0% found this document useful (0 votes)
2K views58 pages

Ee March August

The document contains a 33 question practice exam for registered electrical engineers with multiple choice questions covering topics like circuits, motors, transformers, lighting and electrical engineering fundamentals. The exam includes questions testing knowledge of circuit analysis, motor specifications, transformer configurations, illumination calculations and core electrical concepts. Solutions are provided for several questions.

Uploaded by

REMY CUBERO
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
0% found this document useful (0 votes)
2K views58 pages

Ee March August

The document contains a 33 question practice exam for registered electrical engineers with multiple choice questions covering topics like circuits, motors, transformers, lighting and electrical engineering fundamentals. The exam includes questions testing knowledge of circuit analysis, motor specifications, transformer configurations, illumination calculations and core electrical concepts. Solutions are provided for several questions.

Uploaded by

REMY CUBERO
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
You are on page 1/ 58

MARCH 2018

REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION


MARCH 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

1. In Delta connected circuit, when one resistor is open, then power will be
A. 0 C. decreased by a factor of 3
B. increased by a factor of 3 D. remain unchanged
2. Which does not belong
A. lines per square centimeter C. flux
B. maxwells per square centimeter D. gauss

3. The diameter of a conductor is 0.064 inch. What is the area in circular mils?
A. 5,000 B. 10,000 C. 3,217 D. 4,096

Solution:

d=0.064inch

d= 0.064inch (1000mils/1inch) = 64mils

A = d² = 64² = 4096 CM

4. In a shunt motor, what is the relative connection of the armature winding with
respect to the field winding?
A series B. parallel-series C. series-parallel D. parallel
5. A series RLC circuit has R = 10 ohms , L = 40 x 10^-6 H and C = 60 x 10^-12
F. What is the resonant frequency of the circuit in MHZ?
A. 20.17 B. 3.24 C. 4.49 D 1.62
Solution:
R= 10 ohms

L= 40 x 10^-6 H

C= 60 x 10^-12 F

√ 𝐶 √( )( ) 𝒁

6. A capacitor charged with 10,000 uC. If the energy stored is 1 joule, find the
capacitance.
A 500 uF B. 5 uF C. 5,000 uF D. 50 Uf
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
MARCH 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

Solution:
Q= 10000 uC

W= 1 J

C=?

W = ½ Q^2 / C

C = ½ Q^2 / W = ½ (10000x10^-6)² / 1 = 50 x 10^-6 F or 50 uF

7. A 440-volt synchronous motor is taking 50 amperes at a pf of 0.9 lagging. The


effective resistance of armature is 0.8 ohm. If the iron and friction losses amount
to 600 watts, find the horsepower output.
A 25.7 В 37 1 C. 23.1 D. 40.5
Solution:
VL= 440v

IL= 50A

PF= 0.9

Ra= 0.8 ohms

Pfw= 600W

Pi = √3 VL IL PF = √3 440 50 0 9 =34 3 KW

Pcus= IL²Ra = 50²(0.8) =2000 W

Pout = 32.3 KW – 2000 W – 600 W = 31.7 KW ÷ 746 = 42.5 HP

8. What is considered as the most important value of a sine wave?


A. Effective value C. Average value
B. Peak value D. Instantaneous value
9. For a 3-phase load balanced condition, each phase has the same value of _
A. impedance B. resistance C. power factor D. all of these
10. Three 100 uF capacitors are connected in delta to a 500-volt, three-phase, 50
Hz supply. Find the line current.
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
MARCH 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

A 15.71 A B. 27.21 A C. 31.83 A D 8.66 A


Solution:
C= 100pF

VL= 500v, 50Hz


IL=?
Xc= 1/ 2πfC = 1/ 2π(50)(100uF)= 31.8
Iphase= 500/31.8 = 15.72A

IL= √3 Iphase = √3 15 72 = 27.23 A

11. Assuming the most efficient technologies for the extraction of energy, the fuel
with the largest known reserves is

A. Coal B. Oil C. Natural gas D. Nuclear

12. 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 of the line to neutral if its length is 125 miles.

^ Ω B. 202.4 x 10^3 Ω C. 25.3 x 10^3 Ω D. 16.2 x 10^3 Ω

Solution:
. Given : f = 60 Hz, D = 12m, d = 3.28 = .0328, r = .0164m

L = 125mi = 201.168 km

C = 8.138x10-12 F/m ( 1000m/ km )(201.168 km)

C = 1.637x10-6 F

Xc = 1/wC = 1/ 2πx60x1 637x10-6) = 1.6x103 Ω

13. A circuit has a resistance of 20 ohms and reactance of 30 ohms. What is the
p.f of the circuit ?

A. 0 B. 0.55 C. 0.832 D. 0.99

Solution:
Given: R = 20, X = 30

Z = R + jX
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
MARCH 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

Z = 20 j30 = 36 06∟56 31˚

Pf = cos(56.31) = 0.55

14. A capacitive load always has a __________ power factor

A. Leading B. Lagging C. Zero D. Unity

15. A road is illuminated by two 500 candela lamps place 6.1 m above the center
line of the road and 18.3 m apart. Find the illumination produced on the surface
of the road at a point on the center line midway between lamps

A. 2.29 lux B. 4.58 lux C. 9.16 lux D. 18.32 lux

Solution:
Given: I = 500

h1 = h2 = 6.1m, d = 18.3/2 m = 9.15 m

d2 = 6.12 + 9.152= 120.93

tanƟ = 9 15/ 6 1, Ɵ = 56 31

Ep 1= (I/d2 cosƟ
Ep1 =(500/120.93)cos(56.31) = 2.29 = Ep2
Eptotal = Ep1 + Ep2 = 2.29 + 2.29 = 4.58 lux

16. A magnetizing force of 1000 AT/m will produce a flux density of ___________
in air

A. 1.257 mWb/m^2 B. 0.63 mWb/m^2 C. 1.257 Wb/m^2 D. 0.63 Wb/m^2

Solution:
Given: H = 1000 AT/m

ᵦ = H x uo = 1000 4πx10-7) = 1.257x10-3 mWb/m^2

17. In hydro electric power plant, it is use to prevent the sudden rise and drop of
water pressure.

A. Penstock B. Intake tower C. Dam D. Surge tank

18. The basis where the flow of current is from positive to negative was derived
from __________
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
MARCH 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

A. Superstition B. Tradition C. Scientific D. Convention

19. The following are the given loads of a particular circuit


5 - 15 HP motors
3 - 5 HP motors
8 KW - lighting loads
3.5 KW - miscellaneous loads
Calculate for the maximum demand if the demand factor is 65%

A. 78.64 B. 45.81 C. 51.12 D. 120.28

Solution:
Given : 5 - 15 HP motors = 5( 15x746 ) = 55950
3 - 5 HP motors = 3( 5x746 ) = 11190
8 KW - lighting loads
3.5 KW - miscellaneous loads
DF = PFxTCL = (0.65)(55950 + 11190 + 8000 + 3500)/1000
DF = 51.12
20. By using two wattmeter method, power can be measured in

A. 3-phase, 2-wire B. 3-phase, 3-wire C. 3-phase, 4-wire D. All of these

21. The double energy transient occur in the


A. purely inductive circuit C. R-C Circuit
B. R-L Circuit D. R-L-C Circuit

22. Its primary source is to provide ground source


A. zig-zag transformer C. surge arrester
B. sectionalizer D. tertiary transformer

23. A 20 KV/7.87 KV autotransformer has a secondary line current of 100 A.


What is the current in the common winding?
A. 139.25 A B. 100 A C. 60.65 A D. 39.35 A
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
MARCH 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

Solution:

Given:

