MODULE 6

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EE MODULE VI
MULTIPLE CHOICE
1. A de motor 120 volts has 100 ampere rated current. The allowable voltage drop of
3% of input voltage. Find the maximum distance from the supply if the wire has a
diameter of 316 mils and the resistivity of wire is 10.4 ohm-emf.
A. 508ft. c, 178ft.
B. 316ft. D. 203ft.
VD= Vs-Vt
0.03Vs =Vs -120
120
Vs = - =123.71. volts
0.97
Voltage Drop= 123.71-120;:::: 3.71volts
Voltage Drop= 2Irw
3.71 = 2(100)rw
l
rw =0.018560 = 10.4 (316~)
~

I= 178ft.
Answer is C

2. Determine the per unit impedance of a 3-phase, 50 MVA, 12.8 kv alternator with
30% impedance at 100 MVA and 13.2 KV bases.
A. 0.56 C. 0.48
B. 0.32 D. 0.39
Znew=Z iven(KVgivenY( MVAbase l=o.3(12.8)
- - _g · KVbase) ~MVAgiven) - 13.2)
2

(100)=o.s
50
6u
··- p
Answer is A
3. The capacitance per km of a 3 -wire cable are 0.9 uF between the three bunched
conductor and the sheath, and 0.4 uF between conductor and the other two
connected to the sheath. Determine the line to ground capacitance in J1 F for 30
km length of cable.
A. 9 c. 13.5
B. 4 D. 7
0.9=3Cs
0.4= 2Cc+Cs
Then, Cs = 0.3
Cc=0.05
Cn = 3Cc + Cs =0.45 j1F I km
For 30km, Cn=0.45(30)=13.5pF
AnswerisC
 4. The conductor of transmission line are arranged in the form of equilateral triangle
with sides of 6 m each. If the conductor are 500 mils in diameter . What is its
capacitive reactance of the line in ohm per mile?
A. 128,787 C. 203,105
B. 134,685 D. 125,728
0.5 .
r=-mx-x ft m = 6 .35 x1o-3 m
I 2 12in 3.28ft
I,
tlr.i
il C= 21't&o = 27r(8.85x10-I2 f I m) (1609m) = 0.013061ffarad I mile
11 Ln GMD Ln ?J(6m)(6m)(6m) mile
I; ~: 6.35xl0-3 m
J
I;
I~
l
Xc = -s =203,105 ohm - mile
2Jr60(l.306xl{f )
It
Answer is C
il
I
I 5. A generating station has a maximum demand of 40 MW, annual load factor of
75%, annual plant capacity factor of 65% and a plant use factor of 85%. What is
i the annual energy produced?
I
A. 148.9 x106 lfwh C. 200.9x106 kwh
i
B. 195.6 x10 kwh
6
D. 812.3x106 kwh
LF = TEP O.?S = TEP 1
MDxSr. 40xSr
CF = TEP 0.65 = TEP --~~----2
IC x Pr. ICx8760
PUF = TEP 0.85 =- TEP --------3
IC x Sr. ICxSr
Solving for Sr, equate 2 aREI 3, thus
0.65(8760) =0.85(Sr)
Sr=6699hrs

Solving for TEP, substitute Sr value to equation 1. Thus,

!.'
I TEP=O. 75(40MW)(6699hrs)=200,965MWHrs
,,
!

Answer is C

6. The sending and receiving end voltage of a three phase short transmission line
are Vs= 33 KV and VR=31.2 kv respecti~ely; The per phase line parameter are
R=1 0 ohm and XL=20 ohms. Calculate the maximum power that can be delivered
by the line.
A. 26.57 MW C. 8.86 MW
B. 53.2MW D. 108 MW

P _IVsi~VRI 1vR 1 p
2

max- IZI -lZtcos

 j33j~31.2l
2
·131.2 1 . 0
P
max
= !22.36j I22.36
I
cos63.43 = 26.57 MW
Answer is A

7. The positive, negative and zero sequence reactance's of a 20 MVA, 13.2KV

Synchronous generator are 0.4 pu, 0.3 pu, and 0.2 p.u respectively .The
generator is solidly grounded and is not loaded. A double line to ground fault
occurs at the generator terminals, determine the fault current.
A. 4750 A C. 4500 A
B. 4250 A D. 3030 A
For double line to ground fault,
lpu = . . . 3Xz- = 3(0.3! = 3.46pu
X 1 X 2 +X1X 0 +X2 X 0 0.4(0.3)+0.4(0.2)+0.3(0.2)

