MULTIVECTOR REVIEW AND TRAINING CENTER
DIRECT CURRENT (DC) CIRCUITS
PART I
RESISTANCE/OHM’S LAW/NETWORK THEOREMS
RESISTANCE(R): The property of the material that opposes the flow of an
electric current.
I I2
V
R= p = p =p
A
V A2
for constant temperature only.
Where:l = length of the conductor in m or ft.
A = cross-sectional area in m2 or circular mils(CM)
A = d2 CM; if the diameter (d) of the conductor is in mils
P = specific resistance or resistivity of the conductor material in Ω-m or ΩCM/ft
= 1.77 x 10-8 Ω-m or 10.66 Ω-CM/ft for hard drawn copper(HDC)
at 20oC
= 1.72 x 10-8 Ω-m or 10.50 Ω-CM/ft for annealed copper at 20oC
= 2.83 x 10-8 Ω-m or 17.0 Ω-CM/ft for aluminum at 20oC
V = volume of the conductor in m3 or ft3
Note: 1 inch = 1000 mils
VARIATION OF RESISTANCE WITH THE TEMPERATURE
The resistance of metallic conductors increases w/ rise I temperature.
R1 = R0 [1 + α0 (t1 – t0)]
R1 = R1 [1 + α1 (t1 – t0)]
Where: R0 =resistance of the conductor @to; to = 0oC
MULTIVECTOR REVIEW AND TRAINING CENTER
R1 =resistance of the conductor @t1
R1 =resistance of the conductor @t2
α0 = temperature coefficient of resistance at 0o in oC-1
1
1
αo = __________ = ___________
to – T
│T│
T=inferred absolute zero resistance temperature in
For copper,T= -234.5oC
α1 =temperature coefficient of resistance at t1 in oC
1
1
αo = __________ = ___________
t1 – T
t1 + │T│
α2 = temperature coefficient of resistance at t2 in oC-1
1
1
α2 = __________ = ___________
t2 – T
t2 + │T│
∆T = change in temperature or temperature rise in oC
∆T = final temperature – initial temperature, say (t2 – t1)
OHM’S LAW (Ω): It states that the ratio of potential difference (V) between
any two points on a conductor to the current (I) flowing through them is
constant provided the physical conditions (e.g. temperature) of the conductor
does not change.
V
V
___ = constant or ___ = R
I
I
MULTIVECTOR REVIEW AND TRAINING CENTER
RESISTANCES IN SERIES:
SERIES:
RT = R1 + R2 + R3 + ….. +Rn
1
Series Circuit Characteristics:
____+____+…..+_____
VT = V1 + V2 + V3 + ….. +Vn
R3
Rn
IT = I1 + I2 + I3 + ….. +In
Characteristics:
RESISTANCES IN
1
1
1
1
____= ____+
RT
R1
R2
Parallel Circuit
1
1
1
1
1
+In
____= ____+ ____+____+…..+_____
+Gn
GT
G1
G2 G3
Gn
Where: G = conductance in mho Ω-1
G = 1/R
PT = P1 + P2 + P3 + ….. +Pn
VT = V1 + V2 + V3 + ….. +Vn
IT = I1 + I2 + I3 + …..
GT = G1 + G2 + G3 + …..
PT = P1 + P2 + P3 + ….. +Pn
CURRENT DIVIDERS:
If, in a two-branched parallel circuit, R1 and R2 are the branch resistance,
IT the total current, then branch current are
R2
I1 = IT x _______
R1 + R 2
R1
and
I2 = IT x _______
R 1 + R2
VOLTAGE DIVIDERS:
If, for two resistors R1 and R2 in series, VT is the total voltage, then
individual voltage drops are
MULTIVECTOR REVIEW AND TRAINING CENTER
R1
V1 = VT x _______
R1 + R2
and
R2
V2 = VT x _______
R 1 + R2
WORK, POWER and ENERGY
According to Joule’s Law of Electric Heating, the amount work required
to maintain a current I amperes through a resistance R ohms for t seconds is
given by:
V2t
W = I2 Rt = VIt = _____ = Pt in watt – see or Joules
R
The amount of heat produced electrically is given by:
W
H = _____ in kcal or BTU
J
Where: J = the mechanical equivalent of heat
J = 4.186 KJ/kcal = 778.16 ft-lb/BTU
NOTE: 1 Joule = 107 ergs = 0.24 gm-calorie
THERMAL EFFICIENCY:
It is the ratio of the heat actually utilized to the total heat produced
electrically.
total amount of heat utilized
mCp(∆T)
η = ____________________________= ________
MULTIVECTOR REVIEW AND TRAINING CENTER
total amount of heat produced
VIt/J
where: η = thermal efficiency
m = mass in gm. or lb
Cp = specific heat constant in cal./gm-oC or BTU/lb-oF
∆T = change in temperature or temperature rise in oC or oF
∆T = (tf – ti)
tf = final temperature
ti = initial temperature
Battery – a group of cells that generate electric energy from their internal
chemical reaction.
1. The emf (E) of a battery is the total potential difference established
within the battery between the two electrodes when the battery is not
supplying any current.
2. The terminal potential difference (V) is equal to the emf minus the
internal voltage drop.
That is
V = E – Ir
Where: I – the current supplied by the battery
r – the internal resistance of the battery
Note: E and rare constants.
V and I varies with the load
CHEMICAL EFFECTS OF CURRENT
FARADAY’S LAW OF ELECTROLYSIS:
MULTIVECTOR REVIEW AND TRAINING CENTER
First Law. The mass of an ion set free by a current in the process of
electrolysis is proportional to the quantity of charge that has passed through
the electrolyte.
1
a
m = zlt = (____= ___) It
F
v
Second Law. When the same current passes through the several electrolytes
for the same time, the mass of various ions deposited at each of the
electrolytes are proportional to their chemical equivalents.
m1 z1
___=___
m2 z2
where: m = mass of the ions liberated in gms
a = atomic weight of the substance
z = electrochemical equivalent (ECE) weight in gm/coul.
= the mass of the ions liberated due to the passage of one ampere of
current for one second or one coulomb of charge
v = valency of the substance
Faraday’s constant is defined as the charge required to liberate one – gram
equivalent of any substance.
Chemical equivalent (CE)
F = Faraday ‘s constant =_______________________________= 96,500
coul/gm- equivalent
Electrochemical equivalent (ECE)
CIRCUIT DEFINITIONS:
Circuit. A conducting part or a system of conducting parts through which an
electric current flows or is intended to flow.
Parameters. Various elements of an electric circuit such as Resistance,
Inductance and Capacitance. The parameters may be lumped or distributed.
Linear Circuit. A circuit in which the parameters are constant i.e. do not
change with voltage or current.
MULTIVECTOR REVIEW AND TRAINING CENTER
Non-Linear Circuit. A circuit in which the parameters change with the
voltage or current.
Unilateral Circuit. A circuit whose characteristics change with the direction
of operation. A vacuum tube rectifier is a unilateral circuit.
Bilateral Circuit. A circuit having the same characteristics in either direction.
Transmission line is a bilateral circuit.
Electric Network. It is a combination of any number of electric elements, the
impedances of which may be either lumped or distributed, or both, which are
connected in any manner, conductively, inductively, or capacitively.
Active Network. A network containing one or more sources of energy.
Passive Network. A network containing no source of energy.
Ideal Constant-Voltage Source. Voltage source whose output voltage
remains absolutely constant regardless of any change in load current. Such a
voltage source must possess zero resistance so that internal voltage drop is
zero.
Ideal Constant-Current Source. Voltage source whose internal resistance is
infinite. The resistance of the source is very high compared to the external
load resistance.
Node. A point at which two or more elements have a common connection.
Branch. A single path containing one element which connects one node to
any other node.
DC CIRCUITS PART I
EXERCISES:
REE – April 2007
1. If a certain conductor has an area of 336,400 circular mils, what is the
radius of this conductor in cm?
A. 0.663
B. 0.810
C. 0.737
D.0.921
REE – April 2004
MULTIVECTOR REVIEW AND TRAINING CENTER
2. A copper wire has a diameter of 0.032 inch. What is the area in circular
mils?
A. 1,228
B. 1,024
C.810
D.1,624
REE – Sept. 2003
3. In the American wire gauge, as the number of gauge increases, the
diameter of wire___________.
A .increases
B. decreases
B. Does not change
D. become twice
REE – April 2006
4. An ampere is how many abampere?
A. 0.4
B.0.3
C. 0.1
D.0.2
REE – Sept.2002
5. One (1) kW is equal to _________hp.
A. 0.746
B. 1.34
C. 550
D.1.5
REE – Sept. 2006
6. The specific resistance depends upon
A. The area of cross- section and the length of the conductor
B. The material of the conductor, its area of cross- sectional and length
C. The nature of the material of the conductor only
D. The area of cross- section of the conductor
REE – Oct. 2000
7. A copper bar has a length of 20 ft, width of 4 inches and thickness of 0.5
inch. If the resistivity of copper is 10.37 Ω-CM/ft, what is the resistance
of the bar?
A. 81.4 µΩ
B. 8.14 µΩ
C. 814 µΩ
D. 0.814 µΩ
REE – May 2008
8. A piece of wire 1 foot long with a cross sectional area of 10-5 square
inch has a resistance of 0.75 ohm. What is the resistance of 200ft of wire
of the same material with a cross sectional area of 3 x 10-5 square inch?
MULTIVECTOR REVIEW AND TRAINING CENTER
A. 100 ohms
ohms
B. 200 ohms
C. 150 ohms
D. 50
9. A kilometer of wire having a diameter of 11.7 mm and a resistance of
0.031 ohm is drawn down so that its diameter is 5 mm. What does its
resistance become?
A. 0.013Ω
B. 0.928Ω
C. 0.170Ω
D. 0.072Ω
REE – April 2001
10.
The resistance of the field winding of a DC machine is 0.25Ω at
o
25 C. When operating at full- load, the temperature of the winding is
75oC. The temperature coefficient of resistance of copper is 0.00427 per
o
C at 0oC. Find the resistance of the field winding at full- load?
A. 0.298Ω
B. 0.315Ω
C. 0.512Ω
D. 0.271Ω
DC CIRCUITS PART I
REE – April 2007
11.
A length of copper telephone line has a resistance of 24 ohms at 20
degree C. What is its resistance on hot summer day when its temperature
rises to 36 degree C?
A. 26.5Ω
B. 24.5Ω
C. 25.5Ω
D. 27.5Ω
REE – April 2004
12.
What is the value of Rx in Kohms to be connected in parallel to a
75 Kohms to reduce the value of total resistance to 10 Kohms?
A. 11.7
B.15.7
C.7.5
D.12.7
REE – April 2005
13.
A 100- watt bulb draws 0.833 A and a 200-watt bulb draws 1.666
A when connected across a 120-volt source. If the two bulbs are
connected in series across a 240- volt source, what is the common
current in amperes?
A. 2.51
B. 2.11
C. 1.11
D.1.52
MULTIVECTOR REVIEW AND TRAINING CENTER
REE – Sept. 2005
14.
Five 20- ohm resistors are connected is series across a DC source
voltage. If the voltage across one of the resistors is 25 volts, what is the
source voltage in volts?
A. 150
B. 175
C. 200
D.
125
REE – April 2005
15.
The resistances R1, R2 and R3 are connected in parallel across a
120 volts source. The total current of the circuit is 12 amperes. The
resistance R2 is twice the value of R1 and the resistance R3 is thrice the
value of R1. Find the resistance of R2.
A. 40
B. 33
C. 37
D. 25
REE – April 2007
16.
Three resistors of 3, 6 and Z ohms, respectively are connected in
parallel across a constant current source of 10 A. Find the value of Z if
this resistor draws 2 A current?
A. 3
B. 6
C. 4
D. 8
REE – Sept. 2008
17.
A voltage divider circuit of two resistors is designed with a total
resistance of the two resistors equal to 100 ohms. If the output voltage is
20% of the input, what are the values of the two resistors (ohms) in the
circuit?
A. 15 and 85
B. 20 and 80
C. 10 and 90
D. 2
and 98
REE – April 2007
18.
A 110 v DC motor a current of 30 A and is located 500 ft. from the
power source. What is the supply voltage if the diameter of copper wire
is 0.25 inch?
A. 119
B. 117
C. 121
D.
115
MULTIVECTOR REVIEW AND TRAINING CENTER
19.
A power station supplies 60KW to a load over 2,500 ft of 000 2conductor 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
DC CIRCUITS PART I
REE – Sept. 2003
20.
A switch is connected across a 120 V supply. What is the voltage
across the switch if it is closed?
A. 120 V
B. 0
C. Infinity
D. 60 V
REE – Sept. 2002
21.
What is 1012 ergs/sec in kW?
A. 100 kW
B. 1000 kW
C. 10 kW
D.10,000 kW
22.
What is the work in ergs needed to raise a 10 gram weight 100
meters up?
A. 4.9 x 105
B. 9.8 x 107
C. 98 x 107
D. 1.96 x
107
REE – Oct. 1998
23.
How many calories does an electric heater of 100 watts generate
per second?
A. 10
B. 1,000
C. 23.88
D. 42.25
REE – April 2003
24.
In a certain process water is heated at a rate of 2 liters per minute.
The water enters an electric heater et 25oC and exits at 50oC. Assuming
no heat loss, calculate the wattage of the heater.
A. 2450
B. 3488
C.6520
D. 5125
MULTIVECTOR REVIEW AND TRAINING CENTER
REE – April 2002
25.
Ten (10) kW is equal to _________ gram- cal/sec.
A. 156
B.436
C. 2,388
D. 425
DIRECT CURRENT (DC) CIRCUITS
PART 2
MAGNETISM/ ELECTROMAGNETISM/ ELECTROMAGNETIC
INDUCTION
MAGNETIC HYSTERESIS
MULTIVECTOR REVIEW AND TRAINING CENTER
1. A medium has two permeabilities:
a) Absolute permeability (µ)
b) Relative permeability (µr) with respect to the free space or vacuum
2. The absolute permeability of free space or vacuum (µo) has been given a
value of 4П x 10-7 H/m
3. µ = µr µo
4. Unit of pole strength is Weber (Wb)
5. The law of magnetic force is
m1m2
F=
In a medium
N
4πµ d
2
6. The magnetic field strength magnetizing force (H) at a point distant d
meters from a N pole of m Wb is
m
H=
N/Wb
4πµ d
In a medium
2
7. A magnetic pole strength one weber radiates a magnetic flux of one
weber. In other words, the magnetic flux is also measured in weber. The
flux of magnetic pole of m weber is Φ = m Wb
8. Flux density (B) is given by normal flux per unit area. It is measured in
Wb/ m2
ф
Wb/m2
B=
A
MULTIVECTOR REVIEW AND TRAINING CENTER
B=µH
9. The relation between B and H is given by
10. The relative permeability of a medium is given by
B in medium
µ=
for same H
Bin medium
11. When a current carrying conductor lies in a magnetic field, it is
subjected to a mechanical force is given by
F = BIL
Newton
Where B is the flux density, I is the current and L (in meters) of the
conductor lying within the Magnetic field
12. The axial field of a solenoid having N turns and 1 meters long when
carrying current of I amperes is
H = NI/L
AT/m
13. Field of an infinitely long straight conductor is given by
I
H=
µ0 µr I
AT/m
Wb/m2
B=
2πr
2πr
14. The magnitude of the force between two long straight conductors is
µ0 µrI1I2l
µ0I1I2l
In a medium
F=
In air
N
F=
2πd
N
2πd
15. The ohm’s law for magnetic circuit is
Magnetic force
Flux=
reluctance
16. Flux is measured in weber (Wb)
µ0 µrANI
ф=
Wb
L
MULTIVECTOR REVIEW AND TRAINING CENTER
17. Unit of magneto motive force is ampere turn
18. Reluctance is measured AT/Wb of 1/ henry
19. The ampere Turn required for a certain length of a magnetic path are
given by AT required = HL
20. The leakage coefficient is given by
Total Flux Produce
Useful flux+ leakage flux
λ=
λ=
Useful or air gap Flux
Useful or air gap Flux
21. According to Faraday’s Law of Electromagnetic Induction.
The magnitude of the emf induced in a circuit is equal to rate of change of
flux linkages of the circuit
e = -N dф/dt
22. When a conductor L meters is moved at right angles to the magnetic flux
in a field of flux density B with the velocity of v meters per second, the
dynamically induced emf is given by e = BLv if the conductor moves at
an angle of ɵ with direction of the flux, then
e = BLv sinɵ
23. Coefficient of self- induction is given by:
µ0 µrAN2
H
l
Nф
L=
H
L=
I
24. Coefficient of self- induction is given by:
N2ф1
M=
H
I1
25. Coefficient of coupling is given by:
M
K=
√ L1L2
dI
e=
-L
dt
MULTIVECTOR REVIEW AND TRAINING CENTER
26. When two coils having self- inductances of L1 and L2 and mutual
inductance of M are connected in series, the combined inductance is
given by:
L = L1 +L2 + 2M
L = L1+ L2 – 2M
mmf are additive
mmf are subtractive
27. When two coils having self- inductances of L1 and L2 and mutual
inductance of M are connected in parallel, the combined inductance is
given by:
L1 L2 –M2
L=
Additive
L1+ L2 + 2M
L1 L2 –M2
Subtractive
L=
L1+ L2 - 2M
28. The hysteresis loss is given by the area of the B/H loop.
Wh = area of the hysteresis loop
= (Area of the B/H loop)
joules/m3/cycle provides B is
in Wb/m2 and H in AT/m.
29. If the hysteresis loop has been plotted on a scale such that
1 cm = x AT/m
- for H
And
1 cm = y Wb/m2
- for B
Then
Wh = xy (area of hysteresis loop in cm2) J/m3/cycle.
30. Steinmetz’s empirical law for hysteresis loss is:
-
Wh = η Bx max
Joules/m3/cycle
The value of the index x depends on the value of the maximum flux
density Bmax. Its value is 1.6 whenBmaxlies between 0.1 and 1.2 Wb/m2.
31. If V metre3 of a magnetic material are subjected to f reversals of
magnetization per second, then hysteresis loss is:
- Joules/s or watt
Wh = η B maxf
MULTIVECTOR REVIEW AND TRAINING CENTER
32. The energy stored in the magnetic field of a coil or solenoid of selfinductance L and carrying current of I is:
E = ½ LI2
Joules
33. Energy stored per unit volume of the magnetic field is:
B2
E=
joules/m
B2
3
joules/m3
E=
2µo
2µo µr
In air
in a medium
34. Lifting or tractive force of an electromagnet is given by:
B2
P=
2µo
If the face area of the pole is A m2, then the pulling force is:
B2A
P=
2µo
35. The growth of the current in a coil of resistance R and inductance L is
given by the exponential relation.
I = Im (1 – e-t/λ)
Where Im = V/R
λ = time constant of the coil
λ = L/R seconds
36. The decay of current is given by i = -Im e-t/λ
DC CIRCUITS PART 2
EXERCISES:
MULTIVECTOR REVIEW AND TRAINING CENTER
REE – April 2004
1. How is the storage battery rated in capacity?
A. Volts
B. watts
C. amperes
hour
D. ampere –
REE – April 2002
2. A 12 volts battery has a 50 ampere- hour capacity. The internal
resistance is 0.1 ohm. A 5- ohm load is connected for 5 hours. How
many ampere – hours are still left?
A. 28.51
B. 41.24
C. 38.23
D. 35.92
REE – April 2007
3. Two batteries are connected in series, one of 8 v, 2 kΩ and the other
of 16 v, 3kΩ. the combination supplies a 7 kΩload. Determine the
power dissipated in the 3 kΩ battery.
A. 12 mw
B. 1.2 mw
C. 12 w
D.
1.2 w
REE – April 2002
4. Two 24- volts battery supply power to a 20Ω load. One battery has
0.2Ω internal resistance while the other has 4.0Ω. what power does
the load draws?
A. 28.4 w
B. 22.2 w
C. 30.8 w
D 18.6 w
REE – Sept. 2002
5. It is the unconventional unit of mmf in the CGS system
A. Unit pole
B. statvolts
C. gilberts
turns
D. ampere
REE – April 2002
6. A solenoid has a magnetic reluctance of 2.2 x 10-3. It has 300 turns
and a core area of 5 cm2. What is the flux density when the current
flowing is one ampere?
A.26, 000 gauss
B. 34,200 gauss
C. 12,200 gauss
D. 21,200 gauss
MULTIVECTOR REVIEW AND TRAINING CENTER
7. When the current in a wire is 5A, the flux 3 meters from the wire is
30 mWb. If the current is decreased to 4A, what will be the flux?
A. 24 mWb
B. 37.5 mWb
C. 19.2 mWb
D.
46.88 mWb
REE – May 2008
8. An infinitely long straight conductor located in air is carrying a
current of 100 A. Determine the flux density at point 0.08 m away
from the conductor.
A. 0.50bx 10-3 T
B. 0.25 x 10-3 T C. 0.40 x 10-3 T
D.
-3
0.33 x 10 T
9. Two parallel wires 10 cm apart carry currents in the opposite
direction of 8A. What is the magnetic field halfway between them?
A. 32 x 10-6 T
B. 64 x 10-6 T
C. 16x 10-6 T
D. 8
-6
x 10 T
REE – April 2007
10. A 120 cm long conductor is carrying a current of 3.2 A and is
situated at right angle to a field with flux density of 0.85 T. Calculate
the force on the conductor.
A. 3.2640 N
B. 3.3048 N
C. 3.1396 N
D. 3. 5477 N
DC CIRCUITS PART 2
REE – Sept. 2002
11. A south pole 20 unit pole is placed one inch away from another
south pole 35 unit pole. What is the force between the two poles?
MULTIVECTOR REVIEW AND TRAINING CENTER
A. 108.5 dynes repulsion
C . 70.0 dynes repulsion
B.108.5 dynes attraction
D . 70.0 dynes attraction
REE – Sept. 2004
12. Two opposite poles of strength 160 and 192 units has a force of
1,960 dynes. What is the distance in cm between them?
A. 3
B. 4
C. 5
D. 6
13. A 4 kW electric heater has a resistance of 1 kΩ and an inductance
of 5 mH. If the current that flows through this coil produced a 2 mWb
flux, calculate the reluctance of the coil.
A. 3000
B. 1250
C. 5000
D.
0.0025
14. A 2 – henry inductor has 20 turns. How many turns must be added
to raise its inductance to 3 henry?
A. 147 turns
B. 27 turns
C. 180 turns
D. 60 turns
15. An iron ring electromagnet having a relative permeability of 1250
when it is excited by a coil having 500 turns at 0.8 A. The mean
length of iron parts is 48 cm., the cross section of core is 10 square
cm. and the air gap is 0.5 cm. what is its total reluctance?
A. 306 x 103 AT/Wb
B. 3980 x 103 AT/Wb
C . 4286 x 103 AT/Wb
D. 3674 x AT/Wb
16. A solenoid has an inductance of 1.58 mH. How many turns of wire
does it have if the solenoid is 25 cm long and its radius is 2.5 cm?
A. 160,000 turns
B. 350 turns
C. 400 turns
D. 600 turns
REE – Sept. 2007
17. An air core toroidal coil has a 120 turns and the length is 12 cm.
What is the current required to set up a flux of 1.2 x 10-6 weber, if the
cross- sectional area is 25 square cm?
MULTIVECTOR REVIEW AND TRAINING CENTER
A. 0.8419 A
0.7718 A
B. 0.7960 A
C. 0.8119 A
D.
REE – Sept. 2005
18. Two coils having 30 and 600 turns respectively are wound side by
side on a closed iron circuit of cross- section 100 cm2 and of mean
length 150 cm. A current in first coil grows steadily from zero to 10A
in 0.01 second. Find the emf induced in the coil. Relative
permeability of iron is 2,000.
A. 209.4 v
B. 301.6 v
C. 299.3 v
D. 300.9 v
REE – April 2006
19. Calculate the AT/m to produce flux density of 0.001 Wb/m2 at the
center of along straight solenoid.
A. 783
B.796
C. 799
D.
745
REE – Sept. 2006
20. Wire length of 50 cm is moved in a field density 1 weber per
square meter at a velocity of 30 meters per second. What is the emf
induced if the motion is perpendicular to the field?
A. 1.5 v
B. 150 v
C. 15 v
D.0.15 v
DC CIRCUITS PART 2
21. A DC electromagnet is wound with 1000 turns of wire having a
resistance of 30 ohms. When excited by a voltage of 125 V, a
magnetic flux of 2 mWb links the coil. What is the energy stored in
the magnetic field?
A. 0.48 J
B. 8.34 J
C. 0.96 J
D. 4.17 J
REE – May 2008
MULTIVECTOR REVIEW AND TRAINING CENTER
22. Two coils in a network are positioned such that there is 100%
coupling between them. If the inductance of one coil is 12 mH and
the mutual inductance is 6 mH, find the inductance of the other coil.
REE – May 2008
23. Two identical coupled coils have an equivalent inductance o 40
mH when connected in series aiding and 17.5 mH in series opposing.
What is the value of the mutual inductance?
A. 0.5625 mH
B. 562.5 mH
C. 5.625 mH
D. 56.25 mH
MULTIVECTOR REVIEW AND TRAINING CENTER
DIRECT CURRENT (DC) CIRCUITS
ELECTROSTATICS/ CAPACITOR/ELECTRICAL INSTRUMENTS
1. The permittivity of free space of vacuum is
ϵo = 8.854 x 10 -12 F/m
2. A medium has two permittivities:
(a)Absolute permittivity, ϵ and (b) relative permittivity, ϵr
ϵ
ϵ = ϵo ϵr
ϵr
ϵo
The relative permittivity, being a ratio, has no units. For air (strictly
vacuum), ϵr = 1
3. Coulomb’s law of force between two charges is
Q1Q2
F=
Q1Q2
N
4πϵoϵrd2
In a medium
F=
N
4πϵod2
in air
Where Q1 and Q2 are the two charges in coulombs and d is the
distance between them in meters.
MULTIVECTOR REVIEW AND TRAINING CENTER
4. In the rationalized M.K.S.A system of units, a unit positive charge
radiates out of unit flux. Hence, a positive charge of one coulomb
radiates a unit flux. Electric flux is also measured in coulombs. Hence, a
positive charge of one coulomb radiates a flux of one coulomb. If the
charge is Q coulombs, then
Ψ = Q coulombs
5. The electric field strength (E) at any point within an electric field is
given by
the force experienced by a positive charge of one coulomb placed at
that point.
At any point lying at a distance of d meters within the electric field of a
charge Q coulombs, the electric field strength is given by
Q
E=
Q
N/C
4πϵoϵrd2
In a medium
E=
N/C
4πϵod2
In air
As seen, the unit of E is newton/coulomb (N/C). I t is also known as
volt/meter (V/m).
6. Electric flux density or electric displacement (D) is given by normal flux
per unit area.
If ψ coulombs is the flux passing normally through an area of A m2, then
D= ψ/A = Q/A = σC/m2
MULTIVECTOR REVIEW AND TRAINING CENTER
7. If D is the flux density of an electric field, then the electric field intensity
D
is given by
E
ϵ
Where D = flux density (i.e. ψ/A)
Ψ = absolute permittivity of a medium
σ
E=
or
ϵoϵr ϵoϵr
D
σ
E=
ϵoϵoϵr
D
or
In a medium
8. The relation between D and E is given by
D = ϵE
In air
Also
D
ϵ=
E
Hence, the absolute permittivity of a medium is equal to the
quotient of electric flux density by the electric field intensity.
D = ϵoϵr E
- in a medium
D = ϵoE
- in air
9. Potential at a point lying in an electric field is numerically equal to
Now
The work done (in joules) in bringing a positive change of one coulomb from
infinity to that point.
The potential at a point is 1 volt, if work done in bringing one coulomb from
infinity to that point is one joule.
MULTIVECTOR REVIEW AND TRAINING CENTER
1 volt = 1 joule/coulomb
Similarly, a potential difference of one volt exists between two points if
one joule of work is done in shifting a charge of one coulomb from one point
to the other.
10.
The potential at a point is given by
Q
Q
E=
N/C
4πϵoϵrd
2
In a medium
E=
N/C
4πϵoϵrd
2
In air
11.
(a) The charge resides on the outer surface of a sphere but for
calculating the electric field intensity or potential at a point, it is
considered to be concentrated at the center of the sphere.
(b)
Inside a charged sphere
(i) E = 0
(ii) Q = 0
(iii) ψ = 0
(iv) but potential at any point within it is the same as at its surface
12.
Potential gradient is defined as the rate of change of potential with
distance.
g = dV/dx
13.
At any point within an electric field. The electric field strength (E)
is equal to the potential gradient g at that point
E = g V/m
14.
The dielectric strength of the medium is equal to the potential
gradient necessary to cause the breakdown of the insulation of the
medium. It is usually expressed in kV/mm.
MULTIVECTOR REVIEW AND TRAINING CENTER
15.
The electric field intensity, potential gradient and the dielectric
strength of the medium are all equal.
