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```Board Exam Problems
on
AC MACHINES
(Alternators)
41
1.
The number of cycles generated in a 10 pole alternator in one revolution is
A. 2
B. 5
C. 10
Solution:
𝑛𝑜. 𝑜𝑓
𝑐𝑦𝑐𝑙𝑒𝑠
𝑟𝑒𝑣

2.
1 𝑐𝑦𝑐𝑙𝑒
=(
𝒏𝒐. 𝒐𝒇
2 𝑝𝑜𝑙𝑒
𝒄𝒚𝒄𝒍𝒆𝒔
𝒓𝒆𝒗
)(
10 𝑝𝑜𝑙𝑒𝑠
𝑟𝑒𝑣
.
D. 20
)
= 𝟓 𝒄𝒚𝒄𝒍𝒆𝒔/𝒓𝒆𝒗
What is the number of electrical degrees made per revolution for a special purpose 40 pole
alternator.
A. 28, 800
B. 14, 400
C. 7, 200
D. 1, 600
Solution:
𝑛𝑜. 𝑜𝑓 𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑎𝑙 𝑑𝑒𝑔/𝑟𝑒𝑣 = (
180°
𝑝𝑜𝑙𝑒

40 𝑝𝑜𝑙𝑒𝑠
)(
𝑟𝑒𝑣
)
𝒏𝒐. 𝒐𝒇 𝒆𝒍𝒆𝒄𝒕𝒓𝒊𝒄𝒂𝒍 𝒅𝒆𝒈/𝒓𝒆𝒗 = 𝟕, 𝟐𝟎𝟎°/𝒓𝒆𝒗
3.
Alternators are usually designed to generate
A. variable frequency
B. definite currents
C. definite frequencies
D. definite p.f
4.
A 4-pole, 3-phase, star-connected alternator armature has 12 slots with 24 conductors per slot and
the flux per pole is 0.1 Wb sinusoidally distributed. Calculate the line emf generated at 50 Hz.
A. 1, 066 V
B. 3, 198 V
C.1, 846 V
D. 5, 538 V
Solution:
24 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑜𝑟𝑠
𝑁∅ = (

𝑠𝑙𝑜𝑡
)(12 𝑠𝑙𝑜𝑡𝑠)(
1 𝑡𝑢𝑟𝑛
2 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑜𝑟𝑠
)(
1
3 𝑝ℎ𝑎𝑠𝑒
)
𝑁∅ = 48 𝑡𝑢𝑟𝑛𝑠/𝑝ℎ𝑎𝑠𝑒
1 turn
1 conductor
1 conductor
𝐸∅ = 4.44𝑓𝑁∅ ∅
𝐸∅ = 4.44(50 𝐻𝑧)(48 𝑡𝑢𝑟𝑛𝑠/∅)(0.1

𝑊𝑏
𝑝𝑜𝑙𝑒
)
𝐸∅ = 1, 065.6 𝑉
𝐸𝐿 = √3𝐸∅
𝐸𝐿 = √3(1, 065.6)
 𝑬𝑳 = 𝟏, 𝟖𝟒𝟓. 𝟔𝟕 𝑽
REE – May 2008
5. A three-phase wye-connected 50 Hz, 2-pole synchronous machine has a stator with 2, 000 turns of
wire per phase. What rotor flux would be required to produce a terminal (line to line) voltage of 5
kV?
A. 8.4 mWb
B. 6.5 mWb
C. 5.2 mWb
D. 7.8 mWb
Solution:
𝐸∅ = 4.44𝑓𝑁∅ ∅
𝐸∅ =

𝐸𝐿
√3
=
5,000
√3
𝐸∅ = 2, 886.75 𝑉𝑜𝑙𝑡𝑠
42
𝐸∅
∅=
4.44𝑓𝑁∅
2,886.75
∅=
4.44(50)(2,000)