Find: Icb

24. In a star connected system, the current flowing through the line is
A. Greater than the phase current C. Lesser than the phase current
B. Equal to the phase current D. None of these

25. The capacitive reactance of a transmission line is 90,000 ohms per kilometer.
Find the total capacitive reactance if the transmission line is 50-km long.
A. 4.5 x 106 B. 1800 C. 90,000 D. 180,000

Solution:

26. Three phase induction motor is more suitable than single phase because
A It’s self-starting C. it has better power factor
B.it has better efficiency D. all of these

27. A 100 watts 230 volts gas fitted lamp has a mean spherical candlepower of
92 Find it’s efficiency in lumens per watt
A. 11.56 B. 1.156 C. 115.6 D. 1,156
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
MARCH 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

Solution:

( )

28. What is the speed of an induction motor of six poles if the percent slip is
2.5%?
A. 1462 rpm B.1170 rpm C.877 rpm D.1755 rpm

Solution:

Given:
Find: Nr

( )

( )

( )

29. Two (2), 1 ⌀, 25-KVA transformers are connected in open-delta bank.


Determine the maximum KW the bank can supply without overloading at 0.8 p.f
lagging.
A. 43.3 B. 34.64 C. 40 D. 50

Solution:

Given: ⌀

( )( ⌀)

( )( )
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
MARCH 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

( )

( )

30. The luminous intensity is defined as


A. lumen per square meter C.illumination per square meter
B.luminous flux per unit solid angle D. candela per unit solid angle

31. MHO relay is used for

A. rectifier B. circuit breaker C. transmission lines D. feeders

32. Three units of transformers /γ are connected to supply a 3-phase load from
a 400 volts, 3-phase source. What is the voltage on the load side?

A. 2,000v B. 3,464v C. 693v D. 1,200v

Solution:
VL= √3 (400)
=693V

33. In a serious motor, what is the relation of the connection of motor field
winding with the respect of the motor armature?

A. series B. parallel C. series-parallel D. delta

34. The field circuit of a 4-poledc generator has a resistance of 200Ω and an
inductance of 100H. If it is connected to a 400v supply, find the time taken for the
current to reach ljkjhggfdfd1.5A.

A. 0.125 sec B. 1.5 sec C. 0.369 sec D. 0.693 sec

Solution:
I= 1.5A, R= 200Ω, L= 100H, E= 400v, t=?

I= E/R(1-e^(R/L) t

1.5=400/200(1-e^-(200/100) t
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
MARCH 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

t= 0.693sec

35. Calculate the effective voltage in one phase of an alternator, given the
following particulars; f= 60 cps; turns per phase N = 240; flux per pole =
2.08x10^6.

A. 1,330v B. 2,304v C. 1,225v D. 2,122v

Solution:
f= 60 cps, N= 240rpm, = 2.08x10^6

E= (240) (2.08x10^6) =1324V ≠ 1330V

2 (60)

36. a 60 F capacitor is connected in series with 40,000 Ω resistor If the


capacitor is initially uncharged. Determine the resistor voltage when t= 1.5 times
the time constant for a suddenly applied source emf of 120volts.

A. 93.22v B. 120v C. 80v D. 26.78v

Solution:
I= E/R(1-e^(R/L) t

I= 120 = 3 x 10^-3

40,000

3x10^-3= E/40,000 (1-e^-(40,000/60) (1.5)

E= 120v

37. What is the force on a +5x10^-11 C charge that is 5 cm form each of two
+2x10^-9 C charge that are separated by 8cm?

A. 0.432x10^-6N B. zero C. 0.36x10^-6N D. 0.14x10^-6

Solution:
Q1= 5x10^-11, Q2= 2x10^-9, d= 0.05

5cm x 1m = 0.05m

100cm

F= KQ1Q2
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
MARCH 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

d^2

F= (9x10^9)(5x10^-11)(2x10^-9) = 0.36x10^-6

(0.05) ^2

38. A balanced delta connected load consist of 20 cis 25 Ω impedances The


60Hz line voltage 208v. What is the line current?

A. 6A B.10.4A C. 12A D. 18A

Solution:
R= 20∟25 Ω, V= 208V

IL= 208v = 10.4A

20Ω

39 The impedance of 69 kV transformer is 14 28Ω when referred to primary. The


transformer is a 10-MVA unit. What is the percent impedance of transformer?

A. 4 B. 6.2 C. 3 D. 4.5

Solution:
Zpu= (Zbase x Sbase)

Vbase^2

Zpu= (10 x 14.28) x 100 = 2.99≠3

69^2

40. What is the cost of operating a 70percent 20hp pump for 20 days, 8 hrs per
day? Cost of energy is P0.05/kW-hr.

A. P112 B.P170.45 C. P228.60 D. P83.55

Solution:
Pin= 20Hp, ɳ= .70, Cost of energy= P0.5kW/hr

Pin= Pout = 20x746 =21.31x10^3kW-hr

ɳ 0.70

=20days x 8hrs = 160hrs


REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
MARCH 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

1day

= 160 x 0.5 = 80kW-hr

= 80 kW-hr (21.31 x 10^3)

= 170.45

41. The ratio of the wattmeter readings in three-phase inductive system is 2:1
when power is measured by the two-wattmeter method. What is the power factor
of the load?

A. 0.5 B. 0.866 C.0.75 D. 0.9

Solution:

Given: ratio: 2:1

Find: Pf=?

√ ( ) √ ( )

( )

42. A coil of 100 turns is cut by a magnetic field that increases at the rate of 40
Wb/min. What is the induced voltage?

A. 66.7 v B.16.7 v C. 4,000 v D. 6.67 v

Solution:

Given: N = 100 turns

Find: e =?

( )

43. Find the horsepower rating of a motor running at 1,500 rpm and developing a
torque of 105 lb-ft.
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
MARCH 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

A. 22 B. 28 C. 30 D. 36

Solution:

Given: N = 1,500 rpm

T = 105 lb-ft

Find: Hp=?
( )( )

44. The rms value of a sinusoidal wave is equal to _________ of maximum


value.

A. 0.637 B. 0.707 B. 0.506 D. 1.414

45. The voltage across a 50-µF capacitor rises at a constant rate of 10 V/ms.
Calculate the current through the capacitor.

A. 2 mA B. 5 mA C. 50 mA D. 500 mA

Solution:

Given: C =

Find: I=?

( ) (𝐶) ( )( )

46. One advantage of disturbing the winding in alternator is to

A. reduce harmonics C. reduce the amount of copper

B. improve voltage waveform D. decrease the value of the voltage

47. A series RLC circuit has R = 100 ohms, L = 100µH and C = 100 pF. What is
the frequency in MHZ at unity power factor?
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
MARCH 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

A. 1.59 B. 1.36 C. 15.9 D. 13.6

Solution:

Given:

Find: F=?

(√ ) (√( )( ))

48. In dc machines, lap winding is used for

A. high voltage low current C. high voltage high current

B. low voltage low current D. low voltage high current

49. The ratio of active power to apparent power is known as _______________


factor.

A. demand B. load C. power D. form

50. At ________________ frequencies, the parallel R-L circuit behaves as purely


resistive.