16.
Capacitance of a capacitor is defined as
The charge required to create a unit potential difference between
its two plates.
Or
Q
C=
V
When Q is in coulombs and V in volts, then C is in farads (F).
One farad is the capacitance of a capacitor which requires a charge
of one coulomb to establish a p.d. of 1 volt between its plates.
Submultiples of farad like microfarad (µF) and micro microfarad
(µµF) also called Pico farad(pF) are extensively used.
17.
Capacitance ( in farad) of an isolated sphere of radius r meters is
given by
18.
C = 4πϵoϵr
- when placed in medium
C = 4πϵor
- when placed in air
Capacitance of a spherical capacitor is
ab
C = 4πϵoϵr
F
b-a
19.
(i) The capacitance of a parallel- plate capacitor with a uniform
dielectric medium of absolute permittivity Є is
MULTIVECTOR REVIEW AND TRAINING CENTER
C=
(ii)
ϵA
F=
d
ϵoϵrA
C=
d
ϵoA
F
d
With air electric
The capacitance with a composite dielectric medium is
ϵoA
ϵoA
C=
=
Ʃd/ϵr
F
d1
d2
+ .….
+
ϵr1
20.
ϵr2
In the case of a multiple- plate capacitor.
(n – 1)
ϵoϵrA
C=
F
d
Where n is the number of plates
21.
The capacitance of a cable capacitor is
2πϵoϵrI
C=
F
2.3 log 10b/a
Where
22.
l = length of the cable in meters
a = radius of the inner cylinder or core
b = radius of the outer cylinder
The capacitance (in µF) of a cable 1 km in length is
0.024ϵr
C=
µF/km
log 10
b/a
MULTIVECTOR REVIEW AND TRAINING CENTER
23.
Potential gradient in a cable capacitor is
V
g=
24.
(i)
V/m
2.3 x log 10b/a
The combined capacitance when
Capacitors are connected in series is
1
1
=
Cr
(ii)
1
+
C1
1
+
C2
1
+ …… +
C3 Cn
Capacitors are connected in parallel is
Cr = C1+ C2 + C3 +……+ Cn
25.
The voltage distribution when the two capacitors are joined in
series is
C2
C1
V1 = VT
V1 = VT
C1 +C2
C1 +C2
MULTIVECTOR REVIEW AND TRAINING CENTER
26.
The charge distribution when two capacitors are connected in
parallel is
C1
C2
Q1 = QT
C1 +C2
27.
Q1 = QT
C1 +C2
The insulation resistance of a cable capacitor is given by
2.3p
R=
joules
2πl
Where l = cable length in meters
P = resistivity in Ω-m
28.
Energy stored in a capacitor is
E = ½ CV2
29.
joules
Energy stored per m3 of a dielectric medium is
E = ½ DE
joules
D2
E=
joules
2ϵoϵr
30.
The force of attraction between the two plates of a parallel-plate
capacitor
D2 when
D2 charged is
D2
D2
F=
=
2ϵ
N/m2
2ϵoϵr
F=
=
2ϵ
N/m2
2ϵoϵr
MULTIVECTOR REVIEW AND TRAINING CENTER
31.
(i)
During the charging of a capacitor
Voltage rises according to the equation
ν = V (1 – e -t/λ)
where ν = voltage at any instant during charging
V = final or maximum voltage
λ = time constant
λ = RC seconds
(ii)
Charge increases according to the equation
q = Q (1 – e -t/λ)
where q = maximum charge
Q = CV coulombs
(iii) Charging current decreases according to the relation
i = -Ime -t/λ
where Im = maximum current i.e. at the start of charging
Im = V/R amperes
32.
(i)
During the discharging of a capacitor
Voltage decreases according to the relation
ν = Ve -t/λ
MULTIVECTOR REVIEW AND TRAINING CENTER
(ii)
Charge decreases according to the relation
q = Qe -t/λ
(iii) Discharging current decreases according to the relation
i = -Im e -t/λ
DC CIRCUITS PART 3
EXERCISES:
1. An 800 W electric iron operates on 110 V. How many electrons pass
through the iron per second?
A. 4.54 x 10 19
B. 6.4 x 109
C. 70.4 x 106
D.
19
3.25 x 10
REE – April 2006
2. Two equality charged spheres repel each other with a force of 0.1 kg. If
their centers are 20 cm apart, find the charge on each.
A. 2.09 x 10-6 coul
B. 2.09 x 10+6 coul C. 2.00 x 10-6coul
D.
+6
2.00 x 10 coul
REE – Oct.1997/ Sept. 2002
3. Two point charges 10 cm apart exert a force of 1 x 10-3g on each other.
If the charges are of the same value, what is each charge in stat
coulombs?
A. 8.9 stat coulombs B. 9.90 stat coulombs C. 6.90 stat coulombs D.
7.90 stat coulombs
MULTIVECTOR REVIEW AND TRAINING CENTER
REE – April 2005
4. A charge of 20 units is placed 5 cm from a like charge of 3 units in mica.
The force of repulsion is 0.52 dynes. Find the dielectric constant of
mica.
A. 2.4
B. 4.61
C. 6.6
D.
1.25
5. A pair of point charges with equal magnitude and opposite sign.
A. Couple
B. electric dipole
C. electric field
D. bipolar
6. An electric dipole consists of two point charges, 12 nC, placed 10 cm
apart. Determine the potential at a point in the middle of the electric
dipole.
A. 0 V
B. 2160 V
C. 4320 V
D. 1080 V
REE – Oct. 1998
7. Two metallic plates separated by 1 cm are connected across a 12 V
battery. A certain plastic material is inserted completely filling the space
between the plates and the charge on each plate is observed to double.
What is the dielectric constant of the plate material?
A. 0.5
B. 4
C. 2
D. 6
REE – Sept.2005
8. A two parallel-plate capacitor charged to 15 KV has a capacitance of 4.5
x 10-4µF. If the plates of the capacitor of 200 square centimeter each are
separated by a distance of 1 mm. what is the permittivity of the dielectric
medium?
A. 2.8
B. 2.65
C. 2.55
D.
2.54
REE – Sept.2004
9. A multi-plate capacitor is made up of 501 sheets of aluminum 25 cm x
30 cm with an insulation of paraffin paper 0.0015 cm thick. What is the
capacitance in microfarads of the capacitor?(k for paraffin paper is 3)
A. 50
B. 55
C. 35
D. 67
REE –May 2008
MULTIVECTOR REVIEW AND TRAINING CENTER
10.
Two capacitors C1 = 50 µF and C2 = 30 µF are connected in series.
Find the equivalent capacitance in microfarad.
A. 187.5
B. 18.75
C. 1.875
D. 187
REE- Sept. 2008
11.
If the charge accumulated on two parallel conductors charged to
4V is 500 pC, what is the capacitance of the parallel conductors?
A. 1,000 pF
B. 2,000 pF
C. 250 pF
D.
125 pF
DC CIRCUITS PART 3
REE- April 2004
12.
A 12 µF capacitor charged at 500 volts and 6 µF capacitor charged
at 250 volts are connected in parallel. What is the potential difference of
the combination?
A. 750 V
B. 500 V
C. 375 V
D.417 V
REE- Sept. 2007
13.
A 10- microfarad capacitor is charged by a constant
current source and its voltage is increased to 2 volts in 5 seconds.
Calculate the value of the constant current source in microamperes
A. 5
B. 6
C. 4
D. 3
REE- Sept. 2002
14.
A. Inductance
B. Capacitor
What extends the range of a DC ammeter?
B. current transformer
D. shunt resistance
15.
To increase the range of a voltmeter, a _____________.
A. Low resistance is connected in series
C. high R is
connected in series
B. Low resistance is connected in parallel
D. high R is
connected In parallel
16.
Two meter A and B require 50 mA, respectively, to give
full scale deflection Which is more sensitive?
MULTIVECTOR REVIEW AND TRAINING CENTER
A. A
D. cannot be determined
B. B
C.A and B
REE- Sept. 2001
17.
A 0 – 10 ampere ammeter has an internal resistance of
0.01 ohm. What shunt resistance to the ammeter must be connected to
have a reading of 0 – 100?
A. 1.11 x 10-3 ohm
B. 1.0 x 10-3 ohm
C. 0.111 ohm
D. 0.1 ohm
18.
A DC 0-50 millivolt meter is used to measure 5 amperes
current by means of a shunt. The meter was tested and found to have a
terminal resistance of 4.17 ohms. Determine the resistance to be
connected to indicate a full- scale deflection of 5A.
A. 0.01 Ω
B. 0.042 Ω
C. 1735 Ω
D. 413 Ω
19.
A 75- volt voltmeter has a resistance of 7,500 Ω. What
will be the multiplying factor of the instrument if a resistance of 7,500 Ω
is connected in series?
A. 10
B. 6
C. 4
D. 2
REE-April 2006
20.
An ampere-hour meter is calibrated to give a reading of
energy in kilowatt hour on a 220v circuit. When used on a 250v circuit,
it records 845 KWHR in a certain time. calculate the energy actually
supplied
A. 920
B.960
C.950
D.955
REE- Sept. 2002
21.
To the test accuracy of kilowatt hour meter, an
electrician used a wattmeter. He found out that the meter makes 5
revolutions in 18 seconds. The meter constant in 3.2. What is the
percentage error if the wattmeter reading is 3368?
MULTIVECTOR REVIEW AND TRAINING CENTER
A. 8% slower
5% slower
B. 8% faster
C. 5% slower
D.
ALTERNATING CURRENT (AC) CIRCUITS
PART 1
Single Phase (1ф) System
An alternating quantity either current or voltage is that which acts in
alternative direction and whose magnitude undergoes a definite cycle of
change in a definite interval of time.
Sine Wave/ Sinusoidal Wave
Positive peak ------------------------------------------------------
0
π
Peak to peak
Value
2π
Negative peak ------------------------------------------------- -Alternation
1 cycle
(360o)
Cycle – a complete change in value and direction of an alternating quantity. A
cycle of alternating voltage and current completes in 360 electrical degrees.
There are two (2) alternating per cycle.
MULTIVECTOR REVIEW AND TRAINING CENTER
Frequency (f) – the number of cycles per second expressed in hertz (HZ).
PxN
f=
120
Where:
f – frequency, hertz (HZ) or cps
P- no. of poles
N – speed, rpm
Period (T) – the time it takes to complete one cycle.
1
T=
F
Wave length (λ) – the length of the complete wave or cycle or the distance
traveled by the waveVelocity
in one cycle
ν
λ=
=
frequency
f
Note:
1 . For electromagnetic waves, the velocity (ν) in air or vacuum is
186,000 mi/s or 3 x 1010 cm/s, which
is the speed of light.
MULTIVECTOR REVIEW AND TRAINING CENTER
2 . For second waves, the velocity (ν) in air is 1, 130 ft/s.
ν(t)
-----------------------------------------------Vm
0
π
2π
-----------------------------------------------From the above figure,
ν(t) = Vm sin (1)t volts
where:
ν(t) = instantaneous value of voltage, volts
Vm = maximum value of voltage, volts
(1) = angular frequency
t = time, seconds
(1)t = radian
(1) = 2πf
Similarly, if the above voltage wave is a current wave, then the
equation is
i(t) = Im sin (1)t Amp
where:
i(t) = instantaneous value of current, amperes
Im = maximum value of current amperes
Note:
expressed in degrees.
(1)t whose unit is radian can be replaced by 0
MULTIVECTOR REVIEW AND TRAINING CENTER
ν3(t)
ν1(t)
ν2(t)
--------------------------------------------------------------Vm
θ
θ
π
θθ
2π
θ
--------------------------------------------------------------From the above figure,
ν1(t) = Vm sin 1(t) volts
ν2(t) = Vm sin( 1t +θ) volts
ν2(t) = Vm sin( 1t -θ) volts
where:
+θ = leading or ahead
-θ = lagging or behind
θ = phase angle or angle of displacement
Instantaneous Value – the value of alternating quantity of the instant.
Maximum Value – the maximum value attained by an alternating quantity
during positive or negative half cycle. This is also called the peak value, or
amptitude of alternating quantity.
MULTIVECTOR REVIEW AND TRAINING CENTER
Average Value – the average of all the instantaneous values of half (either
positive or negative) cycle of alternating quantity.
For sinusoidal voltage wave, Vave = 0.636 Vm
For sinusoidal current wave, Iave = 0.636 Im
Effective Value – the value of alternating quantity which when applied to a
given circuit for a given time produces the same expenditure of energy as
when d.c. is applied to the same circuit for the same interval time. The
effective value is also called as the “root-mean-square” (rms) value.
For sinusoidal voltage wave, Vave = 0.707 Vm
For sinusoidal current wave, Iave = 0.707 Im
For other waves, average value of voltage and current is given by
T
Vave =
𝟏
𝑻
∫
T
V(t) dt
Iave =
0
𝟏
𝑻
∫
0
And for RMS of effective value
T
∫V (t) dt
𝟏
𝑻
Vave = √
2
0
i(t) dt
T
Iave = √
∫I (t) dt
𝟏
𝑻
2
0
MULTIVECTOR REVIEW AND TRAINING CENTER
Effective value
Form factor =
Average value
For sine wave, Form factor = 1.11
Maximum value
Crest or peak factor =
Effective value
For sine wave, crest or peak factor 1. 414
Pure Resistance
I
Power Triangle
S
V
R
Q
θ
0
P
V = IR
factor (p.f.) angle or phase angle
Where:
factor = cos θ = P/S
V = effective or RMS voltage
reactive factor = sin θ = Q/S
I = effective or RMS current
Q/P
R = effective or resistance
θ = power
p.f. = power
r.f.=
tan θ =
For pure R, since θ =
0o
MULTIVECTOR REVIEW AND TRAINING CENTER
For sinusoidal voltage supply,
cos θ = 1 or unity
Vm
V=
= 0.707 Vm
P=S;Q=0
√2
p.f. =
o
Im
r.f. = sin θo = 0
Energy expended
I=
= 0.707 Im
W = Pt joules or watt - sec
√2
Vm = Imr
Phasor diagram
0
Pure Inductance
I
I
V
V
θ=0
L
o
I is in phase with V
V = IXL
Where:
XL = inductive
reactance in ohms
1. Real or True or Active or Average Power
= 2πfLΩ
P = VI cos θ watts
henry (H)
VmIm
voltage supply,
P=
cos θ watts
XL = (1)L
f = frequency, Hz
L = inductance,
For sinusoidal
MULTIVECTOR REVIEW AND TRAINING CENTER
2
Im
Vm
V=
; Im
=
P = I2R watts
√2
√2
∴↷ Vm = ImXL
2. Reactive Power
`
Phasor diagram:
Q = VI sin θ volt – ampere reactive (vars)
0
V
o
VmIm
θ = 90
Q=
sinθ vars
2
I
3. Apparent Power
I lags V by 90o
S = VI
volt- ampere (va)
VmIm
S = va
2
∴ For pure L, since θ = 90o
P= VI cos 90o = 0
R and L in Series
Q= VI sin 90o = VI
I
S= VI
∴S=Q
R
VR
Energy expended,
V
f~
2
IL
WL = joules
VL
Im2L
WL=
joules
L
MULTIVECTOR REVIEW AND TRAINING CENTER
2
4
Note: I is constant
XL = (1)L = 2πfL
Ω
Pure Capacitance
I
Phasor diagram:
VL = IXL
V = IZ
V
C
θ
0
I
VR = IR
V = IXC
Where:
XC = capacitive reactance in ohms
1
2
XL
XC=
=
2πfC
(1)C
V = √VR2 + VL2
V = √R2 +
1
V = IZ volts
Where:
C = capacitance, farad (F)
impedance in ohms
For sinusoidal voltage supply
Vm
Im
R/Z
V=
;
I=
VL/V = XL/Z
√2
√2
Z=
Z = √R2 + XL2
p.f. = cos θ = VR/V =
r.f. = sin θ =
tan θ = VL/VR = XL/R
∴Vm = ImXC
Phasor diagram:
I
Impedance Triangle:
Z
MULTIVECTOR REVIEW AND TRAINING CENTER
θ = 90o
V
XL
θ
\0
0
I leads V by 90o
∴ For Pure C, since θ = 90o
P = 0; Q = VI, S = VI
watts
S=Q
R
P = VI(p.f.) = I2R = VR/R watts
Q = VI(r.f.) = I2XL = VL/XL
S = VI = I2Z = V2/Z va
Also,
S = √P2 + Q2 va
Energy expended,
V2C
WC =
2
Vm2C
joules
WC =
4
R and C in Series
RLC in Series
I
I
R
VR
V
f~
joules
VR
R
MULTIVECTOR REVIEW AND TRAINING CENTER
V
C
Vc
1
XC =
1
=
(1)C
2πfC
f~
L
C
VC
Ω
Phasor diagram:
0
VR = IR
θ
I
XL = (1)L = 2πfC Ω
1
XC =
(1)C
VC = IXC
VL
V = IZ
V = √ VR2 + VC2
1
=
2πfC
Ω
V = √ VR2 + (VL – VC)2
V = √R2 + (XL – XC)2
V = IZ volts
Where:
Z = √ R2 + (XL –
XC)2
V = I √ R 2 + X c2
p.f. = cos θ = VR/V = R/Z
V = IZ volts
r.f. = sin θ = (VL – VC)/V =
(XL – XC)/Z
Where:
tan θ = (VL – VC)/VR = (XL – XC)/R
Z = √ R2 + Xc2
Note:
p.f. = cos θ = VR/V = R/Z
Transpose XL and XC if
XC > X L
r.f. = sin θ = Vc/V = XC/Z
tan θ = VC/VR = XC/R
P = VI (p.f.) = I2R = VR2/R
watts
Q = VI (r.f.) = I2 (XL – XC)
Q = VL2/XL = (VC2 – XC)
MULTIVECTOR REVIEW AND TRAINING CENTER
Impedance Triangle:
0
R
θ
XC
Z
S = VI = I2Z = V2/Z va
Also,
S = √ P2 + Q 2 va
P = VI (p.f.) = I2R = VR2/R watts
Q = VI (r.f.) = I2XC = VL2/XCvars
S = VI = I2Z = V2/Z va
Also,
S = √ P2 + Q 2 va
R and L in Parallel
Phasor diagram:
IT
IC
IT
IL
V
IR
f~
R
L
θ
0
V
IR
IT = √ IR2 + IC 2
Note:
V is constant,
√ R2 + XC
XL = (1)L = 2πfLΩ
IR = V/R ; IL = V/XL
IT = V
RXC
2
Phasor diagram:
IR
V
()
θ
IL
IT
√ R2 + X L 2
IT = V/Z amp.
where:
RXC
Z=
√ R2 + XC 2
p.f. = cos θ = IR/IT = Z/R
r.f. = sin θ = IC/IR = Z/XC
MULTIVECTOR REVIEW AND TRAINING CENTER
tan θ = IC/IR =
IT = √ IR2 + IL 2 -V
R/XC
RXL
IT = V/Z amp.
2
2
IR R = V /R watt
Where:
Z = impedance in ohms
RXL
Z=
2
√ R + XL
P = VI(p.f.) =
Q = VI(r.f.) = IC2XC
Q = V2/XC vars
S = VIT = IT2Z = V2/Z va
Also,
S = √ P2 + Q 2
2
p.f. = cos θ = IR/IT = Z/R
r.f. = sin θ = IL/IT = Z/XL
tan θ = IL/IR = R/XL
RLC in Parallel
IT
IR
2
L
2
P = VI(p.f.) = IR R = V /R watt
P = VI(r.f.) = IL2XL = V2/XL vars
C
2
S = VIT = IT Z = V2/Z va
Also,
S = √ P2 + Q 2
R and C in Parallel
IT
IL
V
f~
IC
R
XL = (1)L = 2πfLΩ
1
1
XC =
=
Ω
2πfLΩ
(1)C
IR = V/R ; IL = V/XL ; IC = V/XC
IT = √ IR2 + (IL– IC)2
√ XL2 XC 2 + R2(XC – XL)2
IT = V
RXLXC
IT = V/Z amperes
MULTIVECTOR REVIEW AND TRAINING CENTER
IC
IR
V
f~
R
1
XC =
C
1
=
Ω
2πfLΩ
IR = V/R ; IC = V/XC
(1)C
Where:
Impedances in Series
RXLXC
Z=
ZT = Z1 + Z2 + Z3 + …… + Zn
√ XL2 XC 2 + R2(XC – XL)2
p.f. = cos θ = IR/IT = Z/R
Impedances in
Parallel
r.f. = sin θ = (IL –IC)/IT = Z (XC – XL)/ XLXC
1
1
1
1
1
tan θ = (IL –IC)/IR = R (XC – XL)/ XLXC
=
+
+
+…..+
Note:
ZT Z1 Z2 Z3 Zn
Transpose IL and IC if IL> IC
∴ For two(2) impedances in
parallel
Impedances in Complex Form
Z1Z2
ZT =
1. For Pure R
Z = R + j0 Ω
equations, impedances
Z = ǀ R ǀ∠0o Ω
impedance Z.
2. For Pure L
Z1 + Z2
Note: In the above
must be in complex form.
Admittance (Y) – is the reciprocal of
1
MULTIVECTOR REVIEW AND TRAINING CENTER
Z = 0 + jXL Ω
mho or Siemen
Z = ǀ XLǀ∠90o Ω
or resistance R.
3. For Pure C
Z = 0 + jXC Ω
mho or siemen
Z = ǀ XCǀ∠-90o Ω
reactance X.
4. For Pure RL
Z = R + jXL Ω
Z =√ R2 + XL 2 ∠+θ
Z = ǀ Z ǀ∠+θ Ω
Where: 0o < θ < 90o
BC= 1/ XC
5. For Pure RL
Z = R + jXC Ω
Z =√ R2 + XC 2 ∠-θ
Capacitive susceptance BC
Z = ǀ Z ǀ∠-θ
susceptance BL
Where: 0o < θ < 90o
6. For Pure RLC
Z = R + j(XL + XC) Ω
1
1
2
2
Z = √ R +( XL + XC) ∠ ±θ
+…..+
Y=
Z
Conductance (G) – is the reciprocal
1
Y=
R
Susceptance (B) – is the reciprocal or
1
Y=
mho or siemen
X
∴ Inductive susceptance. BL= 1/ XL
Capacitive susceptance.
In complex form,
Y = G± jB
where
+jB =
-jB = Capacitive
Admittances in Series
1
1
=
+
1
+
MULTIVECTOR REVIEW AND TRAINING CENTER
Z = ǀ Z ǀ∠-θ
YT
Y1
Y2
Y3
Yn
Note: 0o < θ < 90o
∴ For two(2) impedances in
parallel
In general, Z = R ± jX
Where
+ jX = inductive reactance
- jX = inductive reactance
Y1Y2
YT =
Y1 +Y2
* For the complex expression of n impedance,
Admittances in
Parallel
Its angle ranges from 0o to 90o only either
YT = Y1 + Y2 + Y3 +
…… + Yn
Positive or negative. Mathematically, 0o≤θ ≤ 90o
Note: In the above equations,
admittances
Must be
expressed in complex form.
Power in Complex Form
1. Voltage Conjugate Method
2. Current Conjugate
Method
S = I x Conjugate of V with
S = V x Conjugate of
I with
Respect to the horizontal axis
Respect to
*
*
the horizontal axis
S = IV
S = VI
= P ±jQ
= P∓jQ
2
2
S = √P + Q ∠ ±θ va
S = √P2 + Q2∠ ∓θ va
where:
where *
S = apparent power, va
I = conjugate of
the current, amperes
P = real or true or average power, watts
Q = reactive power, vars
note:
I = current, amperes
-j = Q is capacitive
(p.f. is leading)
MULTIVECTOR REVIEW AND TRAINING CENTER
V = voltage, volts
(p.f. is lagging)
*V = conjugate of the voltage, volts
θ = power factor angle
Note:
+j = Q is capacitive (p.f. is leading)
-j = Q is inductive (p.f. is lagging)
+j = Q is inductive
Resonance
A circuit is said to be in resonance when the applied voltage V and the
voltage current I are in phase. Thus, at resonance, the equivalent complex
impedance of the circuit consists of only resistance R. Since V and I are in
phase, the power factor of a resonant circuit is unity.
Series Resonance (For RLC in Series)
Parallel Resonance/
Antiresonance (For RLC in Parallel)
At resonance,
At resonance,
1. VL = VC
1. VL = VC
2. XL = XC
2. XL = XC
1
1
3. L =
3. L =
2
2
(1)o C
(1)o C
1
1
4. C =
4. C =
2
2
(1)o L
(1)o L
1
1
5. (1) =
5. (1) =
√LC
√LC
Where (1)o = angular frequency at
Resonance, rad/sec
1
1
6. fr =
6. Fr =
2π√LC
2π√LC
Where fr= resonant frequency,
Hertz (HZ)
MULTIVECTOR REVIEW AND TRAINING CENTER
7. Supply voltage, V = VR
= VR
8. Z = R → a pure resistance
pure resistance
∴ Z is minimum
9. Total reactance, XT = 0
=0
10. I = V/Z = V/R
V/R
∴ I is maximum
11. p.f.= 1 or unity
12. S = P = VI
13. Q = 0
7. Supply voltage, V
8. Z = R → a
∴ Z is minimum
9. Total reactance, XT
10. IT = IR = V/Z =
∴ IT is maximum
11. p.f.= 1 or unity
12. S = P = VI
13. Q = 0
Parallel Resonance, Two – Branch Circuit
RL
RC
L
C
V
f~
RL2 – L/C
1
At resonance,
B L = BC
XL
(1)o =
√LC RC2 – L/C
XL
=
RL2 + XL2
2
RC + XC
1
2
(1)o =
2π√LC
L/C
RL2 - L/C
RC2 –
MULTIVECTOR REVIEW AND TRAINING CENTER
Quality Factor Q
sinusoidal
Effective Value of a No
Current or Voltage Wave
The quality factor of coils, capacitors
and circuits is defined by
Imn2
I=
+…….+
maximum stored energy
2
Q = 2π
Energy dissipated per cycle
1. For series RL,
+…….+
XL (1)L
2
Q=
=
R
R
2. For series RC,
voltage
XL (1)L
current
Im12 Im22 Im32
Idc2 +
2
+
+
2
2
Vm12 Vm22 Vm32
V = Vdc2 +
+
2πfL
2
2
Vmn2
+
2
=
R
2πfL
where:
I = effective value of the current
V = effective value of
Idc = dc component of the
MULTIVECTOR REVIEW AND TRAINING CENTER
Q=
=
of the voltage
R
R
component
=
R
Vdc = dc component
Im = maximum value of the ac
Of the current, subscript
indicates the
3. For series RLC, at resonance
harmonic (i.e.. 1 for
1
(1)oL
predominant
Qo =
=
second
R
(1)oCR
harmonic and so on)
degree of the
the fundamental or
sinusoidal component, 2 for
harmonic, 3 for third
Vm = maximum value of the ac
component
fr
fr
subscript indicates
Qo =
=
f2- f1 BW
of the voltage,
the degree of the harmonic
Harmonics- components of the
current or voltage in
where:
in which the frequencies
are multiples of
f1 and f2 are the frequencies corresponding
the fundamental.
to the half-power points.
BW = bandwidth which is the distance between
Note:
Half-power points, measured in hertz (HZ)
f=
fundamental frequency
2f = 2nd harmonic
frequency
3f = 3rd
(1)o= √(1)1(1)2
harmonic frequency
(1)1 = 2πf1 ; (1)2 = 2πf2
MULTIVECTOR REVIEW AND TRAINING CENTER
∴ fr = √f1f2
4. For Parallel RLC.
R
Qo =
= (1)oCR
(1)oL
This is the reciprocal of Qo for series RLC
AC CIRCUITS
PART I – SINGLE PHASE SYSTEM
EXERCISES:
1. The length of the time between a points in one cycle to the same point of
the next cycle of an AC wave is the____.
A. Frequency
B. period
C. magnitude
D.
polarity
2. In an experiment, a sinusoidal wave form is observed to complete 8
cycles in 25 msec. Determine the frequency of the wave form.
A. 320 Hz
B. 40 Hz
C. 200 Hz
D. 64 Hz
REE – Sept. 2007
3. Wavelength is the distance traveled by an electronic wave during the
time of the cycle. Given a wavelength of 12 meters, what is the
frequency?
A. 250 KHZ
B. 25 KHZ
C. 250MHZ
D. 25
MHZ
4. If emf in a circuit is given by e = 100 sin 628t, the maximum value of
voltage and frequency is ____________.
MULTIVECTOR REVIEW AND TRAINING CENTER
A. 100 V, 50 Hz
50√2 V, 100 Hz
B. 100 V, 100 Hz
C. 50√2 V, 50 Hz
D.
REE – April 2007
5. What is the complex expression for a given alternating current, i = 250
sin (𝜔t – 25 deg)?
A. 227 + j106
B. 160 – j75
C. 227 – j106
D. 160 + j75
6. A sinusoidal voltage wave has an RMS value of 70.71 V and a
frequency of 60 Hz. Determine the value voltage 0.0014 second after the
wave crosses the 𝜔t axis.