!base= 20xl0 3
::::; 874.77 A
.J3(132)
I 1 == 874.77(3.46) = 3028A
Answer is D

8. In a .1r -network, we have Y1=0.2x10-3/0°S , Y2=0.02x10"3/-90°S and Ys=0.25x1o·

3
/90°5. r-ind Y22 parameter.
A. (0.2-j0.02)x1 o-s c. j0.02x1 o-s
a. j0.23x1 o-s D. -j0.02x1 o-a
3
Y22=Y2+Ys=- j0.02x10-3 + j0.25x10.3= j0.23x10"
Answer is B

9. An electromagnetic device for finding the short circuit

A Hi-pot C. Growler
B. Ohm-:-meter D. Megger
Answer is C

10. A 3-phase 480 velts supply delivers power to a wye-connected lead with
impedance per phase of 1O+j5 and a wye connected pure resistive load of 15+j0.
What is the average power of one phase?
A 11.2 kw C. 12 kw
B. 3.3 kw D. 33.8 kw
2 2
I,
.:!!;'· 480 ] [480]
'
~~
~~
P, =LI';R, = .j 73 (10): .Jj (t5)=11,200watts
:,~
:i
[ 102 +52 15
·ll

I Answer is A
.I!
l
;j
:{
11. A three phase, balanced delta connected load draws 30 ampere phase current
from balance three phase supply. An open circuit fault occurs in one of the lines.
Determine the line current.
A.45A C.90A
8.60A D. 100
IL =IrjJ + 0.5/rjJ ::;: 30 ± 0.5(30) = 45A

Answer is A

12. A device that recloses a circuit or isolates a faulted line.

A. Circuit breaker C. Recloser
B. Sectionalizer D. Disconnect Switch
AnswerisC

13. What is the time at which the load is disconnected when there is fault.
A. Arcing time C. Cut-off time
B. Melting time D. Operating time
Answer is D

14.A 30 kva, 220 volt, wye connected, 3-phase, salient pole synchronous generator
supplies rated load at 0.707lagging power factor. The reactance per phase are
Xd::::2Xq::::4 ohms. Determine the percent voltage regulation.
A. 100% C. 224%
B. 600% D. 10%
30 000
Ia = = 78.73A
v= 2
:.? =127 v
-J3
•
.J3(220)

Solving for power angle, 8

~ laXqcosB 78.73(2)cos45°
tan (1 - - ----'--'-----

- V¢ + IaXqsinfJ -129 + 98.73(2)sin45°

0 =25.03°
E¢ = Vq>coso + I{PXd sin(J +B)= 127 cos25.03° + 78.73( 4)sin(25.03° + 45°) = 411 V
. . 411-127
Voltage Regulatwn = xlOO% = 223.7%
127
Answer is C

15. A transformer has a tum ratio of 400:800. If a 0.5 ohms resistor is connected
across the secondary, what is the resistance referred to primary?
A. 2 ohms C. 0.5 ohm
B. 1 ohm D. 0.25 ohms
Consider Step down transformer (2:1)
Rpri = a 2 Rsec = 2 2 (0.5) = 2
Answer is A
i /
i

16. The electric field between two concentric cylindrical conductors at r=0.01m and
r=0.05m is given by E =(10 5 I r)a.(V I m), fringing neglected. Find the energy
stored in a 0.5m length. Assume free space.
A. 0.224 joule C. 224 joules
B. 2.24 joules D. 22.4 joule

E=
2
1 Je E 2dv =t
0
fJ r+O.S r r.Ol --;:-)2
1< .0

{
105
rdr d¢1 dz = 0.224 joules

Answer is A

17.A 150 candela lamp at a distance of 12.37m gives sufficient illumination on a

certain working plain. How far a 250 candela lamp be placed to have the same
illumination as before.
A. 15.97 m C. 49.5 m
B. 9.58 m D. 8.95 m
r J
L = :::=.yc
d-

---=- d = 15.97m
d22
Answer is A

18. Receptacle shall be installed so that no point along the floor line in any wall space
is more than nim, measured horizontally, from an outlet in that space.
A. 1800 C. 1500
B. 1200 D. 1000
Answer is A