A. very low B. low C. high D. very high

51. One kilowatt equals to _ horsepower.


A. 1.26 B. 1.36 C. 1.46 D. 1.56

Solution:

52. The output of motor in watts when in takes a power of 3 kW and its efficiency
is 75%, is
A. 2000 B. 2250 C. 2500 D. 2750

Solution:
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
MARCH 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

Given:
Find:

53. In a certain Y-Y system, the source phase currents each have a magnitude of
9 A. The magnitude of each load current for a balanced load condition is
A. 12 A B.3 A C. 9 A D. 27 A

Solution:

Given:

54. In a three-phase system, the voltages are separated by


A. 90° B. 45° C. 120° D. 180°

Solution:

55. In a three-phase system, when the loads are perfectly balanced, the neutral
current is
A. one-third of maximum C. two-thirds of maximum
B. zero D. at maximum

56. When the motor runs on no load, then


A. Back emf is almost equal to applied voltage
B. Back emf will be greater than applied voltage
C. Back emf will be less than applied voltage
D. None of these

57. What is the source of heat generation in cables?


REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
MARCH 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

A. Copper loss in conductor


B. Dielectric losses in cable insulation
C. Losses in metallic sheathings and armouring
D. All of these

58. In a 3 core cable, the capacitance between two conductors is 3 µF. What will
be the capacitance per phase?
A. 1.5 µF B. 3 µF C. 6 µF D. 12 µF

Solution:

Given: 𝐶
𝐶 𝐶

59. The process by which a heavy nucleus is splitted into two light nuclei is
known as
A. Splitting B. Fission C. Fusion D. Disintegration

60. Efficiency of a power plant is more in summers or winters?


A. summers B. winters C. same in both D. depends on variation

61. If there are eleven resistors of 33 kΩ each in parallel, what is the total
resistance?
A. 363 kΩ B. 3.3 kΩ C. Ω D. 16 5 kΩ

Solution:

Given:

Find:
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
MARCH 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

62. To reverse the direction of rotation of a three-phase induction motor, it is


necessary to
A. interchange all the three line connections
B. reverse the polarity of the rotor circuit
C. increase the resistance of the rotor circuit
D. interchange any two of the three line connections

63. The law that induced emf and current always opposes the cause producing
them was discovered by
A. Lenz B. Maxwell C. Faraday D. Ohm

64. The power factor of an alternator is determined by its


A. speed B. load C. excitation D. prime mover

65. A heater wire of length 50 cm and 1 mm 2 in cross-section carries a current of


2 A when connected across a 2 v battery. What is the resistivity of the wire?
A. 2 x 10-6 Ω-m B. 2 x 10-8 Ω-m C. 4 x 10-6 Ω-m D. 4 x 10-8 Ω-m

Solution:

Given:

( )

Find: ρ

( )( )

66. A six-pole three-phase 440-v, 60 Hz induction motor develops 10-hp at 1,150


rpm, the power factor being 80% lagging. Stator and core losses amount to
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
MARCH 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

400 w and 350 w, respectively. Frictional losses amount to 0.5-hp. Calculate the
motor line current.
A. 12.24 A B. 13.41 A C. 11.42 A D. 14.64 A

Solution:

Given:

Find:

( )
( )
( )( )

√ √ ( )( )

67. In a circuit containing R, L, and C, power loss can take place in


A. C only B. L only C. R only D. all of these

68. The efficiency of thermal power plant is approximately


A. 10% B. 30% C. 60% D. 80%

69. The time constant of a series RC circuit is given by


A. R/C B. RC2 C. RC D. R2C
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
MARCH 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

70. A power plant consumes 3,600 tons of coal per day. If the coal has an
average energy content of 10,000 BTU/lb, what is the plant’s power output?
Assume an overall efficiency of 15 percent.
A. 132 MW B. 130 MW C. 128 MW D. 126 MW

Solution:

Given:

Find:

( )

71. A resistance of 40 ohms and inductance of 79.6 mH are connected in parallel


across a 240 v, 60 Hz ac supply. Find the total current in amperes.
A. 11 B. 12 C. 10 D. 14

Solution:

Given:

Find:

( )( )
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
MARCH 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

√ √

72. A 230 v, 3-phase source supplies a balanced delta connected load with
impedance of 8-j6 ohms per phase. Calculate the line current in amperes.
A. 33.1 B. 39.8 C. 26.5 D. 29.3

Solution:

Given:

Find:

√ √ ( )

73. Three unbalanced 3-phase currents are given as follows:


Ia=10 -30° A, Ib=0, Ic=10 -150° A
A. 3.34 150° A B. 3.34 -150° A C. 3.34 30° A D. 3.34 -30° A

Solution:

Given:

| | | | | |

Find:
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
MARCH 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

( )

[( ) ( )( ) ( )( )]

74. A three-phase, 6 pole, 50 Hz, induction motor develops a maximum torque of


30 N-m at 960 rpm. Calculate the torque produced by the motor at 6% slip. The
rotor resistance per phase is 0 6 Ω
A. 24.7 N-m B. 25.7 N-m C. 26.7 N-m D. 27.7 N-m

Solution:

Given:

Find:

( )
( ) ( )

( )( )

( )
( ) ( )

( )
( ) ( )
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
MARCH 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

75. A device that repeatedly reverses the polarity of a magnetic field in order to
keep a dc motor rotating is known as
A. a solenoid B. an armature coil C. a commutator D. a field coil

76. The form factor is the ratio of


A. peak value to r.m.s value C. average value to r.m.s value
B. r.m.s value to average value D. none of these

77. The per unit impedance of a circuit element is 0.30. if the base KV and base
MVA are halved, then the new value of the per unit impedance of the circuit
element will be.
A. 0.30 B. 0.60 C. 0.0030 D. 0.0060

Solution:

Given:
⁄ ⁄

Find:

( )( )


( +( +

𝒁 𝒘

78. If two sinusoids of the same frequency but of different amplitudes and phase
angles are subtracted, the resultant is
A. a sinusoidal of the same frequency C. a sinusoidal of double the freq
B. a sinusoidal of half the original frequency D. not a sinusoid
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
MARCH 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

79. A 170-mile, 230 V, 60 Hz, three-phase single circuit transmission line uses a
triangular arrangement with 20 ft, 20 ft, and 36 ft spacing respectively. If the line
conductors have a radius of 0.0217 ft, determine the capacitive susceptance to
neutral per mile.
A. 4.81 x 10-8 mho per mile C. 7.92 x 10-6 mho per mile
B. 12.74 x 10-8 mho per mile D. 2.47 x 10-6 mho per mile

Solution:

Given:

√ √( )( )( )

( )
𝐶

𝐶 ( )

80. An 8-pole lap wound dc generator has 1,000 armature conductors, flux of 20
mWb per pole and emf generated is 400 v. What is the speed of the machine?
A. 1,000 rpm B. 1,800 rpm C. 1,200 rpm D. 1,500 rpm

Solution:

Given:

Find:

( )( )
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
MARCH 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

( )( )( )
( )( )( )

81. In a balanced 3-phase system, wye-connected voltage source with phase


sequence abc, Vab = 230 10o V. Find the phase voltage Vcn.
A. 230 -230oV B. 132.8 -80o V C.132.8 -140o V D. 132.8 -260o V

Solution:

Given:
Find:

82. A three-phase 60 Hz wye-connected wound rotor synchronous generator


rated at 10 kVA, 230V has a synchronous reactance of 1.2 ohm per phase and
an armature resistance of 0.5 ohm per phase. Calculate the percent regulation at
full-load with 0.8 lagging power factor.
A. 18.3% B. 20.8% C. 21.8% D. 22.5%

Solution:

Given:
Find:
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
MARCH 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

√( ) ( )

83. Two equally charged spheres repel each other with a force of 0.1kg. If their
centers are 20 cm apart, find the charge on each.
A. 2.09 x 10-6 coul B. 2.09 x 106 coul C. 2.00 x 10-6 coul D. 2.00 x 106 coul

Solution:

Given:
Find:

√ √

( )( )( )