A. 70.71 V
B. 100 V
C. 50 V
D. 141.42
V
7. An alternating current varying sinusoid ally with frequency of 50 Hz an
RMS value of 20 A. at what time measured from the positive maximum
value will the instantaneous current be 14.14 A ?
A. 1/600 sec
B. 1/200 sec
C. 1/300 sec
D. 1/400 sec
REE – April 2001
8. An alternating rectangular wave has a maximum value of 10 V and a
frequency 1 cycle per second. What is the average value of the wave?
A. 5 V
B. 10 V
C. 0
D.
7.07 V
REE – April 2003
9. What is the rms value of a square wave with an amplitude of 10 V and a
frequency of 1 Hz.
A. 0
B. 10 A
C. 5 A
D.
7.07 A
REE – October 2000
MULTIVECTOR REVIEW AND TRAINING CENTER
10.
A sinusoidal current wave has a maximum value of 20 A. What is
the average value of one – half cycle?
A. 5
B. 12.7
C. 14.14
D.0
REE – April 1997
11.
A wire carries a current, i = 3 cos 314t amperes. What is the
average current over 6 seconds?
A. 0 A
B. 1.5 A
C. 3.0 A
D. 0.532 A
AC CIRCUITS
12.
The average value of the function i = 50 sin 𝜔t is equal to
___________________.
A. 31.8 A
B. 25 A
C. 38.2 A
D. 51.43 A
13.
Determine the rms value of a semi – circular current wave which
has a maximum value of a.
A. 0.816 a
B. 0.237 a
C. 0.886 a
D. 0.707 a
14.
The rms value of a half – wave rectified current is 100 A. Its value
for full wave rectification would be __________ amperes.
A. 141.4 A
B. 200 A
C. 200/π A
D. 40π A
15.
A half – wave rectified sine wave has an average value of 100
amp. What is the effective value?
A. 157 A
B. 444 A
C. 70.71 A
D. 100 A
16.
The form factor of a half – wave rectified alternating current is
__________________.
A. 1.11
B. 1.57
C. 1.73
D.
1.0
MULTIVECTOR REVIEW AND TRAINING CENTER
17.
Three alternating currents are given by i1 = 141 sin (𝜔t + 45o) ; i2 =
30 sin (𝜔t + 90o) ; i3 = 20 cos (𝜔t + 120o). Find the equation of
resultant current.
A. 167.4 sin (𝜔t + 45.66o)
B. 74.6 sin 𝜔t
C . 143.8 sin (𝜔t + 51.4o)
D. 64.7 sin (𝜔t + 30o)
REE – Sept. 2006
18.
When the sole purpose of ac is to produce heat, the selection of
conductor is based on ___________ value of the current.
A. Average
B. Instantaneous C. RMS
D. Peak
19.
The maximum value of sine wave AC voltage which will produce
heat in a resistor at the same average rate as 115 V of direct current is
___________________.
A. 81.2 V
B. 115 V
C. 162.6 V
D.230 V
20.
A sinusoidal voltage source has a peak value of 150 volts. What
equivalent DC voltage source would produce the same heating effect in
a 1 ohm resistor?
A. 15 V
B. 212 V
C. 95 V
D. 106 V
21.
The effective value of v(t) = 100 + sin 𝜔t is known to be 103.1.
The amplitude A of the sine term is ___________.
A. 25
B. 4.85
C. 35.48
D. 100
22.
An alternating current and a direct current flows simultaneously in
the same conductor. If the effective value of the AC is 8 A and DC is 12
A, what will an AC ammeter read when connected in the circuit?
A. 14.42 A
B. 12 A
C. 11.66 A
D. 16.49 A
23.
Find the reading of an AC voltmeter connected across the series
source of 100 sin (𝜔t – π/2) and 100 sin 𝜔t.
A. 100
B. 130.65
C. 170.71
D. 184.78
MULTIVECTOR REVIEW AND TRAINING CENTER
24.
An impedance draws a current i = 10 cos (𝜔t + 30o) A from a v =
220 sin 𝜔t volts. What is the power?
A. 550 W
B. 1100 W
C. 190.5 W
D. 1320 W
AC CIRCUITS
REE – October 2000
25.
A series circuit has an applied voltage of V = 220 sin(𝜔t + 30o)
and draws a current I = 10sin(𝜔t + 30o). What is the average power and
power factor of the circuit?
A. 1,905 W, 86.6% lagging
B. 1,905 W, 86.6% lagging
C . 2, 200 W, 100%
D. 1,100 W, 50% lagging
26.
The effective voltage across a circuit element is (20 + j10) V and
the effective current through the element is 4 – j3 A. Calculate the true
and the reactive power taken by the element.
A. 50 watts & 100 Vars lagging
B. 20 watts & 100 Vars
leading
C . 110 watts & 20 Vars lagging
D. 110 watts & 20
Vars leading
27.
A 110 – volt AC line feeds two circuits in parallel. The current are
(2.2 – j6.5) A and (1.8 – j3.5) A . The power consumed by the two
circuits is approximately_____________________.
A. 1,185 watts
B.440 watts
C. 755 watts
D. 433 watts
28.
Find the average power in a resistance R = 10 ohms if the current
in series form is i = 10 sin𝜔t + 5 sin 3𝜔t + 2 sin 𝜔t amperes.
A. 65.4 watts
B. 645 watts
C. 546 watts
D. 5.46 watts
29.
If e = 100 sin(𝜔t + 30o) – 50 cos 3𝜔t + 25 sin(5𝜔t + 150o) and i =
20 sin(𝜔t + 40o) + 10 sin(𝜔t + 30o) – 5 sin(5𝜔t + 50o). Calculate the
power in watts.
A. 1177
B. 919
C. 1043
D.
1224
MULTIVECTOR REVIEW AND TRAINING CENTER
REE – Sept. 2007
30.
The capacitive reactance XC of circuit varies inversely as the
capacitance C of the circuit. If the capacitance of a certain circuit is
decreased by 25%, by what percentage will XC change?
2
A. 25% increase
B. 20% increase
C. 66 % increase
D. 33
1
3
3
% increase
REE – April 2005
31.
A circuit has resistance of 20 Ω and a reactance of 30 Ω. What is
the power factor of the circuit?
A. 0
B. 0.55
C. 0.832
D. 0.99
REE – Sept. 2006
32.
A two- element series circuit with R = 15 ohms, L = 20 mH has an
impedance of 30 ohms and an unknown angle. What is the frequency in
HZ?
A. 2.0
B. 21
C. 2,067
D. 207
REE – Sept. 2008
33.
The current in a circuit of R = 10 ohms and L = 60 mH lags the
applied voltage by 80 degrees. What is the impedance in ohms?
A. 10 + j0.5671 Ω
B. 10 + j56.71 Ω
C. 10 + j5.671 Ω
D. 10 + j567.1 Ω
REE – April 2007
34.
What are the two elements in a series circuit connection having a
current and voltage of i = 13.42 sin (500t – 53.4o) A and v = 150 sin
(500t + 10o) v?
A. R = 8.8 Ω and L = 35 mH
B. R = 6.3 Ω and L =
25 mH
C . R = 7.5 Ω and L = 30 mH
D. R = 5 Ω and L =
20 mH
MULTIVECTOR REVIEW AND TRAINING CENTER
AC CIRCUITS
REE – Sept. 2006
35.
A two – element series circuit has V = 240 +j0 volts and current
51.96 – j30 A. What is the current in amperes which results when the
resistance is reduced 50% of its former value?
A. 68.57 – j59.39
B. 68.57 – j59.39
C. 59.39 – j68.57
D. 59.39 – j68.57
REE – Sept. 2001
36.
In laboratory experiment, the impedance of a coil was obtained at
60 Hz and 30 Hz. 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
REE – Sept. 2002
37.
A 10 Ω inductive resistor is connected in series with an unknown
capacitance. At 60 Hz the impedance of the circuit is 10 + j11.72. At 30
Hz the impedance of the circuit is 10 – j5. What is the value of L in
millinery’s?
A. 50
B. 500
C. 100
D.
250
REE – April 2007
38.
A series RLC circuit has elements R = 50 Ω, L = 8 mH and C =
2.22 microfarads. What is the equivalent impedance of the circuit if the
frequency is 796 HZ?
A. 50 + j50 Ω
B. 50 + j130 Ω
C. 50 + j50 Ω
D. 50 + j130 Ω
REE – May 2008
39.
A coil of inductance of 1.0 H and resistance of 4 ohms is
connected in series with a condenser of 70.4 microfarad capacitance
across a 60 Hz, 240 V line. How much current will flow on the circuit?
A. 55A
B. 60A
C. 57.5A
D.52.5A
MULTIVECTOR REVIEW AND TRAINING CENTER
REE – Sept. 2004
40.
A capacitor in series w/ a 200 Ω resistor draws a current of 0.3
ampere from 120 volts, 60 Hz source. What is the value of capacitor in
microfarad?
A. 8.7
B. 9.7
C. 6.7
D. 7.7
41.
A series circuit consisting of a 66.2 µf capacitor and a variable
resistor. For what two values of resistance will the power taken by the
circuit be 172.8 watts, if the impressed 60 – cycle emf is 120 volts?
A. 83.33 & 3.33 ohms
B. 53.33 & 30 ohms C. 5.33 & 3.0 ohms
D. 83.33 & 5.33 ohms
REE –April 2004
42.
Reactances are connected in series: XC1 = 100, XC2 = 40, XL1 = 30,
XL2 = 70. What is the net reactance?
A. 80 XC
B. 40 XC
C. 60 XL
D. 50 XL
REE – Sept. 2005
43.
Which is not 180 degrees apart?
A. XL and IC
B. XC and IC
D. XC and XL
C. IL and IC
44.
An impedance of 100 Ω resistance and an unknown inductance is
connected across the capacitor. The resulting impedance is a pure
resistance of 500 Ω if (1) = 105 rad/ sec. Calculate the values of inductor
and capacitor.
A. 1µf & 2 mH
B. 5µf & 1 mH
C. 7µf & 3 mH
D.
0.04 µf & 2 mH
AC CIRCUITS
REE – March 1998
MULTIVECTOR REVIEW AND TRAINING CENTER
45.
A 50 µf and 100 µf capacitors are connected in series and across a
100 sin(𝜔t + 30o) voltage. Write the equation of the current.
A. 1.26 sin(𝜔t + 120o)
B. 1.26 sin(𝜔t + 90o) C. 5.65 sin(𝜔t + 120o)
D. 5.65sin(𝜔t + 90o)
REE – May 2009
46.
Find the value of an impedance which absorbs a complex power of
2300 VA with a 30 degrees angle at 115 V rms.
A. 5 + j7 Ω
B. 7 + j5 Ω
C. 5 + j3 Ω
D. 3
+ j5 Ω
REE – April 2007
47.
Find the value of an impedance which requires – 1, 540 vars at 230
v rms and A rms.
A. 17.40 – j15.00
B. 16.82 –j15.69 C. 17.09 – j15.39
D.
17.62 – j14.78
REE – Sept. 2006
48.
A 10 ohm resistor is connected in parallel to an impedance Z = 3 +
j4 ohms. Find the circuit power factor.
A. 0.600 lagging
B. 0.866 lagging C. 0.609 lagging D. 0.809 lagging
REE – March 1998
49.
A 25 Ω resistor connected in series with a coil of 50 Ω resistance
and 150 mH. What is the power factor of the circuit?
A. 85%
B. 80%
C. 90%
D.
75%
50.
The maximum values of alternating voltage and current are 400 V
and 20 A, respectively, in a circuit connected to 50 Hz supply and these
quantities are sinusoidal. The instantaneous values of voltage and
current are 283V and 10 A respectively at t = 0 both increasing
positively. What is the power factor of the circuit?
MULTIVECTOR REVIEW AND TRAINING CENTER
A. 0.707
0.966
B. 0.83
C. 0.85
D.
51.
Two impedances ZA = 4 + j6 Ω and ZB = 6 + j12 Ω are connected
in parallel. The apparent power for the impedance B is 1490 VA.
Determine the total apparent power.
A. 4250 VA
B. 3290 VA
C. 2652 VA
D. 8031 VA
REE – October 1996
52.
The resistor of 6 Ωand unknown impedance coil in series draw 12
A from a 120 V, 60 Hz line. If the real power taken from the line is 1152
watts, what is the coil inductance?
A. 15.9 mH
B. 10 mH
C. 20 mH
D. 1.59
mH
53.
Two coils A and B are connected in series across a 240 V, 50 Hz
supply. The resistance of A 5 Ω and the inductance of B is 0.015 H. If
the input from the supply is 3 kW and 2 kVAR, find the inductance of A
and resistance B.
A. 0.0132 H & 8.3 Ω
B. 0.215 H & 3.8 Ω C. 0.026 H & 12 Ω
D. 0.031 H & 5.3 Ω
54.
A current of 5 A flows through a non – inductive resistance in
series with a choking coil when supplied at 250 V, 50 Hz. If the voltage
across the resistance is 120 V and across the coil is 200 V, calculate the
power absorbed by the coil in watts.
A. 168.75 W
B. 137.5 W
C. 51.37 W
D. 75.31 W
AC CIRCUITS
55.
A feeder supplies two loads, one at 50 amperes at 50% power
factor, the other 150 amperes at unity power factor. The total current
supplied by the feeder is approximately ________________.
MULTIVECTOR REVIEW AND TRAINING CENTER
A. 180 A
B. 200 A
C. 175 A
D. 150 A
56.
A single phase, 7.46 kW motor is supplied from a 400 V, 50 Hz
AC mains. If its efficiency is 85% and the power factor is 0.8 lagging,
find the reactive component of the input current.
A. 16.46 A
B. 21.96 A
C. 27.43 A
D. 21 A
REE – October 1996
57.
A capacitor is rated 100 KVAR, 380 V, 50 Hz. What will its rating
be at 60 Hz, 220 V?
A. 50 KVAR
B. 40 KVAR
C. 90.9 KVAR
D. 57.7 KVAR
58.
A series RLC circuit consists of 20 ohms resistance, 0.2 H
inductance and an unknown capacitance. What is the value of the
capacitance if the circuit has a leading angle of 45o at 60 Hz?
A. 35.18 µF
B. 47.9 µF
C. 27.8 µF
D. 30.7 µF
59.
A fluorescent lamp and inductive ballast draw a 1.0 A current at
50% lagging power factor from a 120 V, 60 Hz source. What is the
overall power factor when a 26.5 µf capacitor is connected across the
fixture?
A. 0.832 lagging
B. 0.832 leading C. 0.5 leading
D. 0.5
lagging
60.
A 3 HP, 120 V, 60 Hz induction motor operating at 80% efficiency
and 0.866 lagging power factor is to be used temporarily with 240 V, 60
Hz source. What resistance in series with the motor will be required for
the motor to have 120 V across its terminal at full load?
A. 6.68 Ω
B. 4.77 Ω
C. 13.76 Ω
D. 9.54 Ω
61.
A fluorescent lamp taking 80 W at 0.7 pf lagging from a 230 V, 50
Hz supply is to be corrected to unity power factor. What element should
be placed in _____________ for this purpose?
MULTIVECTOR REVIEW AND TRAINING CENTER
A. Series, 59.4 µF
D. parallel, 8.36 µF
B. parallel, 4.95 µF
C. series, 31.2 µF
REE – October 2000
62.
A series circuit consists of a 20 – ohm resistance, a 150 mH
inductance and unknown capacitance. The circuit is supplied with a
voltage v = 100 sin 377t. Find the value of capacitance at resonance.
A. 42 µF
B. 47 µF
C. 34.65 µF
D. 72.57
µF
63.
A 50 Ω resistance, 30 Ω inductive reactance and 25 Ω capacitive
reactance are connected in series across a 100 V, 60 Hz supply. What
will be its resonant frequency?
A. 65.726 Hz
B. 53 Hz
C. 25 Hz
D.
54.77Hz
REE –April 2005
64.
A series RLC circuit has R = 10 Ω, L = 40 x 10-6 H and C = 60 x
10-2 F. What is the resonant of the circuit in MHZ?
A. 20.17
B. 3.24
C. 4.49
D.
1.62
REE – April 2001
65.
A 5 mH pure inductance is connected in parallel with one
microfarad capacitor. What frequency will the circuit be in
antiresonance?
A. 250 Hz
B. 2250 Hz
C. 60 Hz
D. 100 Hz
AC CIRCUITS
REE – October 1998
66.
One leg of a radio tuned circuit has a capacitance of 1 x 10-9 F. It is
tuned at 200 kHz. Hat is the inductance of the other leg in Henry?
A. 6.33 x 10-4
B. 20 x 10-3
C. 8.25 x 10-5
D. 129 x 10-3
MULTIVECTOR REVIEW AND TRAINING CENTER
67.
A coil having a Q – factor of 5 is connected in series with an ideal
capacitor across an AC source of 60 V. Calculate the voltage across the
capacitor at resonance?
A. 150 V
B. 300 V
C. 12 V
D. 65 V
REE –April 2004
68.
A series RLC circuit has a resistance of 15 Ω and an inductive
reactance of 1,500 Ω. What is the Q – factor of the circuit at resonance?
A. 120
B. 100
C. 140
D. 150
REE –April 2004
69.
A tuned circuit is resonant at 4 MHz. Its Q is 100. What is its
bandwidth?
A. 4 kHz
B. 400 kHz
C. 400 Hz
D. 40 kHz
REE –Sept. 2007
70.
A circuit has load impedance Z = 50 + j80 Ω. Determine the
parallel impedance required to correct the power factor to unity.
A. 32 µF
B. 36 µF
C. 24 µF
D. 28 µF
71.
A circuit consisting of a capacitor in series with a resistance of 10
ohms is connected in parallel with a coil having a reactance and
resistance 0f 17.32 ohms and 10 ohms, respectively. What is the
reactance of the capacitor that will draw minimum current from a 230 V,
60 Hz supply?
A. 17.32 Ω
B. 10.32 Ω
C. 5.78 Ω
D. 22.18 Ω
REE –April 2007
72.
An impedance Z3 is n series to the two impedances Z1 = 10 + j10
Ω and Z2 = 20 – j30 Ω that are connected parallel. What is the value of
the impedances Z3 in ohms that will produce resonant circuit?
A. J13.1
B. – j5.4
C. j5.4
D. – j13.1
MULTIVECTOR REVIEW AND TRAINING CENTER
AC Circuits Part 2
Three Phase Circuits / Systems
Balanced 3ф Circuits / Systems:
1. Wye (star) Connected System
IL = Iф
Vф
Zф
VL
VL
IL = Iф
Zф
Zф
VL
IL = Iф
Relationships:
Iф = IL
VL
;
√3
Vф =
MULTIVECTOR REVIEW AND TRAINING CENTER
2. Delta (Mesh) Connected System
IL
Vф
Iф
VL
VL
Iф
Zф Zф
Zф
IL
VL
IL
Iф
Relationships:
IL
;Iф =
√3
Vф = VL
Where:
Vф – phase voltage
Iф – phase current
VL – line voltage
IL – line current
Power in Balanced 3ф Circuits/ systems
(Whether Wye or Delta connected)
PT = √3 VLIL(p.f.) = 3VфIф (p.f.) = 3 IфRф watts
QT = √3 VLIL(r.f.) = 3VфIф (r.f.) = 3 IфXф vars
ST = √3 VLIL= 3VфIф = 3 Iф2Zф = √PT2 + QT2 va
Power Triangle:
Impedance Triangle:
MULTIVECTOR REVIEW AND TRAINING CENTER
ST
Zф
QT
Xф
θT
0
θT
PT
0
ST = √PT2 + QT2 va
Ω
From the above Triangles:
PT KW
p.f. = cos θT =
=
ST KVA
Zф = √Rф2 + Xф2
Rф
=
Zф
QT
r.f. = cos θT =
ST
KVAR
Xф
=
=
KVA
Zф
QT
KVAR
Xф
=
=
KW
Rф
tan θT =
PT
Rф
Where
PT = total real power drawn by the balanced 3ф load
QT = total reactive power drawn by the balanced 3ф load
ST = total apparent power drawn by the balanced 3ф load
p.f. = power factor of the balanced 3ф load
r.f. = reactive factor of the balanced 3ф load
θT = power factor angle of the balanced 3ф load
Power Sequence = the order in which the generated voltages in the phase
windings of an alternator reach or attain their peak or maximum value.
a. Positive Phase Sequence
ABC → BCA → CAB
AB→BC → CA → BC → CA → AB →CA – AB – BC
AN → BN → CN → BN → CN → AN
→ CN – AN – BN
MULTIVECTOR REVIEW AND TRAINING CENTER
Examples of Vector Representations
VC = ǀ VCǀ∠ - 240o
VA = ǀ VAǀ∠ - 240o
VBǀ∠ - 240o
120o
120o
120o
VB = ǀ
120o
VA = ǀ VAǀ∠0o120oVB = ǀ VBǀ∠0o120o
120o
VB = ǀ VBǀ∠120o
120o
VC = ǀ VCǀ∠120o
VC = ǀ VCǀ∠0o
120o
VA = ǀ VAǀ∠120o
Sequence ABC
Sequence BCA
Sequence CAB
b. Negative Phase Sequence
ACB → CBA → BAC
AB → CA → BC → CA → BC → AB
→BC – AB – CA
AN → CN → BN → CN → BN → AN
→ BN – AN – CN
Examples of Vector Representations
VB = ǀ VBǀ∠ - 240o
VA = ǀ VAǀ∠ - 240o
VCǀ∠ - 240o
120o
120o
VBǀ∠0o
VC = ǀ
120o
120o
VA = ǀ VAǀ∠0o120oVC = ǀ VCǀ∠0o120o
120o
VC = ǀ VCǀ∠120o
120o
VB = ǀ VBǀ∠120o
VB = ǀ
120o
VA = ǀ VAǀ∠120o
MULTIVECTOR REVIEW AND TRAINING CENTER
Sequence ACB
Sequence BAC
Sequence CBA
If the phase sequence is not given, assume a positive phase sequence.
Three phase (3ф) alternators are designed to operate with positive phase
sequence voltages.
Power Measurements in three Phase (3ф) Systems
1. One – wattmeter Method – used to measure the total real power in a
balanced three – phase system with a single wattmeter.
Method employed:
a. Potential – lead shift method
b. Artificial – neutral method
c. T method
d. Current – transformer method
2. Two – wattmeter Method – usually used to measure the real power
being drawn by a three – phase (3ф), 3 – wire system.
Current Coil1
Potential Coil1
IL
±
W1
Balanced
3ф
Source
IL
VL
VL
VL
W2
IL
±
Potential Coil2
Current Coil2
Balanced
3ф
Load
MULTIVECTOR REVIEW AND TRAINING CENTER
W1 = VLIL cos ǀ30o±0ǀ
W2 = VLIL cos ǀ30o∓0ǀ
PT = W1 + W2→added algebraically
QT = √3 (W1 – W2), if W1> W2
QT = √3 (W2 – W1), if W2> W1
tan θ =
√3 (W1 – W2)
, if W1> W2
W 1 + W2
tan θ =
√3 (W2 – W1)
, if W2> W1
W 2 + W1
Where:
θ = power factor of the balanced 3ф load
PT = total real or true power drawn by the balanced 3ф load
QT = total reactive power drawn by the balanced 3ф load
Notes:
1. At unity p.f., the two wattmeter’s have equal reading. That is
W1 = W2
2. At 0.866 p.f., one wattmeter reading is twice the other.
3. At 0.5 p.f., one wattmeter reads zero while the other registers
the total circuit power.
4. At less than 0.5 p.f., one wattmeter gives a negative reading and
the other positive.
Symmetrical Components
Fortes cue’s Theorem: A set of anunbalanced related phasors may be
resolved into n systems of balanced phasors called the symmetrical
components.
MULTIVECTOR REVIEW AND TRAINING CENTER
For three – phase systems, we have:
a. Positive sequence component
b. Negative sequence component
c. Zero sequence component
For voltages:
𝟏
VA = VA0 + VA1 + VA2
VA0 = (VA + VB + VC)
VB = VA0 + a2VA1 + aVA2
VA1 = (VA + aVB + a2VC)
2
VC = VA0 + aVA1 + a VA2
𝟑
𝟏
𝟑
𝟏
VA2 = (VA + a2VB + aVC)
𝟑
Where:
a = 1∠120o = -0.5 + j0.866
a2 = 1∠240o = -0.5 + j0.866
VA, VB, VC = unbalanced voltages
V0, V1, V2 = zero, positive and negative symmetrical components of
voltage respectively
For currents:
IA = IA0 + IA1 + IA2
2
IB = IA0 + a IA1 + aIA2
2
𝟏
IA0 = (IA + IB + IC)
𝟑
𝟏
IA1 = (IA + aIB + a2IC)
𝟑
𝟏
IC = IA0 + aIA1 + a IA2
IA2 = (IA + a2IB + aIC)
𝟑
IN = IA + IB + IC
Where:
IA, IB, IC = unbalanced currents
I0, I1, I2 = zero, positive and negative symmetrical components of
current respectively
MULTIVECTOR REVIEW AND TRAINING CENTER
IN = neutral current
MULTIVECTOR REVIEW AND TRAINING CENTER
AC CIRCUITS
PART 2 – POLYPHASE SYSTEM
EXERCISES:
1. In a balanced three – phase Y – connected system, the line voltage and
the corresponding phase voltage are displaced from each other
by_____________.
A. 0o
B. 30o
C. 90o
D. 120o
2. In balanced star (wye) connected system, the line voltage is
______________.
A. 0.707 times the phase voltage
B. 1.414 times the phase voltage
C . phasor sum of two phase’s voltages
D. phasor difference of two
voltages
3. The phase sequence of a three- phase system, the line voltage is
____________.
A. CBA
B. CAB
C. ACB
D. none of
these
REE – May 2008
4. A three – phase system has line to line voltage Vab = 1,500 V rms with
30 deg angle with a wye load. Determine the phase voltage.
A. – 433 + j750 V rms
C. 750 + j433 V rms
C . j866 V rms
D. 866 V rms
REE – Sept.2007
MULTIVECTOR REVIEW AND TRAINING CENTER
5. The wye connected three – phase voltage source has V sub c = 115 v
rms with – 240 degrees angle. Find the line to line.
A. – 99 – j173 v rms
C. 199 + j0 v rms
C . 0 – j199 v rms
D. 99 – j173 v rms
REE – April 2002
6. Three 10 – ohm resistors connected in wye are supplied from a balanced
three phase source where phase a line voltage is given by 230 sin 377t.
What is the phase a line current?
A. 13.28 sin 377t
C. 13.28 sin ( 377t – 30o)
C . 23 sin (377t – 30 o)
D. 40 sin (377t + 30o)
REE – Sept. 2008
7. In a balanced three – phase wye – wye system, the source is an abc
sequence set of voltages. The load voltage on a phase A is Van = 110 V
rms with 80 degrees angle. Zline = 1 + j1.4 ohms and Zload= 10 + j13
ohms. Determine the input sequence of the line to neutral voltage V rms
between phase B and neutral.
A. – 78.1 – j41.9
C. 93.4 – j77.8
C . 120.2 – j20.7
D. – 114.1 – j41.9
REE – April 2002
8. In a balanced three phase system, the phase A voltage is 132.8 cis 0o,
what is the line to line voltage VCA ?
A. 230 cis 30o
B. 230 cis( - 30o)
C. 230 cis( - 60o)
D.
o
132.8 cis 120
9. Line B of a 230 V undergrounded – wye system touches the ground.
What is the voltage between line A and the ground?
A. 230 V
B. 115 V
C. 0 V
D. 132.79 V
AC CIRCUITS
MULTIVECTOR REVIEW AND TRAINING CENTER
10.
A system consist of three equal resistors connected in delta and is
fed from a balanced three – phase supply. How much power is reduced
if one of the resistors is disconnected/
A. 33%
B. 50%
C. 25%
D. 0%
REE – April 2005
11.
Three heater units each taking 1, 500 watts are connected delta to a
120 volt three phase line. What is the resistance of each unit in ohms?
A. 9.6
B. 5.4
C. 8.6
D.
7.5
12.
The resistance between any two terminals of a balanced delta connected load is 12 ohms. The resistance of each phase is
_______________.
A. 12 ohms
B. 18 ohms
C. 6 ohms
D. 36
ohms
REE – Sept. 2006
13.
A three – phase balanced load has 10 ohms resistance in each of its
phases. The load is supplied by a 229 V, three – phase source. The load
is connected in wye. What is the power absorbed by the load?
A. 19.36 KW
B. 14.52 KW
C. 9.68 KW
D.
4.84 KW
REE – March 1998
14.
Three impedances, - j10, j10 and 100 Ω are wye connected.
Determine the impedance of an equivalent delta.
A. 12.5, j12.5, - 12.5 Ω
B. 10, j10, - j10 Ω
C . j8.5, j12.5, 8 Ω
D. 5, 5j, - j5 Ω
REE – Sept. 2005
15.