19. What is the maximum output of cellular transmitter? •

A.3W C.6W
B. 30W D. 60W
Answer is

20. The frequency variation of FM transmitter is express as a function of _ __

A. AM C. Amplitude
B. FM D. Phase
Answer is D

21. A short transmission line with an impedance of 3+j4 is supplying a balanced load
at receiving end voltage of 13.2 kv. Under this condition, the voltage regulation is
8%. Calculate the sending end voltage.
A 13.8 kv C. 13.5 kv
B. 14.3 kv D. 15.2 kv

f
,I
'1
}
!
 %VR = Vs- Vr xlOO%
Vr
8% = Vs-l3. 2 xlOO% Vs =14.256KV
13.2
Answer is D

22. The power taken by a 500 ft resistive coil made of copper wire is 250 watts at 110
volts. Resistivity of copper is 10.4 ohm-emil per ft. Calculate the cross sectional
area of the coil in circular mil.
A.146 C.107
'· B. 168 D. 175
li V 2 1102
~~ R=-=--=48Aohms
~~
p 250
!l
l
II R=p-
'.~,
A
,,,,
500
48.4 =10.4
!
A
!I A = 107 .44cmil
'i '
AnswerisC
i
23. What is the equivalent area in sq. inch for a conductor 336,400 circular mil?
A. 0.428 C. 0.264
:I B. 0.678 D. 0.768
!I
,t d == .r;;;;izxi 0_,; =.Jo.3364 =0.58inch .
I A = Jr d 2 ""'0.264in 2
I 4
I Answer is C

I
'I
24. Which of the following does not belong for grid design?
A. Capacity C. Voltage
'l' B. Timing D. Circuit Design
':

't Answer is 8

25. What is the minimum surfaee area exposed fer plate electmde?
2
A. 1/6 m2 C. 1/5 m
B. % m2 D. 1/3 2 m
AnswerisC

26.A source of 100 volts rms is supplying a resistance 80 ohms in parallel with a
200 mH inductor. To what angle is the·current lag or lead?
A. 46.7 degrees lag C. 46.7 degrees lead
B. 43.3 degrees lag D. 43.3 dE!grees lead
 XL= 27l'60(0.2) = 15.4ohms

Y= I + l = 0.01822L -46.7°
80+j0 0+j75.4
Answer is A

27. For differential protection, When are the current transformers of the distribution
transformer are connected wye-delta?
A. Delta -wye C. wye-wye
B. Delta-delta D. Open Delta-wye
Answer is A

28. Current through the coil increases to 20A in 1/1000 sec. If its inductance is 100 P
H. What is the induced voltage at this instant?
A. 20V C. 0.2 V
B.200V D.2V
e =L di = lOOxl0-6 ~ = 2volts
dt 0.001
Answer is D

29.1n metering, kwhr meter in conjunction with varhr-meter is used' to

determine_ __
A. Load Factor C. Power Factor
. B. Use Factor D. Demand Factor
Answer is C

30. What is the maximum power that can be transmitted over a three phase short
transmission line having a per phase impedance of 0.3 +j 0.4 ohm if the receiving
end voltage is 6351 volts per phase and the voltage regulation of the line is not to
exceed 5%.
A 108.9 MW C. 266.9 MW
B. 165,6 MW D. 312.3 MW
21 1 2
P¢
v NI'V I .jvRN
= l s i RN·---cosP= !6668.5'1·163511
. - 16351 1
cos53.13=36.3MW/phase
max IZI !ZI !051. 1° 5 1
Pmax = 3(36.3MW) = 108.9MW
Answer is C

31. A series loop contains the following circuit elements in order: a 8-volt source, a 2
kO resistor, a 3 kO resistor, a 16-V source and a 7 kO resistor. What is the
voltage across the 2 kO resistor? Source voltage area additive.
A. 6V C.14V
8. 8V D. 4V
I= 8+16 =2mA
2k+3k+7k
Voltageacross2k0 ::::= 2mA(2kO) =:= 4volts

-------- --~------
 Answer is D

32. The illumination on a certain object 2.5m from the light source is 6648 lux, using
the same light source. what is the illumination on the object at 5 m?
A. 1,226 lux C. 3,324 lux
B. 1662 lux · D. 4700 lux
I
E==-
d2
E1d\ = E2 d 2 2
6648(2.5) 2 = E 2 (5) 2
E1 =1662/ux
Answer is B

