August 2013
August 2013
August 2013
AUGUST 2012
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS
Solution:
XCT = 1 = 1
2πfC 2π (60) (6x10-6)
XCT = 442.1 Ω
XCT2 = VT = 240 = 0.54
ZT 442.1
V1 = I1 x Xc1
V2 = V T - V1
V2 = 240-59.68
V2 = 180 v
P 10+ P 30 P0 = 42.2 KW
P0 = =
2
3,030.7+81,830.1
2
GIVEN:
Solution: d = 325mils
A. 363 kΩ B. 3.3 kΩ C. 3 kΩ
D. 16.5 kΩ
A. KW B. KV C. KWHR D. KVA
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
AUGUST 2012
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS
GIVEN:
Solution:
2 2
IRMS = √ (Idc)^2+ ( 5 m ) + ( ℑ1 ) +¿ 2
IRMS = 1.89
A. 25 A B. 50 A C. 500 A D. 1250
GIVEN: Solution:
I =?
10. In a series RL circuit, the inductor current _____ the resistor current.
11. A 120v dc shunt motor has an armature resistance of 200 mΩ. If the
full-load armature current is 75 A, find the starting resistance needed to
limit the armature starting current to 150% of the full-load value.
13. The law that induces emf and current always opposes the cause
producing them was discovered by
15. Solution:
COD = pi/ pr
1.2 = 1000/x
X = 0.833 kw
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
AUGUST 2012
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS
2,400/240
PRIMARY =2,400
SECONDARY =240
18. A 400 v series motor working with unsaturated field is taking 60 A and
running at 840 rpm. The total resistance of the motor is 0.1 ohm. At what
speed will the motor run when developing half-full load torque.
A. 35 µF B. 70 µF C. 115 µF D. 140
µF
GIVEN: C = Σo Σr / d (n-1)
25. The pitch factor for 3rd harmonic in an alternator having 18 slots per
pole and colt spam 5/6 of pole pitch is
26. A certain power plant has reserve capacity above the peak load of
MW. The annual factors are load factor = 59%, capacity factor = 41% and
use factor 46%. Determine hours per year not in service.
LF
= L/ PL ∆L = LE (PL)
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
AUGUST 2012
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS
= 752 HRS
GIVEN: d = 1x10-3 m2
L = 50 x 10 -2 m I = 2∆
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
AUGUST 2012
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS
29. A string insulator has 4 units. The voltage across the bottom most unit
is 30% of the total voltage. What is the string efficiency?
30. When a line to ground faults occurs, the current in a faulted phase is
100 A. What is the zero sequence current in this case?
100.3 = 3Iao /3
A. 12 MW B. 10 MW C. 9.8 MW D. 11 MV
35. 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%.
Ia1(pu) = 1/ Z1 +Z2+Z3
= 1 / j0.3+j0.2+j0.1
= -j1.67
VØ = 220/√8 = 127.02 v
Pt= 3,495.56
39. A coil of wire of 100 turns enclosing an area of 0.09 m 2 has its plane
perpendicular to a uniform magnetic field of flux density 0.5 T. the coil is
turned to position wherein its plane is parallel to the magnetic field in 0.05
sec. Find the average emf generated in the coil in volts.
A. 15 B. 10 C. 12 D. 9
Ø=ß∆ e = N dd/dt
Ø = 0.0045 e = 9v
e= 0.0045
A. 200 v B. 100 V C. 75 v D. 50 v
Solution:
QFactor = EL / ET
20 = EL/10
EL = 200 V
42. One advantage of distributing the winding in alternator is to
GIVEN: Xc
Q=
R
Q=5
F = 100 X 103Hz P 100
P = 10W R = 2 = 2 =156.25
I 0.8
I = 0.8∆ Xc
C =? S= ; Xc = 781.25
156.25
1
Xc = ; 782.25
2 πFC
=
SOLUTION: 1
3
; C=2.04
2 π (100 x 10 )(C)
47. An iron ring of circular cross-section of 5 x 10 -4 m2 has a mean
circumference of 2 m. It has a saw-cut of 2 x 10 -3 m length and is wound
with 800 turns of wire. Determine the exciting current when the flux in the
air gap is 0.5 x 10-3 Wb. Given µr of Iron = 600 and leakage factor is 1.2.
Assume areas of air gap and iron are same.
A. 11 B. 12 C. 10 D. 14
A. 11 B. 12 C. 10 D. 14
SOLUTION:
IL= (√3)Vp = (√3)(230)
Z 8 – j6
IL = 39.8
SOLUTION:
Ia1 = 1 {(Ia+a(Ib)+a2(Ic)}
3
= 1{(10<-30)+(1<120(0)+(1<240)(10))}
3
= 3.34<300∆
IC1= a(Ia1)
=(1<120)(3.34<30)
=3.34<1500A
SOLUTION:
L
F=(2X10-7) (I1I2) ( )
D
2
F=(2X10-7)(8X8)( )
3 X 10−3
F= 8.53X10-3N
57. A 40 miles, 60 cycle single phase line consists of two 000 conductors
spaced 5 ft apart. Determine the charging current if the voltage between
wires is 33 kV. The diameter of 000 conductors is 410 mils.