A balanced wye – connected load has a phase impedance of 15 +
j10 ohms. What is the impedance per phase of the equivalent delta
connected load?
MULTIVECTOR REVIEW AND TRAINING CENTER
A. 5 + j3.33 Ω
8.66 + j5.77 Ω
B. 26 + j17.32Ω
C. 45 + j30 Ω
D.
16.
Three identical capacitances, each of 150 µF are connected in star.
The value of capacitance in each phase of the equivalent delta connected
load would be _____________.
A. 150 µF
B. 450 µF
C. 50 µF
D. 300 µF
17.
Three identical wye – connected resistances consume 1,000 watts.
If the resistances are connected in delta across the same supply. The
power consumed will be _______________.
A. 3,000 W
B. 6,000 W
C. 1,000 W
D.
333 W
REE – Sept. 2004
18.
A 480 – volt three – phase system supplies 85 A at a power factor
of 0.8. What is the power factor of the angle?
A. 60o
B. 45o
C. 37o
D.
15o
REE – Sept. 2007
19.
A balanced wye connected load of 8 + j6 ohms per phase is
connected to a three – phase 480 – v system. Find the real power in
KW?
A. 23
B. 13.8
C. 18.4
D. 10.6
AC CIRCUITS
REE – Sept. 2007
20.
A balanced three – phase, wye – connected load of 150 kw takes a
leading current of 100 A, when the line is 2,400 v, 60 Hz. What is the
capacitance per phase?
A. 21 mF
B. 21 µF
C. 205 mF
D. 205 µF
MULTIVECTOR REVIEW AND TRAINING CENTER
21.
A 50 – HP, three – phase induction motor with full load efficiency
of 85 % and power factor of 0.80 is connected to a three phase, 480 V
system. The equivalent star connected impedance that can replace this
motor is __________.
A. 7.3∠ - 36.87o Ω
B. 7.3∠36.87o Ω
C . 4.2∠ - 36.87o Ω
D. 4.2∠36.87o Ω
REE – April 2004
22.
It is required to increase the power factor of a 750 KW three –
phase balanced load from 70% lagging to 90 % lagging. The line voltage
is 6, 900 volts, 60 Hz. Specify the capacitor required to increase the
power factor of this wye – connected load in microfarad per phase.
A. 20.45
B. 22.39
C. 18.58
D. 17.22
REE – Sept. 2007
23.
A balanced wye – connected load of 30 + j10 ohms per phase is
connected to a 3 – phase 60 HZ 208 volts supply. Determine the value of
each capacitor in µF which will yield unity power factor.
A. 9.63
B. 8.39
C. 9.28
D. 8.84
24.
Find the average power absorbed by a balanced three – phase load
in an ACB circuit in which one line voltage is VAC = 480 ∠30o and one
line current to the load is Ib = 2.1∠80o A.
A. 1337 W
B. 1122 W
C. 1719 W
D.112 W
REE – Sept.2001
25.
The phase B line voltage and the phase A line current of a
balanced three – phase systems are V = 220 sin (𝜔t + 210o) volts and I =
10 sin (𝜔t + 180o) amperes, respectively. What is the power of the
system?
A. 1,905 W
B. 3,300 W
C. 5,716 W
D.
3,810 W
REE – April 2005
MULTIVECTOR REVIEW AND TRAINING CENTER
26.
The loads on 230 – volt three – phase generator consist of (a) three
groups of incandescent lamps, each group containing twenty 100 watts
lamps and (b) a 9 hp motor operating at 82 percent power factor and 87
percent efficiency. Find the total load current in amperes.
A. 30
B. 37
C. 40
D. 45
27.
Two balanced loads are connected in parallel to a three – phase
460 – volt source. Load A is 90 kVA at a power factor of 0.6 lagging
and load B is 25 kVA at unity power factor. What is the new power
factor of the system?
A. 0.74 lagging B. 0.74 leading
C. 0.80 lagging
D. 0.80
leading
28.
A three – phase, 440 V, 60 Hz supply is connected to a three –
phase, we – connected induction motor and synchronous motor. The
impedance per phase of the induction motor is (1.5 + j2.5) Ω. The
synchronous motor is over - excited and draws 100 A at 0.8 power
factor. What is the combined power factor of the load?
A. 0.99 lagging B. 0.99 leading
C. 0.51 lagging
D. 0.51
leading
AC CIRCUITS
29.
Two – wattmeter method is applied to three – phase motor running
at full load. The two wattmeters indicate 85.5 kW, respectively. What is
the operating power factor of the motor?
A. 87.45 %
B. 80.69%
C. 89.49 %
D. 94.76%
REE – April 2005
30.
The total power consumed by a balanced 3 ф load is 3,000 watts at
80% power factor. Two wattmeters. WA and WB are connected to the
line. What is the reading of WA?
A. 2,150 watts
B. 2,000 watts
C. 1,500 watts
D.
2,500 watts
MULTIVECTOR REVIEW AND TRAINING CENTER
REE – Sept. 2004
31.
The power drawn by a balanced 3∅ load is 3,000 watts at 80 %
power factor. If two wattmeters WA and WC are used, what is the
reading of WC?
A. 800
B. 950
C. 900
D.
850
REE – Sept. 2008
32.
In a balanced three – phase 230 V circuit, the line current is 90 A.
The power is measure by two – wattmeter method. If the power factor is
100 % and the line current is the same, what is the reading on each
wattmeter?
A. 25.6 kW
B. 30.4 kW
C. 17.9 kW
D.
20.7 kW
33.
The two – wattmeter method is applied to a three – phase, three –
wire, 100 V, ABC system with the meters in lines B and C, WB = 836
watts and WC = 224 watts. What is the impedance of the balanced delta
– connected load?
A. 10 ∠45o Ω
B. 10 ∠ - 45o Ω
C. 20 ∠ - 45o Ω
D. 20
o
∠45 Ω
34.
Two wattmeters are connected for two – wattmeter method with
current coils in line A and B of a 208 V, ABC circuit that has a balanced
delta load. If the meter readings are 6 kW and – 3 kW, respectively, find
the load impedance per phase.
A. 18.8 ∠ - 35.2o Ω / phase
B. 18.8 ∠ - 79.1o Ω / phase
C . 22.3 ∠ - 36.2o Ω / phase
D. 32.2∠ - 36.2o Ω / phase
35.
Three impedance Za = 6 ∠20oΩ, Zb = 8 ∠40oΩ, Zc = 10 ∠0oΩ are
connected in wye and are supplied by a $*0 V balanced three phase
source. The current in line b is ___________.
A. 46.19 ∠ - 50o A
B. 33.11 ∠ 169.45o A C. 27.71 ∠ - 90o A
D.
o
43.2 ∠ - 90 A
MULTIVECTOR REVIEW AND TRAINING CENTER
36.
Three impedances Zab = 4 – j3 ohms, Zbc = 3 + j4 ohms and Zca = 10
+ j0 ohms are connected in delta across a 220 volts, three phase source.
The total power of the circuit is_____________.
A. 1,839 W
B. 18, 392 W
C. 1,936 W
D.
1,849 W
37.
A 480 V, three – phase, three – wire source supplies to a
combination of industrial loads. The line currents drawn from the source
are: Ia = 55.78 ∠ - 59.4oA, Ib = 43.85 ∠ 176. 7oA, Ic = 48.0 ∠ 71.3oA.
Determine the total power drawn from th source.
A. 17.31 kW
B. 36.24 kW
C. 3.1 kW
D. 38.29
kW
38.
A three – phase, four – wire system has the following unbalanced
loads: Za = 10 +j10 Ω, Zb = 13 - j2 Ω, Zc = 7.5 +j10 Ω. Determine the
total power delivered to the load if the line voltage is 208 V.
A. 4.48 kW
B. 1.48 kW
C. 2.49 kW
D.
2.90 kW
AC CIRCUITS
REE – April 2005
39.
A 3 – phase 4 – wire system has the following unbalanced loads:
Z1 = 10 +j5 Ω, Z2 = 8 +j4 Ω, and Z3 = 20 +j0 Ω. The line to neutral
voltage of the system is 120 volts. What is the reading of the wattmeter
in watts in Z3?
A. 900
B. 720
C. 1,000
D. 1, 050
REE – March 1998
40.
The three unbalanced currents are: Ia = 10 cis (-30o), Ib = 0, Ic = 10
cis 130o. Find the negative sequence of phase A current.
A. 8.66 cis 30o A
B. 5.7 cis (-60o) A
C. – 5.77 A
D.
5.77 A
REE – Oct. 1998
MULTIVECTOR REVIEW AND TRAINING CENTER
41.
The load of a wye connected transformer are: Ia = 10 cis (-30o), Ib
= 12, Ic = 15 cis 82o. What is the neutral current?
A. 1.04 cis 72.8o B. 0.92 cis 62.5o C. 2.21 cis (- 30o)
D. 3.11 cis 72.8o
REE – Oct. 1998
42.
For the following phase currents Ia = 34.6, Ib= -10 – j17.32, Ic = 10 + j17.32. What is the positive sequence for phase C?
A. 12.44 – j21.55
B. 12.44 + j21.55
C. -12.44 + 21.55
D. –
12.44 – j21.55
43.
A wye – connected, three – phase system has the following
sequence components of current.
Zero sequence current = 12.5 cis 30o
Positive sequence current = 28 cis (-45o)
Negative sequence current = 20 cis (- 32o)
Determine the current following in the neutral wire.
A. 4.17 cis 30o
B. 37.5 cis 30o
C. 35.7 cis( -60o)
22.5 cis 10o
D.
REE – April 2007
44.
Given the following sets of symmetrical components, determine
the phase current for phase B.
Ia0= 1 at 75 deg
Ia1 = 1 at 15 deg
Ia2 = 1 at -45 deg
A. 0.259 + j0.966
B. 0.516 +j1.931
C. 1.932 + j0.518
D.1.414 + j1.414
REE – Sept. 2002
45.
The phase currents of a three – phase system are :
IA = 100 cis 0o A
IB = 80 cis 240o A
IC = 91.8 cis13 0.9o A
What is the zero sequence current?
A. 0 A
B. 270.7 cis 3.68o A
MULTIVECTOR REVIEW AND TRAINING CENTER
C . 34.68 cis (-30.24o) A
D. 90.23 cis 3.68o A
REE – Sept. 2006
46.
One conductor of a three – phase line is open. The current
following to the ∆- connected load through line a is 20 A. Which the
current in line a as reference and assuming that line c is open, what is the
negative sequence current of phase a?
A. 10 + j5.78
B. 10 – j5.78
C.j11.55
D. – 10 +
j5.78
ELECTRICAL TRANSIENT
Electrical Transient – is the study of characteristics of current potential drop,
power charged, energy dissipation and energy stored across different load
parameters as energized by a D.C or A.C. source by means of the activation of
a switch.
In a general, transient disturbances are produced whenever:
a. An apparatus or ckt. Is suddenly connected to or disconnected from the
supply.
b. A ckt. Is shorted
c. There is a sudden change in the applied voltage from one finite value to
another.
Transient Disturbances are classified as follows:
1. Initiation Transients – these are produced when a ckt; which is originally
dead is energized.
2. Subsidence Transients – these are produced when an energized circuit is
rapidly de – energized and reaches an eventual steady – state of zero
current or voltage.
3. Transition Transients – these are due to sudden but energetic changes
from one steady state to another.
4. Complex Transients –these are produced in a circuit which is
simultaneously subjected to two transients due to two independent
disturbances.
5. Relaxation Transients –transition occurs cyclically towards states, which
when reached, become unstable themselves.
MULTIVECTOR REVIEW AND TRAINING CENTER
DC Transient
Series RL Circuit
Case 1 Initiation Transient:
Given initial condition: at t = 0, i = 0
Model Circuit:
k3
E – Ri = 𝑒
𝑅
− 𝑡
𝐿
E
− 𝑡
+ k 2= 𝑒
𝑅
− 𝑡
𝐿
𝑒
𝑘2
= k3𝑒
SW
i=
R
VR
R
-
let k =
∴
k3
𝑒
𝑅
𝐿
R
−k3
R
𝐸
𝑅
i = + k𝑒
𝑅
−𝐿𝑡
→ general
solution
E
VL
L
where k = constant of integration to
find k,
At t = 0, i = 0
i
When the switch is closed:
VR + V L = E
general solution
di
Ri + L − E → L.D.E. of the first order
dt
By variable separable
di
E−Ri
1
0=
𝑅
k=
+ke0
−𝐸
𝑅
, subs’t back in
Hence,
1
E
E
R
R
R
i = - e− L t
= dt
L
𝐸
1
- 1n (E – Ri) = t + k1k2
R
L
component
−R
1n (E – Ri) =
t - Rk1
L
component
E
where: = steady state current
R
E
R
- e− L t = transient current
R
𝑅
𝐿
− 𝑡
MULTIVECTOR REVIEW AND TRAINING CENTER
1n (E – Ri) =
e
R
−Lt
−R
L
E
t +k2
Finally: i = (1 R
→ particular solution
)
Current growth eq ‘ n. from i = 0
to i = Imax =
E
R
Open at t = 0 →i = 0
Note: for inductor with no initial current
Shorted at t = ∞ →i = Imax =
E
R
ELECTRICAL TRANSIENT
Voltage Across R (VR):
Power Across L (Pt) :
VR = Ri = E (1 - e
R
−Lt
PR = VR∙I =Ee
)
Voltage Across L (VL):
VL = L
PR =
di
R
(e
)∙
R
L
− t
E
R
-e
(1 - e
−
R
−Lt
)
2R
t
L
t = 0, PL = 0
dt
E
R
−R
− t( L )
= L{ [0 − e L
R
 VL = E e
E2
R
−Lt
]}
t = ∞, PL = 0
R
−Lt
PT = PR + PL = Ei
E
=E ∙ (1 - e
Or
R
 VL= E - VR
*PR =
= E – E (1 - e
(WL):
R
−Lt
)
E2
R
R
−Lt
)
R
(1 - e− L t )2
Energy Stored in the Magnetic Field
R
−Lt
 VL = E e
L i di
Power across R (PR)
𝑽𝟐𝑹
PR = VR∙i = i2R =
𝑹
Using:
PR = VR∙i
dWL = PLdt = VLi dt = L
di
dt
when: t = 0, i = 0, WL = 0
t = 0, i = 1, WL = WL
Hence
WL
1
∫0 dW2 = L∫0 idi
∙i dt =
MULTIVECTOR REVIEW AND TRAINING CENTER
R
=E (1 - e− L t ) ∙
E
R
R
1
(1 - e− L t )
* WL = L I2
2
R
−Lt 2
E2
 PR = (1 - e )
R
(t = ∞), I = Imax
When t = 0, PR = 0
t = ∞PR =
at maximum time condition
1
Note: t = ∞(steady – state) WL = L I2max
2
E2
R
CASE 2 Subsidence Transient:
Given initial condition: when t = 0, i = Io
switch is in position 2
Model Circuit:
1
2
when the
at t = 0, i = Io
t = t, i = i
t 𝑑𝑖
𝑅 1
=- ∫0 dt
∫𝑖
𝑜
E
R
𝑖
𝐿
R
−Lt 2
o
i=I e )
or from gen. solution
R
i
E
R
i = + k e− L t = k
e
R
R
−Lt
i
Io
L
L
Io
When the switch is in position l
Io
at t = 0, i = Io
Io = keo
VR + VL = 0
R
E
i= (1 - e− L t )
R
R
−Lt
Ioe
Ri + L
𝑑𝑖
𝑑𝑡
=0
∴k=
i=
→ current decay equation
𝑑𝑖
𝑡
𝑅
= dt
𝐿
from i = Ioto i =
0
Note: when the switch transfer from 1 to 2 during steady – state condition,
E
Io=
R
MULTIVECTOR REVIEW AND TRAINING CENTER
ELECTRICAL TRANSIENT
Time Current (TC):
The time constant of any equation dealing with transient is in the value
of time (t, sec) which makes the magnitude or absolute value of the exponent
of the nos. e equivalent to unity or 1.
 In both current growth and current decay curves the current tends to
approach the steady – state values after 5TC.
Thus, the transition interval is:
0 ≤ t ≤ 5 TC
For RL Circuit
TC = L/R
For RC Circuit
TC = RC
Time constant at Initiation:
-Is the time for the current to reach 63.2% of its final value.
Proof:
R
E
−Lt
E
R
i= (1 - e )
R
When
t = 1TC = L/R
i= (1 - e− L t ) = 0.632
R
E
R
= 63.2% Imax
i
E
R
0.632
E
R
E
i= (1 - e
R
t(sec)
0
1TC
5TC
Time constant at Subsidiency:
R
−Lt
)→ exponential rise
MULTIVECTOR REVIEW AND TRAINING CENTER
-is the time elapsed for the current to reach 36.8% of its initial value.
Proof:
R
L
− t
i = Ioe
L
When t = 1 TC =
R
−t
i = Ioe = 0.368% Io = 36.8% Io
i
R
i = Ioe− L t → exponential decay
Io
0.368 Io
t(sec)
0
1 TC
5 TC
ELECTRICAL TRANSIENT
Transient in Series RC Circuit (D.C.)
E
i
When the switch is closed
E = VR +VC
E = Ri + q/C
E=R
E-
q
C
dq
dt
+
=R
q
C
dq
dt
dq
R
VR = Ri = R
C
VL = q/C = ∫ 𝑖 dt
C
dt
1
to find i,
t
dq
1
i = = CE [0 − e ⁄RC (− )]
dt
E t⁄
i = e RC
R
but E/R = Io
it means,
RC
MULTIVECTOR REVIEW AND TRAINING CENTER
CE – q = RC
dq
=
CE−q
1
t
dq
dt
R
dt
voltage Across R:
RC
-In (CE – q) =
In (CE – q) =
t
E t
i = Ioe ⁄RC = e ⁄RC
t
VR = Ri = R ∙ e ⁄RC
+ k1
t
VR = Ee ⁄RC
RC
−t
RC
E t
+ k1
k2
t
CE – q = e ⁄RC k1 = e ⁄RC ∙ e .k1
R
Voltage Across C:
t
−t
q = CE – k2e ⁄RC
VC = q/C =
−t
q = CE – ke ⁄RC → gen. solution
CASE 1 (Charging Process)
Given Condition: at t = 0, q = 0
CE (1−e ⁄RC )
C
t⁄
VC = E(1 − e RC )
or
t
VC = E - VR = E - Ee ⁄RC
t
VC = E(1 − e ⁄RC )
t
q = CE + ke ⁄RC
0 = CE + k
k = - CE
t
q = CE (1 − e ⁄RC )
t
q = Qmax(1 − e ⁄RC )
t
E t
i = Ioe ⁄RC = e ⁄RC
R
shorted at t = 0, Io = E/R
Note:
If the capacitor is initially uncharged
Open at t = ∞, I∞ = 0
t
E t
Proof: of i = Ioe ⁄RC = e ⁄RC
R
E = VR + VC= Ri +q/C
At t = 0, i = Io, q = 0
E = RIo + 0/C
Io = E/R
ELECTRICAL TRANSIENT
Power Across R:
Energy Stored in C due to charge Q
in “t” sec:
MULTIVECTOR REVIEW AND TRAINING CENTER
2
PR = VR∙i =i R
V2R
dWC= PC dt = VCi dt =
R
t⁄
E t⁄
= Ee RC ∙ e RC
R
E2 t⁄
PR =
e RC
R
E2
q dq
C dt
dt =
Q
1 q2 Q
WC = ∫0 q dq =
] o
C
C 2
1 Q2
1
1
WC=
= Q VC= C VC2
2 C
2
2
q
C
dq
1
At t = 0, PR =
R
At t = ∞, PR = 0
since VC = Q/C; Q = CVC
at maximum condition (t = ∞)
Q = Qmax = CE
Hence,
Power Across C:
PC = VCI
1 Q2max 1
1
2
2
WC=
t
E t
PC= E(1 − e ⁄RC ) e ⁄RC
R
C
= Qmax E = CE2
2
At t = 0, PC = 0
At t = ∞, PC = 0
CASE 2: if at t = 0, q = Qo
(charging with Qo)
CASE 3: (discharging with Qo)
R
E
R
E
Qo
Qo
C
i
i
From,
t
q = CE + ke ⁄RC
At t = 0, q = Qo
Qo = CE + keo
K = Qo – CE
t
q = CE + (Qo – CE)e ⁄RC
t
i = Ioe ⁄RC =
C
E− Qo/C t⁄
e RC
R
From,
t
q = CE + ke ⁄RC
At t = 0, q = - Qo
- Qo = CE + keo
K = -(Qo + CE)
hence,
MULTIVECTOR REVIEW AND TRAINING CENTER
t
Proof: E = VR+ VC = Ri + q/C
q = CE - (Qo + CE) e ⁄RC
t
i = ioe ⁄RC =
E+ Qo/C t⁄
e RC
R
At t = 0, q = Qo, i = Io
E = RIo +Qo/C
Io=
E+ Qo/C
Proof:
At t = 0
E− Qo/C t⁄
e RC
R
E = RIo + (- Qo)/C; Io =
R
Hence,
t
i = Ioe ⁄RC =
E− Qo/C t⁄
e RC
R
ELECTRICAL TRANSIENT
CASE 4 (For discharging of initially charged capacitor):
1
1
2
V
2
C
R
CASE 5:
C1
V
i
R
i
C2
R
Q
R
C
C1
i
Qo
i
C2
MULTIVECTOR REVIEW AND TRAINING CENTER
t
q = Qoe ⁄RC
t
i = Ioe ⁄RC =
− Qo/C t⁄
e RC1
R
− Qo/C t⁄
e RC
R
t
i = Ioe ⁄RC =
CT=
C1 C2
C1 +C2
RLC Circuit (Double Energy Transient)
t
q=
Qo CT C1
C1
(
CT
–1+
e ⁄RC1 )
Series RLC Circuit (DC):
E
R
VR
L
VL
C
VC
I
By KVL,
VR + VL + VC = E
di
1
Ri + L + ∫ 𝑖 dt = E
dt C
Differentiate “i” with respect to t
d𝑖
d2𝑖
𝑖
1
[R + L + = 0]
dt
dt C
L
d21
dt2
Or
+
R di
L dt
+
𝑖
LC
= 0→L.D.E. of the second order with three possible solutions.
MULTIVECTOR REVIEW AND TRAINING CENTER
(D2 +
R
L
D
1
LC
)i = 0
ELECTRICAL TRANSIENT
Auxiliary Equation
Case II – Critically – damped case:
R
1
m2 + m + = 0
-when the roots are real and
L
LC
equal/repeated
R
4
By Quadratic Equation Formula:
Test: ( )2 - = 0
L
m=
R
R
4
− ±√( )2−
L
L
LC
=−
or b2 – 4ac = 0
2 (1)
R
±
2L
m= −
m= −
R
2L
R
2L
R
L
√( )2−
4
LC
or if (
√4
±
R
L
√( )2−
LC
4
LC
𝛽=0
√4
R
4
L
LC
± √( ) 2 −
R
1
2L
)2 = ( )
LC
0
m=∝+𝛽
m = m1 = m2 = ∝
(D - ∝ )(D - ∝) i = 0
𝑡
𝑡
Gen. Solution: i = k1e𝑚1 + k2 te𝑚2 =
𝑡
𝑡
k1e𝑚1 + k2 te𝑚2
i = e∝𝑡 ( k1 + k2t)
Let ∝ = −
R
R
4
L
LC
, 𝛽 √( ) 2 −
2L
dumped CASE: (Oscillatory Case)
Therefore,
imaginary part.
m = ∝ ±𝛽
Roots;
m1 = ∝ +𝛽
m2 = ∝ +𝛽
imaginary, 𝛽 = j𝜔
CASE III – Under
- when the roots contain a real and
test: (
R
2L
4
)2 - ( )< 0
LC
or
b2 – 4ac < 0
R
1
or if ( )2 < ( )
2L
LC
𝛽 = is
MULTIVECTOR REVIEW AND TRAINING CENTER
Damping Cases of RLC in Series:
m = ∝ ±𝛽
Case 1: Over – damped Case:
m= −
-when the roots are real and unequal
Test: (
R
2L
4
)2 - ( )> 0
m= −
LC
b2 – 4ac> 0
Or if (
R
2L
m= −
∝=−
1
R
2L
R
LC
𝛽 is a real number
m1 ≠m2
m1 = ∝ + 𝛽, m1 = ∝ − 𝛽
[D – (∝+𝛽)][D – (∝ − 𝛽)]i = 0
− j𝜔)]i = 0
Gen. Solution
𝑡
𝑡
i = k1e𝑚1 + k2e𝑚2 , amps.
i = k1e(∝+𝛽)𝑡 + k2 te(∝− 𝛽)𝑡
= k1e∝𝑡 ∙ e𝛽𝑡 + k2e∝𝑡 ∙ e−𝛽𝑡
i = e∝𝑡 (k1e𝛽𝑡 + k2e−𝛽𝑡 )
2L
R
2L
R
1
L
1
LC
R 2
± √( ) 2 −
± √[
± 𝑗√
1
LC
−(
2L
)]
R
LC
− ( )2
2L
2L
let 𝜔 = √
)2 > ( )
R
1
LC
R
− ( )2
2L
m = ∝ ± j𝜔
or m = ∝ + j𝜔
m = ∝ − j𝜔
[D – (∝+j𝜔)][D – (∝
Gen. Solution
𝑡
𝑡
i = k A e𝑚 1 + k B e𝑚 2
= kAe(∝+𝑗𝜔)𝑡 + k2 te(∝− 𝑗𝜔)𝑡
= kAe∝𝑡 ∙ e𝑗𝜔𝑡 + kBe∝𝑡 ∙ e−𝑗𝜔𝑡
i = e∝𝑡 (kAe𝑗𝜔𝑡 + kBe−𝑗𝜔𝑡 )
ELECTRICAL TRANSIENT
From Euler’s Formula:
e±𝑗𝜔𝑡 = cos 𝜔t ± j sin𝜔t
i = e∝𝑡 [ kA(cos 𝜔t ± j sin𝜔t) + kB (cos 𝜔t ± j sin𝜔t)]
= e∝𝑡 [ kA + kB ] cos 𝜔t + (jkA - jkB) sin𝜔t]
i = e∝𝑡 [ kA(cos 𝜔t ± j sin𝜔t)
Where:
𝜔 = angular velocity of oscillation, rad/ sec
f = frequency of oscillation, cps or hertz
MULTIVECTOR REVIEW AND TRAINING CENTER
𝜔 = 2𝜋𝑓= √
f=
1
2𝜋
√
1
LC
1
LC
R
− ( )2
2L
R
− ( )2
2L
AC Transients
Series RL Circuit:
Series RC Circuit:
R
R
e j
j
j
j
j
h
h
e
L
i
𝜃 z)
𝜙 = position of sinusoidal voltage
i
ǀZC ǀ
tan 𝜃 Z =
𝜃 Z = tan
For ex,
At t = 0, e = Em
1 = sin ϕ
ϕ = 90o
Em
sin (𝜔t + 𝜙 +
ZC = R – j XC= R – j
At the instant of switching/ application
0 = Em sin (0 + ϕ)
ϕ =0
C
i = Cet⁄RC +
e = Em sin ( 𝜔t + 𝜙)
At t = 0, e = 0
j
j
j
j
j
h
h
-1XC
R
XC
R
tan
1
At t = 0, e = Em
1
Em= Emsin ϕ
ϕ = 30o
2
\
After the switch is closed,
e = eR + eL
di
Emsin ( 𝜔t + 𝜙) = Ri + L
dt
i = iT + iSS
Steady - state component
Transient component
2
=
1
ωCR
-1 1
ωCR
1
ωC
MULTIVECTOR REVIEW AND TRAINING CENTER
Resultant
𝑅
i = Ce−𝐿𝑡 +
Em
ǀZL ǀ
sin (𝜔t + 𝜙 − 𝜃 z)
ZL = R + j 𝜔L=ǀZL ǀ∠ 𝜃z
X
ωL
tan 𝜃 Z = C =
R
𝜃 Z = tan
-1XL
R
R
tan -1
ωL
R
ELECTRICAL TRANSIENTS
EXERCISES:
1. The field circuit of a 4 – pole DC motor has a resistance of 200 Ω and an
inductance of 100 H. If it is connected to a 400 – volt supply, find the
time taken for the current to each 1.5 A.
A. 0.125 sec
B. 1.5 sec
C. 0.369 sec
D. 0.693
sec
REE – Sept. 2003
2. In an RL circuit having a source voltage of 100 V, the resistance is 400
Ω and the inductance is 20 H. What is the time constant?
A. 20
B. 0.05
C.0.2
D.0.25
3. A coil has a time constant of 1 second and an inductance of 8 H. If the
coil is connected to a 100 V DC source, determine the rate of rise of
current at the instant of switching.
A. 8 amp/sec
B. 12.5 amp/sec
C. 0.25 amp/sec
D.