∅ = 𝟔. 𝟓 × 𝟏𝟎−𝟑 𝑾𝒃
REE – April 2006
6. A 3-phase, 8-pole generator is delta-connected. The terminal voltage is 2, 400 Volts while the line
current is 500 Amperes. If the machine is converted to wye, what will be the terminal voltage in
volts?
A. 4, 800 V
B. 4, 157 V
C. 4, 600 V
D. 3, 800 V
Solution:
∗ 𝑤ℎ𝑒𝑛 𝑐𝑜𝑛𝑛𝑒𝑐𝑡𝑒𝑑 𝑖𝑛 𝑑𝑒𝑙𝑡𝑎
 𝐸∅ = 𝐸𝐿 = 2, 400 𝑉𝑜𝑙𝑡𝑠
∗ 𝑤ℎ𝑒𝑛 𝑐𝑜𝑛𝑛𝑒𝑐𝑡𝑒𝑑 𝑖𝑛 𝑤𝑦𝑒
𝐸𝐿 = √3𝐸∅
𝐸𝐿 = √3(2, 400)
 𝑬𝑳 = 𝟒, 𝟏𝟓𝟔. 𝟗𝟐 𝑽𝒐𝒍𝒕𝒔
REE – September 2005
7. A three-phase, 8-pole, 2,400 Volts delta connected generator has a line current of 500 Amperes. If
converted in wye, what is the tolerable current?
A. 371.5 A
B. 288.7 A
C. 245.1 A
D. 315.2 A
Solution:
∗ 𝑤ℎ𝑒𝑛 𝑐𝑜𝑛𝑛𝑒𝑐𝑡𝑒𝑑 𝑖𝑛 𝑑𝑒𝑙𝑡𝑎
𝐼∅ =
𝐼𝐿
√3
=
500
√3
 𝐼∅ = 288.68 𝐴𝑚𝑝𝑒𝑟𝑒𝑠
∗ 𝑤ℎ𝑒𝑛 𝑐𝑜𝑛𝑛𝑒𝑐𝑡𝑒𝑑 𝑖𝑛 𝑤𝑦𝑒
 𝑰𝑳 = 𝑰∅ = 𝟐𝟖𝟖. 𝟔𝟖 𝑨𝒎𝒑𝒆𝒓𝒆𝒔
8.
If the alternator winding has fractional pitch of 4/5, the pitch factor 𝑘𝑝 is
A. 0.833
Solution:
B. 0.966
4
𝑝° = (180°)
5
C. 0.972
.
D. 0.951
full-pitch coil span or 1 pole span
 𝑝° = 144
𝑝°
𝑘𝑝 = 𝑠𝑖𝑛( )
𝑘𝑝 = 𝑠𝑖𝑛(
2
144
2
)
 𝒌𝒑 = 𝟎. 𝟗𝟓𝟏
9. A six pole, three-phase alternator has 72 slots and a coil span of 1 to 10, what is the
pitch factor?
A. 0.924
B. 0.939
C. 0.966
D. 0.985
Solution:
43
10-1= 9 slots
1
𝑝° =

10
(10−1)𝑠𝑙𝑜𝑡𝑠
72 𝑠𝑙𝑜𝑡𝑠
6 𝑝𝑜𝑙𝑒
(180°)
𝑝° = 135
𝑝°
𝑘𝑝 = 𝑠𝑖𝑛( )
𝑘𝑝 = 𝑠𝑖𝑛(

2
135
2
)
𝒌𝒑 = 𝟎. 𝟗𝟐𝟒
10. The power factor of an alternator is 75%. The operator is ordered to increase the power factor to
80%. What shall he do?
A. increase the voltage
B. operate the governor
C. increase the excitation
D. decrease the excitation
11. Calculate the distribution factor for a 36-slot, 4-pole three phase winding.
A. 0.924
B. 0.951
C. 0.960
Solution:
𝑛=
D. 0.975
36 𝑠𝑙𝑜𝑡𝑠
(4 𝑝𝑜𝑙𝑒)(3 𝑝ℎ𝑎𝑠𝑒𝑠)
𝑠𝑙𝑜𝑡𝑠

𝑑° =
𝑛=3
𝑝𝑜𝑙𝑒
/𝑝ℎ𝑎𝑠𝑒
180°
(
)(4 𝑝𝑜𝑙𝑒𝑠)
𝑝𝑜𝑙𝑒

36 𝑠𝑜𝑡𝑠
𝑑° = 20°
𝑑°
𝑘𝑑° =
𝑘𝑑° =

𝑠𝑖𝑛 𝑛( 2 )
𝑑°
2
20°
𝑠𝑖𝑛 3( 2 )
20°
3 𝑠𝑖𝑛( )
2
𝑛 𝑠𝑖𝑛( )
𝒌𝒅° = 𝟎. 𝟗𝟔𝟎
12. A 144-slot stator has a whole-coiled 12 pole three-phase winding. What is the number of coils per
phase and per group, respectively?
A. 12 and 4
B. 24 and 4
C. 36 and 4
D. 48 and 4
Solution:
𝑛𝑜. 𝑜𝑓 𝑐𝑜𝑖𝑙𝑠 𝑝𝑒𝑟 𝑝ℎ𝑎𝑠𝑒 =
1 𝑐𝑜𝑖𝑙
)(144 𝑠𝑙𝑜𝑡)
𝑠𝑙𝑜𝑡
(
3 𝑝ℎ𝑎𝑠𝑒
44

