1.

1Kwhr = 3413 BTU Solution :

2. C. Philippine Electrical Code V 1 26.59− j 4.69 V
I 1= =
3. C. Outlet Z1 2.2= j2.2 Ω
4. A. 1800 mm
¿ 8.68 ⎳−5 5
5. B. 200 Lux
Solution : V =I 1 ( Z 1+ Z 2+ Z 3 )
2
I2 d1
= ( )
I1 d2 ¿ 8.68 ⎳−55+ ( ( 2.2+ j2.2 ) + ( 10+ j10 ) + (− j5 ) )
2
800 d1
=(
2d 2)
¿ 111.908− j50.894 ¿ 1.8+10.37 ohm CM / ft
E2
I 2=200 Lux 12. B. 18
6. D. 16.25 Solution :
Solution: 240 V
Rt = =60 Ω
I ampacity=( 1.25 ) ( 5 ) +10=16.25 A 4A
60 V
R 1= =15Ω
7. B. doubled 1A
Solution: 60=15+1.5 R 3+ R 3
R2 L2 R 3=18 Ω
=
R1 L1
R 2 2 L1 13. A. 3556 kW, 4.444kW
= =2
R 1 L1 Solution :
R 2=2 R 1 @58Hz
8. A. 7, 194 W 60−58 60−fr
=
Solution: Pb 8000
( 120 ) ( 60 ) @57.5Hz
Ns= =1200 rpm
6 60−57.5 60−fr
Nr=1,150 rpm =
Pa 8000
1200rpm−1,150 rpm
Sr= =0.04166 8000 kW =Pa+ Pb
1200 rpm
Pb=800−Pa
Pcur 300
Pinr= = =7200W 60−58 60−57.5
S 0.04166 = −Pb
9. C. 156 mm pb 8000
Solution: Pb=4.444 kW
ρa ρc Pa=800−4444=3556 kW
=
Aa Ac 14. A. 2
Solution ;
17 ΩCM /ft 10.37 Ω Cm/ ft L1 N 12
= =
−3 2 2
π ( 2 x 10 ) /4 π ( x ) /4 l 2 N 22
X =1.56 x 10−3 m N1 2

10.
N2
=
−5 √
=2

11. D. 112-j51 V 15. D. 72

 Solution: −5 2
Q=IT
¿ 10 ( )
1
=2.5 mH
Q= (10 )( 60 x 60 )( 2 ) 25. B. 0.12 v
Q=72 kC Solution :
16. C. 18Ω e=BlV
Solution : ¿ ( 10 x 103 ) ( 10 ) ( 120 ) ( 1 x 10−8 ) e=−0.12 v
36 R 26.
=12
36+ R 27. D. 500 w
R=18Ω Solution ;
17. rT =40+40=80
18. C. RA 9136
2002
19. C. Collecting Efficiency Pnew= =500 W
80
20. A. 0.3174Ω
28. D. 3.48
21. C. 178µC
Solution :
Solution:
Vϕ
Q Iϕ= =¿ 220/15+j20= 5.28+j 7.04
C= Zϕ
E
C=( ( 4 x 10 ) ¿ ¿−1+ ( 5 x 10 ) + ( 6 x 10 ) )S=3
¿ IϕVϕ=3 ( 5.28+ 7.04 ) ( 220 )
−6 −6 −1 −6 −1

C=1.62 x 10−6 F S=P+ jQ=3484.8+ J 4646.4

Q=( 1.6 x 10−6 ) ( 110 )
Q=178 µC 29. B. Inductance
22. D. 17.89 30. D. 360 watts
Solution: V2 1202
Pin= =
1 BTU =252 cal 4 R 4 ( 9.5+ 0.5 )
7.1 x 1010 x 252 cal=17.89 Gcal ¿ 360 Watts
23. D. 28.76 31. B. 1 182.5
Solution: Solution :
2 πNT ( 120 x 60 )
20= =¿ N s= =1200 rpm
33000 6
T =75.03 105
fr= =0.875
2 x 60
T =( X −dea dweight ) (length of arm ) 0.875
S= =0.0145
60
75.03= ( x −3.75 )( 3 )
Nr=( 1200 ) ( 1−0.0145 )
¿ 28. 76lb Nr=1182.6 rpm