0.04amp/sec
4. A series RL circuit with R = 50 ohms and L = 10 H has a constant
voltage E = 100v applied at t = 0 by the closing of a switch. Find the
time at which vR = vL.
A. 0.6931 sec
B. 0.1386 sec
C. 2.3026 sec
D. 0.
2773 sec
MULTIVECTOR REVIEW AND TRAINING CENTER
5. A constant voltage is applied to a series RL circuit at t = 0 by closing a
switch. The voltage across L is 25 volts at t = 0 and drops to 5 volts at t
= 0.025 second. If L = 2 H, what must be the value of R in ohm?
A. 17.85
B. 128.8
C. 1288
D.178.5
6. A coil having L = 150 H, R = 200 ohms us connected in series with 100
ohms resistor. A 240 – V dc source is connected to the circuit at t = 0.
The voltage across the coil at t = 0.5 sec is
A. 151.8 V
B. 189.4 V
C. 88.2 V
D.101.2 V
7. A circuit of resistance R and inductance L has a direct voltage of 230 V
applied to it 0.3 second after switching on, the current was found to be 5
amperes. After the current had reached its final steady value, the circuit
was suddenly short – circuited. The current was again found to be 5
amperes at 0.3 second after short circuiting the coil. Find the value of R
and L.
A. 25 ohms, 5 H B. 5 ohms, 10 H C. 23 ohms, 10 H
D. 20 ohms, 5 H
8. The field winding of a separately – excited dc generator has an
inductance of 60 H and a resistance of 30 ohms. The discharge
resistance of 50 ohms is permanently connected in parallel with the
winding which is excited from a 200 V supply. Find the value of the
decay current 0.6 sec after the supply has been switched off.
A. 4 A
B. 3 A
C. 2 A
D. 1
A
9. A coil of 10 H inductance and 5 ohms resistance are connected in
parallel with 20 ohm resistor across 100 V dc supply which is suddenly
disconnected. The voltage across the switch contacts at the instant of
separation is,
A. 500 V
B. 400 V
C. 300 V
D. zero
REE – April 2006
MULTIVECTOR REVIEW AND TRAINING CENTER
10.
A capacitor of 8 µF is to be charged by a voltage of 400 V through
a resistor of 100 ohms. How long will it takes for the voltage across the
capacitor from its initial zero value to reach 300 V?
A. 1.05
B. 1.10
C. 0.99
D. 1.00
ELECTRICAL TRANSIENTS
11.
A resistance R and a 4 µF capacitor are connected in series across
a 200 V dc supply. Across the capacitor is a neon lamp that strikes
(glows) at 120 V. Calculate the value R to make the lamp strike(glow) 5
seconds after the switch have been closed.
B. 2.45 MΩ
B. 1.98 MΩ
C. 1.36 MΩ
D.
3.62 MΩ
REE – April 2004
12.
A series circuit with 10 – ohm resistor and 50 µF capacitor is
suddenly connected to a DC source of 120 volts. What is the current at
one time constant?
A. 12 A
B. 2.44 A
C. 7.6 A
D. 4.4 A
REE – April 2001
13.
A 200 microfarad capacitor is connected in series with a 10 Ω
resistor. The circuit is connected to a 24 V battery. What is the initial
current when putting the switch on?
A. 12 A
B. 2.4 A
C. 0.225 A
D. 0.0125 A
14.
When a capacitor charged to a p.d of 400 V, is connected to a
voltmeter having a resistance of 25 MΩ, the voltmeter reading is
observed to have fallen to 50 V at the end of an interval of 2 minutes.
Find the capacitance of the capacitor.
A. 2.81 µF
B. 2.65 µF
C. 2.31 µF
D. 32 µF
15.
A 100 µF capacitor charged to 24 V, is connected with a 200 µF
uncharged capacitor, a 1 kΩ resistor and a switch. What is the current
0.1 sec after the switch is closed?
MULTIVECTOR REVIEW AND TRAINING CENTER
A. 5.36 mA
4.88 mA
B. 14.56 mA
C. 8.83 mA
D.
16.
A time of 10 ms is required for the current in an RL circuit to reach
90 percent of its final value. If R is 10 ohms, find the value of C to be
inserted in series with the RL circuit so that the frequency of oscillation
of the resulting transient currents is 1,000 cycles per second.
A. 58.3 µF
B. 0.748 µF
C. 0.583 µF
D. 5.83 µF
REE – May 2008
17.
A series RLC circuit contains a resistor R of 2 ohms and a
1
capacitor C = farad. Determine the value of the inductor so that the
2
circuit is critically damped,
A. 2 H
B. 1⁄2H
C. 1 H
D. 3⁄2H
REE – May 2008
18.
A parallel RLC circuit contains a resistor R = 1 ohm and an
inductor L = 2H. Find the value of the capacitor so that the circuit I
critically damped.
A. 1.0 F
B. 0.85 A
C. 0.64 A
D. 0.96 A
19.
A series RL circuit with R = 50 ohms and L = 0.2 H has a
sinusoidal voltage source e = 150 sin (377t + ф) volts applied at t = 0
when passing to its maximum value. Find the resultant transient current
0.001 sec after switching on.
A. 0.52 A
B. 0.85 A
C. 0.64 A
D. 0.96 A
REE – Sept. 2006
20.
What is the current through a 0.02 microfarad capacitor at t = 0 if
the voltage across it is 2sin2pi x 10 to the 6th x t?
A. 0.3142 A
B. 0.2513 A
C. 0.1885 A
D.
0.1256 A
MULTIVECTOR REVIEW AND TRAINING CENTER
ELECTRICAL TRANSIENTS
REE – April 2007
21.
Find the current thru a 0.10 microfarad capacitor at t = 0 if the
voltage across it is – 2 e raised to – 10 raised to 7 x t V.
A. 200 mA
B. 20 mA
C. 2 A
D. 20 A
REE – Sept. 2008
22.
The current to a capacitor is given by i = 2t + 3. The initial charge
on the capacitor is 8.13 coulombs. Find the charge when t = 1 second?
A. 16 coulombs B. 12.13 coulombs C. 13 coulombs
D. 4
coulombs
MULTIVECTOR REVIEW AND TRAINING CENTER
DIRECT CURRENT (DC) MACHINES
PART 1
DYNAMO – is a rotating electrical machine that converts mechanical energy
into electrical energy or vise – versa.
Two (2) General Type of a Dynamo:
1. DC Generator
2. DC Motor
Two (2) Major Part of a Dynamo
1. Stator –that portion which is stationary; the most important part of
which is the field winding.
2. Armature– the portion which rotates; the most important part of which
is the armature winding.
Two (2) General Types of a DC Armature Winding
1. Lap Winding - is one which the coil ends are connected to commutator
segments that are near one another; adjacent in simplex – lap winding.
MULTIVECTOR REVIEW AND TRAINING CENTER
2. Wave Winding - is one which the coil ends are connected to
commutator segments that are of some distance from one another, nearly
360 electrical degrees.
Coil pitch – the distance between the two sides of a coil (the no. of slots
enclosed by the two coil sides).
COIL PITCH
𝑺
Ys = - K
𝑷
Where:
Ys = coil pitch, in slots
S = total no. of armature slots
P = no. of poles
K = any parts of S/P that is subtracted to make Ys an integer
Commutator Pitch –the distance between the two ends of a coil (the no. of
commutator bars between the two coil ends).
For Lap Winding,
Yc =± m
For Wave Winding,
Yc =
𝑪 ±𝒎
𝑷/𝟐
Where:
YC = commutator pitch, bars or segments
S = total no. of commutator bars or segments
P = no. of poles
K = plex or degree of municipality of the winding.
MULTIVECTOR REVIEW AND TRAINING CENTER
For simplex winding,
For duplex winding,
For triplex winding,
For quadruplex winding,
m = 1 bar or segment
m = 2 bars or segments
m = 3 bars or segments
m = 4 bars or segments
Note: (+) – for progressive winding
( - ) – for retrogressive winding
Parallel Paths in Lap and Wave Windings
1. For Lap Winding
a=mxP
2. For Wave Winding
a=2xm
where:
a = no. of parallel paths in the armature (always an even no.)
m = plex of the winding
P = no. of poles
Dead or Dummy Element – an electrically an connected coil element in the
armature that serves only to keep the revolving structure mechanically
balanced.
Frog – Leg Winding – a type of armature winding that combines the features
of ordinary lap and wave windings. It was Allis – Chalmers Manufacturing
Company who developed this type of winding.
Average Generated Voltage in a Conductor
ф
Eave = x 10-8
𝒕
Volts
Where: Eave = average generated voltage in a conductor, volts
Ф = total flux cut, lines or Maxwell’s
MULTIVECTOR REVIEW AND TRAINING CENTER
t = time, seconds, during which cutting takes place.
General Voltage Equation for DC Generator
Eg =
ф 𝒙 𝑷 𝒙 𝑵 𝒙 𝒁 𝒙 𝟏𝟎−𝟖
𝟔𝟎 𝒙 𝒂
Volts
Where:
Eg= total generated voltagein the armature, volts
Ф = flux per pole, lines/pole or Maxwell’s/ pole
P = no. of poles, an even no.
N = speed of the armature conductors
Z = total no. of active armature conductors
a = no. of parallel paths in the armature
DC Generator – a type of a dyno that converts mechanical energy into
electrical energy.
Two (2) General classifications of a DC generator
1. Self – excited DC Generator – is one which the field winding is
energized by its own armature.
2. Separated – excited DC Generator – is one which the field winding
is energized from a source of direct current external to the machine
called an “exciter”. An exciter may be a battery or another de
generator of smaller capacity.
General Types of Self – Excited DC Generator
1. Series Wound DC Generator – its field winding called series field
winding is connected in series with the armature. It is used for constant
application like in series street lighting.
Schematic Wiring Diagram
Equivalent Circuit Diagram
MULTIVECTOR REVIEW AND TRAINING CENTER
Formulas:
Ia = Is = IL
Eg = Vt + Va + Vbc + Vs
Eg = Vt + Ia (Ra + Rbc) + IsRs
Pg = EgIa
Pg = po + electrical loses
Po = VtVL = IL2RL
Electrical Losses or Copper Losses:
a) Armature winding copper loss, Pa = Ia2 Ra
b) Brush contract resistance loss, Pbc = Ia2 Rbc
c) Series field winding copper loss, Ps = Is2 Rs
Pin = Pg +spL
Pin = Po + electrical losses + spL
Pin = Po + total losses
𝜼=
𝑷𝒐
𝑷𝒊𝒏
x 100
𝜼𝒄 =
𝑷𝒐
𝑷𝒈
x 100
Where: Ia = armature current
winding resistance
Is = series field current
or
IL = load current
Eg = generated or developed or induced
power
Voltage in the armature (also
Called as open circuit or no loss
𝜼𝒎 =
𝒑𝒈
𝑷𝒊𝒏
x 100
Rs = series field
Pg = power generated
developed in the armature
Pin = input
Po = output power
spL = stray power
MULTIVECTOR REVIEW AND TRAINING CENTER
𝜂 c = electrical
Load voltage of the generator)
efficiency
Vt = terminal or line or output voltage of the generator 𝜂 m= mechanical
efficiency
Va = armature winding resistance drop
𝜂 = over – all or
conventional or
Vbc = brush contract resistance drop
commercial
efficiency
Vs = series field winding resistance drop
Ra = armature winding resistance
Rbc = brush contact resistance
2. Shunt Wound DC Generator – its field winding called shunt field
winding is connected across the armature. It is used for constant voltage
application like in electrical winding.
Schematic Wiring Diagram
Equivalent Circuit Diagram
Formulas:
Ia = IL = If
If = Vt /Rf
Eg = Vt + Va + Vbc
Eg = Vt + Ia (Ra + Rbc)
Pg = EgIa
Pg = po + electrical loses
Po = VtVL
Electrical Losses or Copper Losses:
a) Armature winding copper loss, Pa = Ia2 Ra
MULTIVECTOR REVIEW AND TRAINING CENTER
b) Brush contract resistance loss, Pbc = Ia2 Rbc
c) Series field winding copper loss, Pf = If2 Rf
Pin = Pg +spL
Pin = Po + electrical losses + spL
Pin = Po + total losses
𝜼=
𝑷𝒐
𝑷𝒊𝒏
x 100
𝜼𝒄 =
𝑷𝒐
𝑷𝒈
x 100
𝜼𝒎 =
𝒑𝒈
𝑷𝒊𝒏
x 100
Where:
If = shunt field current
Rf = shunt field winding resistance
3. Compound Wound DC Generator –it has both the series and shunt
field windings which are connected either long shunt or short shunt.
a) Cumulative Compound Generator – the series field flux is aiding
field to supply a power and lighting loads.
a.1) Under Compounded – the full – load terminal voltage is less
than the no – load voltage. It is used when the load is located near
from it.
a.2) Flat Compounded – the full – load terminal voltage is the same
as the no – load voltage. It is used when the load is at in medium
distance from it.
a.3) Over Compounded – the full – load terminal voltage is greater
than the no – load voltage. It is used when the load is far from it.
Percent Voltage Regulation
% V.R. =
𝑽𝑵𝑳 ÷ 𝑽𝑭𝑳
𝑽𝑭𝑳
x 100
MULTIVECTOR REVIEW AND TRAINING CENTER
Where:
%V.R. = percent voltage regulation
VNL = no – load voltage
VFL = full – load voltage
a) Differential Compound Generator – the series field flux opposes
the shunt field flux. It is used in electric winding.
Long Shunt Compound Generator
Schematic Wiring Diagram
Equivalent Circuit Diagram
Formulas:
Ia = IS = IL + If
If = Vt /Rf
Eg = Vt + Va + Vbc+ Vs
Eg = Vt + Ia (Ra + Rbc+ Rs)
Short Shunt Compound Generator
Schematic Wiring Diagram
Equivalent Circuit Diagram
MULTIVECTOR REVIEW AND TRAINING CENTER
Formulas:
Ia = IL + If
If = Is
𝑉𝑡+ 𝑉𝑠
If =
𝑅𝑓
Eg = Vt + Vs + Va + Vbc
Eg = Vt + IsRs + Ia(Ra + Rbc)
4. Separately – Excited DC shunt Generator – its field winding also
called shunt winding is connected to an outside DC source ( a battery or
another DC generator). It is used for laboratory experiment purpose
where voltages are likely to be varied.
Schematic Wiring Diagram
Formulas:
Ia = IL
If = Vf/Rf
Eg = Vt + Va + Vbc
Eg = Vt + Ia(Ra + Rbc)
Equivalent Circuit Diagram
MULTIVECTOR REVIEW AND TRAINING CENTER
Note: The following may be included in the design of DC machines;
a) Interpole/ Commutating pole winding – used to correct the
objectionable commutation effects of armature reaction. This winding
is permanently connected in series with the armature.
b) Compensating Winding –used for the purpose of neutralizing the
effects of armature reaction in the zones outside the influence of the
interpoles. This winding is permanently connected in series with the
armature.
c) Diverter – a low resistance shunt connected directly across the series
field of a compound generator for the purpose of adjusting the degree
of compounding.
DC MACHINES – PART 1
MULTIVECTOR REVIEW AND TRAINING CENTER
EXERCISES:
1. A 4 – pole simplex wave – wound generator has 57 slots, what is the coil
pitch?
A. 7
B. 14
C. 4
D. 14.25
2. What is the commutator pitch of a 4 – pole, duplex wave – wound
armature having 120 commutator bars?
A. 122 or 118 bars
B. 30 or 29 bars C. 61 or 59 bars D. 60 or 58 bars
REE – April 2001
3. The difference between the front pitch and the back pitch of a
retrogressive lap winding is 2. If the front pitch is 21, what is the back
pitch?
A. 19
B. 23
C. 21
D. 21.5
REE – Sept. 2008
4. How many parallel paths there will be in the armature of a 10 – pole
machine if the armature is simplex lap wound?
A. 4
B. 12
C. 8
D. 10
5. How many parallel paths does a 6 – pole quadruplex wave – wound
connected armature winding of a DC generator has?
A. 24
B. 12
C. 6
D. 8
6. What is the average voltage generated in a moving conductor if it cuts
2.5 x 106Maxwell’s in 1⁄40 second?
A. 1 volt
B. 2 volts
C. 10 volts
D. 100 volts
REE – Sept. 2007
7. A DC machine has 8 poles and a rated current of 100 A. How much
current will flow in each path at rated condition if the armature is
simplex lap wound?
A. 2 A
B. 12.5 A
C. 6.25 A
D. 50 A
MULTIVECTOR REVIEW AND TRAINING CENTER
8. A six – pole dynamo, with interpoles, requires 360 inductors on its
armature with 3 turns per coil and 2 coil sides per slot. Determine the
number of coils, slots, and commutator bars.
A. 120, 120, 120 B. 60, 60, 60
C. 60, 60, 120
D. 120,
60,60
REE – Sept.2008
9. When a generator is being driven at 1,200 rpm, generated emf is 230 V.
What is the generated emf if the flux is decreased by 10 % with the
speed remains unchanged.
A. 205 V
B. 207 V
C. 200 V
D.203 V
REE – October 2000
10.
A four –pole DC generator has an armature containing 4 elements
per slot and 48 slots per pole. The armature is simplex lap – wound. The
flux per pole is 2,500,000 Maxwell’s and the speed of the armature is
1500 rpm. Calculate the total voltage generated in the armature?
A. 120 V
B. 480 V
C. 960 V
D. 240 V
DC MACHINES – PART 1
REE – Sept. 2005
11.
What is the maximum voltage generated in a drum armature with
concentrated windings consisting of 300 series conductors in each path?
The armature speed is 1,200 rpm. Each conductor cuts twice through a
field of 1.5 x 106 lines of force per revolution.
A. 282 V
B. 267 V
C. 273 V
D. 256 V
12.
A shunt generator delivers 195 A at a terminal voltage of 250 V.
The armature resistance and the shunt field resistance are 0.02 and 50
ohms, respectively. The iron and the friction losses are equal to 950 W.
Find the commercial efficiency of the generator.
A. 94.2%
B. 98.2%
C. 95.9%
D. 93.5%
MULTIVECTOR REVIEW AND TRAINING CENTER
13.
In a problem No. 12, find the mechanical efficiency.
A. 93.5%
B. 95.9%
C. 97.35%
D. 98.2%
14.
In Problem No. 12, find the electrical efficiency.
A. 93.5%
B. 97.4%
C. 97.3%
D. 94.2%
15.
A DC generator has a mechanical efficiency at 98.2 % and an
electrical efficiency of 95.9%. Find the commercial efficiency.
A. 96y.5%
B. 97.4%
C. 98%
D. 94.2%
REE – Oct. 1998
16.
The pulley of an old gen – set has a diameter of 20 inches. The belt
exerts a pull or 353 lbs. on the pulley. The gen – set runs at 900 rpm.
What I the approximate kW of the gen –set?
A. 75
B. 200
C. 250
D. 35
17.
A short shunt compound DC generator delivers a load current of
30 A at 220 V and has armature, series field and shunt fields resistances
of 0.05 ohm, 0.03 ohm and 200 ohms, respectively. Calculate the
generated voltage (allow 1 V for brush contact drop).
A. 295.2 V
B. 223.45 V
C. 221.05 V
D. 301.56
V
18.
A 4 – pole, wave wound, 750 rpm shunt generator has an armature
and field resistances of 0.4 ohm and 200 ohms, respectively. The
armature has 720 conductors and the flux per pole is 28.95 mWb. If the
load resistance is 10 ohms, what is the terminal voltage of the machine?
A. 521 V
B. 510 V
C. 505 V
D. 500 V
19.
The voltage of a 100 – kW, 250 – volt shunt generator rises to 260
volts when the load is removed. What is the full – load current does the
machine deliver, and what it’s percent regulation?
A. 400 A, 4%
B. 385, 3.85%
C. 400 A, 2%
D.
500 A, 5%
MULTIVECTOR REVIEW AND TRAINING CENTER
20.
Each of the series field’s coil of a 4 – pole 50 – kW 250 – volt
short shunt compound has 6 1⁄2 turns of wire. The resistance of the
entire series field is 0.012 ohm, and the diverter resistance is 0.036 ohm.
What is the number of ampere – turns of each series – field coil at full –
load?
A. 975 AT
B. 1300 AT
C. 325 AT
D. 2275 AT
DC MACHINES
PART 2
Maximum Efficiency of a DC Generator
The efficiency of a DC generator is maximum.
When
Variables Losses = Constant Losses
For DC shunt generator,
Variable losses = Ia2 (Ra + Rbc)
Constant losses = If2Rf + spL
Hence,
𝛈𝐦𝐚𝐱 =
𝐏𝐨 ′𝐱 𝟏𝟎𝟎
𝑷𝒐 ′ + 𝐕𝐚𝐫𝐢𝐚𝐛𝐥𝐞 𝐥𝐨𝐬𝐬𝐞𝐬 + 𝐂𝐨𝐧𝐬𝐭𝐚𝐧𝐭 𝐥𝐨𝐬𝐬𝐞𝐬
𝛈𝐦𝐚𝐱 =
𝐕𝐭 𝐈𝐋 ′𝐱 𝟏𝟎𝟎
𝟐
𝐕𝐭 𝐈𝐋 ′ 𝐈′𝐚 (𝐑 𝐚 + 𝐑 𝐛𝐜 ) + 𝐈𝐟𝟐 𝐑 𝐟 + 𝐬𝐩𝐋
MULTIVECTOR REVIEW AND TRAINING CENTER
Where:
𝛈𝐦𝐚𝐱 = maximum efficiency
Po’ = output power at maximum efficiency
Vt = terminal or output voltage (assumed to be practically
constant)
IL’ = load current at 𝛈𝐦𝐚𝐱
Ia’ = armature current at 𝛈𝐦𝐚𝐱
Since for DC shunt generator, Ia = IL + If
But If is negligible compared to IL, therefore neglecting If makes Ia = IL.
Hence at 𝛈𝐦𝐚𝐱 .
Variables losses = Constant losses
Ia’2 (Ra + Rbc) = If2 Rf + spL
Where Ia’ = IL’ by neglecting If
∴
IL’ 2 (Ra +Rbc) = If2Rf + spL
IL’ =√
𝑰𝟐𝒇 𝑹𝒇 +𝒔𝒑𝑳
𝑹𝒂 +𝑹𝒃𝒄
Advantages of Parallel Operation of DC Generator
1. For generator efficiency of operation
2. For continuity of service
3. For repair and maintenance problems
4. Serves as an added capacity as the service demands changed
MULTIVECTOR REVIEW AND TRAINING CENTER
Conditions to be met for ideal parallel operation of
I.
DC shunt Generator
1. Same voltage
2. Same polarities
II. DC Compound Generator
1. Same voltage rating
2. Same polarities
3. With equalizer connection
If condition (1) is not met, the generators will not deliver power in
proportion to KW ratings.
If condition (2) is not met, there will be big current that will circulate the
armature windings of the generators which may damage the windings.
If condition (3) is not met, the generation will operate in unstable
equilibrium. One may become a motor to be supplied by the generator.
There will be “motorization”.
DC Generator – a type of a dynamo that converts electrical energy into
mechanical energy.
General Types of DC Motor
I.
Series Wound DC Motor – its field winding also called series
field winding is connected in series with the armature.
Speed Characteristic:
Variable Speed
Torque Characteristic:High Starting Torque
Uses:
For elevator, crane, conveyor, hoist,
gear, drive, etc.
MULTIVECTOR REVIEW AND TRAINING CENTER
To reverse the direction of rotation of this motor, interchange the
brushes.
Caution: never start this motor without load or remove the load while it
is operating for it will “race” or “run away”.
Schematic Wiring Diagram
Equivalent Circuit Diagram
Formulas:
IL = Is = Ia
Vt = Eb + Va + Vbc + Vs
Vt = Eb + Ia (Ra + Rbc) + IsRs
Pb = EbIa
Pin = Pb-spL
Pin = VtIL
Pin = Po + total losses
Pin = Pb + electrical losses
𝜼=
𝑷𝒐
𝑷𝒊𝒏
x 100
𝜼𝒄 =
𝑷𝒃
𝑷𝒊𝒏
Where: IL = line current
in the armature
Is = series field current
Ia = armature current
Eb = back or counter emf
Ra = armature winding resistance
conventional or
x 100
𝜼𝒎 =
𝒑𝒐
𝑷𝒃
x 100
Pb = power developed
Po = output power
Pin = input power
spL = stray power loss
𝜂 = over – all or
MULTIVECTOR REVIEW AND TRAINING CENTER
Rbc = brush contact resistance
Rs = series field winding resistance
efficiency
commercial efficiency
𝜂 c = electrical
𝜂 m= mechanical efficiency
II.
Shunt Wound DC Motor – its field winding also called shunt
field winding is connected in across with the armature.
Speed Characteristic:
Nearly constant or adjustable
speed
Torque Characteristic: Medium Starting Torque
Uses:
For fan, blower, pump, grinder, etc.
To reverse the direction of this motor, interchange the brushes or reverse
the connection of the shunt field winding terminals with respect to the
armature but not both.
Caution: never open the field circuit while this motor operating for it
will “race” or “runaway”.
Schematic Wiring Diagram
Equivalent Circuit Diagram
Formulas:
IL = Ia + If
Pb= EbIa
Vt = If Rf
Po= Pb- spL
2
Vt = Eb + IaRa + Rbc + Ia Rbc
Pin = Po + total losses
Vbc = IaRbc
Pin = Vt IL
Where:
Vbc = brush contact resistance drop
MULTIVECTOR REVIEW AND TRAINING CENTER
If = shunt field current
Rs = shunt field winding resistance
III.
Compound Wound DC Motor – it has the series field and the
field windings which are connected either long shunt or short
shunt.
Speed Characteristic:
Variable Speed or adjustable
speed
Torque Characteristic:High Starting Torque
Uses:
For elevator, conveyor crane, milling
machine, punching machine, etc.
To reverse the direction of rotation of this motor, interchange the
brushes.
A. Long Shunt – adjustable speed.
Schematic Wiring Diagram
Equivalent Circuit Diagram
Formulas:
IL = Ia + If
Vt = If Rf
Vt = Eb + IaRa + Rbc + IaRs
Vbc = IaRbc
B. Short Shunt – variable Speed.
Schematic Wiring Diagram
Equivalent Circuit Diagram
MULTIVECTOR REVIEW AND TRAINING CENTER
Formulas:
IL = IS = Ia + If
Vt = If Rf + ILRs
Vt = Eb + IaRa + IaRbc + ILRs
Vbc = IaRbc
DC MACHINES – PART 2
EXERCISES:
1. A 60 kW, 250 V DC shunt generator has an armature circuit resistance
of 0.05 ohm, field resistance of 50 ohms and maximum efficiency of
91.2%. The approximate stray power loss is
A. 1785 W
B. 160.25 W
C. 1660 W
D. 2135 W
2. A 230 V shunt generator has an armature circuit resistance of 0.05 ohm
and shunt field resistance of 57.5 ohms. When the generator operates at
rated voltage it has a total core and mechanical losses of 1880 W. At
what load does the generator operate so as to achieve maximum
efficiency?
A. 78.5 W
B. 53.51 kW
C. 50.61 kW
D. 60. 15
kW
MULTIVECTOR REVIEW AND TRAINING CENTER
3. The output of a shunt generator is 24 kW at a terminal voltage of 200
volts. Armature resistance = 0.05 ohm, shunt field resistance = 40 ohms.
If iron and friction losses equal the copper losses at this load, the output
o the engine driving the generator is
A. 27.562 kW
B. 25.672 kW
C. 22.762 kW
D.
26.257 kW
4. A shunt generator has an external characteristics whose terminal voltage
given by the equation Vt = (250 – 0.4 IL), where IL is the load current
and power will be supplied by the machine to a load resistor of 5.85
ohms?
A. 40 A, 936 W B. 2.5 A, 3 V
C. 10 A, 585 W
D. 253.3 V
5. Two shunt generators A and B, with ratings of 250 and 400 W,
respectively, and having identical straight – line voltage vs. percent
kilowatt – output external characteristic, are connected in parallel. If no load voltage is 260 and the full load voltage is 240, calculate the
terminal voltage when the total output is 575 kW.
A. 242.3 V
B. 245.3 V
C. 248.3 V
D. 253.3 V
6. Two 220 V dc generators, each having a linear external characteristics
operate in parallel. One machine has a terminal voltage of 270 V on no –
load and 220 V at a load current of 35 A, while the other has a voltage of
280 V on no – load and 220 V at 50 A. Calculate the output current of
one machine is the total load current is 60 A.
A. 23.6 A
B. 19.5 A
C. 31.7 A
D. 28 A
REE – Sept. 2006
7. Which DC motor cannot be used without a load connected to it?
A. Shunt
B. Series
C. Compound
D. Universal
REE – Sept. 2003
8. In a DC compound motor, what is the relation of the connection of
motor field windings with respect to the motor armature?