𝒏𝒐. 𝒐𝒇 𝒄𝒐𝒊𝒍𝒔 𝒑𝒆𝒓 𝒑𝒉𝒂𝒔𝒆 = 𝟒𝟖 𝒄𝒐𝒊𝒍𝒔 𝒑𝒆𝒓 𝒑𝒉𝒂𝒔𝒆
𝑛𝑜. 𝑜𝑓 𝑐𝑜𝑖𝑙𝑠 𝑝𝑒𝑟 𝑔𝑟𝑜𝑢𝑝 =

48 𝑐𝑜𝑖𝑙𝑠/𝑝ℎ𝑎𝑠𝑒
12 𝑔𝑟𝑜𝑢𝑝𝑠/𝑝ℎ𝑎𝑠𝑒
𝒏𝒐. 𝒐𝒇 𝒄𝒐𝒊𝒍𝒔 𝒑𝒆𝒓 𝒈𝒓𝒐𝒖𝒑 = 𝟒 𝒄𝒐𝒊𝒍𝒔 𝒑𝒆𝒓 𝒈𝒓𝒐𝒖𝒑
13. A part of an alternator winding consists of six coils in series, each coil having an e.m.f of 10 Volts
(r.m.s) induced in it. The coils are placed in successive slots and between each slot and the next, there
is an electrical phasor displacement of 30°. Find the e.m.f of the six coils in series.
A. 77.28 V
B. 9.66 V
C. 38.64 V
D. 19.32 V
Solution:
𝐸𝑅 = 𝐸𝑐1 + 𝐸𝑐2 + 𝐸𝑐3 + 𝐸𝑐4 + 𝐸𝑐5 + 𝐸𝑐6
𝐸𝑅 = 10∠0° + 10∠30° + 10∠60° + 10∠90° + 10∠120° + 10∠150°
 𝑬𝑹 = 𝟑𝟖. 𝟔𝟒∠𝟕𝟓° 𝑽𝒐𝒍𝒕𝒔
14. The following information is given in connection with an alternator: 𝑠𝑙𝑜𝑡𝑠 = 144; 𝑝𝑜𝑙𝑒𝑠 = 8; 𝑟𝑝𝑚 =
900; 𝑡𝑢𝑟𝑛𝑠 𝑝𝑒𝑟 𝑐𝑜𝑖𝑙 = 6; ∅ = 1.8 × 106 ; 𝑐𝑜𝑖𝑙 𝑠𝑝𝑎𝑛 = 𝑠𝑙𝑜𝑡𝑠 1 𝑡𝑜 16; 𝑤𝑖𝑛𝑑𝑖𝑛𝑔 𝑐𝑜𝑛𝑛𝑒𝑐𝑡𝑖𝑜𝑛𝑠 = 𝑠𝑡𝑎𝑟.
What is the voltage generated between terminals?
A. 1, 276 V
B. 2, 210 V
C. 635 V
D. 1, 100 V
Solution:
(16−1)𝑠𝑙𝑜𝑡𝑠
𝑝° =
144 𝑠𝑙𝑜𝑡𝑠
8 𝑝𝑜𝑙𝑒𝑠

(
180°
𝑝𝑜𝑙𝑒
)
𝑝° = 150°
𝑝°
150°
2
2
𝑘𝑝° = 𝑠𝑖𝑛 ( ) = 𝑠𝑖𝑛(

𝑛=
𝑘𝑝° = 0.966
144 𝑠𝑙𝑜𝑡𝑠
(8 𝑝𝑜𝑙𝑒𝑠)(3 ∅)
𝑠𝑙𝑜𝑡𝑠

𝑑° =
𝑛=6
𝑝𝑜𝑙𝑒
/𝑝ℎ𝑎𝑠𝑒
180°
(
)(8 𝑝𝑜𝑙𝑒𝑠)
𝑝𝑜𝑙𝑒
144 𝑠𝑙𝑜𝑡𝑠

𝑑° = 10°/𝑠𝑙𝑜𝑡
𝑑°
𝑘𝑑° =
𝑠𝑖𝑛 𝑛( 2 )
𝑑°
𝑛 𝑠𝑖𝑛( )
2

𝑓=
)
10°
=
𝑠𝑖𝑛 6( 2 )
10°
6 𝑠𝑖𝑛( )
2
𝑘𝑑° = 0.956
𝑝×𝑟𝑝𝑚
120

=
2(900)
120
𝑓 = 60 𝐻𝑧
6 𝑡𝑢𝑟𝑛𝑠
𝑁∅ = (
𝑐𝑜𝑖𝑙
)(
1 𝑐𝑜𝑖𝑙
𝑠𝑙𝑜𝑡
)(144 𝑠𝑙𝑜𝑡𝑠)(
1
3 𝑝ℎ𝑎𝑠𝑒)
 𝑁∅ = 288 𝑡𝑢𝑟𝑛𝑠/𝑝ℎ𝑎𝑠𝑒
𝐸∅ = 4.44(𝑘𝑝°)(𝑘𝑑°)𝑓𝑁∅ ∅ × 10−8
𝐸∅ = 4.44(0.966)(0.956)(60)(288)(1.8 × 106 )(10−8 )
 𝐸∅ = 1, 275.36 𝑉𝑜𝑙𝑡𝑠
𝐸𝐿 = √3𝐸∅ = √3(1, 275.36)
 𝑬𝑳 = 𝟐, 𝟐𝟎𝟗 𝑽𝒐𝒍𝒕𝒔
15. The disadvantage of a short-pitch coil is that
.
A. harmonics are introduced
B. waveform becomes non-sinusoidal
C. voltage around the coil is reduced D. both a and b
45
16. A 4-pole alternator, on open circuit, generates 200 Volts at 50 Hz when the field current is 4
Amperes. Determine the generated e.m.f at a speed of 1, 200 rpm and a field current of 3 Amperes,
neglecting saturation in the iron parts.
A. 40 V
B. 240 V
C. 60 V
D. 120 V
Solution:
𝐸 = 𝑘𝑓𝐼𝑓
𝐸1 = 𝑘𝑓1 𝐼𝑓
1
𝐸1
𝑘=
𝑓1 𝐼𝑓
(50)(4)
1