33. An armature conductor 100 em long moves perpendicular to magnetic flux of

0.33 Tesla and while carrying 30 amperes when supplied at 120 volts. What is the
Force experienced by the conductor?
A.20N C.50N
B.10N 0.60N
F = Bllsin(} = 0.33T(30A)(l.Om)sin90° = 9.9N
Answer is B

34.A copper bus has a cross section of 0.5inx 6in. Calculate the area in circular mil.
A. 3.82x106 C. 3.82x105
7
B. 3.0x10 D. 3.0x10 5
Sq. mil= 500mils(6000mils) =3000000

Circular mil= isq.mil = i(3000000) = 3.82xl0 6

7! 7!
Answer is A

35. A coil connected in series with a resistance are supplied from a 240 volt 60Hz
source draw 1.0 ampere current. The voltage across the series resistance and
across the coil are both equal to 160 volts. Find the value of coil resistance.
A.180 C.20
8.30 0.60
By cosine law:
Vc01L 2=VR2+Vl-2 VRVreosfh
1602=1602+2402-2( 160)(240)cos eT
BT=41.4°
v 240 .
Zr =....L=-=240L41.4=180+ ;158.75
I 1
v 160
R 3 =_!i.=-=160 ohms
I 1
RcoiL = Rr- R 3 = 180-160 = 20 ohms
Answer is C

- ---
----- ---------- - __:_______~-=---
 36. A balanced delta connected load of 12 + j 15 is connected at the end of 3-phase
line with impedance of 0.1 +j0.2 ohms per phase. The line is supplied from a three
phase source with a line to line voltage of 240 V rms. What is voltage at the load?
A. 232 volts C. 230 volts
B. 228 volts D. 235 volts
Equivalentwye-connected load, Zv = z~3 =4+ j5

By voltage divider method:

vLoad = J3[ 2~ r/
4
: + s' 2 ] ] = 232 volts
""3' 4.1 +5.2
Answer is A
37. Two identica1606 volts generators are connected in parallel, one delivers 400A,
0.95 power factor lagging while the other delivers 150A at 0.75 power factor
lagging. What is the overall load power factor?
A.88% C. 93%
B. 91% D. 90%
1
Ir =400L -cos- 0.95 + 150L -cos- 0.75
1

Ir=400L -18.2° + 150L -41.4° = 541L -24.5°

Pfr := cos24.5 =0.91 lagging
Answer is B
38. A core of annealed sheet steel is wound with 1500 turns of wire through which a
current of 40 mA is flowing. If the length of the coil is 20cm, calculate the
magnetic strength in ampere-turn per meter.
A.300 . C.400
8.350 0.450
H = NI = 1500turns(0.04amp) =300
l 0.2m
Answer is A
39.A simple power system composed of a generator, an ideal transformer(T1) 1:10,
transmission line of impedance 30+j210 ohms, transformer (T2) 20:1 three phase
load of impedance 7.07+j7.07 ohm per phase at line voltage of 460 volts. Bases
are 10 MVA, 460 V. What is the required generator voltage?
A. 13,426 volts C. 960 volts
B. 1197 volts D. 1343 volts
Base Voltages:
460volts @ the load Vpu=1.0
9200volts @ the line
920volts @ the generator
Per unit current:
Base Current:

,I
'~t.'
K

l
l
 460
10000
Jbase = r::; = 12551A factual =.J 73 =26.56A
-v3(0.46)
7.07 2 + 7.07 2
26.56
Jpu = - - = 0.002116pu.
12551

Base Impedance: Per unit Impedance:

~22 30+ j210 .
@ the line, Zbase = -·- = 8.464 ohms ZuNE = =3.544+ ;24.8Ip.u.
10 8.464
Using per unit values;
Vspu = ILO + 0.002116L- 45° (3.544 + j24.81) =1.0429Ll.75° p.u.
At the generator, Actual voltage=1.0429{920)=959.5 volts
Answer is C c

40.A 4 pole, 60Hz induction has a full load slip of 5% slip. What is the full load rotor
speed? ·
A. 1780 rpm C. 1760 rpm
B. 1730 rpm D. 1710 rpm
AT _ 120/ _ 120(60) _ ,.
J.V s - - - - - 1800 :pm
p 4