MULTIVECTOR REVIEW AND TRAINING CENTER
A. Series
B. Parallel
C. Series- parallel
D. delta – wye
REE – April 2002
9. A certain shunt motor armature has a resistance of 0.07 Ω and current of
50 amperes at 120 volts. Determine the counter emf.
A. 108.5 V
B. 116.5 V
C. 112.2 V
D. 118.3 V
10.
A 440 – volt shunt motor has an armature resistance of 0.8 Ω and a
field resistance 200 Ω. What is the back emf when giving an output of
7.46 kW at 85% efficiency?
A. 465.8 V
B. 425.8 V
C. 400 V
D. 440 V
DC MACHINES – PART 2
REE – April 2001
11.
A DC shunt motor develops 15 HP at 120 V. The armature
efficiency is 95%. What is the armature effective resistance?
A. 0.10 Ω
B. 0.06 Ω
C. 0.08 Ω
D. 0.60 Ω
12.
A 200 – volt shunt motor is taking a current of 30 amperes.
Armature resistance is 0.2 Ω and shunt field resistance is 100 Ω. Iron
and friction losses amount to 500 watts. What is the B.H.P output of the
motor?
A. 6.63 HP
B. 66.3 HP
C. 3.66 HP
D. 36.6 HP
MULTIVECTOR REVIEW AND TRAINING CENTER
DC MACHINES
PART 3
Force Exerted in a Motor Conductor
F
BI𝑙
10
dynes
F BIl dynes
MULTIVECTOR REVIEW AND TRAINING CENTER
Where:
F = force exerted, dynes
B = flux density, gauss (lines/cm2)
I = current through the conductor, amperes
l = length of the conductor, cm.
Iab = current through the conductor, abampere
Note:
Iab = I/10
Total Force Exerted in the Motor Armature
FT = FZ
𝐙𝐁𝐈 𝑙
FT =
10
But I = Ia/a
FT =
𝐙𝐁Ia 𝑙
10
Where:
FT = total force exerted, dynes
Z = total no. of armature conductor
Ia = total armature current
a = no. of parallel paths in the armature
Motor Developed Torque/ Armature Torque
Td = FT x r
𝐙𝐁Ia /𝑟
Td =
10a
dynes- cm
MULTIVECTOR REVIEW AND TRAINING CENTER
Where:
Td= developed torque, dyne – cm
r = average distance of Z conductors from the center of rotation,
cm.
Note:
445,000 dynes = 1lb
30.48 cm = 1 ft.
Motor Torque Relations
Pb = EbIa watts
But
Eb =
𝝓𝒙𝑷𝒙𝒁𝒙𝑵
𝟔𝟎 𝒙 𝒂
x 10-8 volts
Therefore
Eb =
𝝓𝒙𝑷𝒙𝒁𝒙𝑵
𝟔𝟎 𝒙 𝒂
x 10-8 x Ia volts
Also
2 T
Hpdev = π dN
33,000
Where:
Hpdev= developed horsepower in the armature, hp
Td = developed torque, lb – ft
N = speed of the armature, rpm
Note:
MULTIVECTOR REVIEW AND TRAINING CENTER
33,000 ft−lb
1 hp =
min
Pb= Hpdev (746)
Pb =
2π TdN
33,000
(746) watts
2
Equating 1 and 2
ϕxPxZxN
x 10-8x Ia
60 x a
2π TdN
33,000
(746)
33,000 x10−8 P x Z
Td =(60)(2 )(746)x
x 𝜙 x Ia
a
π
0.1173
PxZ
Td =( 8 ) (
)x 𝜙 x Ia lb – ft
10
a
Where: k = proportionally constant
0.1173
PxZ
K =( 8 ) (
)
10
a
Td = k𝜙Ia lb – ft
1) For a series motor,
At light loads, 𝜙 is directly proportional to Ia(𝜙 = klIa)
∴
Td= k (klIa) Ia
Td = k2Ia2
→ a parabola
At heavy loads, 𝜙 is practically constant(𝜙 = kl)
∴
Td= k (kl) Ia
Td = k2Ia
→ a straight line
2) For a shunt motor,
Shunt fields current (If) is practically constant for all conditions of
loading.
MULTIVECTOR REVIEW AND TRAINING CENTER
∴
𝜙 remains substantially constant also (𝜙 = kl)
Td= k (kl) Ia
Td = k2Ia
→ a straight line
Motor Speed Relations
For a given DC motor,
𝜙𝑥𝑃𝑥𝑍𝑥𝑁
Eb =
x 10-8 volts
60 𝑥𝑎
Where:
𝑃𝑥𝑍𝑥𝑁
k=
60 𝑥𝑎
k = proportionally constant
Eb = k𝜙N
N = Eb/k𝜙
Percent Speed Regulation
% = S.R. =
No−load speed−Full−load speed
Full−load speed
N
% = S.R. = NL
− NFL
NFL
x 100
x 100
Acceleration of DC Motors
At the instant of starting, Eb = 0 since N = 0
∴
Eb = Vt – Ia (Ra + Rbc)
0 = Vt – Ia ( Ra + Rbc)
Vt
Ia=
Ra + Rbc
From the above equation, since the armature circuit resistance (Ra + Rbc)
is very small, the armature tends to draw high and excessive value of armature
current. To prevent this, an external resistance (called starting resistance) must
be inserted in series with the armature. As the motor accelerates the starting
resistance may be cut gradually, because Eb rises, until the entire external
MULTIVECTOR REVIEW AND TRAINING CENTER
resistance is ultimately cut out (or short – circuited), under which condition
the motor is connected across the line and running at full speed.
Prony Brake – a device used to determine the output KW and output torque
of an electronic motor.
DC MACHINES – PART 3
EXERCISES:
REE – April 2001
1. 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. 95%
D. 94%
2. In a 5 – HP, 230 – volt, 1500 rpm shunt motor, the resistance of the
armature including brushes is 0.175 Ω, and that of the shunt field is
610 Ω. The stray power losses when the motor delivers rated load at
rated voltage are 305 watts. What is the efficiency of motor at the
rated load?
A. 85.32%
B. 89.32 %
C. 87.32 5
D. 91.32
%
3. A 5 – KW, 220 – V DC shunt motor has a full – load line current of
30 A. The armature circuit resistance is 0.3 Ω and the shunt field
resistance is 200 Ω. The constant loss for this motor is
A. 1,600 W B. 1,350 W
C. 250 W
D. 1,500 W
REE – May 2008
4. A 240 V shunt motor delivers 50 HP at the shaft of 1,200 rpm. The
motor has an efficiency of 85 % at this load. The torque loss due to
friction and wind age is 9% of the shaft torque. What is the motor
developed torque in N-m?
A. 296.8
B. 314.6
C. 307.4
D. 323.5
5. A compound motor develops a torque of 271 lb – ft when it is
operating at 1,200 rpm, under which condition its armature current Ia
MULTIVECTOR REVIEW AND TRAINING CENTER
is 215 amperes. What will be the developed horsepower of the motor
if the load increased so that it slows down to 1.120 rpm in which case
Ia changes to 238 amperes and the total flux increases by 8 percent?
A. 69.1
B. 61.8
C. 65.1
D. 67.8
6. A series motor develops 164 lb – ft torque when the current is 94
amp. If the load increases the current 50 percent with a flux increase
of 20 percent, what is the torque at the second value of current?
A. 225.9 lb –ft
B. 259.2 lb –ft
C. 229.5 lb – ft
D.
295.2 lb – ft
REE – Sept. 2006
7. Which of these is not a factor in determining the speed of a DC
motor?
A. Strength of armature circuit magnetic field
B. Strength of magnetic field produced by field coils
C. The load on the motor
D. The thickness of the commutator bars
REE – Sept. 2005
8. A DC shunt motor runs at 900 rpm from a 460 V supply when taking
an armature current of 25 amperes. Calculate the speed at which it
will run from a 230 V supply when taking an armature current of 15
amperes. The resistance of the armature circuit is 0.8 Ω. Assume the
flux produced per pole at 230 volts to have decreased to 75% of its
value at 460 volts.
A. 495 rpm B. 585 rpm
C. 485 rpm
D. 59 rpm
DC MACHINES – PART 3
REE – Sept. 2007
9. A shunt motor with an armature resistance of 0.15 ohm is running
1,200 rpm for a load which requires an armature current of 50 A from
a 230 V source. At no – load, the armature current is 5 A. if the effect
MULTIVECTOR REVIEW AND TRAINING CENTER
of armature reaction has reduced the air – gap flux 2% from no – load
to full –load speed?
A. 1,213 rpm B. 1,210 rpm
C. 1,208 rpm
D. 1,212
rpm
REE – Sept. 2006
10. A 240 V shunt motor driving its normal load draws an armature
current of 50 A from 240 V source. The armature resistance of this
machine, including brushes is 0.25 ohm. If the speed of operation is
1,200 rpm, what armature current will this machine draw from 220 V
source when driving the same load, with the field adjusted to maintain
a speed of 1,200 rpm
A. 50 A
B. 60 A
C. 57.5 A
D. 55 A
11. The no – load speed of a compound motor is 1,650 rpm. How fast
will it operate when it delivers rated horsepower output, if the speed
regulation is 12.5 percent?
A. 1,444 rpm B. 1,856 rpm
C. 1,467 rpm
D. 1,500
rpm
12. A 10 – HP, 220- V, DC shunt motor has an armature and field
resistances of 0.25 Ω and 100 Ω, respectively. The full – load
efficiency is 83 %. Determine the value of starting resistance in order
that the armature starting current will not exceed 200 percent of the
full – load value
A. 2.6 Ω
B. 3.6 Ω
C. 4.6 Ω
D. 5.6 Ω
13. In a brake test of an elevator door motor, the armature and
voltmeter measuring the input read 34 amperes and 220 volts,
respectively. The speed of the motor is found to be 910 rpm and the
balance on a 2 ft. brake arm reads 26.21 lb. The dead weight of the
arm is found to be 2.4 lb. The HP output of the door motor at this
particular load is
A. 9.91 HP B. 5.28 HP
C. 8.25 HP
D. 7.25 HP
MULTIVECTOR REVIEW AND TRAINING CENTER
AC MACHINES
Part 1
AC GENERATORS (ALTERNATORS)
GENERAL TYPES OF ALTERNATOR
1. Synchronous generator – its speed is called synchronous speed ant it is
used in almost all types of application.
2. Induction generator – (a synchronous generator) – it is an induction
motor which is run as a generator with a speed above the synchronous
speed. Its power factor is leading and usually connected in parallel with
a synchronous generator in order to supply lighting and power loads.
3. Inductor alternator – it generates voltages at higher frequencies (500
HZ to 10,000 HZ). It is used to supply power to induction furnaces in
order to heat and melt alloys and steel.
MULTIVECTOR REVIEW AND TRAINING CENTER
Two Possible Constructions of an Alternator
1. Stationary field and a revolving armature
Stator – as field
Rotor – as armature
2. Stationary armature and a revolving field
Stator – as armature
Rotor – as field
Note: The latter (2nd) arrangement is more preferable.
PRIME MOVERS FOR ALTERNATORS
A. For Large AC Generators
Generators
1. Steam turbine
engine
2. Hydraulic turbine
3. Gas turbine
4. Internal combustion engine
Frequency of General Voltage
f=
𝐏𝐍𝐒
𝟏𝟐𝟎
Where:
f = frequency, HZ or cps
P = no. of poles
NS = synchronous speed, rpm
Generated Voltage in an Alternator
𝛟
Eave = N x 10-8 volts
𝐭
Where:
Eave = average generated voltage, volts
B. For Small AC
1. Internal combustion
MULTIVECTOR REVIEW AND TRAINING CENTER
N – no. of turns in coil
ϕ = flux per pole, lines/poles or maxwells/ pole
t = time for flux to change by ϕ, seconds
Or
𝛟
Eave = N volts (if ϕ is in webers/pole)
𝐭
Eave = 4 f N 𝛟 x 10-8 volts
1
For sinusoidal flux – density distribution, effective voltage is
E = 1.11 Eave
Eave = 4.44 f N 𝛟 x 10-8 volts
For a 3ϕ alternator
𝐄𝛟 = 4.44 f 𝐍𝛟 𝛟 x 10-8 volts
Where:
Eϕ = generated voltage per phase
Nϕ = no. of turns per phase
2
3
Note: Equations 1, 2, and 3 are used for alternators having concentric
armature windings and full pitch coils.
ARMATURE WINDINGS FOR ALTERNATORS
Alternator windings are very similar to the dc lap windings. Coils in ac
armatures are joined together by merely connecting the proper coil ends in the
correct sequence.
Two (2) General Arrangements of Coils Employed:
1. Half – coiled – (single layer) – one coil side/slot
2. Whole – coiled – (double layer) – two coil side/ slot.
MULTIVECTOR REVIEW AND TRAINING CENTER
Note: The whole – coiled winding is the more usual of the two
arrangements.
Coil Pitch – the distance between the two sides of a coil. It is usually
express as a percent of full pitch.
Coil Pitch
Coil Sides
A. Full – Pitch coil – a coil having a distance between its two sides exactly
equal to 10 electrical degrees.
Note: for a full – pitch coil,
generated voltages in both coil
sides are exactly in phase.
180o
A. Fractional Pitch coil – a coil having a distance between its two sides
exactly equal to 10 electrical degrees.
Note: for a fractional pitch coil,
generated voltages in two (2) coil
sides are not in phase.
180o
MULTIVECTOR REVIEW AND TRAINING CENTER
Pitch Factor – the ratio of the voltage generated in the fractional – pitch coil
to the voltage generated in the full – pitch coil.
Kp = sin
𝐏𝐨
𝟐
Where:
Kp= pitch factor in decimal
Po = span of the coil in electrical degrees
For a full – pitch coil, Kp = 1 (unity)
For a fractional – pitch coil, Kp< 1
Distribution Factor – the factor by which the generated voltage E must be
multiplied because the coils are distributed in several slots under the poles
instead of being concentrated in single slots under the poles.
𝐝𝐨
Kd=
𝐬𝐢𝐧 𝐧( 𝟐 )
𝐝𝐨
𝐧 𝐬𝐢𝐧 ( 𝟐 )
Where:
Kd = distribution factor, decimal
n = no. of slots per pole per phase
do= no. of electrical degrees per slot
For concentric winding, Kd= 1 (unity)
For distributed winding, Kd< 1
Corrected Voltage of an Alternator
𝐄𝛟 = 4.44 Kp Kd f 𝐍𝛟 𝛟 x 10-8 volts
ALTERNATORS SCHEMATIC WIRING DIAGRAM
a
a
F.R.
+
MULTIVECTOR REVIEW AND TRAINING CENTER
DC source
(Exciter)
To 3𝜙
Load
b
b
c
DC Field
Winding
armature windings
EQUIVALENT CIRCUIT DIAGRAMS
c
1. For WYE connected armature windings:
a
a
c
IL
c
+
c
Iф
If
Zs
Vf
Rf
-c
+
~
Eф
Eф - = Eф
+ ~~ +
Zs
c
VL
VL
Zs
Iф
Iф b
c
VL = √3 Vф
IL
c
;
IL = Iф
IL
bc
VL
C
c
;
EL = √3 Eф
2. For DELTA connected armature windings:
+c
a
If
c
a
-
c
MULTIVECTOR REVIEW AND TRAINING CENTER
Zs
Vf
Eф
~+
IL
Rf
+ Iф
Iф
Eф ~
Iф
c
ZS
IL
-
C
-
c
VL
VL
Zs
~
+
Eф
c
b
-
bc
IL
VL
c
C
VL = Vф ;
IL = √3 Iф ;
EL = Eф
Per phase Equivalent Circuit Diagram:
(Wye or Delta Connected)
Eф = Vф + Ia Zs →added vector ally
Iф = Ia
Zs = Ra + j Xs
Zs
Xs = Xar + XL
Vф
L
+
Eф~
Where:
Eф = generated or open circuit or no – load voltage per phase
EL = generated or open circuit or no – load line to line voltage
Vф= terminal (output) voltage per phase
VL = line to line terminal voltage
Iф= Ia = armature current per phase
IL = line current
Ra = Re = effective or ac resistance of the armature per phase
Xs= synchronous reactance per phase
XL = armature leakage reactance per phase
Xar = reactance due to armature reaction per phase
Zs = synchronous impedance per phase
MULTIVECTOR REVIEW AND TRAINING CENTER
Rf = field resistance
If = field current
Vf = voltage drop across the field
Alternator Voltage Regulation
% V.R. =
VNL −VFL
VFL
x 100
Or
% V.R. =
Eϕ − Vϕ
Vϕ
x 100
Factors affecting the drop or rise of the alternator terminal voltage:
1. Magnitude of the load
2. Actual over – all power factor of the combined loads
Effect of various types of a load on the alternator terminal voltage:
1. Resistive loads (i.e. incandescent lamps, heating devices) or loads
with unity power factor.
-8% to 20% drop in terminal voltage below its no – load value
2. Inductive loads (i.e. induction motors, electrical welders, fluorescent
lighting) or loads with lagging power factor.
-25% to 50% drop in terminal voltage below the no - load value
3. Capacitive loads (i.e. capacitor devices or special types of
synchronous motor) or loads with leading power factor.
-tend to raise or increase the terminal voltage of the alternator above
the no – load value.
Three (3) factors responsible for the change in voltage of an alternator:
1. Resistance drop in the armature circuit (IaRa)
2. Change in the flux or armature – reaction effect (IaXar)
3. Armature reactance drop (IaXL)
Alternator Phasor (Vector) Diagrams:
a. For lagging power factor
b. For unity power factor
c.
For leading power factor
MULTIVECTOR REVIEW AND TRAINING CENTER
Eф
IaZs
Eф
Eф
IaXs Ia
IaXs
IaZs
IaZs
0
0
θ
IaXs
Vф
0
IaRa
θ
Vф
Ia
Vф
IaR
IaRa
Iф = Ia
θ=0
From the phasor diagrams:
1. In complex form, Eф= Vф+ IaZs
2. In magnitude, Eф= √(Vϕ cos θ + Ia R a )2 + (Vϕ sin θ ± Ia X s )2
Where:
θ = power factor angle of the load
= for lagging power factor
= for leading power factor
Three (3) tests necessary to perform in order to obtain data for the
calculation of the regulation an alternator:
1. Armature dc resistance test
2. Open –circuit or no – load test
3. Short – circuit test
The armature DC Resistance Test
With the DC field winding open, measure the DC resistance between
each pair of terminals. The average of the three sets of resistance values is
called Rt.
Adc
1
2
Open DC field wdg.
DC source
Vdc
MULTIVECTOR REVIEW AND TRAINING CENTER
3
Rt =
R12 + R23 + R31
3
; (i.e. R12 =
Vdc rdg
Adc rdg
High current rheostat
R
For wye – connected armature windings. Radc= t
2
3R
For delta – connected armature windings, Radc= t
2
To get the equivalent effective or AC resistance armature, use a factor of
1.25 to 1.75. Say,
Ra or Re = 1.25 Radc
The Open – Circuit Test or No – load Test
With the armature – winding circuit open, the alternator is driven at
synchronous speed. A dc source is connected to the field, making provision to
adjust the field current so that, starting at zero, it may be raised until the ac
voltage between any pair terminals of the armature winding is somewhat
above the rated voltage.
If
F.R
DC source
Open Armature
Windings
V~t
Prime Mover
AC Voltmeter
V
Open Circuit voltage per phase, Eoc = t
√3
MULTIVECTOR REVIEW AND TRAINING CENTER
The Short – Circuit Test
The alternator is driven at synchronous speed. Starting with a very low
direct field current, progressively increase its value as the ac ammeters
increase their deflections to rated current and above.
Highly Resistive
If F.R
A1
A2
DC source
armature
Shorted
Windings
Prime Move r
A3
AC Voltmeter
Average short circuit per phase, Isc =
A1 + A2 + A3
Synchronous impedance per phase, Zs =
3
Eoc
Isc
a. For wye – connected armature windings
Zs =
Eoc
Isc
=
𝑉 𝑡 / √3
Isc
b. For delta – connected armature windings
Zs =
Eoc
I sc
√3
=
Vt
I sc
√3
MULTIVECTOR REVIEW AND TRAINING CENTER
Synchronous reactance per phase, Xs = √(Zs )2 − (R a )2
ALTERNATOR EFFICIENCY:
P
Po x 100
η = 0 x 100 =
Pin
Po +Total losses
Alternator losses include:
1. Rational losses
a. Friction and wind age
b. Brush friction at the field collector rings (often neglected; quiet small)
c. Ventilation to cool th machine (if necessary)
d. Hysteresis and eddy currents in the stator
2. Electrical losses
a. Field winding
b. Armature winding
c. Brush contacts (often neglected; quiet small)
3. Losses in the exciter used for field excitation
4. Stray – load loss (negligible for small alternators)
OPERATION OF ALTERNATORS IN PARALLEL
Requisites for connecting alternators in parallel
1. Their voltages must be the same.
2. Their wave forms (frequencies) must be the same.
3. Their phase sequences must be the same.
If any of the above requisites is not met, there will be cross current
between the alternator windings thru the common bus bars. With the
cross – current, the alternator cannot be synchronized.
Alternators Under Faults
MULTIVECTOR REVIEW AND TRAINING CENTER
1. Three phase (3ϕ) fault at the alternator terminals (either wye or delta
connected alternator)
Note:
In a 3ϕ fault, the fault currents are symmetrical (balanced).
IscL = ILrated (
100
% Z1
)
If alternators resistance is neglected, then
IscL = ILrated (
100
% X1
)
2. Phase to phase fault at the alternator terminals (wye connected
alternators)
In a phase to phase fault, the fault currents are asymmetrical
(unbalanced). One fault current is zero. The other two fault currents will
have a magnitude of
100
√3 [ Iϕrated ( % X +%X )]→ alternator resistance is neglected
1
2
3. Line to ground or phase to ground fault (wye connected alternator)
In a line to ground fault, the fault currents are asymmetrical
(unbalanced). Two fault currents are zero. The remaining one will have a
magnitude of
100
√3 [ Iϕrated ( % X +%X
1
2+%𝑋3
)] → alternator resistance and neutral
wire impedance re neglected.
4. Double to ground fault (wye connected alternator)
In a double line to ground fault, the fault currents are asymmetrical
(unbalanced). One fault current is zero. The other two fault currents will
have a magnitude of
1
Iϕrated [
%X %X2 %X0
1
%X2 +%X0
]
MULTIVECTOR REVIEW AND TRAINING CENTER
Where:
IscL = Three phase fault line current
ILrated = rated line current of the alternator
Iϕrated = rated phase current of the alternator
%Z1 = percent positive phase sequence impedance
%x1 = percent positive phase sequence reactance
%X2 = percent negative phase sequence reactance
%X0 = percent Zero phase sequence reactance
AC MACHINES PART 1 - ALTERNATORS
EXERCISES:
1. The no. of cycles generated in a 10 pole alternator in one revolution is
_______________.
A. 2
B. 5
C. 10
D. 20
2. What I 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
3. In huge alternators, the moving part is
A. Brushes
B. armature
these
C. poles
D. none of
4. Alternators are usually designed to generate
A. Variable frequency B. definite currents C. definite frequencies
D. definite p.f
5. 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 sinusoid ally
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
6. In a 3 – phase, 4 – pole, star connected alternator the e.m.f per phase is
1,000 V at a frequency of 50 Hz. Calculate the speed and line voltage.
MULTIVECTOR REVIEW AND TRAINING CENTER
A. 750 rpm, 3000V
B. 1500 rpm, 1732V C. 300 rpm, 1414V
D.1500 rpm, 2000V
REE – May 2008
7. 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. 6.5mWb
C. 5t.2 mWb
D.
7.8 mWb
REE – April 2006
8. A 3 – phase, 8 – pole generator is delta connected. The terminal voltage
is 2,400 V 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
REE – Sept. 2005
9. A 3 – phase, 8 – pole, 2,400- volt 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
10.
If the alternator winding has a fractional pitch of 4⁄5, the pitch
factor kp is ____________.
A. 0.833
B. 0.966
C. 0.972
D. 0.951
11.
A six pole, three phase alternator has 72 slots and a coil span of 1
to 10, what is the Fitch factor?
A. 0.924
B. 0.939
C. 0.966
D.0.985
12.
Calculate the distribution factor for a 36 – slot 4 pole three phase
winding?
A. 0.924
B. 0.951
C. 0.960
D. 0.975
MULTIVECTOR REVIEW AND TRAINING CENTER
13.
The voltage generated across one inductor of a three phase
alternator is 6 volts. There are 120 slots and 4 turns in every coil pitch is
short by 2 slots, number of poles is 10 and frequency is 60 Hz. The
armature winding correction factor of the alternators is nearest to
______________.
A. 0.957
B. 0.925
C. 0.987
D. 0.967
AC MACHINES PART 1 – ALTERNATORS
14.
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 & 4
B. 24 & 4
C. 36 & 4
D. 48 & 4
15.
The following information is given in connection with an
alternator; slots = 144; poles = 8; rpm = 900; turns per coil = 6; ф = 1.8
x 106; coil span = slots 1 to 16; winding is whole – coiled three – phase;
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
16.
A part of an alternator winding consists of six coils in series, each
coil having an e.m.f of 10 V (r.m.s.) induced in it. The coils are placed in
successive slots and between each slot and the next, there is an electrical
phase displacement of 30o. 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
17.
A 4 – pole alternator on open circuit, generates 200 V at 50 Hz
when the field current is 4 A. Determine the generated e.m.f. at a speed
of 1200 r.p.m. and a field current of 3 A, neglecting saturation in the
iron parts.
A. 40 V
B. 240 V
C. 60 V
D. 120 V
18.
A 3 ф, wye connected alternator having a sinusoidal line potential
with a 5% 5th harmonic content, 3% 7th harmonic content, 1.5% 11th
harmonic content and 0.85% 13th harmonic content. What is the total
harmonic distortion (THD) of the system line voltage?
MULTIVECTOR REVIEW AND TRAINING CENTER
A. 11%
B. 9%
C. 6%
D. 3%
19.
A 100 MVA, 13.8 kV, 3 ф, Y – connected alternator will have a
per phase nominal impedance of ___________________.
A. 3.5 Ω
B. 2.9 Ω
C. 1.9 Ω
D. 7.5 Ω
REE – March 1998
20.
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 DC 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
REE – April 2006
21.
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
REE – Sept. 2004
22.
A 3,600V 500 - KVA, 60 HZ, 3 – phase Y – connected generator
on test yields the following results: Mechanical and iron losses are 10
KW; field current at full – load 10% p.f. is 50 A; field current at fullload 80% p.f. is 70 A; 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
AC MACHINES PART 1 – ALTERNATORS
REE – April 2006
MULTIVECTOR REVIEW AND TRAINING CENTER
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 – April 2005
24.
A single – phase 2,400 – volt 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
REE – May 2005
25.
A three – phase wye connected wound rotor synchronous generator
rated at 20 KVA, 240 V has a synchronous reactance of 1.5 ohms/ phase
and an armature resistance of 0.6 ohm/phase. What is the percent voltage
regulation at full – load with 80% lagging p.f.?
A. 38
B. 34
C. 47
D. 51
26.
A 3 – phase, star – connected alternator is rated at 2,000 kVA, 13.5
kV. The armature effective resistance and synchronous reactance are 1.3
Ω and 20 Ω respectively per phase. Calculate the percentage regulation
at full – load for power factors of 0.8 lagging and 0.8 leading.
A. 7.76%, - 5.06%
B. 15.52 %, -10.12% C. 2.91%, - 2.1%
D.
6.56%, - 4.3%
27.
A 1,000 kVA, 3,000 V, 50 Hz, 3 – ф 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 1040 V (line value). Calculate the full- load percentage regulation at a
power factor of 0.8 lagging.
A. – 21.43%
B. 21.43%
C. -24.31%
D. 24.31%
28.
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 it’s percent
efficiency.
A. 90.48%
B. 92.32%
C. 91.58%
D. 89.12%
MULTIVECTOR REVIEW AND TRAINING CENTER
29.
A 25 kVA, 220 volt 3 ф alternator delivers rated kVA at a power
factor of 0.84. The effective ac resistance between armature winding
terminals I 0.18 Ω. The field takes 9.3 amp. At 115 volts. If friction and
wind age loss is 460 watts and the core loss is 610 watts. Calculate the
percent efficiency.
A. 87.82%
B. 82.27%
C. 86.41%
D. 88.33%
30.
Adjustment of field excitation on one of the two alternators
operating in parallel will
A. Increase its load
B. change its frequency
C . decrease its load
D. change its power factor
31.
Two alternators running in parallel supply the following loads:
(i) 1500 kW at 0.9 lagging p.f.
(ii) 1000 kW at 0.8 lagging
p.f.
( iii)
800 kW at unity p.f.
(iv) 600 kW at 0.8 leading
p.f.
The load on one machine is adjusted to 2,000 kW at p.f. 0.95 lagging. Find
the load and p.f of the other machine.