𝑓2 =
200
=
𝑘=1
𝑃×𝑟𝑝𝑚
120
=
4(1,200)
120
 𝑓2 = 40 𝐻𝑧
𝐸2 = 𝑘𝑓2 𝐼𝑓
2
𝐸2 = 1(40)(3)
 𝑬𝟐 = 𝟏𝟐𝟎 𝑽𝒐𝒍𝒕𝒔
17. A 3-phase. Wye-connected alternator having a sinusoidal line potential with a 5% 5 th harmonic
content, 3% 7th harmonic content, 1.5% 11th harmonic content and 0.85% 13 th harmonic content.
What is the total harmonic distortion (THD) of the system line voltage?
A. 11%
B. 9%
C. 6%
D. 3%
Solution:
𝑇𝐻𝐷 = √
𝑇𝐻𝐷 = √

𝐸5 2 +𝐸7 2 +𝐸11 2 +𝐸13 2
𝐸1
(0.05𝐸1 )2 +(0.03𝐸1 )2 +(0.015𝐸1 )2 +(0.0085𝐸1 )2
𝐸1
𝑻𝑯𝑫 = 𝟔. 𝟎𝟖%
18. A 100 MVA, 13.8 kV, three-phase, wye-connected alternator will have a per phase nominal
impedance of
.
Ω
A. 3.5 ⁄∅
B. 2.9 Ω⁄∅
C. 1.9 Ω⁄∅
D. 7.5 Ω⁄∅
Solution:
𝑍∅ =
𝑍∅ =
𝑘𝑉 2
𝑀𝑉𝐴
13.82
100
𝒁∅ = 𝟏. 𝟗 Ω⁄∅

REE – March 1998
19. A generator is rated 100 MW, 13.8 kV and 90% power factor. The effective resistance to ohmic
resistance is 1.5. The ohmic resistance is obtained by connecting two terminals to a d.c source. The
current and voltage are 87.6 amperes and 6 Volts, respectively. What is the resistance per phase?
A. 0.0617 Ω
𝑩. 𝟎. 𝟎𝟓𝟏𝟑 Ω
C. 0.0685 Ω
D. 0.342 Ω
Solution:
𝑅𝑡 =
𝑉𝑑𝑐
𝐼𝑑𝑐

𝑅𝑎 𝑑𝑐 =
=
6
87.6
𝑅𝑡 = 0.0685 Ω
𝑅𝑡
2
=
→ dc ohmic resistance between terminals
0.0685
2
𝑅𝑎 𝑑𝑐 = 0.03425 Ω
46

𝐴𝐶 𝑓𝑎𝑐𝑡𝑜𝑟 =
𝑅𝑎
𝑅𝑎 𝑑𝑐
𝑅𝑎 = 1.5(0.03425)
 𝑹𝒂 = 𝟎. 𝟎𝟓𝟏𝟑 Ω/∅
REE – April 2006
20. A single-phase alternator gives 250 Amperes at 1, 200 Volts. What is the output of the machine in
KVA?
A. 300
B. 450
C. 400
D. 350
Solution:
𝐾𝑉𝐴𝑜𝑢𝑡𝑝𝑢𝑡 =
𝐾𝑉𝐴𝑜𝑢𝑡𝑝𝑢𝑡 =

𝑉𝐿 𝐼 𝐿
1,000
(1,200)(250)
1,000
𝑲𝑽𝑨𝒐𝒖𝒕𝒑𝒖𝒕 = 𝟑𝟎𝟎
21. High speed alternators have a rotor construction
A. similar to d.c. machines
B. of non-salient type
C. of salient type
D. none of these
22. The imaginary or fictitious part of synchronous reactance on alternator takes care of
A. inductive reactance
B. leakage reactance
C. armature reaction
D. copper losses
REE – April 2006
23. A single-phase generator delivers 80 A at 240 V and 75% lagging p.f. What kind of load is the
generator supplying?
A. capacitive
B. resistive
C. inductive
D. reactive
REE – September 2004
24. A 3,600 Volts, 500 kVA, 60 Hz, three-phase, Y-connected generator on test yields the following
results: Mechanical and iron losses are 10 kW; field current at full load 100% p.f is 50 Amperes; field
current at full load 80% p.f is 70 Amperes; resistance per phase of armature winding is 0.4 ohm. The
exciter voltage is constant at 120 Volts and voltage control is done by means of rheostat. Determine
the full load armature current at 80% lagging p.f
A. 96.5 A
B. 99.8 A
C. 64.2 A
D. 80.2 A
Solution:
𝐼𝑎 = 𝐼𝐿 =
𝐼𝑎 =
𝐾𝑉𝐴𝑜𝑢𝑡𝑝𝑢𝑡 (1,000)
√3
(500)(1,000)

√3
𝑰𝒂 = 𝟖𝟎. 𝟏𝟗 𝑨𝒎𝒑𝒆𝒓𝒆𝒔
REE – April 2005
25. A single-phase 2, 400 Volts synchronous generator delivers 450 Amperes at unity power factor. The
synchronous impedance of the generator is 0.08 + 𝑗2.8 Ω. What is the regulation in percent?
A. 14.2
B. 16.3
C. 20.1
D. 18.4
Solution:
𝐸𝑔 = 𝑉𝑡 + 𝐼𝑎 𝑍𝑠
𝐸𝑔 = 2, 400∠0° + (450∠0°)(0.08 + 𝑗2.8)
47