84. The area of one plate of a two-plate mica capacitor is 0.0025 m2 and the
separation between plates is 0.02 m. If the dielectric constant of mica is 5, what
is the capacitance of the capacitor?
A. 5.53 µF B. 5.53 pF C. 7.74 µF D. 7.74 pF

Solution:

Given:
Find: 𝐶

𝐶
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
MARCH 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

( )( )( )
𝐶
𝐶

85. For coil having a span of 2/3 of pole pitch, the coil span factor is
A. 0.8 B. 0.866 C. 0.786 D. 0.966

Solution:

Given:
Find:

( )
( *

86. The number of parallel paths in a simplex lap winding is equal to


A. 2 B. number of poles C. number pair of poles D. 1

Solution:

Given:
Find:

( )

87. A 240 V DC series motor develops a shaft torque of 200 N-m at 92%
efficiency while running at 600rpm. Calculate the motor current.
A. 62.09 A B. 52.36 A C. 48.17 A D. 56.91 A

Solution:

Given:
Find:
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
MARCH 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

( )( )

88. A voltage of e = 70.7sin50t V is applied to a series circuit containing a


resistance of 4 ohms and an inductance of 60 mH. What is the impedance of the
circuit in ohms?
A. 5 30˚Ω B. 5 60˚Ω C. 5 53 13˚Ω D. 5 7˚Ω

Solution:

Given:
Find:

( )

89. An office, 17m by 8m, is lighted by 10 lamps and each lamp has a luminous
intensity of 200 cd. Allowing an absorption loss of 5,000 lm for the reflectors,
walls, and ceilings, calculate the average illumination on the working plane.
A. 177 lm/m2 B. 136 lm/m2 C. 118 lm/m2 D. 148 lm/m2

90. A 480-V rms line feeds two balanced three-phase loads.


If the two loads are rated as follows,
Load 1: 5 kVA at 0.8 p.f. lagging Load 2: 10 kVA at 0.9 p.f. lagging
Determine the magnitude of the line current from the 480-V rms source.
A. 17.97 A B. 14.94 A C. 13 A D. 7.36 A

Solution:
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
MARCH 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

Given:

Find:

√ ( )

91. A colliery workshop 20 m x 10 m requires an average illumination of 135


lm/m2 on the working plane. Calculate the electrical power and current required
for the purpose. Allow a coefficient of utilization of 0.45 and a depreciation factor
of 1.4. The supply is 230-V at 50 Hz and the efficiency of each of the lamps is 12
lm/W.
A. 8kW, 34.8 A B. 9kW, 39.1 A C. 6kW, 26.1 A D. 7kW, 30.4 A

Solution:

92 A series RL circuit has R = 6 Ω and XL = 8 Ω What is the circuit admittance?


A. 0.06 mho B. 0.1 mho C. 0.08 mho D. 0.125 mho

Solution:

Y = Z-1
Y = (6 + j8)-1
Y = 0.06 – j0.08 0.1 - 59.03
Y = 0.1 mho

93. A balanced 6,600-volt, 3-phase, 3-wire, 60 Hz transmission line is designed


to transmit power to a 450 kVA inductive load at 0.8 power factor. The line has a
resistance of 3.5 ohms and a reactance of 10.5 ohms per wire. Find the voltage
regulation of the line.
A. 7.6% B. 5.6% C. 9.6% D. 11.6%
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
MARCH 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

Solution:

94. A balanced Y-connected load of 300 j100 Ω s supplied by a 3-phase line


40 km long line with an impedance of 0 6 j0 7 Ω per km line to neutral Find
the voltage at the receiving end when the voltage at the sending end is 66 kV.
A. 60 kV B. 62 kV C. 58 kV D. 64 kV

Solution:

ZT = 300 j100 40 0 6 j0 7 = 324 j128 Ω

Is = = = 189.46 -21 56˚ = IR


( )

VR = IR ZT = (189.46 -21 56˚ (300 + j100)


VR = 59.9 kV -3 12˚
VR 60 kV

95. A plant has a total generating capacity of 800 kW. The coal consumption is
1,900 lbs. per hour. Heating value of the coal is 9,500 BTU per lb. What
approximate percent in heat in the coal is converted into useful energy?
A. 8.7 % B. 17.5 % C. 12.3 % D. 15%

Solution:

9,500 x 1900 = 18.05 x 106 x = 5288.6 kW (100%)

x = 15.13% 15%

96. A 100 km transmission line has a 1,200 ohms shunt reactance. What is the
shunt reactance per km?
A 1,200 Ω B 120,000 Ω 2Ω D 120 Ω

Solution:
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
MARCH 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

= 12

97. A relay is used to


A. receive a signal and record it C. connect a line to the source
B. protect a circuit D. relay a message to a remote place

98. A power plant consumes 100,000 lbs. of coal per hour. The heating value of
the coal is 12,000 BTU per lb. the overall plant efficiency is 30%. What is the kW
output of the plant?
A. 175,000 kW B. 142,500 kW C. 105,500 kW D. 205,000 Kw

Solution:

100 x 12,000 = 1.2 x = 351. 597 kW

Po = 105, 500 kW

99 The condition in Ohm’s law is that


A. ratio V/I should be constant
B. current should be proportional to voltage
C. the temperature should remain constant
D. the temperature should vary

100. How is the storage battery rated in capacity?


A. volts B. watts C. amperes D. ampere-hour
AUGUST 2018
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
AUGUST 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

1. In the transformers
A. high tension winding is used near the core
B. high tension winding is used away from the core
C. low tension winding is used near the core
D. low tension winding is used away from the core

2. Which of the following parameters can be neglected for a short line?


A. resistance B. inductance C. reactance D. capacitance

3. Find the frequency of a radio wave whose wavelength is 600m.


A. 500Hz B. 500kHz C. 150Hz D. 150kHz

Given: L= 600m Req’d: f

Solution:

L= v ; f = v = 3 x 105 m/s = 500kHz


f L 600m

4. The curve representing Ohm’s law is


A. a parabola B. linear C. a sine function D. a hyperbola

5. A series connected circuit has R= 4Ω and L=25mH Calculate the value of C that will
produce a quality factor of 50.
A. 62.5µF B. 6.25µF C. 0.625µF D. 25µF

Given: R= 4Ω Req’d: C

L=25mH

Qfactor = 50

Solution:

Q.F. = 1/ R (√L/C)
50 = ¼ (√25 x 10-3 / C)
C= 0.625µF

6. The overcurrent protective device is caused to operate by ___________________.


A. voltage surge B. current flow C. voltage drop D. impedance
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
AUGUST 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

7. What is the maximum number of bands permitted in rigid conduits between outlets?
A. 4 quarter bends C. 2 quarter bends
B. 6 quarter bends D. 3 quarter bends

8. Which of the following insulators will be selected for high voltage application?
A. strain type B. disc type C. suspension type D. pin type

9. At a 115kV substation, the PT ratio is 1000 and the CT ratio is 1200/5. The potential
going into the wattmeter is 115volts. What is the MW indicated when the wattmeter
reads 800 watts?
A. 150 MW B. 192 MW C. 15 MW D. 19.2 MW

Given: V=115kV Req’d: Pactual in MW


PTR=1,000
CTR=1200/5
VW= 115V
Prdg= 800W
Solution:

Pactual = Prdg (PTR x CTR)


= (800) (1000 x 1200/5)
= 192 x 106
= 192 MW

10. An inductive coil consumes active power of 500W and draws 10A from a 60Hz ac
supply of 110V. What is the coil inductance?
A. 26mH B. 28mH C. 30mH D. 32mH

Given: V= 110Vf = 60Hz Req’d: L


I = 10A
P= 500W
Solution:

R= P/ I2 = 500 / (10)2 = 5Ω
Z= V / I = 110 / 10 = 11Ω

Z= √R2 + XL2 L= XL / 2лf


XL= √Z2 – R2 = 9 78 / 2л 60
XL= √112 - 52 = 25.94 x 10-3
XL= 9 78Ω L=26mH
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
AUGUST 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

11. If the torque angle continuously increases, system will become


A. stable C. oscillatory stable
B. unstable D. asymptotic stable

12. The condition in Ohm’s Law is that


A. ratio V/I should be constant
B. current should be proportional to voltage
C. the temperature should remain constant
D. the temperature should vary

13. Which of the following shall be included in the calculation of the reliability standard
indices of the distributor asset by the Philippine Distributor Code?
A. Outages that occur on the secondary lines of the distribution system
B. Planned outages where the customers have been notified at least three(3)
days prior to the loss of power
C. Outages that occur on the primary lines of the distribution system
D. Outages that occur in the transmission line

14. If 25 ohms and 125 amperes are the base impedance and base current, respectively
for a given system, what is the base KVA?
A. 781.25 B. 39.1 C. 390.625 D. 76.12

Given: Zbase= 25Ω Req’d: kVAbase

Ibase = 125A
Solution:
kVAbase= Ibase2 x Zbase
= (125)2 (25)
kVAbase= 390.625

15. A type of electronics communication in which only one party transmits at a time.
B. Full duplex B. Half duplex C. Bicom D. Simplex

16. An equal number of electrons is placed on two metal spheres 3cm apart in air. How
many electrons are on each sphere if the resultant force is 4kN?
A. 1.47 x 1014 electrons C. 1.25 x 1014 electrons
B. 1.33 x 1014 electrons D. 1.40 x 1014 electrons
Given: d= 3cm Req’d: e
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
AUGUST 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

F= 4kN

Solution:

F= kQ1Q2 / d2 Qe = 1.6x10-19
4x103 = (9x109)(Q)2 / (3x10-2)2 e = 2x10-5 / 1.6x10-19
Q= 2x10-5 e = 1.25 x 1014 electrons

17. Base load plants are designed to supply power at


A. low capital cost and low operating cost
B. high capital cost and low operating cost
C. low capital cost and high operating cost
D. high capital cost and high operating cost

18. Peak load plants are designed to supply power at


A. low capital cost and low operating cost
B. high capital cost and low operating cost

C. low capital cost and high operating cost


D. high capital cost and high operating cost

19. A symmetrical three-phase 240V supplies a balanced delta connected load of 6-j8
ohms per phase. What is the line current?
A. -25 + j32A B. 25 – j5.78A C. -25 - j32A D. 25 + j32A

Given: V= 240V Req’d: IL


Z= (6-j8 Ω

Solution:

Iɸ = V / Z = 240 / 6-j8 = (14.4+j19.2) A


IL = Iɸ x √3 = (14.4+j19.2) (√3)
= 24.9+j33.3 A
IL = 25 + j32A

20. A 440V, 60Hz, 6 pole, 3 phase induction motor has a rotor input power of 33kW at
rated voltage and frequency. Under this condition, the rotor emf makes 126
complete cycles/minute. Find the rotor resistance in ohms/phase if the rotor current
is 60A/phase.
A. 0.0118 B.0.0107 C. 0.107 D. 0.118
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
AUGUST 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

21. If the effect of earth is taken into account, then the capacitance of line to ground
A. decreases C. remains unchanged
B. increases D. becomes infinite

22. The effective resistance of conductor is increased by


A. skin effect C. corona effect
B. proximity effect D. both (a) and (b)

23. Surge impedance of transmission line is given by


A. √C/L B √LC C 1/√LC D √L/

24. A 100 watts 230 volts gas fitted lamp has a mean spherical candle power of 92. Find
its efficiency in lumens per watt.
A. 9.56 B. 10.56 C. 11.56 D. 12.56

Given: P= 100W
I = 92mscp
V=230V
Solution:

n = Im / P = 4лI / P = 4л 92 / 100 = 11.56

25. The overload KVA of each transformer of an open-delta bank is 11.56. If each
transformer is rated 37.5 KVA, then what is the KVA of the balanced 3ɸ load?
A. 70 B. 80 C. 85 D. 90

Given: Req’d: Sload

S1ɸ = (37.5 + 11.55) KVA


Solution:

S1ɸ = Svv / √3 = Sload / √3


S1ɸ = Sload / √3
Sload = S1ɸ √3 = (37.5 + 11.55) √3 = 84 96 85

26. In a 6-pole alternator how many mechanical degrees is the equivalent of 180
electrical degrees?
A. 60 B. 30 C. 120 D. 90
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
AUGUST 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

Given: Req’d: Θm

P= 6
Θe= 180
Solution:

Θe= P / 2 Θm

Θm= Θe / (P / 2) = 180 / (6 / 2) = 60
27. A two terminal pair of network of a transmission line can be represented by a
A. Л- network B. T-network C. either (A) or (B) D. tree network

28. In television system, channels 2 and 13 are classified as


A. VHF B. UHF C. SHF D. EHF

29. An aluminum cable has 10 identical strands, a length of 1,500m and a diameter of
2.50 mm each. What is the resistance of the cable?
A. 86 3Ω B Ω C 56 3Ω D 0 563Ω

Given: Req’d: R

n = 10 identical strands
L = 1, 500m
d = 2.50mm = 2.5 x 10-3 m
ρ = 2 83 x 10-8 Ωm @ 20 0C

Solution:

R=ρ L/A

= ρ { L / л / 4 d2 n) }

= 2.83 x 10-8 { 1, 500 / л / 4 2 5 x 10-3 )2 x 10 ) }


R= Ω

30. An ACSR conductor has (n-1) layers around its single center strand, the total
number of strands will be
A. 3n2 + 3n +1 B. 3n2 - 3n -1 C. 3n2 + 3n -1 D. 3n2 - 3n +1
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
AUGUST 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

31. Three resistors A, B and C are connected in series to a 120V supply. If the the
resistor A=50 ohms and the voltage across resistor B=40V when the current is 1/2A,
what is the resistance of resistor C in ohms?
A. 105 B. 90 C.110 D.100

Given: Req’d: Rc

RA = 50Ω
VT = 120V
VB = 40V
I = 1/2A
Solution:

RB = VB / I = 40 / 1/2 = 80Ω

RT = VT / I = 120 / 1/2 = 240Ω

RT = RA + RB + Rc

240 = 50 + 80 + Rc

Rc = 240 - 50 – 80 = Ω

32. An inductor motor with 8 poles, 60Hz is operated with a slip of 3%. Calculate the
rotor speed.
A. 1,684 B.1,024 C. 873 D. 927

Given: Req’d: Nr

P=8 poles
f= 60Hz
%s= 3%
Solution:

Ns= 120f / p = 120(60) / 8 = 900rpm

%s = (Ns-Nr) / Ns
0.03 = (900- Nr) / 900
Nr = 900 – 0.03(900) = 873rpm

33. Which of the following methods of generating electric power from the sea water is
more advantageous?
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
AUGUST 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

A. water power B. ocean power C. tidal power D. none of these


34. When installing two grounding electrodes, the minimum distance between them is
A. 1,800 mm B. 2,000 mm C. 1,500 mm D. 2,400 mm

35 A 36 resistor is connected in parallel with an unknown resistor R Their


combination is them connected in series with a 12 resistor Find the value of R such
that the power drawn by the parallel combination of 36 and R is equal to the power in
the 12 resistor
A 6 B C 24 D 48

Solution:

P1 = Peq V1=V2

V1 = V1

R1 Req

R1 = Req

R1 = R (36)/R + 36

R = 18 Ω

36. The current flowing in L and C at parallel resonance are


A. zero B. equal C. infinite D. Difference

37. What does PEMC in EPIRA stands for


A. Philippine Energy Market Corporation
B. Private Energy Marketing Company
C. Philippine Energy Market Corporation
D. Philippine Electrical Marketing Corporation

38. For two alternators operating in parallel, if the load shared by one of the, to be
increased, as field excitation is
A. to be strengthened keeping input torque the same
B to be weakened keeping input torque the same
C. to be kept constant but input torque should be increased
D. to be kept constant but input torque should be decreased
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
AUGUST 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

39. A 7.5 MVA, 69/13.8 kV, 3-phase transformer has 8% impedance. What us the per
unit impedance at 10 MVA base?
A. 0.24 B. 0.106 C. 0.06 D. 0.096

Given: Zpuold = 0.08 pu

Sold = 7.5 MVA

Snew =10 MVA

Solution:

Zpunew = Zpuold (Zpuold / Sold)

= 0.08 ( 10 / 7.5 )

= 0.106 pu

40. Open circuit test on transformer is conducted to determine


A. hysteresis loss B. copper loss C. core loss D. eddy current loss

41. Which of the following power plant is a conventional source of energy?


A. Solar energy B. Geothermal energy C. Coal energy D. Wind energy

42. A transformer is rated 10kVA, 440/110V and 60 Hz when operated as conventional


2 winding transformer. This transformer is to be used as 550/440 V step-down
autotransformer in a distribution system. What is the new kVA rating of the transformer
in this manner?
A. 20 B. 10 C. 37.5 D. 50

43. In a synchronous motor, the magnitude of stator back emf depends on


A. load on the motor C. both the speed and rotor flux
B. dc excitation only D. none of theses

44 A dc ammeter has an internal resistance of 0 1 A shunt of 1 01 x 10 -3 is


connected to be ammeter. What is the multiplier of the set up?
A. 10 B. 50 C. 80 D. 100

Given: Rg = 0.1 Ω
Rsh = 1.01x10-3 Ω
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
AUGUST 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

Solution:
Multiplier = (Rg + Rsh) / Rsh
= (0.1 + 1.01x10-3) / 1.01x10-3
= 100
45. A single-phase, 10-km transmission line has 16.65 mH total inductance. If the
distance between the conductors is 1 m, what is the conductor radius?
A. 2 cm B. 1.5 cm C. 1 cm D. 0.75

Given : Lt = 16.65 mH

D=1

Solution:

Lt = 4x10-7 ln (d/ 0.7788r)

Lt = 16.65 mH x (1 / 10,000 m)

Lt = 1.665x10-6 H/m

1.665x10-6 = 4x10-7 ln (d/ 0.7788r)

r = 0.0199 m or 2 cm

46 4-pole wave wound generator has 220 coils of 10 turns each. The resistance of each
turn is 0 02 Find the resistance of armature winding.
B 22 C 02 D 0 44

Solution:

Ra = (220 coils)(10 turns x 0.02 Ω) / 4 Poles

Ra = 11 Ω

47. PDC in EPIRA means


A. Philippine Distribution Commission C. Philippine Distribution Corporation
B. Philippine Distribution Code D. Philippine Distribution Company

48. What is the maximum plug and cord rating of a 30 A branch circuit?
A. 24 A B. 30 A C. 26 A D. 28 A
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
AUGUST 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

Solution:

Rating = 80 % of OCPD

= 0.8 ( 30 A )

Rating = 24 A
49. Wavelength is the distance travelled by an electronic wave during the time of one
cycle. Given a wavelength of 12 meters, what is the frequency?
A. 250 kHz B. 25 kHz C. 250 MHz D. 25 MHz

Given: λ = 12 m

Electronic wave = 3x108 m/sec.

Solution:

λ = V/f

f = (3x108) / 12

f = 25 MHz

50. Alternators installed in a hydro power station, are of


A. low speed B. high speed C. medium speed D. very high speed

51. A sine wave has a frequency of 60 Hz. Its angular frequency is


A. 120 π B 60 π C 30 π D 6π

Solution:

f= 60 Hz

ω= 2π (60)

ω 2 π

52. In a loss-free RLC circuit, the transient current is


A. oscillating B. Square wave C. sinusoidal D. non-oscillating

53. What is the maximum flux in mWb of a transformer with 200 primary turns when
connected to a 60 Hz, 200 V supply?
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
AUGUST 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

A. 10.35 B. 6.25 C. 50.5 D. 3.75

Given: N= 200 turns

f= 60 Hz

E= 200V

⌀= ?
Solution:

E= 4.44 Nf⌀

⌀ = E / 4.44Nf
⌀= 200 / (4.44x200x60)
⌀= 3.75 mWb
54. A 60 Hz, 4 pole, 3 phase induction motor has a slip of 2.5 %. What is the slip speed
in rpm ?
A. 55 B. 40 C. 60 D. 45

Given: f = 60 Hz

P=4

S= 2.5%

NSLIP =?

Solution:

Ns = 120f/ P

Ns = 120(60) / 4

Ns = 1800 rpm

Nr = Ns (1-s)

= 1800 (1 – 0.025)
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
AUGUST 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

Nr = 1755 rpm

NSLIP = Ns-Nr

NSLIP= 1800-1755

NSLIP = 45 rpm

55. A 15 hp, 460 V, 60 Hz, 6-pole, 3 phase induction motor has full-load slip of 4%. Find
the full-load torque in N-m.
A. 85 B. 88 C. 93 D. 90
Given: POUT = 15 HP

f = 60 Hz

P = 6 Poles

S=4%

T=?

Solution:

HP = 2πNrT / 44,760

Nr = (120f/P) / (1-s)

= ((120x60)/6) / (1-0.04)

Nr = 1152 rpm

T = HP (44,760)/ 2πNr

= (15) ( 44,760)/ 2π(1152)

T= 92.76 N-m or 93 N-m

56. At very low frequencies a series RC circuit behaves as almost purely


A. resistive B. inductive C. capacitive D. none of these
57. A power plant gets water from a dam from a height of 122.45 m at the rate of 1,000
cubic meters per minutes. If the output of the plant is 15,000 kW, what is the plant
efficiency?
A. 80% B. 75% C. 70% D. 65%
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
AUGUST 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

Given: h = 122.45 m

Q = 1000m3 / min. x (1 min/ 60 sec.) = 16.67 m3 / sec.

Pout = 15,000 kW or 15 MW

Solution:

Pin = Qwh

Specific weight of water W = 9810 N/m3

Pin = (16.67)(9810)(122.45)

= 20 MW

n = (Pout / Pin) x 100

= (15 MW / 20 MW) x 100

n = 75 %

58. A 500 kVA, 34.5/13.8 kV, 3-phase transformer is delta-wye connected. The primary
and secondary winding resistances are 35 ohms and 0.876 ohm per phase,
respectively. Calculate the percent efficiency at full-load and unity power factor id core
loss is 3 kW.
A. 97.8 B. 98.7 C. 96.4 D. 97.2

Given: S = 500 kVA rp = 35 ohms

34.5/13.8 kVA rs = 0.876 ohms

Pcore = 3 kW

n=?