A. 1900 kW, 0.98 lagging
B. 950 kW, 0.96 lagging
C . 1900 kW, 0.98 leading
D. 950 kW, 0.96 leading
AC MACHINES PART 1 – ALTERNATORS
32.
Two exactly similar turbo- alternators are rated 20 MW each. They
are running parallel. The speed – load characteristics of the driving
turbines are such that the frequency of alternators one drop 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
MULTIVECTOR REVIEW AND TRAINING CENTER
REE – Sep.2001
33.
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
34.
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 displayed 30o from a position of
phase opposition.
A. 915 A
B. 1,830 A
C. 457.4 A
D. 119A
35.
In prob. No. 34, when the voltage 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
36.
A 600 kVA, 2,400 V, 3ф alternator has a zero phase sequence
reactance of 12% and a positive and negative phase sequence
reactance’s of 8%. If the alternator resistance is negligible, determine the
fault currents that the alternator can sustain if a 3ф occurs at its
terminals.
A. 1,042 A
B. 1,804 A
C. 3,125 A
D. 602 A
37.
In Prob. No. 36, determine the fault currents delivered if a phase
fault occurs at the alternator terminals.
A. 1,562 A
B. 902 A
C. 1,804 A
D. 3,125 A
MULTIVECTOR REVIEW AND TRAINING CENTER
38.
A 3ф 11 kV, 10 MVA alternator has a sequence reactance’s of
positive sequence, X1 = j0.15 p.u., negative sequence, X2 = j0.15 p.u.,
and zero sequence, X0 = j0.05 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. 4499 A, 3499 A B. 2814 A, 2187 A C. 4499 A, 3030 A D.
1842 A, 854 A
AC MACHINES
PART 2
TRANSFORMER
Transformer – it is an AC device that transfers power from one circuit to
another without a rotating part and change of frequency.
Transformer construction
1. Core Type – the coil surround a considerable part of the magnetic
circuit (core).
Laminations
Coils
MULTIVECTOR REVIEW AND TRAINING CENTER
2. Shell Type – the magnetic circuit (core) surrounds a considerable
portion of the windings.
Laminations
Primary and
Secondary Coils
Elementary Diagram of a simple transformer
Laminated silicon
Steel sheets (Core)
Primary Windings
Mutual Flux (Useful
Flux)
-flux that links the primary
And secondary windings
Primary Circuit
± ~
AC
Circuit
Secondary
Circuit
L
MULTIVECTOR REVIEW AND TRAINING CENTER
Secondary leakage flux
(wasted Flux)
-it links the secondary turns
Primary leakage flux (Wasted Flux)
-it links the primary turns only.
TRANSFORMER VOLTAGES
EP = 4.44 f NPфm (10-8)volts
8
)volts
Secondary Windings
ES = 4.44 f NSфm (10-
Where:
EP = primary induced voltage
фm = BmAn
ES = secondary induced voltage
Bm = maximum flux
density, line/cm2 or
f – frequency, Hz
Maxwell’s/cm2
NP= number of primary runs
An = core net cross –
2
section: cm
NS = number of secondary turns
фm = maximum value of the mutual flux, lines or maxwells
Voltage and Current Ratios in Transformers
𝐄𝐩
𝐄𝐬
=
𝐍𝐩
𝐍𝐬
=a
Where:
a =turn ratio or the induced voltage ratio,
also called the ratio of transformation
if there is no – loss (for ideal transformer only).
Input (primary) power = output (secondary) power
VPIP (p.f.p) = VSIS (p.f.s)
Where
p.f.p = p.f.s
∴VPIP= VSIS
MULTIVECTOR REVIEW AND TRAINING CENTER
𝐕𝐩
𝐕𝐬
𝐈
= 𝐒 = a → approximate ratio of transformation
𝐈𝐩
Where:
VP = primary line (impressed) voltage
VS = secondary line (impressed) voltage
IP = primary line current
IS = secondary line current
p.f.p= primary power factor
p.f.S = secondary power factor
rp
XP
IP
INIb
Iф
BO GO
IS
L
ES
VS
VP
EP
Iє
Transformer Exact Equivalent Circuit Diagram
EpIb = EsIs
𝐄𝐩
𝐄𝐬
=
𝐈𝐒
𝐈𝐛
Is = aIb
= a→ exact ratio of transformation
Where:
IN = no – load current
Iф = magnetizing current
I
Ib s
a
MULTIVECTOR REVIEW AND TRAINING CENTER
Iε = no – load energy current
Ib = component of IP that supplies IS
rp = primary resistance
rs = secondary resistance
XP = primary leakage reactance (due to primary leakage flux)
Xs = secondary leakage reactance (due to secondary leakage flux)
rp
a2rs
Xp
a2Xs
IS
Ip
Ib
L
Vp
Iф
B0G0
Iє
aVs
Transformer exact equivalent ckt. Diagram referred to the
primary
Note: IN<< IP, hence IP≈ Ib
rp
XP a2rs a2xs
Ze-p
Re-p Xe-p
IP
IP
L
VP
L
aVs VP
aVs
MULTIVECTOR REVIEW AND TRAINING CENTER
Transformer approximate equivalent circuit diagram referred to the
primary. (no- load current IS is neglected).
R e-p = rp + a2rs
X e-p = Xp + a2rs
Ze-p =
√𝐑 𝐞−𝐩𝟐 + 𝐗 𝐞−𝐩𝟐
In complex form,
Ze-p = Re-p + j Xe-p
VP = aVS + IP Ze-p
VP = √(𝐚𝐕𝐬 𝐜𝐨𝐬𝛉 + 𝐈𝐩 𝐑 𝐞−𝐩 )𝟐 + (𝐚𝐕𝐬 𝐬𝐢𝐧𝛉 +
In magnitude,
𝐈𝐩 𝐗 𝐞−𝐩 )𝟐
Where:
Re-p = total or equivalent resistance of the transformer referred to the
primary.
Xe-p = total or equivalent reactance of the transformer referred to the
primary.
Ze-p = total or equivalent impedance of the transformer referred to the
primary.
VP = no – load primary line voltage
θ= p.f. angle of the load
= for lagging p.f.
= for leading p.f.
Ze-s
𝑟𝑝
𝑋𝑝
𝑎2
𝑎2
rs
xs
Re-s Xe-s
IP
IP
L
𝑉𝑝
𝑎
L
Vs
𝑉𝑝
𝑎
Vs
MULTIVECTOR REVIEW AND TRAINING CENTER
Transformer approximate equivalent circuit diagram referred to the
secondary. (no- load current IS is neglected).
R e-s =
𝑟𝑝
𝑎
2 + rs
X e-s =
𝑋𝑝
𝑎2
+ rs
Ze-s =
√𝐑 𝐞−𝐬𝟐 + 𝐗 𝐞−𝐬𝟐
In complex form,
𝑉𝑝
𝑎
Ze-s = Re-s + j Xe-s
= VS + IS Ze-S
In magnitude,
𝑉𝑝
𝑎
= √(𝐕𝐬 𝐜𝐨𝐬𝛉 + 𝐈𝐬 𝐑 𝐞−𝐬 )𝟐 + (𝐕𝐬 𝐬𝐢𝐧𝛉 + 𝐈𝐬 𝐗 𝐞−𝐬 )𝟐
Where:
Re-s = total or equivalent resistance of the transformer referred to the
secondary.
Xe-s = total or equivalent reactance of the transformer referred to the
secondary.
Ze-s = total or equivalent impedance of the transformer referred to the
secondary.
𝑉𝑝
= no – load secondary line voltage
𝑎
Note the fallowing relationships:
R e-p = a2 Re-s
X e-p = a2 Xe-s
Z e-p = a2 Ze-s
Transformer Regulations
𝐕 − 𝐕𝐅𝐋
%V.R = 𝐍𝐋
𝐕𝐅𝐋
x 100
a. As referred to the primary,
secondary
b. as referred to the
MULTIVECTOR REVIEW AND TRAINING CENTER
𝑽𝒑
|𝐕𝐩 |−|𝐚𝐕𝐬 𝐫𝐚𝐭𝐞𝐝 |
%V.R =
%V.R =
𝐚𝐕𝐬 𝐫𝐚𝐭𝐞𝐝
| 𝒂 |−|𝐕𝐬 𝐫𝐚𝐭𝐞𝐝 |
𝐚𝐕𝐬 𝐫𝐚𝐭𝐞𝐝
Note: The transformer regulation must be the same whether referred to the
primary or to secondary.
Transformer Phasor Diagrams:
(As referred to the primary)
a. For Lagging power factor b. For Leading power factor
Leading power factor
c. For
IPXep
VP
Vp
IP
IpZe-p
IpXe-p
IPRe-p
0
IPZe-p
IPXe-p
0
Θ
VP
θ
IPZe-p
0
aVs
IP
aVS IPRe-p
aVS
IpRe-p
0 = 0o
Percentage, Resistance, Reactance and Impedance
These quantities are usually measured by the voltage drop at full – load
current expressed as a percentage of the normal voltage of the windings on
which calculations are made.
a. Percent resistance
%R =
%R =
𝐈𝐩 𝐫𝐚𝐭𝐞𝐝 𝐑 𝐞−𝐩
𝐕𝐩𝐫𝐚𝐭𝐞𝐝
x 100 =
𝐑𝐚𝐭𝐞𝐝 𝐂𝐮 𝐋𝐨𝐬𝐬 𝐢𝐧 𝐤𝐖
𝐫𝐚𝐭𝐞𝐝 𝐊𝐕𝐀
𝐈𝐬 𝐫𝐚𝐭𝐞𝐝 𝐑 𝐞−𝐩
𝐕𝐬 𝐫𝐚𝐭𝐞𝐝
x 100
= % Cu. Loss
MULTIVECTOR REVIEW AND TRAINING CENTER
b. Percent Resistance
c.
%X =
𝐈𝐩 𝐫𝐚𝐭𝐞𝐝 𝐗 𝐞−𝐩
𝐕𝐩𝐫𝐚𝐭𝐞𝐝
x 100 =
𝐈𝐬 𝐫𝐚𝐭𝐞𝐝 𝐗 𝐞−𝐩
𝐕𝐬 𝐫𝐚𝐭𝐞𝐝
x 100
d. Percent Impedance
%Z =
𝐈𝐩 𝐫𝐚𝐭𝐞𝐝 𝐙𝐞−𝐩
𝐕𝐩𝐫𝐚𝐭𝐞𝐝
x 100 =
𝐈𝐬 𝐫𝐚𝐭𝐞𝐝 𝐙𝐞−𝐩
𝐕𝐬 𝐫𝐚𝐭𝐞𝐝
x 100
%Z = √(%𝐑)𝟐 + (%𝐗)𝟐
Transformer Losses
1. Core (Iron) loss – constant loss, its value is always rated
𝟏.𝟔
𝐄
a. Hysteresis loss, Ph = kh f Bm1.6 = k1( 𝟎.𝟔 )
2
𝐟
Bm2= k2 E2
b. Eddy current loss, Pe = kh f
Core loss, Pc = Ph + Pe
2. Copper (electrical loss) – varies as the square of the load
a. Primary copper loss = Ip2rp
copper loss, Pcu = Ip2
rp + Is2rs
b. Secondary copper loss = Is2rs
copper loss, Pcu = Ip2 Re-p
copper loss, Pcu = Is2 Re-s
Note:
1. If the transformer is connected to the source, whether loaded or not
loaded, there is core loss in the transformer
2. If the transformer is loaded, there is copper loss in the transformer. At
no – load, copper loss is negligible.
𝐆𝐢𝐯𝐞𝐧 𝐥𝐨𝐚𝐝 2
Copper loss at any given = (
) (Rated or full – load copper loss)
𝐅𝐮𝐥𝐥 𝐥𝐨𝐚𝐝
Where given load & full – load must be apparent powers.
Transformer Efficiency
MULTIVECTOR REVIEW AND TRAINING CENTER
η=
Po
η=
x 100
Pin
Po x 100
Po x 100
η=
Po +total Losses
Po +copperl Losses+core loss
Condition for maximum Efficiency
Copper Loss = core loss
If IP’= primary current at maximum efficiency, then
IP’2 Re-p = core loss
IP’ = √
core loss
Re−p
Similarly, the secondary current at maximum efficiency is
IP’ = √
core loss
Re−s
KVA load at 𝛈max = full – load KVA = √
𝐜𝐨𝐫𝐞 𝐥𝐨𝐬𝐬
𝐅𝐮𝐥𝐥 𝐥𝐨𝐚𝐝 𝐂𝐮 𝐥𝐨𝐬𝐬
All – day Efficiency (also known as “ Operational Efficiency”)
It is the ratio of the energy (kilowatt – hours) delivered by the
transformer in a 24 – hour period to the energy input in the same period of
time.
η=
Output energy/dayx 100
energy
Output day +Energy losses/day
This efficiency is always less than the commercial efficiency of a transformer.
TRANSFORMER TESTS
1. The open – circuit or no – load Test
The purpose of this test is to determine the no – load loss or core loss
and no – load current IN which is helpful in finding GO and BO.
Voltage Regulator
Adjust voltage across
Low Side Accurately
to
W
±±
V
MULTIVECTOR REVIEW AND TRAINING CENTER
To Low voltage
High Voltage
AC source
Low voltage
windings
Open
Windings
Open ckt. Test ckt. Diagram
Wrdg = Rated core loss + Negligible copper loss
Wrdg = Rated core loss
2. The short circuit or Impedance Test
This is an economical method for determining the following:
a. Equivalent impedance (Ze-p or Z e-s), leakage reactance (Xe-p or X e-s)
and total resistance (Re-p or R e-s) of the transformer as referred to the
winding in which the measuring instruments are placed.
b. Rated or full- load copper loss (and at any desired load). This loss is
used in calculating the efficiency of the transformer.
c. Knowing Ze-p or Z e-s, the total voltage drop in the transformer as
referred to primary or secondary can be calculated and hence
regulation of the transformer is determined.
Adjusting Rheostat
±±
W
Irated
A
V
Low voltage
Low Voltage
To Low voltage
Windings
AC Source
windings
Short
MULTIVECTOR REVIEW AND TRAINING CENTER
High voltage windings
Short ckt. Test ckt. Diagram
Wrdg = rated copper loss + negligible core loss
∴ Wrdg= Rated Copper loss, Irated2Re
Ze =
Ardg = Irated
Vrdg = Irated Ze;
Vrdg
Ardg
Re =
Wrdg
Ze = √Ze2 − R2e
(Ardg )2
Transformer Priority
-it refers to the relative directions of the induced voltages in the primary and
secondary windings of a transformer with respect how the terminal leads are
brought out and marked accordingly.
In practiced, transformers are marked as shown:
H1
±
X1
±
H1
X1
a. Subtractive polarity marking
marking
H1
±
H1
Where:
H1&H2 are the high voltage side terminals
instantaneous polarity
X1&X2 are the low voltage side terminals
instantaneous polarity
X2
±
X2
b. Additive polarity
H1&X1 have the same
H2&X2 have the same
MULTIVECTOR REVIEW AND TRAINING CENTER
Voltmeter Test to determine Transformer Polarity
a. if Vrdg> V1, then polarity is
additive.
V
Vrdg = V1 + 1
a
impressed
Voltage
b. if Vrdg< V1, then polarity is
subtractive
V
Vrdg = V1 + 1
a
V
Why does the polarity of a transformer must be known?
1. In order to know what terminals are to be connected together if windings
are to be connected in series or in parallel.
2. In order to know what terminals are to be connected together if
transformers are to be connected in bank or in group.
Autotransformer – it is a transformer with one winding only, part of this
being common to both primary and secondary. Obviously, in this transformer
the primary and secondary windings are electrically connected to each other.
Three(3) Principal Types of autotransformers
1. Step down autotransformer
2. Step – up autotransformer
3. Variable autotransformer
Step – down Autotransformer(a>1)
𝐕𝐩
𝐍
𝐄
𝐈
a. 𝐚𝐜
= 𝐚𝐜 = 𝐬 =
𝐍𝐜𝐛
a
IP
VP
b
IS
IcbVS
L
𝐄𝐜𝐛
𝐈𝐩
𝐕𝐬
MULTIVECTOR REVIEW AND TRAINING CENTER
c
The autotransformer acts exactly like a two – winding transformer if
from the standpoint of transformer action, it is considered that the portion of
the windings ab is the primary and the portion of the windings bc is the
secondary. To prove this
Nab
Nbc
=
= cb =
I
Is −Ip
Ip
Ip
Re-p
Xe-
Eab
Ebc
I
= s –1=a–1
IP
Ip
Re-p
L
VP
Xe-p
IP
(a – 1)Vs
𝑉𝑝
L
Approximate equivalent circuit diagram
equivalent circuit diagram
Referred to the primary
secondary.
Re-p = rp + (a – 1)2 rs
Approximate
referred to the
Re-s =
Xe-p = Xp + (a – 1)2 Xs
VS
(𝑎−1)
𝐫𝐩
(𝐚−𝟏)𝟐
+ rS
Xe-s =
𝐗𝐩
(𝐚−𝟏)𝟐
+
XS
Power transformed inductively (thru ab)
inductively (thru bc)
Ptrans = (VP - VS)IP = VS (Is - IP)
1
Ptrans = VPIP (1 - )
a
the secondary
1
Ptrans = Power Input (1 - )
a
IS
Power transformed
1
Pcon = Pm(1 - )
a
Total Power transferred to
PS = Pcom + Ptrans
MULTIVECTOR REVIEW AND TRAINING CENTER
Nab
a
Nbc
Nab
b
V
VP IP
Nbc
VS
=
=
Eab
Ebc
Eab
Ebc
I
= s
Ip
Is
=
Ibc
1
=
Ip −Is
Is
I𝑠
Is −Ip
1
=1
𝑎
−1
=
1
1−𝑎
Ibc
C
Variable Autotransformer
IS
IP
VP
IC
movable
brush
L
prepared VS
Winding
surface
Instrument Transformer
Two (2) kinds of instrument transformers
1. Current transformer(CT) – it is a special transformer used in
conjunction with ordinary AC measuring instruments to measure high
current, power at high current and energy at high current. Its
secondary is almost always designed for 5 A maximum.
2. Potential Transformer (PT) – it is a special transformer used in
conjunction with ordinary ac measuring instruments to measure high
voltage, power at high voltage and energy at high voltage. The
transformation ratios are usually designed for a nominal secondary
voltage of 120 V.
Note: Loads of instruments transformers are called ”burdens”.
Parallel Operation of Transformers
MULTIVECTOR REVIEW AND TRAINING CENTER
Conditions to be met for ideal operation
I– of two (2) or more single (1ф) transformers in parallel:
1. Same voltage ratings (same ratios)
2. Same polarities
1
1
3. Ze-1∝
, Ze-2∝
, etc.
Re−1
Xe−1
rated KVA1
Re−2 Re−3
=
Xe−2
=
Xe−3
rated KVA2
= etc.
If only two (2) transformers are connected in parallel, then
𝐙𝐞−𝟏 𝐑𝐚𝐭𝐞𝐝 𝐊𝐕𝐀𝟐
=
𝐙𝐞−𝟐
𝐑𝐚𝐭𝐞𝐝 𝐊𝐕𝐀𝟏
II- of two (2) or more three phase (3ф) transformers in parallel:
1.
2.
same conditions in single phase (1ф) transformers in parallel.
3.
4. Same windings connections (i.e. Y – Y to be connected in parallel
with Y – Y).
Exception: ∆ − ∆ can be connected in parallel with Y – Y.
Effects / Consequences if these conditions are not met.
If condition 1 is not met (1ф or 3ф). There will be current that will
circulate thru the transformer secondaries which will contribute to the
transformer losses.
If condition 2 is not met (1ф or 3ф). There will be big current that will
circulate thru the transformer secondaries which may damage the windings.
If condition 3 is not met (1ф or 3ф). The transformers will not share the
common load in proportion to their KVA ratings.
If condition 4 is not met (3ф). There will be current that will circulate
thru the transformer which will contribute to the transformer losses.
No – load Operation
For two (2) transformers in parallel.
Let a1 = the ratio of transformation of transformer 1
MULTIVECTOR REVIEW AND TRAINING CENTER
a2 = the ratio of transformation of transformer 2.
As referred to their secondaries at no – load, the circulating current is
a. In complex form.
(approximate)
IC =
b. In magnitude
Vs1 −Vs2
IC =
Ze−s1 +Ze−s2
(a1 −a2 )Vs
a1 Ze−s1 +a2 Ze−s2
Load Operation – Equal ratios of Transformation
Ze−s1
IS1
IS2
ILT
Ze−s2
𝐕𝐩
L
𝐚
VS
Equivalent circuit of two transformers as referred to their secondaries,
having equal ratios of transformation. Connected in parallel and supplying
power to a common load.
𝐊𝐕𝐀 𝟏
𝐊𝐕𝐀 𝟐
=
𝐈
𝐙 𝐞−𝐬𝟏
𝐈𝐬𝟐
𝐙 𝐞−𝐬𝟐
; 𝐬𝟏 =
IS1 Z e-s1 = IS2 Z e-s2
𝐙 𝐞−𝐬𝟏
;
𝐙 𝐞−𝐬𝟐
Load current of each transformer,
IS1 = ID1(
𝑍𝑒−𝑠2
𝑍𝑒−𝑠1 + 𝑍𝑒−𝑠2
)
KVA load of each transformer,
IS2 = ID1(
𝑍𝑒−𝑠1
𝑍𝑒−𝑠1 + 𝑍𝑒−𝑠2
)
MULTIVECTOR REVIEW AND TRAINING CENTER
S1 = ID1(
𝑍𝑒−𝑠2
𝑍𝑒−𝑠1 + 𝑍𝑒−𝑠2
)
S 2 = ST (
𝑍𝑒−𝑠1
𝑍𝑒−𝑠1 + 𝑍𝑒−𝑠2
Where: IDT = total load current
ST = total KVA of the common load
Load Operation – Unequal Ratios of Transformation
Load current of each transformer
a. In complex form,
Es1 Ze−s2 +(Es1 − Es2 )ZL
IS1 =
;
Ze−s1 Ze−s2 + ZL (Ze−s1 + Ze−s2 )
Es2 Ze−s1 +(Es1 − Es2 )ZL
IS2 =
Ze−s1 Ze−s2 + ZL (Ze−s1 + Ze−s2 )
Where:
Es1 = secondary induced voltage of transformer 1.
Es2 = secondary induced voltage of transformer 2.
ZL = load impedance
b. In magnitude,
IS1 =
(a1 −a2 )Vs (a1 Ze−s1 IDT )
(a2 −a1 )Vs (a2 Ze−s2 IDT )
a1 Ze−s1 + a2 Ze−s2
;
IS2 =
a1 Ze−s1 + a2 Ze−s2
THREE PHASE (3ф) TRANSFORMERS
Three Phase (3ф) Transformer Ratings
1. The KVA rating of a 3ф transformer is a 3фKVA.
2. The voltage rating 3ф transformers are line to line voltages.
)
MULTIVECTOR REVIEW AND TRAINING CENTER
3. The percent impedance (%Z) or equivalent impedance (ZC) of a 3ф
transformer is phase.
Three Phase Transformer connections
There are various methods for transforming 3 – phase voltages to higher
or lower 3- phase voltages. The most common connections are
1. Y – Y – it has the advantage that the insulation is stressed only to the
extent of the line to neutral voltage, which is 57.7% of the line
voltage.
2. ∆ − ∆ - This arrangement is generally used in systems of moderate
voltages but high current.
3. ∆ −Y – it is generally employed to step up the voltage, as, for
example, at the beginning of a high – tension transmission system. It
is also used for stepping down the voltage. The ratio of the line to line
voltage, high to low, is 1.73a.
4. Y - ∆ - it is used primarily for steeping down the voltage, as, for
example, at the end of a transmission line.
5. V – V or Open Delta- it is employed
a. When the three – phase load is comparatively small, so that the
installation does not warrant a ∆ − ∆ bank.
b. When one of the transformers in a ∆ − ∆ bank fails.
c. When it is anticipated that the future load will increase to
warrant the closing of the open ∆ at some later time.
Note: Total capacity of the V – V (open delta) bank is 57.7% of the
∆ − ∆ bank
6. Three – Phase Scott Tee or T – T – used to transform power from
a. Three – phase to three – phase
b. Three – phase to two – phase and vice versa
This connection used two special transformers, the main transformers
whose primary and secondary windings are provided with the center (50%)
taps and the teaser transformer whose windings are provided with 86.6%
taps.
MULTIVECTOR REVIEW AND TRAINING CENTER
Note: Total Capacity of the two Scott – connected transformers is also
57.7% of the ∆ − ∆ bank.
MULTIVECTOR REVIEW AND TRAINING CENTER
AC MACHINES PART 2 – TRANSFORMERS
EXERCISES:
1. The high – voltage coil of a transformer is wound with 700 turns of wire,
and the low – voltage coil is wound with 292 turns. When used as a step
– up transformer (the low – voltage coil used as the primary), the load
current is 10.5 A. Find the component of the primary current.
A. 43.5 A
B. 4.38 A
C. 25.18 A
D.2.518 A
REE – May 2008
2. A transformer has a primary winding of 2,000 and of 2,400 v and a
current of 8.66 – j5 A with impedance Z2 connected across the
secondary winding. If the secondary winding has 500 turns, what is the
value of the secondary current?
A. 20 – j34.64 A
B. 34.64 – j20 A C. 34.64 + j20 AD. 20 + j34.64
A
3. A 4,600/ 230 V, 60 Hz step –down transformer has core dimension of
76.2 mm by 111.8 mm. A maximum flux density of 0.93 Wb/ m2 is to be
used. Assuming 9 percent loss of area due to stacking factor of
laminations, calculate the primary and secondary turns required.
A. 2,395 & 120
B. 120 & 2,395
C. 2,180 & 109
D.
109 & 2,180
4. A 120 V to 27.5 V, 400 Hz step – down transformer is to be operated at
60 Hz. What is the highest safe input voltage?
A. 220V
B. 400 V
C. 120 V
D. 18 V
REE – Sept. 2006
5. The primary of transformer has 200 turns and is excited by a 240 V, 60
Hz. What is the maximum value of the core flux?
MULTIVECTOR REVIEW AND TRAINING CENTER
A. 4.04 mWb
D. 4.32 mWb
B. 4.40 mWb
C. 4.13 mWb
REE – Sept. 2008
6. A transformer is rated 1 kVA, 220/110 V, 60 Hz. Because of an
emergency this transformer has to be used on a 50 Hz system. If the flux
density in the transformer core is to be kept the same as at 60 Hz and
220 V, what is the kilovolt – ampere rating at 50 Hz?
A. 0.890 kVA
B. 0.833 kVA
C. 0.909 kVA
D. 0.871 kVA
7. A 1∅ transformer has a no – load power input of 250 watts, when
supplied at 250 – volt; 50 Hz has a p.f. of 0.25. What is the magnetizing
component of the no – load current?
A. 4.00 A
B. 3.87 A
C. 1.00 A
D. none of these
REE – Oct. 1997
8. A small single – phase transformer has 10.2 watts no –load loss. The
core has a volume of 750 cu. Cm. The maximum flux density is 10,000
gauss and the hysteresis constant of the core is 5 x 10 -4. Using the
Steinmetz law to find the hysteresis, determine the eddy current loss.
A. 4.55 watts
B. 5.55 watts
C. 3.55 watts
D. 2.55 watts
9. The eddy – current loss in a 2,300 volt, 60 – cycle transformer is 280
watts. What will be this loss if the transformer is connected
i)
To a 2,300 – volt 50 – cycle source?
A. 233 W
B. 280 W
C. 194 W
D. 403 W
ii) To a 2,400- volt 60 – cycle source?
A. 292 W
B. 268 W
C. 257 W
D. 305 W
AC MACHINES PART 2 – TRANSFORMERS
iii) To a 2,200 – volt 25 – cycle source?
A. 268 W
B. 306 W
C. 256 W
D. 293 W
MULTIVECTOR REVIEW AND TRAINING CENTER
10.
The hysteresis loss in a 6,600 volt 60 cycle transformer is 480
watts. What will be this loss when the transformer is connected:
i)
To a 6,900 – volt 60 – cycle source?
A. 515 W
B. 502 W
C. 447 W
D. 459 W
ii) To a 6,600 – volt 60 – cycle source?
A. 720 W
B. 320 W
C. 376 W
D. 612 W
iii) To a 6,900 – volt 40 – cycle source?
A. 570 W
B. 350 W
C. 657 W
D. 306 W
11.
A 100 – kVA distribution transformer has a full – load copper of
1,180 watts. For what kilowatt, at a power factor of 0.71, will the copper
losses in the transformer be 1,500 watts?
A. 90.25
B. 71
C. 112.75
D. 80.05
REE – Sept. 2004
12.
In an ideal transformer, what is the efficiency?
A. 100%
B. 90%
C. 80%
70%
D.
13.
Given a 10 – kVA transformer with full –load losses amounting to
70 watts in the iron and 140 watts in the copper. Calculate the efficiency
at half – load unity power factor.