𝐸𝑔 = 2, 742.57∠27.35° 𝑉𝑜𝑙𝑡𝑠
%𝑉. 𝑅 =
%𝑉. 𝑅 =

𝐸𝑔 −𝑉𝑡
× 100
𝑉𝑡
2,742.57−2,400
2,400
× 100
%𝑽. 𝑹 = 𝟏𝟒. 𝟐𝟕%
26. A 25 kVa alternator has a total loss of 2, 000 watts when it delivers rated kVA to a load at a power
factor of 0.76. Calculate its percent efficiency.
A. 90.48%
B. 92.32%
C. 91.58%
D. 89.12%
Solution:
𝑃𝑜𝑢𝑡 = 𝑘𝑉𝐴(𝑝. 𝑓) = 25(0.76)
 𝑃𝑜𝑢𝑡 = 19 𝑘𝑊
ɳ=
ɳ=
𝑃𝑜𝑢𝑡
× 100
𝑃𝑜𝑢𝑡 +𝑡𝑜𝑡𝑎𝑙 𝑙𝑜𝑠𝑠
19,000
19,000+2,000
× 100
 ɳ = 𝟗𝟎. 𝟒𝟖%
REE – September 2010
27. A three-phase, wye-connected wound rotor synchronous generator rated at 10 kVA, 230 V has a
synchronous reactance of 1.2 ohms/phase and an armature resistance of 0.5 ohm/phase. What is the
percent voltage regulation at full-load with 85% lagging power factor?
A. 18.3%
B. 24.7%
C. 20.8%
D. 22.5%
Solution:
𝑉∅ =
𝑉𝐿
√3

∠0° =
230
√3
∠0°
𝑉∅ = 132.79∠0° 𝑉𝑜𝑙𝑡𝑠
𝐼𝑎 = 𝐼𝐿 =
𝑟𝑎𝑡𝑒𝑑 𝑘𝑉𝑎 ×1,000
√3 𝑉𝐿𝑟𝑎𝑡𝑒𝑑
=
10,000
√3 (230)
𝐼𝑎 = 25.1∠ − 31.79° 𝐴𝑚𝑝𝑒𝑟𝑒𝑠
𝐸𝑔 = 𝑉∅ + 𝐼𝑎 𝑍𝑠
𝐸𝑔 = 132.79∠0° + (25.1∠ − 31.79° )(0.5 + 𝑗1.2)

𝐸𝑔 = 160.45∠6.8° 𝑉𝑜𝑙𝑡𝑠
%𝑉. 𝑅 =
%𝑉. 𝑅 =

𝐸𝑔 −𝑉∅
× 100
𝑉∅
160.45−132.79
132.79
100
%𝑽. 𝑹 = 𝟐𝟎. 𝟖%
REE – September 2010
28. A three-phase, wye-connected wound rotor synchronous generator rated at 10 kVA, 230 V has a
synchronous reactance of 1.2 ohms/phase and armature resistance of 0.5 ohm/phase. What is the
A. – 0.92%
B. – 5.14%
C. – 3.08%
D. – 3.91%
Solution:
𝑉∅ =

𝑉𝐿
√3
∠0° =
230
√3
∠0°
𝑉∅ = 132.79∠0° 𝑉𝑜𝑙𝑡𝑠
𝐼𝑎 = 𝐼𝐿 =
10,000
√3 (230)
∠𝑐𝑜𝑠 −1 0.80
 𝐼𝑎 = 25.1∠36.87° 𝐴𝑚𝑝𝑒𝑟𝑒𝑠
𝐸𝑔 = 𝑉∅ + 𝐼𝑎 𝑍𝑠 = 132.79∠0° + (25.1∠36.87° )(0.5 + 𝑗1.2)
48

𝐸𝑔 = 128.69∠14.22° 𝑉𝑜𝑙𝑡𝑠
%𝑉. 𝑅 =

𝐸𝑔 −𝑉∅
𝑉∅
× 100
%𝑽. 𝑹 = −𝟑. 𝟎𝟖%
REE – May 2010
29. A three-phase wye-connected wound rotor synchronous generator rated at 10 kVA, 230 V has
synchronous reactance of 1.2 ohms per phase and armature resistance of 0.5 ohm per phase. What is
the power factor such that the voltage regulation at full load is zero?
Solution:
𝐼𝑎 = 𝐼𝐿 =
10,000
√3 (230)
∠0°
 𝐼𝑎 = 25.1∠0° 𝐴𝑚𝑝𝑒𝑟𝑒𝑠
𝐼𝑎 𝑍𝑠 = (25.1∠0° )((0.5 + 𝑗1.2)
 𝐼𝑎 𝑍𝑠 = 32.63∠67.38° 𝑉𝑜𝑙𝑡𝑠
𝑉∅ =