Solution:

Refer to secondary side

Rs = rs + rpa2

= 0.876 + 35 ((13.8/ √3)/34.5)2

= 2.743 ohms
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
AUGUST 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

Is = S / √3 Vs)

= 500x103 / √3x13.8103 )

= 20.92

@ Unity pf

Po = S = 500kW

Pcu = 3Is3Rs

= 3 (20.92)2 (2.743)

= 3601.39 W

n = (Po/Pi)x100 = (Po/Po+Pcu+Pcore)x100

= 500x103 / (500x103 + 3x103 + 3601.39)

n = 98.7 %

59. One of the advantages of distributing the winding in alternators is to


A. reduce harmonics C. improve voltage waveform
B. reduce the amount of copper D. decrease the value of the voltage

60. The effect of corona is


A. increased reactance C. Increased energy loss
B. increased inductance D. all of these

61. The rms value of pure cosine function is


A. 0.5 of peak value C. same as peak value
B. 0.707 of peak value D. zero

62. Calculate the use factor of a power plant if the capacity factor is 35% as it operates
8,000 hours during the year.
A. 38.32% B. 33.82% C. 36.82% D. 32.54%

63. How many commutator bars does a 6-pole dc generator with 4 coil elements per slot
and 20 slots per pole has?
A. 60 B. 120 C. 240 D. 480
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
AUGUST 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

Solution:

Z = (4 Coil/ Slot x 20 Slot/Pole) ( 6 Pole )

= 480 Coils

Nc = Z/2

= 480 / 2

Nc = 240 Coils

64. A coil having a resistance of 10 and an inductance of 4 H is switched across a 20


V dc source. Calculate the rate of current change when the current reaches 50% of its
final steady value.
A. 2.5 A/s B. 3.5 A/s C. 4.5 A/s D. 5.5 A/s

Solution:

XL = 2π(60)(4)

= 480π

ISV = V/Z

IFSV (50%) = 20 / (√102 + 480π2)

IFSV = 0.026 (100)

IFSV = 2.65 A or 2.5 A/s

65. Calculate the torque in N-m developed by a 440-V dc motor having an armature
resistance of 0 25 and running at 750 rpm when taking a current of 60 A
A. 240 B. 325 C. 285 D. 342

Given: V = 440 N = 750 rpm


Ra = 0.25 IL = 60 A
Solution:

Eg = VL – IaRa

= 440 – (60)(0.25)
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
AUGUST 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

= 425 V

Pd = (425)(60)

= 25,500 N

HP = 25,500/746 = 34.182 HP

HP = 2πNrT / 44,760

34.182 = (2πx750xT / 44760

T = 324.67 or 325 T-N

66. In a 13.8 kV, 10 MVA base three-phase system, a phase to phase fault occurs. If
the Thevenin’s equivalent impedance of the system is 2 63%, what is the per unit fault
current?
A. 19.01 pu B. 32.93 pu C. 57.03 pu D. 98.78 pu
Solution:

ILL = √3/2 10 MVA/ √3 13 8 kV 0 0263

ILL = 13776.38

Ipu = I FAULT/ I BASE

= 13776.38 / ( ( 10MVA/ √3 13 8kVA

= 32. 93 pu

67. A single-phase transmission line having a length of 15 miles is composed of two


copper conductors spaced 7ft apart having a diameter of 0.0575inch. Determine the line
to line capacitance in farad if the system frequency is 60Hz.
A. 4.9 x10-12 B. 9.8 x10-12 C. 11.8 x10-9 D. 118 x10-9

68. What is the equivalent IEEE device function number for directional power relay?
A. 21 B. 32 C. 37 D. 87

69. Energy radiated continuously in the form of lighting waves is called?


A. luminous intensity B. lumen C.luminousflux D. illumination
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
AUGUST 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

70. Efficiency is secondary consideration in case of


A. base load plants C. both base load and peak plants
B. peak load plants D. none of these

71. A capacitor is charged with 10,000 µC. If the energy stored is 1 joule. Find the
capacitance
A. 5µF B. 50µF C. 500µF D. 5,000µF
Solution :

W= 1/2 (Q2/C)

1j = 1/2 ( (10,000 x10-6)2/C)

C= 50µF
72. To a series RLC circuit, a voltage of 10 v is applied. if Q of the coil at resonant
frequency is 20., the voltage across the inductor at resonant frequency will be
A. 200 v B. 100 v C. 75 v D. 50 v
Solution:

Q-factor = EL / Ep

EL = ( Qfactor ) ( Ep )

= ( 20 ) (10 )

EL = 200 V

73. A 75 KVA, 13,200/240, 60Hz, single-phase transformer has 1% resistance and 3%


reactance. Find the copper loss of the transformer at 3/4 of full-load and unity power
factor.
A.155 W B. 422 W C. 368 W D. 508 W
74. The starting current of a 15 hp, 460 v, 3 phase induction motor is 130 A. What
voltage in volts should be applied to the motor so that the starting current will not
exceed 84.8 A.
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
AUGUST 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

A. 238 B. 295 C. 300 D. 310

75. The condition of a liquid electrolyte is measured in terms of its


A. current value B. specific gravity C.acid content D.voltage output

76. If the of a dc shunt motor is opened


A. it will continue to run at its rated speed
B.the speed of the motor will become very high
C.The motor will stop
D.the speed of the motor will decrease

77. An ideal full-wave rectifier with non-inductive load has zero transformation losses.
What is the efficiency?
A. 70% B. 80% C. 90% D. 100%
78. In an ACB sequence balanced 3 phase system, the angle between the line voltage
and the line current is 50°, what is the lagging power factor of the system?
A. 0.30 B. 0.50 C. 0.64 D. 0.94
79. A power station supplies 60 KW to the load over 2,500 ft of 000 2-conductor copper
feeder the resistance of which is 0.078 ohm per 1,000 ft. The bus-bar voltage is
maintained constant at 600 volts. Determine the maximum power which can be
transmitted.
A. 60 KW B. 230.7 KW C. 120 KW D. 150 KW

80. IF a certain conductor has an area of 336.400 circular mils, what is the radius of this
conductor in cm?
A. 0.0663 B. 0.810 C. 0.737 D. 0.921
Solution:

Acm = d2 d = 2r r = [290 x (1/100 in)] [2.54 cm/1in]

d2 = 336,400 r=d/2 r = 0.737


REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
AUGUST 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

d = 580 r = 580 / 2

81. In laboratory experiment, the impedance of a coil was obtained at 60Hz and 30Hz.
These are 75,48 ohms and 57.44 ohms , respectively . What is the inductance of the
coil?
A. 150 mH B. 182.5 mH C. 42.5 mH D. 2.1 mH
Solution:

Z2 = R2 + XL2 XL = 2π fL

60 Hz , ( 75.48 )2 = R2 2π 60 L 2
-------- eq. 1

30 Hz , ( 57.44 )2 = R2 2π 30 L 2
-------- eq. 2

Solve 2 equations

L = 150 mH

82. The dummy coil in dc machines is used to


A. eliminate reactance voltage
B. eliminate armature reaction
C. bring about mechanical balance of armature
D. eliminate harmonics developed in the machine

83. A half wave rectified sine wave has a averange value of 100 amperes. What is its
effective value ?
A. 157 A B. 70.71 A C. 100 A D. 141.4 A
Solution:

Half wave , Vave = 0.318 Vmax Vrms = 0.5 Vmax

100 = 0.318 Vmax Vrms = (0.5)(314.465)

Vmax = 100 / 0. 318 Vrms = 157.23 A

Vmax = 314.465 v
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
AUGUST 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

84. A 3-phase transmission line has two aluminum conductors in bundle. The self GMD
of each conductor is 8.90 mm and the distance between the conductors and phase is
40cm and 9m, respectively. What is the inductance of the line per phase ?
A. 1 µH/m B. 2 µH/m C. 3 µH/m D. 4 µH/m
Solution:

L = 2x10-7 ln ( Dm / Ds )

