒𝑟/𝑚𝑖𝑙𝑒
 𝛽 → 𝑝ℎ𝑎𝑠𝑒 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 → 𝑟𝑎𝑑/𝑚𝑖𝑙𝑒
𝑍𝑠𝑒
𝛿 = √𝑍𝑌 = √
𝑍𝑠ℎ
0.42+𝑗0.65
𝛿=√
𝑗(0.12×106 )
𝛿 = √6.45 × 106 ∠163.48°
𝜃
√/𝑍/∠𝜃 = (/𝑍/∠ 2 )
𝐴
√/𝑍/∠𝜃 = √/𝐴/∠𝐴𝑅𝐺 2
 𝛿 = 7.18 × 10−4 + 𝑗2.44 × 10−3
 𝛼 = 7.18 × 10−4 𝑚𝑒𝑡𝑒𝑟/𝑚𝑖𝑙𝑒
 𝛽 = 2.44 × 10−3 𝑟𝑎𝑑/𝑚𝑖𝑙𝑒
𝜔
𝑣=
𝑣=
𝛽
2𝜋(60)(𝑟𝑎𝑑/𝑠𝑒𝑐) 1.61 𝑘𝑚
𝑟𝑎𝑑
2.44×10−3
𝑚𝑖𝑙𝑒
(
1 𝑚𝑖𝑙𝑒
)(
3,600 𝑠𝑒𝑐
1 ℎ𝑟
)
 𝒗 = 𝟖𝟗𝟓 × 𝟏𝟎𝟔 𝒌𝒑𝒉
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
𝛴𝑀𝐴 = 𝑊𝑋 ( ) − 𝑇(𝑦)
2
𝑊𝑥 2
2
= 𝑇𝑦
𝑊𝑥 2
𝑦=
2𝑇
 𝑑1 =
𝑊𝑥1 2
2𝑇
 𝑥1 = 𝑙 −
𝑊(𝑙−𝑥1 )
and
𝑑2 =
and
𝑥2 = 𝑙 +
ℎ𝑇
2𝜔𝑙
2𝑇
ℎ𝑇
2𝜔𝑙
𝐶 = ℎ1 − 𝑑1 + 𝑑2
19 = 20.5 −
−1.5(2𝑇)
0.8
𝑊𝑥1 2
2𝑇
2
+
𝑊(𝑙−𝑥1 )
2𝑇
= −𝑋1 + (150 − 𝑋1 )2
−3.75𝑇 = −𝑋1 2 + (1502 − 300𝑋1 + 𝑋1 2 )
−3.75𝑇 = 22, 500 − 300𝑋1
𝑋1 = 150 −
3.75𝑇+22,500
300
(10)(𝑇)
2(0.8)(150)
= 150 −
10(𝑇)
2(0.8)(150)
 𝑻 = 𝟏𝟑𝟖𝟒. 𝟔𝟐
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:
𝑆𝑏 = 220 𝑘𝑉𝐴
𝑉𝑏 = 11 𝑘𝑉
𝑍𝑏 =
𝑍𝑏 =

𝑉𝑏 2
𝑆𝑏
11,0002
220,000
𝒁𝒃 = 𝟓𝟓𝟎
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/𝑐𝑚2 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 =
3=
𝑇=
𝐴.𝑆
8,070
𝐴.𝑆→𝑇
8,070
3
 𝑇 = 2, 690
𝑊𝑤 = 𝑃𝑑
𝑊𝑤 = (
3.91 𝑔
𝑐𝑚2
)(
1 𝑘𝑔
1,000 𝑔
)(
100 𝑐𝑚 2
) (19.66
1𝑚
𝑚𝑚)(
1𝑚
1,000 𝑚𝑚
)
 𝑊𝑤 = 0.766 𝑘𝑔/𝑚
𝑊𝑇 = √𝑊𝑤 2 + 𝑊𝐶 2
𝑊𝑇 = √0.7662 + 0.8652
 𝑊𝑇 = 1.155 𝑘𝑔/𝑚
𝑦=
𝑊𝑇 𝑋 2
2𝑇
1.155𝑋 2
6.28 =
𝑋=√
2(2,690)
6.28(2)(2,690)