𝑉𝐿
√3
∠0° =
230
√3
∠0° and
%𝑉. 𝑅 =
𝐸𝑔 −𝑉∅
𝑉∅
× 100 = 0
𝐸∅ = 𝑉∅ = 132.79∠0° 𝑉𝑜𝑙𝑡𝑠
𝐼𝑎 𝑍𝑠
𝐸∅
𝐼𝑎 𝑋𝑠
𝐼𝑎 = 𝐼𝐿
67.38°
𝛼
𝜃
𝐼𝑎 𝑅𝑎
𝑉∅
∗ 𝑏𝑦 𝑐𝑜𝑠𝑖𝑛𝑒 𝑙𝑎𝑤:
𝐸∅ 2 = 𝑉∅ 2 + (𝐼𝑎 𝑍𝑠 )2 − 2(𝑉∅ )(𝐼𝑎 𝑍𝑠 )𝑐𝑜𝑠𝛼
(132.73)2 = (132.73)2 + (32.63)2 − 2(132.73)(32.63)𝑐𝑜𝑠𝛼
𝛼 = 𝑐𝑜𝑠 −1 [
32.632
]
2(132.73)(32.63)
 𝛼 = 82.94°
𝜃 = 180° − 82.94° − 63.78°
 𝜃 = 29.68°
𝑝. 𝑓 = 𝑐𝑜𝑠 𝜃 = cos 29.68°
 𝒑. 𝒇 = 𝟎. 𝟖𝟔𝟗 𝒍𝒆𝒂𝒅𝒊𝒏𝒈
30. A 1, 000 kVA, 3kV, 50 Hz, thee-phase star-connected alternator has an armature effective resistance
of 0.2 Ω. A field current of 40 A produces a short-circuit current of 200 A and an open-circuit e.m.f of
1, 040 V (line value). Calculate the full load percentage regulation at a power factor of 0.8 lagging.
49
A. –21.43%
Solution:
B. 21.43%
C. –24.31%
D. 24.31
3,000
𝑉∅ =
√3

𝑉∅ = 1, 732.1∠0° 𝑉𝑜𝑙𝑡𝑠
𝐼𝑎 = 𝐼𝐿 =

√3 (3,000)
∠ − 𝑐𝑜𝑠 −1 0.80
𝐼𝑎 = 192.45∠ − 36.87° 𝐴𝑚𝑝𝑒𝑟𝑒𝑠
𝐸𝑂𝐶
𝑍𝑠 =
1,000(1,000)
=
𝐼𝑆𝐶

1,040/√3
200
𝑍𝑠 = 3 𝑜ℎ𝑚𝑠
𝑋𝑠 = √𝑍𝑠 2 − 𝑅𝑎 2 = √32 − 0.22
 𝑋𝑠 = 2.99 𝑜ℎ𝑚𝑠
𝐸𝑔 = 𝑉∅ + 𝐼𝑎 𝑍𝑠 = 1, 732.1∠0° + (192.45∠ − 36.87° )(3 + 𝑗2.99)

𝐸𝑔 = 2, 152.92∠11.72° 𝑉𝑜𝑙𝑡𝑠
%𝑉. 𝑅 =

𝐸𝑔 −𝑉∅
𝑉∅
× 100 =
2,152.92−1,732.1
1,732.1
× 100
%𝑽. 𝑹 = 𝟐𝟒. 𝟑%
31. Adjustment of field excitation on one of the two alternators operating in parallel will
B. change its frequency
D. change its power factor
REE – September 2001
32. A generator is being synchronized to a large system. The actual system voltage and frequency are
13.7 kV and 60 Hz, respectively. The generator voltage and frequency are 13.6 kV and 60 Hz,
respectively. When the generator is switched to the system, choose which one happens,
A. generator delivers MVAR
B. generator takes MVAR
C. generator delivers MW
D. generator delivers both MW and MVAR
Solution:
𝑠𝑖𝑛𝑐𝑒 𝑡ℎ𝑒 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑖𝑠 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 𝑣𝑜𝑙𝑡𝑎𝑔𝑒
𝑡ℎ𝑒 𝒈𝒆𝒏𝒆𝒓𝒂𝒕𝒐𝒓 𝒕𝒂𝒌𝒆𝒔 𝑴𝑽𝑨𝑹 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚
33. A 25 kVA, 220 Volts, three-phase alternator delivers rated kVA at a power factor of 0.84. The effective
ac resistance between armature winding terminals is 0.18 Ω. The field takes 9.3 Amperes at 115
Volts. If friction and windage loss is 460 watts and the core loss is 610 Watts. Calculate the percent
efficiency.
A. 87.82%
B. 84.27%
C. 86.41%
D. 88.33%
Solution:
𝑃𝑜𝑢𝑡 = 25(0.84)
 𝑃𝑜𝑢𝑡 = 21 𝑘𝑊
𝐼𝑎 =
25,000
√3 (220)
 𝐼𝑎 = 65.61 𝐴𝑚𝑝𝑒𝑟𝑒𝑠
∗ 𝑌 − 𝑐𝑜𝑛𝑛𝑒𝑐𝑡𝑒𝑑
𝑅𝑎 =