24. C. 2.5 mH 32.

Solution ; 33. B. Copper
2
n2
L 2=L1 ( )
n1
34. A. 0.0043⁰C
1
ἀ 0= =0.00430 C
T
 R 2 T +t 1 45. B. Stator
=
R 1 T +t 2 46. D.35.71
35. C. They Producing less arching Solution :
36. 50 x 7=35
37. A. 0.0124 50 x .95=47.5
Solution: 47.5−35
%= =35.71 %
110 47.5
a= =0.25
440
1 2 47. D. DC Source
x ' =( 0.06 )( )
.25
=0.96
48. A. 80 %
4402 Solution:
Xb= =77.44
2.5 x 103 0.25−0.05
0.96 NS %= x 100=80 %
Xpu= =0.0124 0.25
77.4
49.
38. A. 5.98 A 50. B. Conductor
( 3 x 10−6 ) ( 5 )=( 1.5 x 10−6 ) 51. B. 2.74Ω
1 Solution :
X= = j 2122 −8
(2 π ) (50 ) ( 1.5 x 10−6 ) ρL ( 1.72 x 10 ) ( 2000 )
R= = 2
11 x 103 A π ( 4 x 10−3 )
I= =5.98 A
j2122 4
R=2.74 Ω
39. D. half 52. A. 1.25 mJ
40. Solution :
2
41. C. 31mH 1 1 50 x 10−6
Solution: W = C E2= ( 1 x 10−6 )
2 2 (
1 X 10−6 )
M =K √ L1 L2=0.98 √ ( 20 x 10 ) ( 50 x 10 )
−3 −3
W= 1.25 mJ
¿ 31 mH 53. B. 1.56⎳-38.7 A
42. C. 3600 watts Solution :
Solution: V 10
2 2
I= = =1.56 ⎳−38.7 A
V 12 Z 5+ j 4
Pmax= = =3600 watts
4 R 4 ( 0.01 ) 54. C. It decreases by 50 %
43. A. 5.5 kg 55. D. 0.001 F
Solution : Solution :
Q 1 Q2 ( 2 x 10−3 ) ( 3 x 10−3 ) a=10
F= = =54
4 πɛoɛr d 2 4 πɛoɛr ( 100 x 10−2 )2
F=mg=¿
Cp=( .1 x 10−6 )
( a1 )
2

44. A. 400:4000 Cp=( .1 x 10−6 ) ( 1 )=0.

2 001µF
Solution: 10
VL=√ 3 ( 4000 ) =6928.20V 56. D. 1387.5 rpm
 Solution :