1.155
 𝑋 = 171 𝑚
𝑚𝑎𝑥 𝑠𝑝𝑎𝑛 = (𝑥)(2) = 2(171𝑚)
 𝒎𝒂𝒙 𝒔𝒑𝒂𝒏 = 𝟑𝟒𝟐 𝒎
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 ′
𝐶𝑒
𝑉1
𝐼1
a
𝐼2 ′
𝐼2
𝑉2
b
𝐼3
𝑉3
c
𝐶𝑒 = 0.1 𝐶
𝐾𝐶𝐿 @ 𝑎:
𝐼2 = 𝐼1 + 𝐼1′
𝑉2
𝑋𝑐
=
𝑉1
𝑋𝑐
+
𝑉1
𝑋𝑐𝑒
; 𝜔𝐶𝑉2 = 𝜔𝐶𝑉1 + 𝜔𝐶𝑒 𝑉1
𝐶𝑉2 = 𝐶𝑉1 + (0.1𝐶)(𝑉1 )
 𝑉2 = 1.1𝑉1
𝐾𝐶𝐿 @ 𝑏:
𝐼3 = 𝐼2 + 𝐼2′
𝐶𝑉3 = 𝐶𝑉2 + (0.1𝐶)(𝑉1 + 𝑉2 )
𝑉3 = 1.1𝑉2 + 0.1𝑉1
𝑉3 = (1. 1)2 𝑉1 + 0.1𝑉1
 𝑉3 = 1.31 𝑉1
ɳ=
ɳ=
𝑉1 +𝑉2 +𝑉3
3(𝑉3 )
𝑉1 +1.1𝑉1 +1.31𝑉1
3(1.31𝑉1 )
 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:
𝐶𝑃𝐿𝛼(𝑉∅ −𝑉0 )2
@ 𝑐𝑎𝑠𝑒 𝐼
𝐶𝑃𝐿 = 𝑘(𝑉∅1 −𝑉0 )2
@ 𝑐𝑎𝑠𝑒 𝐼𝐼
𝐶𝑃𝐿 = 𝑘(𝑉∅2 −𝑉0 )2
𝐶𝑃𝐿1
𝐶𝑃𝐿2
=

𝐶𝑃𝐿3
𝐶𝑃𝐿1
𝑘(𝑉∅1 −𝑉0 )2
𝑘(𝑉∅2 −𝑉0 )2
42
=
500
−𝑉0 )2
√3
200
𝑘( −𝑉0 )2
√3
𝑘(
𝑉0 = 34.25 𝑘𝑊
=
𝑘(𝑉∅3 −𝑉0 )2
𝑘(𝑉∅1 −𝑉0 )2
𝐶𝑃𝐿3 = 280

→
280
385
−34.25)2 ]
√3
500
[( −34.25)2 ]
√3
[(
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:
𝑍𝑏𝑎𝑠𝑒 =
𝑍𝑏𝑎𝑠𝑒 =
𝑉𝑏𝑎𝑠𝑒 2
𝑆𝑏𝑎𝑠𝑒
34,5002
100(106 )
 𝑍𝑏𝑎𝑠𝑒 = 11.9 𝑜ℎ𝑚𝑠
𝑍𝑎𝑐𝑡𝑢𝑎𝑙 = 𝑍𝑝.𝑢 𝑍𝑏𝑎𝑠𝑒
𝑍𝑎𝑐𝑡𝑢𝑎𝑙 = (0.06)(11.9)
 𝒁𝒂𝒄𝒕𝒖𝒂𝒍 = 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 + 𝑗5 Ω and the transmission line
impedance is 1 + 𝑗3 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
𝑍𝑝.𝑢 = 𝑎2 (2 + 𝑗5)
𝑍𝑝.𝑢 = 102 (2 + 𝑗5)
 𝑍𝑝.𝑢 = 200 + 𝑗500
480
𝑉𝐼𝐼
=
1
2
 𝑉𝐼𝐼 = 960 𝑉𝑜𝑙𝑡𝑠
𝐼𝐿 =
𝑉𝐼𝐼
𝑍𝑝.𝑢 𝑍𝑏𝑎𝑠𝑒
=
960
(200+𝑗500)(1+𝑗3)
 𝐼𝐿 = 1.77∠ − 68.22°
𝐼𝑏𝑎𝑠𝑒 =
𝑆𝑏𝑎𝑠𝑒
𝑉𝑏𝑎𝑠𝑒
=
20,000
480
𝐼𝑏𝑎𝑠𝑒 = 20.83 𝐴𝑚𝑝.