NR = N 5 (1-S) =1800(1-0.05) =1710 rpm

Answer is D

41. In dwelling unit, floor receptacle/outlets shall not be counted as part ofthe
required receptacle outlets unless located within mm of the wall.
A. 1800 C. 450
B. 1600 D. 600
Answer is C PEC 2000 page 73

42.A 4 pole 60Hz squirrel cage induction motor has a full load speed of 1754 rpm.
What is the full load slip?
A 2.3% C. 3%
B. 1.8% D. 2.6%
N s_120/
-P-
_120(60) -1800
- rpm
4
1800-1754 xl 00% = 2.56%
1800

43.1n a 3-phase wye-connected 50 MVA, 13.2 kv 60Hz generator, What is the base
current using the ratings as base values.
------------------------------------------------------------------------

A. 2187A C. 3800A
B. 4311A D. 5600A
50000
lbase = r:; =2,187 A
-v3(13.2)
Answer is A

44. Determine the per unit impedance of a 3-phase, 60 MVA, 13.2 kv Generator with
20% impedance at 100 MVA and 13.2 KV bases.
A. 0.54 C. 0.28
B. 0.12 D. 0.33
Znew =Zgiven(KVgiven)2( MVAbase
KVbase MVAgiven)
j = 0.2(13.2)2(100)
· 13.2 60
= 0.33pu
-
Answer is D

45. A three phase, wye- delta connected, 50 MVA, 345/34.5 KV transformer is

protected by CT's. Determine the CT ratios for differential protection such that the
circulating current through the transformer delta does not exceed 5 A.
A. 500:5 c. 1000:5
B. 2000:5 D. 1200:5
TraAsfarmer Fated seoondafY 8urreAt is
50000
!base ::;;: r::; =837 A
-v3(34.5)
· Use, 1000:5 Wye-connected Current Transformers
Answer is C

46. A transmission line cable consist of 12 identical strands of aluminum, each of 3

mm in diameter., The resistivity of aluminum is 2.8x1 o-s
ohm-meter. Find the 50°C
ac resistance per km of cable .. Assume correction factor of 1. 02 at 60 Hz.
A 0.3774 C. 0.4774
B. 0.2337 D. 0.5668
i
R=p-:-
A
-8 1000 '
RDC@20 = z.8xl0- ( ) ::; 0.33Q
12 !.o.oo3 2
4

RLJC@jO
236 50
=0.33[ 236+20
+ ]= 0.36867

RAe =1.02(0.36867) = 0.376Q

Answer is A

47. A 20 Hp, 240 V shunt motor takes 72 Amperes when operating at full load. The
shunt field resistance is 240 ohms and the armature resistance is 0.22 ohm.
Calculate the starter resistance if the starting torque is limited to 1.5 times the full
load torque {take a 3-V brush drop).
 A. 40 c. 60
B. 20 D. 1.80
240
laFL =IL - !1 =72--=71A
240
lasT =1.5/aFL
Rst = Vt- Vbc - r == 240- 3 0.22 =20
laST a f.5(71)
Answer is B

48. Two coils of inductance L1=1.16mH, L2=2mH are connected in series. Find the
total energy stored when the current is 2 amperes.
A. 1.75mJ C. 17.5mJ
B. 6.32mJ D. 63.2mJ
WL =}_i 2 Ll + }_i 2 L2 =_!_ 2 2 (1.16xl0-3 ) + .!.2 2 (2x1 o-3 )::::: 6.32xl0-3 Joules
2 2 2 2
Answer is B

49. A load of 100 kw, 0.707 lagging power factor is connected in parallel to a
capacitive load of 150'~var of 0.707 power factor. What is the minimum kw rating
of generator?
A. 180 c. 200
8.300 0.250
Pc = ___!;k_ = lSO = 150k»>
· tan Be tan 45°
Pr ;;;;;pL +Pc =100+150=250kw
Answer is D

50. The current through a 60 mH inductor is given by i =15 sin 377t. Determine the
induced emf.
A. 900 sin 377t C. 339.3 sin 377t
B.goo cos 377t o. 339.3 cos snt
e = L di = 60xl0-3 (15)(377)cos377t = 339.3cos377t volts
dt
Answer is 0

----~-----·---------