110
ST =3 ( 10 √3 ) ( 240
√3 )
=7200

Ia 1=10− =9.5 63. C. 2.25⎳-14.3

220
Solution :
E 1=110−( 9.5 )( .8 ) =102.4 1
Ia 2= ( Ia +a2 Ib +aIc )
3
110 1
Ia 2=10+ =10.5 ¿ ¿
220 3
Ia 2=2.25 ⎳105.7
E 2=( 110 ) + ( 10.5 ) .8 ¿=118.4
64. B. 1.08 x10^6 V
E 2 N 1 (118.4 ) ( 1200 ) Solution :
N 2= =
E1 102.4 Q 1.2 x 10−6
V= =
4 πɛoɛr d 2 4 πɛoɛr ( 0.1 )
N 2=1387.5
V =1.08 x 106
57. B. Statically induced emf 65. C. 3000
58. ( 3 x 103 ) (1.257) = 3,000
Tair=
59. A. 1890 rad/sec µo
Solution : 66.
wL 67. B. 94.92
tanθ=
R Solution :
( X ) ( 0.06 ) Po=50 x 10 3 ( .25 ) (−8 )
tan80 0= X
20 ¿ 10,00 W
¿ 1890 rad /sec Po
N=
60. C. 727.5 ( Po+ Plosses )
Solution : Pcu=( 0.25 )2 ( 560 )=35 W
NP ( 500 )( 12 ) 10 x 10 3
f= = =50 Hz N= =94.92 %
120 120
10 x 103 +35+500
@motor
68. D. Fault MVA
120 f ( 120 ) ( 50 )
Ns= = =750 Hz 69. D. Base load Plant
P 8 70. B. 1000
S=Ns ( 1−s )=( 750 ) ( 1−.03 ) Solution :
S=727.5rpm 2 2
I =E d =( 10 )( 10 ) =1000 cp
71.
61. D. is an extremely large number of
72. D. 215
charge carriers
Solution :
62. B. 7200
AveLoad
2400 Lf =
Peakload
Iϕ= √( )
8
3
=10 √3 A ¿
(100 )( 20 ) ( 3.05 ) + ( 50 ) ( 8 ) ( 305 )
100 x 24 x 305
Lf =25 %
73. B. 100 A 85. D. 0337
Solution : Solution :
It=3 Iao W 1=−220
If 300 W 2=940
Ia 0= =
3 3 −1 ( √ 3−220+940
(− 220−940 )
)=−¿70.28
Iao=100 A θ=tan
74. B. 220
75. pf =cosθ=cos (−70.28 )=0.337 ¿ ¿
76. 86.
77. 87. B. 1200 A
78. B. 5 Solution :
Solution : Ia=Iap ( mP )=( 10 ) ( 2 )( 12 )
Ia=1200 A
Iave= (√ 12 ) ( 5 =5 )=5 A
2 2
88. D. 3300A
79. B .93.38 % Soltuion :
Solution : −( 5 x 106 ) ( 0.5 )
Zpu= 2
=0.01313
Pco=x 2 Pcufl ( 13.8 x 103 )
75 Zt=0.01313+ 0.05=0.06313
Pcufl= 2 =208.33
6 5 x 106
If =
( 5 ) ( 103 ) ( .8 ) ( √3 ) ( 13.8 x 103 ) ( .6313 )
N=
( 5 x 10 3 ) ( .8 ) + ( 288.33 ) +75 If =3313 A
N=93.38 % 89. B. 36
Solution :
80. C. 3.46 Ω E 1 2 ( ) 0.8 2
Solution :
T ' =T
E2 ( ) ( )
=1
1
=64

Dm= √3 ( 4 ) ( 4 ) ( 8 )=5.04 Treduced=1−0.64=.36

L=2 x 10−7 ¿ Treduced=36 %
90. D. can have either a positive or
(377 ) ( 1.139 x 10−6 ) ( 580 )( 5 )
X= =3.46 negative charge
3.28
91. D .2 x 10−5 N
81. C. 75 %
Solution : Solution :

Kwh ( 15 x 103 ) ( 60 ) ( 2 x 10−7 ) ( I 1 I 2 L )

N= = F=
Qδh ( 1000 ) ( 9.81 ) (122.45 ) d
N=75 % ( 2 x 10 ) ( 5 ) ( 2.5 x 10−2 )
−7
¿
82. A. Bandwith is 200 Hz 10 x 10−2
Solution : ¿ 2 x 10−¿ 5
fr 10 x 103 92. D. 0.25 H
Bw= = =200 Hz Solution :
Qf 50
83. A. Moving electric charge di
e=N
84. dt
 6−2
100= X =0.25 H
10−3
93. B. two winding transformer with the
short circuit secondary
94. A. 25, 653
Solution :
( 120 )( 50 )
Nr= ( 1−0.05 ) =1440
4
( 2 π )( 1140 )( 160 )
Po= =32.42 x 746=24127.43
44760
Pm=Po+ Spl=24127.43+ 500=24627.43
RCL=( Rcl + Pm )=( 0.04 )( RCL+24627.43 )
RCl=1026.143
Pr=Pm+ Rcl=( 24267.43+1026.143 )
Pc=25653.57
95. D. 127 cis (-120)
Explanation : since it positive sequence
the rotation of Vbc is -120
96.
97.
98. B. 1500 ohms
Solution:
90 x 103
=1500 Ω
60 km
99. A. 37.18 km
Solution :
15 x 10 6
I=
( .8 ) ( √ 3 ) ( 132 x 103 )
I =82
Plosses=( 15 x 106 ) ( 0.05 )
Pl=750 x 103
Pl+ 3 I 2 R
750 x 103
R= =37.18 Ω
822 x 3
L=3718 x 1 x 10 3=37.18 km
100.A. 836.7 A
Sb 50 x 106
3 3
Ib= = =836.7 A
Vb 34.5 x 103
√3 √3