𝐼𝑝.𝑢 =

1.77∠−68.22°
20.83
𝑰𝒑.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:
𝑋𝑝.𝑢 =
𝑋𝑝.𝑢 =
𝑆𝑏𝑎𝑠𝑒
𝑆𝑠𝑐
15
400
 𝑋𝑝.𝑢 = 0.0375 𝑝. 𝑢.
𝑋𝑝.𝑢2 = 𝑋𝑝.𝑢1 + 𝑋𝑝.𝑢𝑇
𝑋𝑝.𝑢2 = 0.0375 + 0.025
 𝑋𝑝.𝑢2 = 0.0625 𝑝. 𝑢.
𝑆𝑠𝑐2 =
15(106 )
0.0625
 𝑺𝒔𝒄𝟐 = 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:
𝐼𝐹 𝑝.𝑢. =
𝐼𝐹 𝑝.𝑢. =
1
𝑋1 𝑝.𝑢
1
10 𝑝.𝑢.
 𝐼𝐹 𝑝.𝑢. = 0.1 𝑝. 𝑢
𝑆𝑠𝑐 =
𝑆𝑠𝑐 =
𝑆𝑏𝑎𝑠𝑒
𝑋𝑝.𝑢
50,000,000
0.1
 𝑺𝒔𝒄 = 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 𝑍(+) = 𝑗0.2577 𝑝. 𝑢.,
𝑍(−) = 𝑗0.2085 𝑝. 𝑢., and 𝑍(0) = 𝑗0.14 𝑝. 𝑢, determine the fault current at phase A
for a single line to ground fault at A.
A. – 𝑗6.4 𝑝. 𝑢
B. – 𝑗3 𝑝. 𝑢
C. – 𝑗11 𝑝. 𝑢
D. – 𝒋𝟓 𝒑. 𝒖
Solution:
𝐼𝐹𝐶 =
3
(𝑗0.2577)(𝑗0.2085)(𝑗0.14)
 𝑰𝑭𝑪 = −𝒋𝟓 𝒑. 𝒖
REE – April 2011
50. Consider a system with sequence impedances of 𝑍1 = 𝑗0.2577 𝑝. 𝑢., 𝑍2 =
𝑗0.2085 𝑝. 𝑢., and 𝑍0 = 𝑗0.14 𝑝. 𝑢., determine the fault current 𝐼𝑎1 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. 𝟔𝟓 𝒑. 𝒖.
Solution:
𝐼𝑎1 = 𝐼𝑎2 = 𝐼𝑎0
𝐼𝑎 = 𝐼𝑎1 + 𝐼𝑎2 + 𝐼𝑎0
 𝐼𝑎 = 3𝐼𝑎1
3
𝐼𝑎 = (𝑗0.2577)+(𝑗0.2085)+(𝑗0.14)
 𝐼𝑎 = −𝑗4.9
𝐼𝑎1 =
𝐼𝑎1 =
𝐼𝑎
3
−𝑗4.9
3
 𝑰𝒂𝟏 = −𝒋𝟏. 𝟔𝟓 𝒑. 𝒖
REE – September 2002
51. At a certain point in a 69 kV transmission line, the positive sequence impedance is
𝑗0.15 𝑝. 𝑢., and the zero sequence impedance is 𝑗0.55 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:
𝐼𝐹𝐶 =
𝐼𝐹𝐶 =

𝑆𝑏𝑎𝑠𝑒
(𝑉𝑏𝑎𝑠𝑒 )(𝑋1 𝑝.𝑢 +𝑋2 𝑝.𝑢 )
50,000,000
(69,000)(−𝑗0.15+𝑗0.15)
𝑰𝑭𝑪 = 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:
𝑋𝑇 𝑝.𝑢 = (𝑗0.2)(
10 𝑀𝑉𝐴
100 𝑀𝑉𝐴
)
 𝑋𝑇 𝑝.𝑢 = 𝑗0.02
𝑋1 𝑝.𝑢 = 𝑗0.02 + 𝑗0.05
 𝑋1 𝑝.𝑢 = 𝑗0.07
1
𝐼𝐹 𝑝.𝑢 =
𝑋1 𝑝.𝑢
=
1
𝑗0.07
𝐼𝐹 𝑝.𝑢 = 14.29 𝑝. 𝑢
𝐼𝐵 𝑝 =
𝑆𝐵
√3𝑉𝐵 𝑝
=
𝑆𝐵
√3(115,000)
 𝐼𝐵 𝑝 = 50.20 𝐴𝑚𝑝.