Dm = ? d1 = 40 cm = 0.4 m

Ds = 8.9 mm d2 = 9 m

for bundled conductor

Dm = Dmxy = squareroot d1d2

= (0.4) (9) = 1.879m x (1000mm / 1m) = 1879mm

L = 2x10-7 ln (1879 / 8.9)

L = 1.07x10-6 H / m

L = 1.07 µH/m
85. The purpose of insulating oil when used on power circuit breaker are the following.
EXPECT:
A. Interrupter B. Coolant C. Insulation D. Quencher
86. A wind generator with an efficiency of 80% has a blade diameter of 20 m. If the wind
velocity is 30 Km/hr, how much power in KW is obtained from the generator?
A. 58.47 B. 45.78 C. 48.75 D. 54.87
Solution:

For wind plant

Pin = 0.2962 D2 V3 P = Watts

D = Blade diameter (m) = 20m V = wind velocity ( m/s )

V = ( 30 Km / hr ) ( 1 hr / 3600 sec ) ( 1000 / 1 km )

V = ( 25/3)m/s
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
AUGUST 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

Pin = ( 0.2962 ) ( 20 )2 ( 25/3 )

Pin = 68.56 KW

PRating = n ( Pin )

= 0.8 ( 68.56 )

PRating = 54.85 KW
87. The field system of a 50 Hz alternator has a sinusoidal flux per pole of 0.1 Wb .
Calculate the emf generated in one turn which span two-thirds of a pole pitch.
A. 17.4 v B. 19.2 v C. 12.2 v D. 9.6 v
Solution:

E = 4.44 Kd Kp f Φ N

Kd = 1 ( since not stated / assumed )

Kp = sin ( 90deg x pitch )

Kp = sin ( 90 x 2/3 )

Kp = 0.866

E = (4.44)(1)(0.866)(50)

E = 19. 2 V
88. A dc shunt motor develops 15 hp at 120 v . If the effective armature resistance is
0.061 ohm and the field winding draws 2 amperes, what is the overall efficiency?
A. 90% B. 93% C. 94% D. 95%
Solution:

Pd = 15hp x 746 = 11.19 KW PspL = 0 since not given

Pd = Pout + PspL PLoss = Pcu + PspL

= 11.19 + 0 PLoss = Ia2 Ra + Vsh Ish

Pd = 11.19 KW PLoss = (98.15)2 (0.061) + (120)(2)

PLoss = 827.60 W
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
AUGUST 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

η= ( Pout / Pin ) = ( Pout / Pout + PLoss )


P d = E b Ia , E b = V T - Ia

11.19x103 = ( 120 - Ia(0.061)) ( Ia )

Ia = 98.15 A

η= ( 11.19 KW / ( 11.19 KW + 0.8276 KW ))


η= 93.11 % = 93%

89. An 8-pole triplex wave-wound armature carries a total of 660 amperes. If there are 8
brush arms, calculate the current in each armature conductor.
A. 110 A B. 66 A C. 22 A D. 11 A
Solution:

Ia = 660 A

a = 2m

a = (2) (3)

Ia/conductor = ( Ia / a ) = 660/6 = 110 A

90. Inside a hollow spherical conductor


A. electric field is zero
B. electric field is constant
C. electric field changes with distance from the center of the sphere
D. electric field is unity

91. Residual magnetism is necessary in a


A. separately excited generator C. both of these
B. self-excited generator D. none of these
92. Three impedance Zab=4-j3 ohms, Zbc=3+j4 ohms and Zac=10+j0 ohms are
connected in delta across a 220 volts, three phase balanced source. What is the total
power of the circuit?
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
AUGUST 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

A. 1,839 W B. 1,836 W C. 6,130 W D. 18,392 W


Solution:

PT = ? assume ABC sequence

PT = Iab2 ( Rab ) + Ibc2 ( Rbc ) + Ica2 (Rca)

Iab = | ( Vab / Zab ) | = | (220<0) / (4-j3) | = 44 A

Ibc = | ( Vbc / Zbc ) | = | (220<-120) / (3+j4) | = 44 A

Ica = | ( Vca / Zca ) | = | (220<120) / (10+j0) | = 44 A

PT = (44)2 ( 4 ) + (44)2 (3) + (22)2 (10)

PT = 18,392 W

93. A light source located 2.75 m from a surface produces an illumination of 528 lux on
the surface.Find the illumination if the distance is chaged to 1.55m.
A. 298 lux B. 1,662 lux C. 937 lux D. 168 lux
Solution:

E = K / d2

E1 d 1 = E 2 d 2

E2 = E1 d12

E2 = E1 d12 = (5.28)(2.75)2 / 1.552

E2 = 1662 lux

94. In case of delta connected circuit, when one resistor is open, power will be
A. unaltered B. reduced to 1/9 C. reduced to 1/16 D. reduced by 1/3

95. Two similar poles, each of 1 Wb, placed 1m apart in ar will experience a replusive
force of_____.
A. 63,000 N B. 796 N C. 8 x1012 N D. 63 x10-3 N

Solution:
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
AUGUST 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

F = Km1m2 / d2

for MKS

K = 1 / 4πµoµr

K = 1/4π 4π x10-7 ) (1)

K = 63325.74

F = 63325.74 (1)(1) / 12

F = 63 ,328 N = 63 000 N

96. In a Transformer, the core loss is found to be 60 watts at 30 Hz and 90 watts at


40Hz measured at the same maximum flux density. Compute the eddy current and
hysteresis losses at 60 Hz in watts.
A. 90,75 B. 85,60 C. 95,69 D. 88,72
Solution:

Pcore = Phrs + PEddy

@30Hz

60 = Phrs1 + PEddy1

@40Hz

with respect to frequency change

P = K f2 ---- eddy

K = PEddy1 / f12 = PEddy2 / f22 = PEddy3 / f32

Phrs = K f2

( PEddy1 / PEddy2 ) = ( f1 / f2 )2

( PEddy1 / PEddy2 ) = ( 30 /40 )2

PEddy1 = 9/16 PEddy2

( Phrs1 / Phrs2 ) = ( 30 /40 )

Phrs1 = 3/4 Phrs2


REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
AUGUST 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

90 = Phrs2 + PEddy2
60 = Phrs1 + PEddy1
60 = (9/16)PEddy2 + (3/4) Phrs2 -------- eq. 1
90 = Phrs2 + PEddy2 -------- eq. 2
PEddy2 = 40
Phrs2 = 50
( PEddy3 / PEddy2 ) = ( f3 / f2 )2 = ( 60 / 40 )2
PEddy3 = ( 40 ) ( 60 / 40 )2
PEddy3 = 90 W --------@60Hz
( Phrs3 / Phrs2 ) = ( f3 / f2 ) = ( 60 / 40 )
Phrs3 = ( 50 ) ( 60 / 40 )
Phrs3 = 75 W ---------60Hz
97. Transient current in an RLC circuit is oscillatory when
A. R=0 B. R= 2√L/C C. R>2√L/C D. R<2√L/C

98. The current from neutral to ground connection is 12 A. What is the zero sequence
component in phase?
A. 12 A B. 36 A C. 4 A D. 3 A
Solution:

In = 12

In = Ia + Ib + Ic

from fortescues theorem

Ico = Ibo = Iao = 1/3 ( Ia + Ib + Ic )

= 1/3 ( In )

= ( 1/3 ) ( 12 )

= 12/3
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
AUGUST 2018
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS

Iao = 4 A
99. With reference to armature windings, wave windings, wave windings are often
called_______ windings.
A. Cascade B. Series C. parallel D. ring

100. This is the compact arrangement of switching devices.


A. Circuit protector B. Switchboard C. Switchgear D. Panelboard

You might also like