𝑅𝑡
2
=
0.18
2
𝑅𝑎 = 0.09 𝑜ℎ𝑚𝑠
𝑳𝒐𝒔𝒔𝒆𝒔:
1. 𝑎𝑟𝑚𝑎𝑡𝑢𝑟𝑒 𝑤𝑖𝑛𝑑𝑖𝑛𝑔
50
𝑃𝑎 = 3𝐼𝑎 2 𝑅𝑎 = 3(65.612 )(0.09)
𝑃𝑎 = 1, 162.26 𝑊𝑎𝑡𝑡𝑠
2. 𝑓𝑖𝑒𝑙𝑑 𝑤𝑖𝑛𝑑𝑖𝑛𝑔𝑠
𝑃𝑓 = 𝑉𝑓 𝐼𝑓 = (115)(9.3)
𝑃𝑓 = 1. 069.5 𝑊𝑎𝑡𝑡𝑠
3. 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝑎𝑛𝑑 𝑤𝑖𝑛𝑑𝑎𝑔𝑒 𝑙𝑜𝑠𝑠
𝑃𝑓𝑤 = 460 𝑊𝑎𝑡𝑡𝑠
4. 𝑐𝑜𝑟𝑒 𝑙𝑜𝑠𝑠
𝑃𝑐 = 610 𝑊𝑎𝑡𝑡𝑠
𝑡𝑜𝑡𝑎𝑙 𝑙𝑜𝑠𝑠𝑒𝑠 = 𝑃𝑎 + 𝑃𝑓 + 𝑃𝑓𝑤 + 𝑃𝑐 = 1, 162.26 + 1, 069.5 + 460 + 610

ɳ=
𝑡𝑜𝑡𝑎𝑙 𝑙𝑜𝑠𝑠𝑒𝑠 = 3, 301.76 𝑊𝑎𝑡𝑡𝑠
𝑃𝑜𝑢𝑡
𝑃𝑜𝑢𝑡 +𝑃𝑙𝑜𝑠𝑠𝑒𝑠

=
21
21+3.302
ɳ = 𝟖𝟔. 𝟒𝟏%
34. When the speed of an alternator is changed from 3, 600 rpm to 1, 800 rpm, the generated emf/phase
will become
A. one half
B. twice
C. four times
D. one fourth
REE – September 2009
35. Two alternators operating in parallel supply 2, 500 kW to a load at a power factor of 80% lagging. If
one machine delivers 1, 200 kW at a power factor of 95% lagging, what is the power supplied by the
other machine?
A. 1, 400 kW
B. 1, 600 kW
C. 1, 300 kW
D. 1, 500 kW
Solution:
𝑃𝑡 = 𝑃1 + 𝑃2
𝑃2 = 𝑃𝑡 − 𝑃1
𝑃2 = 2, 500 − 1, 200
 𝑷𝟐 = 𝟏, 𝟑𝟎𝟎 𝒌𝑾
36. Two exactly similar turbo-alternators are rated 20 MW each. They are running in parallel. The speed
load characteristics of the driving turbines are such that the frequency of alternator one drops
uniformly from 50 Hz on no-load to 48 Hz on full-load, that of the alternator two from 50 Hz to 48.5
Hz. How will the two machines share a load of 30, 000 kW?
A. 14.56 MW, 15.44 MW
B. 12.80 MW, 17.20 MW
C. 17.47 MW, 12.53 MW
D. 16.92 MW, 13.08 MW
Solution:
∗ 𝒇𝒐𝒓 𝒂𝒍𝒕𝒆𝒓𝒏𝒂𝒕𝒐𝒓 𝟏
50−𝑓
50−48
=
𝑀𝑊1
20
 𝑀𝑊1 = 500 − 10𝑓
∗ 𝒇𝒐𝒓 𝒂𝒍𝒕𝒆𝒓𝒏𝒂𝒕𝒐𝒓 𝟐
50−𝑓
50−48.5
=
→ 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝟏
𝑀𝑊2
20
𝑀𝑊2 = 666.67 − 13.33𝑓
30 − 𝑀𝑊1 = 666.67 − 13.33𝑓
 𝑀𝑊1 = 13.33𝑓 − 636.67 → 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝟐
∗ 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝟏 = 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝟐
500 − 10𝑓 = 13.33𝑓 − 636.67
51
𝑓=
500+636.67
13.33+10

𝑓 = 48.72 𝐻𝑧
𝑀𝑊1 = 500 − 10𝑓
𝑀𝑊1 = 500 − 10(48.72)
 𝑴𝑾𝟏 = 𝟏𝟐. 𝟖 𝑴𝑾
𝑀𝑊2 = 666.67 − 13.33𝑓
𝑀𝑊2 = 666.67 − 13.33(48.72)
 𝑴𝑾𝟐 = 𝟏𝟕. 𝟐 𝑴𝑾
37. Each of two single-phase alternators has an armature winding whose resistance and synchronous
reactance are respectively, 0.025 and 0.06 ohm. If the machines are operating without load, calculate
the circulating current in the windings if the alternators are paralleled, when the emfs are equal at
230 Volts, but are displaced 30° from a position of phase opposition.
A. 915 A
B. 1, 830 A
C. 457.5 A
D. 119 A
Solution:
𝑍𝑠 1 = 𝑍𝑠 𝐿 = 𝑅𝑎 + 𝑗𝑋𝑠
𝑍𝑠 1 = 0.025 + 𝑗0.06

𝑍𝑠 1 = 0.065∠67.38° Ω
𝐵𝑦 𝐾𝑉𝐿:
𝐸1 − 𝐼𝑐 (𝑍𝑠 1 + 𝑍𝑠 2 ) − 𝐸2 = 0
𝐼𝑐 =
𝐼𝑐 =
𝐸1 −𝐸2
𝑍𝑠 1 +𝑍𝑠 2
230∠0°−230∠−30°
2(0.065∠67.38°)