𝐼𝐹𝐶 𝑝𝑟𝑖 = 𝐼𝐹 𝑝.𝑢 + 𝐼𝐵 𝑝 = (14.29)(50.2)
 𝑰𝑭𝑪 𝒑𝒓𝒊 = 𝟕𝟏𝟕. 𝟑𝟔 𝑨𝒎𝒑.
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:
𝑍𝑏𝑎𝑠𝑒 =
𝑉2
𝑆𝑏𝑎𝑠𝑒
=
4802
5,000,000
 𝑍𝑏𝑎𝑠𝑒 = 0.046 𝑝. 𝑢
𝑋𝑝.𝑢 =
0.032
0.046
𝑋𝑝.𝑢 = 0.7 𝑝. 𝑢
𝑋𝑝.𝑢 𝑇 = 0.7 + 0.05
 𝑋𝑝.𝑢 𝑇 = 0.75 𝑝. 𝑢
𝐼𝐹𝐶 =

𝑆𝐵
√3𝑉𝐵 𝑋𝑝.𝑢 𝑇
=
𝑆𝐵
√3(480)(0.75)
𝑰𝑭𝑪 = 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:
𝑋1 𝑝.𝑢 =
𝑆𝐵
𝑆𝑆𝐶
=
𝑆𝐵
𝑆𝑆𝐶 𝑝𝑟𝑖
=
50 𝑀𝑉𝐴
600 𝑀𝑉𝐴
 𝑋1 𝑝.𝑢 = 0.083 𝑝. 𝑢
𝑋1 𝑝𝑟𝑠 =
𝑋1 𝑝𝑟𝑠 =
𝑆𝐵
𝑆𝑆𝐶
50 𝑀𝑉𝑎
200 𝑀𝑉𝐴
 𝑋1 𝑝𝑟𝑠 = 0.25 𝑝. 𝑢
𝑋𝑅 𝑝.𝑢 = 𝑋1 𝑝𝑟𝑠 − 𝑋1 𝑝.𝑢 − 𝑋𝑝.𝑢 𝑇
𝑋𝑅 𝑝.𝑢 = 0.25 − 0.083 − 0.04
 𝑿𝑹 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:
𝑗0.12
𝑗0.2
+
-
𝑗0.2
+
N
𝑉𝑇𝐻 =
12,800
13,200
 𝑉𝑇𝐻 = 0.97
𝐼𝐹 𝑝.𝑢 =
𝐼𝐹 𝑝.𝑢 =
𝑉𝑇𝐻
𝑍𝑇𝐻
0.97
𝑗0.123
 𝐼𝐹 𝑝.𝑢 = −𝑗7.89 𝑝. 𝑢
𝐼𝐹 = 𝐼𝐹 𝑝.𝑢 + 𝐼𝑏𝑎𝑠𝑒
𝐼𝐹 = (7.89) + (
30,000,000
√3(13,200)
 𝑰𝑭 = 𝟏𝟎, 𝟑𝟓𝟐. 𝟗𝟒 𝑨𝒎𝒑.