SOLUTION:
%Vr=√(cosθ+(%IR))2+(SINθ+-(21%))2-1
=√(0.8+0.04)2+(SIN(36.87)+(0.06))2-1
=0.068X100
=6.8%
GIVEN: SOLUTION:
PZ 0 IA
I=100A P=4 T=? T=
6.283 a
( 4 )( 500 ) ( 0.003 ) (100)
V=230V T=
6.283( 4)
R=0.03Ω T=238.73N-M
Z=500Ω
60. A 0.5 µF capacitor, an 80 mH inductance and a 500-ohm resistor are
connected in parallel across a voltage source of V = 12 -j6 v. Find the
current as a phasor if the angular velocity is 500 rad/sec.
C. 0.333∠122° A D. 0.333∠-122° A
62. A mode of communications in which either party can hear while taking,
thus allowing one party to instantly interrupt the other at any time, is called
63. suppose a cell of 1.5 v delivers 100 mA for 7 hours and 20 minutes,
and the it is replaced. How much energy is supplied during this time?
20
W(energy) = Pt =(1.5v)(100x10-3)(7+ ) = 1.1Whr
60
65. Find the cross-sectional area of the core of a 10 turns transformer for
a voltage of 50 v at 50 HZ. The flux density is 0.9 Wb/m 2.
66. The rotor resistance of an 8-pole, 50 HZ, wound rotor induction motor
is 0.5 Ω per phase. The speed of the motor is 720 rpm at full load.
Determine the external resistance to be connected with the rotor circuit to
reduce the speed to 680 rpm for full-load torque.
GIVEN: SOLUTION:
Sb
Sb=600KVA IF=
¿ 3Vb( Xfu)
600 X 10(3)
V=2,400V I F=
√ 3(2400)(0.1)
f=60HZ IF=1443 A
P=6
IF=?
A. 2 mH B. 3 mH C. 4 mH D. 5 mH
SOLUTION:
K=0.75
M=K√L1L2
=0.75√(8X10-3)(2X10-3)
=3X10-3H = 3MH
71. Wind mils in addition to proper gearing is the best driver for
A. 0.2 c B. 50 kc C. 20 c
D. 5 c
SOLUTION:
W=Pt
W 500 J
W=Ivt ∴ Q= = = 5C
V 100V
74. A 220 v, 3-phase supply has a balanced delta connected load with
impedance of 15 + j20 ohms/phase. What is the total real power of the
load in watts?
GIVEN:
V=220
ZØ=15tj20Ω=25<53.13
SOLUTION:
V2
P=3( COSØ)
Z
220
P=3( cos (53.13)¿
25
P=3,484.8W
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
AUGUST 2012
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS
75. In a 3-phase, 4-wire system, the current in line A has I a1 = 200∠0° A Ia2
=100∠60° A, the current through the neutral conductor is 300∠-60° A. Find
the line current Ib in amperes.
GIVEN: SOLUTION:
Ia2=100<60 Iao=100<-60
In+300<-60 Ib=100<-60+(1<240)(2)(1<120)(100)
Ib=? Ib=300<-120+360
Ib= 300<240
76. Calculate the no-load current taken by a 100 kW, 460 v, shunt motor
assuming the armature and field resistance to remain constant and equal
to 0.03 Ω and 46 Ω, respectively. The efficiency at full load is 88%.
A. 2,232,435 m3 B. 2,345,456 m3
C. 2,293,560 m3 D. 2,371,402 m3
A. series B. Shunt
81. 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,
Vb1 2 Sb 2
Xpu2=Xpu1( ) ( )
Vb2 Sb 1
2 1
Xpu2=0.3 ( )2 ( ¿ ; xpu2=0.60
1 2
A. 94 B. 96 C. 97 D. 98
SOLUTION:
√ 3(64−36.5)
Ø=tan-1( ¿
(64+ 36.5)
Ø=25.360
W=VLILCOS(30-Ø)
64X103=(600)(IL)COS(30-
25.36)
IL=107
REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
AUGUST 2013
PROFESSIONAL ELECTRICAL ENGINEERING SUBJECTS
84. A power plant has an overall efficiency of 30%. If this plant can
consume 4,200 kilograms of coal per hour, estimate the total electric
energy produced in one day. Assume the calorific value of the coal being
used is 8,000 kcal per kilogram.
85. Four lamps are suspended 6 m above the ground at the corners of a
lawn 4 m on each side. If each lamp emits 250 cd, calculate the
illumination at the center of the lawn.
d2=(2)2+(2)2+(6)2=44
d=6.63
J 250 6
=4( ¿COSØ=4( ¿( ¿
D2 44 6.63
=20.56LUX
A. 4.81 x 10-5 mho per mile C. 7.92 x 10-5 mho per mile
B. 12.74 x 10-6 mho per mile D. 2.47 x 10-6 mho per mile
A. 50 Ω B. 25 Ω C. 400Ω D. 100 Ω
89. A 500 MCM ACSR cable has 37 strands. Determine the diameter in
mils of each strand.
90. A contact device installed at the outlet for the connection of a single
attachment plug.
91. 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?
92. An air cored coil has 500 turns. The mean length of magnetic flux path
is 50 cm and the area of cross-section is 5 x 10 -4 m2. If the exciting current
is 5 A, determine the flux density.
A. 100 B. 150 C. 50 D. 0
A. 30 B. 60 C. 300 D. 600