𝑰𝒄 = 𝟗𝟏𝟓. 𝟖𝟐∠𝟕. 𝟔𝟐° 𝑨𝒎𝒑𝒆𝒓𝒆𝒔
38. A 3-phase 11 kV, 10 MVA alternator has a sequence reactances of positive sequence, 𝑋1 = 𝑗0.15 𝑝. 𝑢 ;
negative sequence, 𝑋2 = 𝑗0.15 𝑝. 𝑢 and 𝑋0 = 𝑗0.15 𝑝. 𝑢, with negligible neutral wire reactance. If the
alternator is on no-load, calculate the line to ground and 3-phase fault currents.
A. 4, 499 A and 3, 499 A
B. 2, 814 A and 2, 187 A
C. 4, 499 A and 3, 030 A
D. 1, 842 A and 854 A
Solution:
∗ 𝑓𝑜𝑟 𝑙𝑖𝑛𝑒 𝑡𝑜 𝑔𝑟𝑜𝑢𝑛𝑑 𝑓𝑎𝑢𝑙𝑡
𝐼𝐹 = 3𝐼𝐿 𝑟𝑎𝑡𝑒𝑑 (
𝐼𝐹 = 3 [
1
𝑋1 𝑝𝑢 +𝑋2 𝑝𝑢 +𝑋0 𝑝𝑢
10×106
](
√3 (11×103 )
)
1
)
0.15+0.15+0.15
 𝑰𝑭 = 𝟒, 𝟒𝟗𝟗 𝑨𝒎𝒑𝒆𝒓𝒆𝒔
∗ 𝑓𝑜𝑟 𝑡ℎ𝑟𝑒𝑒 − 𝑝ℎ𝑎𝑠𝑒 𝑓𝑎𝑢𝑙𝑡
𝐼𝐹 = 3𝐼𝐿 𝑟𝑎𝑡𝑒𝑑 (
𝐼𝐹 = 3 [

1
𝑋1 𝑝𝑢
10×106
)
](
√3 (11×103 )
1
0.15
)
𝑰𝑭 = 𝟑, 𝟒𝟗𝟗 𝑨𝒎𝒑𝒆𝒓𝒆𝒔
39. In Problem No.37, when the voltages are in phase opposition, but one voltage is 230 and the other is
200 Volts.
A. 462 A
B. 231 A
C. 400 A
D. 800 A
52
Solution:
∗ 𝑵𝒐𝒕𝒆: 𝑖𝑓 %𝑋2 𝑖𝑠 𝑛𝑜𝑡 𝑔𝑖𝑣𝑒𝑛 𝑡ℎ𝑒𝑛 𝑎𝑠𝑠𝑢𝑚𝑒 𝑡ℎ𝑎𝑡 𝑖𝑡 𝑖𝑠 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 %𝑋1
𝐼𝑐 =
𝐼𝑐 =
𝐼𝑐 =
𝐸1 −𝐸2
𝑍𝑠 1 +𝑍𝑠 2
230∠0°−200∠0°
2(0.065∠67.38°)
230∠0°−200∠0°
2(0.065∠67.38°)

𝑰𝒄 = 𝟐𝟑𝟎. 𝟕𝟕∠ − 𝟔𝟕. 𝟑𝟖° 𝑨𝒎𝒑𝒆𝒓𝒆𝒔
40. A 600 kVA, 2.4kV, 3-phase alternator has a zero phase sequence reactance of 12%, and a positive and
negative phase sequence reactances of 8%. If the alternator resistance is negligible, determine the
fault currents that the alternator can sustain if a three-phase fault occurs at its terminals.
A. 1, 042 A
B. 1, 804 A
C. 3, 125 A
D. 602 A
Solution:
∗ 𝑓𝑜𝑟 𝑡ℎ𝑟𝑒𝑒 − 𝑝ℎ𝑎𝑠𝑒 𝑓𝑎𝑢𝑙𝑡
𝐼𝐹 = 𝐼𝐿 𝑟𝑎𝑡𝑒𝑑 (
𝐼𝐹 = [

100
)
%𝑋1
600(1,000) 100
√3 (2,400)
](
8
)
𝑰𝑭 = 𝟏, 𝟖𝟎𝟒. 𝟐𝟐 𝑨𝒎𝒑𝒆𝒓𝒆𝒔
41. In Problem No. 40, determine the fault currents delivered if a phase to phase fault occurs at the
alternator terminals.
A. 1, 562 A
B. 902 A
C. 1, 804 A
D. 3, 125 A
Solution:
/𝐼𝑎 /=/𝐼𝑏 /= √3 [𝐼∅ 𝑟𝑎𝑡𝑒𝑑 (
100
%𝑋1 +%𝑋2
600(1,000) 100
/𝐼𝑎 /=/𝐼𝑏 /= √3 [
√3 (2,400)
](
8+8
)]
)
 /𝑰𝒂 /=/𝑰𝒃 /= 𝟏, 𝟓𝟔𝟐 𝑨𝒎𝒑𝒆𝒓𝒆𝒔
a
𝐼𝑎
𝐼𝑏
fault
b
𝐼𝑐
c
53
```