A. 98.62%
B. 95.97%
C. 97.28%
D. 97.94%
14.
A 100 – kVA 2,400/240 volt 60 cycle transformer has the
following constant: rp = 0.42Ω; Xp = 0.72Ω; rs = 0.0038 Ω; Xs =
0.0068Ω. What is the equivalent impedance in primary terms?
A. 0.016Ω
B. 1.612Ω
C. 0.161Ω
D. 16.12Ω
REE – April 2006
15.
A 50 – kVA, 5,000/500 – v, 50 Hz, 1 – phase transformer has the
high voltage winding with a resistance of 8Ω and low voltage winding
with a resistance of 0.06Ω. The no – load losses transformer amount to
MULTIVECTOR REVIEW AND TRAINING CENTER
1,000 watts. Calculate the efficiency of the transformer when delivering
its full rated output at a power factor of 0.8. With the same power factor,
what will be the efficiency if the output drops down to 80% of the rated
value?
A. 92.27%
B. 93.45%
C. 95.33%
D. 94.39%
16.
Calculate the all – day efficiency of a 100 –kVA transformer
operating under the following conditions: 6 hours on a load of 50 kW at
0.73 power factor; 3 hours on a load of 90 kW at 0.82 power factor; 15
hours with no – load on secondary. The iron loss is 1,000 watts and the
full – load copper loss is 1,060 watts.
A. 96.31%
B. 94.87%
C. 95.33%
D. 95.29%
REE – Sept. 2005
17.
A 50 – KVA, single – phase transformer has a 96% efficiency
when it operates at full – load unity power factor for 8 hours per day.
What is the all – day efficiency of the transformer if the copper loss is
60% of full – load losses?
A. 92%
B. 90%
C. 89.5%
D. 93%
AC MACHINES PART 2 – TRANSFORMERS
Asst. EE – Oct. 1991
18.
A 10 kVA, 2,400/240 V, single – phase transformer has the
following resistances and the leakage reactance’s;
rp = 3Ω
rs = 0.03Ω
Xp = 15Ω
Xs = 0.15Ω
Find the primary voltage required to produce 240 V at the secondary
terminals at full – load, when the load power factor is 0.8 lagging.
A. 2,400 V
B. 2,496.5 V
C. 2,348 V
D. 2,445.5
V
19.
A 500 KVA, 1ϕ, 13,200/2,400 V transformer has 4% reactance
and 1% resistance. The leakage reactance and resistance of the high
voltage (primary) winding are 6.34 Ω and 1.83Ω, respectively. The core
MULTIVECTOR REVIEW AND TRAINING CENTER
loss under rated condition is 1,800 watts. Calculate the leakage reactance
and resistance of the low voltage (secondary) winding.
A. 7.56 Ω, 1.66 Ω
B. 13.69 Ω, 3.42 Ω C. 0.25 Ω, 0.055 Ω
D. 13.9Ω, 3.48 Ω
20.
In prob. #19, calculate the %V.R. and efficiency of the transformer
at full – load, 0.85 p.f. lagging and 2,400 V.
A. 4%, 97.8%
B. 6%, 95.4%
C. 5%, 96.8%
D. 3%, 98.4%
21.
A short circuit test was performed upon a 10 – kVA, 2,300/230 –
volt transformer with the following result: ESC = 137 volts; PSC = 192
watts; ISC = 4.34 amp. Calculate in secondary terms the transformer
equivalent reactance.
A. 29.88Ω
B. 2.988Ω
C. 0.2988Ω
D. 298.8Ω
REE – Sept. 2003
22.
A short circuit test was performed on a 100 KVA, 2,200/220V
transformer. Its low side was shorted. A wattmeter, voltmeter and an
ammeter were connected to its high side. If the windings of the
transformer were carrying rated current and the wattmeter and voltmeter
read 750 W and 83V, respectively, find the resistance of the transformer
in ohms.
A. 1.826Ω
B. 1.790Ω
C. 0.363Ω
D. 0.018Ω
23.
An 11,000/230 V, 150 – kVA, 1 – phase, 50 –Hz transformer has a
core loss of 1.4 kW and a full load copper loss of 1.6 kW. What is the
value of maximum efficiency at unity p.f.?
A. 98.17%
B. 96.38%
C. 97.22%
D. 97.64%
24.
A 300 – kVA, single – phase transformer is designed to have a
resistance of 1.5% and a maximum efficiency occurs at a load of 173.2
kVA. Find its efficiency when supplying full – load at 0.8 p.f. lagging at
normal voltage and frequency.
A. 97.56%
B. 96.38%
C. 98.76%
D. 95.89%
MULTIVECTOR REVIEW AND TRAINING CENTER
REE – Sept. 2002
25.
A 20 kV/ 7.87 kV auto transformer has 200 A current in the
common winding. What is the secondary line current?
A. 143.52
B. 200
C. 56.48
D. 329
26.
An auto transformer is adjusted for an output voltage of 85.3 volts
when operated from a 117 volts line. The variable power load draws
3.63 kW at unity power factor at this setting. Determine the transformed
power and the conducted power from the source to the load.
A. 980 & 2,650 watts
C. 1,815& 1,815 watts
B. B. 1,343 & 2,287 watts
D. 1,210 & 2,420 watts
AC MACHINES PART 2 – TRANSFORMERS
REE – April 2007
27.
A transformer is rated 500 kVA, 4,800/480 V, 60 Hz when it is
operated as a conventional two winding transformer. This transformer is
to be used as a 5.28/4.8 kV step down auto transformer in a power
distribution system. In the auto transformer, what is the transformer
rating when used in this manner?
A. 5MVA
B. 6MVA
C. 5.5 MVA
D.
6.5
MVA
REE – Sept. 2008
28.
Two identical transformer bank on open delta serve a balanced 3 –
phase load of 26 kVA at 240 V, 60 Hz. What is the minimum size of
each in kVA needed to serve this load?
A. 25
B. 10
C. 30
D. 15
29.
Two (2), 1ϕ, 100 – kVA transformers are connected in V (open
delta) bank supplying a balanced 3ϕ load. If the balanced 3ϕ load is 135
kW at 0.82 p.f. lagging and 0.823 efficiency, determine the overload
kVA on each transformer.
A. 10.5
B. 5.5
C. 15.5
D. 15
30.
In prob. #29, determine the p.f. of each transformer secondary.
A. 0.820 lag & 0.820 lag
C. 0.996 lag & 0.424 lag
MULTIVECTOR REVIEW AND TRAINING CENTER
B. 0.996 lag & 0.424 leading
lag
D. 0.410 lag & 0.410
31.
In Prob. #29, what is the minimum size in KVAR of a capacitor
bank to be connected across the load so that each transformers is loaded
96% of its rated capacity?
A. 87 KVAR
B. 114 KVAR
C. 27 KVAR
D. 66 KVAR
REE – Sept. 2003
32.
What type of transformer bank is used to convert 2 – phase to 3 –
phase power?
A. Open – delta
B. Scott – T
C. wye – delta
D.
delta – wye
33.
Two T – connected transformers are used to supply a 440 V, 33 –
kVA balanced load from a balanced 3 – phase supply of 3,300 V.
Calculate the kVA rating of the main and teaser transformer.
A. 33 &33
B. 33 & 28.6
C. 28.6 & 24.8
D.
19.05 & 16.5
34.
A polarity test is performed upon a 1, 150/115 – volt transformer.
If the input voltage is 116, calculate the voltmeter reading if the polarity
is subtractive.
A. 127.6 V
B. 106 V
C. 126 V
D. 104.4 V
REE – April 2005
35.
What is the normal secondary circuit current of a current
transformer?
A. 15 A
B. 20 A
C. 5 A
D. 10
A
REE – May 2008
36.
A three – phase, Y/ ∆ connected 50 MVA d345/ 34.5 kV
transformer is protected by differential protection. The current
MULTIVECTOR REVIEW AND TRAINING CENTER
transformer on the low – side of the transformer for differential
protection is 900: 5. What is the current on the secondary side of the
CT?
A. 4.25 A
B. 4.65 A
C. 3.65 A
D. 5.65 A
37.
A 20: 1 potential transformer is used with a 150 volt voltmeter. If
the instrument is 118 volts, calculate the line voltage.
A. 3,000 V
B. 2,850 V
C. 2,360 V
D. 2, 242 V
AC MACHINES PART 2 – TRANSFORMERS
REE April 2004
38.
Instrument transformers are used in indicating and metering and
with protective devices, they are used for ________________.
A. Measuring
B. detecting
C. relaying
D.
sensing
REE – Oct. 2000
39.
The CT ratio and PT ratio used to protect a line are 240 and 2000,
respectively. If the impedance of each line is 10 Ω, what is the relay
impedance to protect the line from fault?
A. 83.33 Ω
B. 1.2 Ω
C. 48, 000 Ω
D. 12Ω
40.
Two transformers 1 and 2 are connected in parallel supplying a
common load of 120 kVA. Transformer 1 is rated 50 kVA, 7,620/240 V
1ϕ and has an equivalent impedance of 8.5 Ω while transformer 2 is
rated 75 kVA, 7,620/240 V 1ϕ and has an equivalent impedance of
5.1Ω. The two transformers operate with the same power factors. What
is the kVA load of each transformer?
A. 48 & 72
B. 45 & 75
C. 42 & 78
D. 40 & 80
41.
Two - 1ϕ transformers are connected in parallel at no – load. One
has a turn ratio of 5,000/440 and rating of 200 kVA, the other has a ratio
of 5,000/480 and rating of 350 kVA leakage reactance of each is 3.5%.
The no – load circulating current is __________.
A. 207 A
B. 702 A
C. 720 A
D. 270 A
MULTIVECTOR REVIEW AND TRAINING CENTER
Transformer – 1 is in parallel with transformer – 2.
T -1
T–2
150 kVA,1ϕ
300 kVA,1ϕ
6600/240 V
6600/240 V
o
Ze−s1 = 0.02425 ∠ 62.9 Ω
Ze−s2 = 0.02425 ∠ 62.9o Ω
Determine the maximum kVA load the bank can carry without
overloading any of the two transformers, assuming that the two
transformers operate at the same power factors.
A. 450 kVA
B. 432 kVA
C. 420 kVA
D. 412 kVA
42.
REE – Oct.1997
43.
A power transformer rated 50,000 kVA, 34.5/13.8 kV is connected
Y – grounded primary and delta on the secondary. Determine the full –
load phase current at the secondary side.
A. 2092 A
B. 1725 A
C. 1449 A
D. 1208 A
REE – April 2006
44.
A 2,000 KW, 2,400 – volt, 75% p.f. load is to be supplied from a
34,500 – volt, 3 – phase line through a single bank of transformers. Give
the primary and secondary line currents in amperes for the wye – wye
connections.
A. 50/700
B. 48/650
C. 60/800
D. 45/642
REE – April 2005
45.
A 3,000 KVA, 2,400 V, 75 percent power factor load is to be
supplied from a 34,500 – volt, 3 – phase line through a single bank of
transformers. What is the voltage rating of each transformer if the
connection is wye- wye?
A. 20,000/ 1,380
B. 18,500/1,350
C. 18,000/1,850
D.
19,000/ 1,350
AC MACHINES PART 2 – TRANSFORMERS
MULTIVECTOR REVIEW AND TRAINING CENTER
REE – March 1998
46.
A 13.8 kV/480 V, 10 MVA three phase transformers has 5%
impedance. What is the impedance in ohms referred to the primary?
A. 0.952 ohm
B. 0.03 ohm
C. 5.125 ohm
D. 9.01 ohm
REE – April 2004
47.
A transformer rated 2,000 KVA, 34,500/240 volts has 5.75%
impedance. What is the per unit impedance?
A. 0.0635
B. 0.0656
C. 0.0575
D. 34.2
REE – April 2001
48.
A 69 KV/13.8 KV, 7.5 MVA transformers have 8% impedance.
What I it’s impedance at 100 MVA base?
A. 0.6%
B. 8%
C. 800%
D. 106.7%
49.
Three 5: 1 transformers are connected in ∆ - Y to step up the
voltage at the beginning of a 13,200 – volt three – phase transmission
line. Calculate the line voltage on the high side of the transformers.
A. 114,300 V
B. 66,000 V
C. 7,815 V
D.
198,000 V
50.
Three (3) – 1ϕ, 50 kVA, 2,400/277 V transformers are connected∆
- Y bank. Each transformer has %Z of 4%. Calculate the fault current
delivered if a 3 ϕ fault occurs at the secondary terminals.
A. 4,512 A
B. 3,908 A
C. 7,815 A
D. 1,504A
51.
A 150 kVA, 2,400/480 volt 3 ϕ transformer with an equivalent
impedance of 4% is connected to an infinite bus and without load. If a 3
phase fault occurs at the secondary terminals, the fault current in
amperes is ______________.
A. 4,512 A
B. 3,908 A
C. 7,815 A
D. 1,504A
MULTIVECTOR REVIEW AND TRAINING CENTER
AC MACHINES PART 3 – AC MOTORS
1. Induction Motor (1𝛟, 𝟑𝛟) - is one which the rotor field is induced by
currents flowing thru the stator. The rotor circuit has no connection what
so ever to the supply circuit. The speed of this motor is less than the
synchronous speed and it operates with a lagging p.f.
General Types of 𝟑𝛟 Induction Motors;
1. Squirrel cage rotor type – its rotor is self – contained and resembles
a birdcage. The rotor conductors are permanently short circuited by
end rings.
Speed characteristics – nearly constant or adjustable or variable
speed
Torque characteristics – in general, it has a medium starting torque
MULTIVECTOR REVIEW AND TRAINING CENTER
Uses – fan, blower, pump, compressor, prime mover of DC
generator
2. Wound rotor type – its rotor has insulated coils of wire and
resembles a DC armature. The rotor conductors are connected to slip
rings and short circuited thru the brushes.
Speed characteristics – variable speed
Torque characteristics – high starting torque
Uses – hoist, crane, conveyor, escalator, elevator, pump, etc.
2. Synchronous motor (1𝛟, 𝟑𝛟) – is one in which the rotor is in step or in
unison with the phase of the alternating current that operates it. It speed
is constant called synchronous speed.
Under – excited SM – lagging power factor
Normally – excited SM – unity power factor
Over – excited SM – leading power factor
The rotor of this motor is called salient pole rotor. Salient pole means
shaped pole.
Uses:
1. For constant speed drives like fan, blower, pump, compressor, prime
mover of DC generators, etc.
2. For power factor correction
Synchronous condenser – a SM without mechanical load, used
for power factor correction only.
3. For voltage regulation of long transmission lines.
To reverse the direction of rotation of a 3ϕ motor, you interchange
the connection to the lines of any 2 motor leads.
Methods employed to start 𝟑𝛟 motors
I.
Full Voltage starting - usually used to start small size 3ϕ motors
MULTIVECTOR REVIEW AND TRAINING CENTER
Example – across the line magnetic starter
II.
Reduced Voltage starting
1. Wye – delta starter – usually used to start medium size 3ϕ
motors
2. Auto – transformer or
Starting compensator
3. Line reactance starter
usually used to start large 3ϕ
motors
4. Line resistance starter
AC MACHINES PART 3 – AC MOTORS
a. Primary line resistance starter
b. Secondary line resistance starter → usually used to start 3ϕ
III. Part Winding Method – usually used to start dual voltage 3ϕ
motors
Reasons why the voltage is reduced during starting large 𝟑𝛟
motors.
In order to reduce the starting line current
1. So that the motor windings line current
2. So that disturbances to nearby load will be avoided (like
flick of nearby lamps).
Advantages of the 𝟑𝛟 IM over the 𝟑𝛟 SM and vise – versa.
1. The 3ϕ IM rugged and simple, therefore, it is easier to
operate and maintain
2. The speed of the 3ϕ IM can be varied over a wide range
or can be adjusted easily
3. The 3ϕ IM is cheaper as to first cost
Vice Versa
MULTIVECTOR REVIEW AND TRAINING CENTER
1. The 3ϕ SM can be constructed with a wider air gap,
which will make it better mechanically (better ventilation)
2. The 3ϕ SM can be operated with a leading power factor
which improve the system power factor (most systems
have lagging power factors).
3. The 3ϕ SM is more efficient especially at higher ratings
because it has no slip.
AC MACHINES PART 3 – AC MOTORS
MULTIVECTOR REVIEW AND TRAINING CENTER
EXERCISES:
REE – April 2006
1. Find the synchronous speed of the revolving field of an induction motor
who’s number of poles Is 12 and operating at 60 Hz frequency.
A. 1,200
B. 1,800
C. 600
D. 900
REE – May 2009
2. A 4 – pole induction motor, running with 5% slip, is supplied by a 60 Hz
synchronous generator. Determine the speed of the motor.
A. 1,674 rpm
B. 1,728 rpm
C. 1,710 rpm
D. 1,692 rpm
REE – Sept. 2004
3. In an induction motor, what is the percent slip if the speed is 3,510?
A. 1
B. 1.5
C. 2
D. 2.5
4. A 6 – pole, 50 Hz, 3 – phase induction motor has a full – load speed of
950 rpm. Its speed in RPM at half load is __________.
A. 975
B. 500
C. 475
D. 1,000
5. A three – phase 60 – cycle six – pole 220 – volt wound – rotor induction
motor has a stator that is connected in ∆ and a rotor that is connected in
Y. the rotor has half as many turns as the stator. For a rotor speed of
1,110 rpm, calculate.
i)
The blocked rotor voltage per phase EBR
A. 110 V
B. 220 V
C. 440 V
D. 16.5 V
ii) The rotor generated voltage per phase ER
A. 14.25 V
B. 16.5 V
C. 18.25 V
D. 8.25 V
iii) The rotor voltage between terminals
A. 8.25 V
B. 14.29 V
C. 16.5 V
D. 4.76 V
MULTIVECTOR REVIEW AND TRAINING CENTER
iv) The rotor frequency
A. 60 Hz
B. 45 Hz
C. 4.5 Hz
D. 30 Hz
The rotor current, if the rotor resistance per phase is 0.1 Ω and the
rotor reactance at standstill is 0.5 Ω/ phase
A. 77.25 A
B. 7.725 A
C. 16.18 A
D. 28.02 A
v)
REE – Sept. 2004
6. An induction motor runs at 1,152 rpm. What is the frequency of the rotor
current?
A. 60 HZ
B. 6.4 HZ
C. 2.4 HZ
D. 4 HZ
REE – March 1998
7. A 50 Hip three phase motor, rated 220 volts, 75% power factor has 6 –
poles. The slip at full load is 3%. What is the frequency of the rotor
current at full load?
A. 1.8 Hz
B. 60 Hz
C. 5.4 Hz
D. 58.2 Hz
AC MACHINES PART 3 – AC MOTORS
8. The nameplate speed of a 60 Hz induction motor is 1750 rpm. If the
speed at no – load is 1795 rpm, what is the percent regulation?
A. 2.78
B. 2.86
C. 2.57
D.
4.24
9. A 3 – phase induction motor running with 2.5% slip takes 100 KW from
the mains. If the stator copper loss amounts to 1.8 KW. What is the rotor
copper loss?
A. 1.8 kW
B. 2.2 kW
C. 2.5 kW
D. 2.45 kW
10.
In Prob. #9, the mechanical power developed is ____________.
A. 95.74 kW
B. 98.2 kW
C. 92.8 kW
D. 97.54 kW
MULTIVECTOR REVIEW AND TRAINING CENTER
REE – Sept. 2007
11.
A 6 – pole, three – phase, 60Hz wound rotor induction motor
delivers 25 hips at 1,150 rpm. If the rational losses of the motor are 300
watts, what is he rotor power input in kW?
A. 21.66
B. 18.95
C. 19.77
D. 22.74
12.
A 3ϕ induction motor draws 4.5 A from its lines at 230 V line to
line at a power factor of 0.153 while running at no – load. Its DC
resistance line to line between two phases of the stator is 0.863Ω. What
is its rational loss?
A. 274.28 W
B. 32.77 W
C. 52.43 W
D. 241.51 W
13.
A 3 – phase induction motor having a star – connected rotor has an
induced elf of 80 V between slip – rings at standstill an open circuit. The
rotor has a resistance and a reactance per phase of 1Ω and 4Ω,
respectively. Calculate the rotor current per phase and power factor
when slip rings are short – circuited.
A. 19.4, 0.422 lag
B. 11.2 A, 0.242 lag C. 1.94 A, 0.422 lag
D. 1.12 A, 0.242 lag
14.
In Prob. #13, calculate the rotor current per phase and p.f. when
slip rings are connected to a star – connected rheostat of 3Ω per phase.
A. 8.16 A, 0.242 lag
B. 11.2 A, unity
C. 8.16 A, 0.707 lag
D. 11.2 A, 0.242 lag
15.
A 220 V, 10 kW,3ϕ 60 Hz, 6 – pole induction motor operates at
full – load with an efficiency of 85%, p.f. of 80% and slip of 10%. What
is the full –load torque of the motor?
A. 59.36 N- m
B. 65.96 N –m
C. 32.80 N- m
D. 88.42 N – m
16.
In Prob. #15, what is the full – load current of the motor?
A. 28.79A
B. 38.59 A
C. 32.80 A
D. 26.24 A
MULTIVECTOR REVIEW AND TRAINING CENTER
REE – Sept. 2002
17.
A3ϕ, 220 – volt, 10 – HP induction motor having a power factor
of 0.8 has 3% losses. Determine the current drawn by the motor.
A. 25.23 A
B. 43.69 A
C. 20.18 A
D. 14.56 A
REE – Sept. 2005
18.
A 50 – HP, 440 V,3ϕ 60 cycle induction motor operates at full –
load with 87% power factor and 90.6% efficiency. What is the power
factor input to the motor in horsepower?
A. 47
B. 50
C. 45
D. 55.2
AC MACHINES PART 3 – AC MOTORS
19.
A 5 – hip four – pole 60cycle three – phase induction motor takes
4,500 watts from the power factor supply. The rotor and stator copper
losses are 92 and 292 watts, respectively. The friction and wind age
losses are 72 watts, and the core loss is 225 watts. Calculate the
efficiency/
A. 87.48%
B. 84.87%
C. 88.47%
D. 89.74%
20.
A 3ϕ , 4 – pole, 230 V, 60 cycle wound – rotor induction motor
has both its stator and rotor connected in Y. The rotor has half as many
turns per phase as the stator. If the rotor is turning at 1725 rpm, R R =
0.075 Ω and XBR = 0.375 Ω, determine the torque developed by the
motor.
A. 2.7543 lb. – ft.
B. 275.43 lb. – ft.
C. 23.745 lb. – ft.
D. 27.543 lb. – ft.
21.
In Prob. #20, what is the maximum torque that can be developed
by the motor?
A. 22.99 lb. – ft.
B. 68.96 lb. – ft.
C. 137.93 lb. – ft.
D. 119.44 lb. – ft.
MULTIVECTOR REVIEW AND TRAINING CENTER
22.
In Prob. #20, what is the speed at which maximum torque will
occur?
A. 1,725 rpm
B. 1,710 rpm
C. 1,620 rpm
D. 1,440 rpm
23.
In Prob. #20, calculate the starting torque of the motor?
A. 26.525 lb. – ft.
B. 79.575 lb. – ft.
C. 45.943 lb. – ft.
D. 53.050 lb. – ft.
24.
An induction motor is tested in the blocked – rotor test. Its rated
line current of 8.5 A is drawn when the line voltage is 16.6 V and the
total wattage is 48.8 watts. Under this condition, what is the equivalent
reactance per phase in stator terms?
A. 1.128Ω
B. 0.225Ω
C. 1.105Ω
D. 0.552Ω
25.
In Prob. #24, if the AC resistance of the stator is 0.127 Ω per
phase, what is the resistance and the reactance of the rotor per phase
under blocked rotor conditions?
A. 0.98Ω & 5.53Ω
C. 0.98Ω & 0.553Ω
B. 0.098Ω & 0.553Ω
D. 0.098Ω & 5.53Ω
26.
A squirrel cage motor develops a maximum torque of 2.25 times
it’s full – load torque and has a standstill reactance which is four times
its resistance per phase. Determine the slip when working on full –load.
A. 0.0586
B. 0.0685
C. 0.0865
D. 0.075
REE – April 2006
27.
A 3,300 – V, 24 – pole, 50 – HZ, 3 – phase star connected
induction motor has a slip – ring rotor resistance of 0.016Ω and
standstill reactance of 0.265Ω per phase. Calculate the ratio of full –
load torque to maximum torque if full – load torque is obtained at 247
rpm.
A. 0.298
B. 0.288
C. 0.375
D. 0.382
REE – Sept. 2003
MULTIVECTOR REVIEW AND TRAINING CENTER
28.
In a wound rotor induction motor, where do you connect the
voltage limiting device for reduced voltage starting?
A. Stator
B. rotor
C. field
D.
panel board
29.
An induction motor has a starting torque of 20 N – m at full –
voltage starting. If the starting voltage is reduced to 80% of rated
voltage, the starting torque is __________ N- m
A. 20
B. 16
C. 25
D12.8
AC MACHINES PART 3 – AC MOTORS
30.
A 10 – Hip, 550 – volt 60 cycle three – phase induction motor has
a starting torque of 160 percent of full – load torque and a starting
current of 425 percent of full –load current. What voltage is required to
limit the starting current to the full –load current value?
A. 129.41 V
B. 74.72 V
C. 224.14V
D.
275 V
REE – April 2002
31.
A three – phase squirrel cage motor is started by a wye – delta
starter. What shall be its starting torque relative to its rated voltage
starting torque?
A. 33.3%
B. 42.3%
C. 57.7%
D. 100%
REE – April 2006
32.
A 3 – phase, 6 – pole, 50 – HZ induction motor drives a single –
drum haulage with a speed of 12 km/hour. The effective diameter of the
haulage drum is 2 m and the motor slip at the started speed is 5%.
Calculate the speed of haulage drum in rpm.
A. 28.1
B. 29.8
C. 29.4
D.
971 rpm
33.
Two 60 – cycle motors are connected in concatenation to drive a
load. If machine 1 has six poles and machine 2 has eight poles, calculate
the speed of the combination if the slip is 0.075.
MULTIVECTOR REVIEW AND TRAINING CENTER
A. 1,110 rpm
D. 971 rpm
B. 832 rpm
C. 475 rpm
REE – April 2004
34.
A synchronous motor that is used for power factor correction only.
A. Synchronous reactor
C. generator
B. Alternator
D. synchronous condenser
35.
For 10 – pole 60 Hz synchronous motor, what is the normal
operating speed?
A. 720 rpm
B. 1200 rpm
C. 1500 rpm
D.
1800 rpm
REE – Sept. 2004
36.
What is the percent slip of the synchronous motor when running?
A. 3
B. 2.5
C. 1
D. 0
REE – Sept. 2008
37.
A three – phase 2,300 V, Y – connected wound rotor synchronous
motor has XS = 2 ohms / phase and Ra= 0.1 ohms/phase. The motor
operates at 0.866 leading power factor while taking a line current of 350
amperes. Find the power angle in degrees.
A. 19.5
B. – 22.9
C. 22.1
D. – 20.7
38.
The full –load losses in a 5,000 kVA three – phase 6,600 – volt
synchronous condenser are 160 KW. Calculate the full – load current
and p.f.
A. 438 A, 0.032 ldg
B. 43.8 A, 0.32 ldg C. 483 A, 0.64 ldg
D. 483 A, 0.064 ldg
39.
A factory load of 900 kW at 0.6 power factor lagging is to be
increased by the addition of a synchronous motor that takes 450 kW. At
what power factor must this motor operate if the over – all power factors
is to be 0.9 lagging?
MULTIVECTOR REVIEW AND TRAINING CENTER
A. 0.673 ldg
D. 0.637 ldg
B. 0.763 ldg
C. 0.736 ldg
ILLUMINATION
1. Luminous flux (ф of F) is the light energy radiated out per second from
the luminous body in the form of light waves.
Its unit is lumen (lm). One lumen is defined as the flux emitted per unit
solid angle from a standard candle.
2. Solid angle is given as
𝛚=
𝐚𝐫𝐞𝐚
(𝐫𝐚𝐝𝐢𝐮𝐬)𝟐
Its unit is steradian.
3. Luminous intensity (or candle power) of a source in any direction is
given by the luminous flux radiated out per unit solid angle in that
direction.
If ф is the luminous flux emitted by a source in a solid angle of ω
steradian, then l = ф/ 𝜔 lm / steradian or candela.
Hence, it is the angular flux density emitted by the source.
4. A source having a uniform luminous intensity of 1 candela radiates out a
total flux of 4πl. Lumens.
5. The mean spherical candle power (M.S.C.P) is given by
M.S.C.P =
𝐭𝐨𝐭𝐚𝐥 𝐟𝐥𝐮𝐱 𝐞𝐦𝐢𝐭𝐭𝐞𝐝 𝐢𝐧 𝐥𝐮𝐦𝐞𝐧𝐬
𝟒𝛑
candela
6. Candela (Cd) is the unit of luminous intensity. One candela is equal to
1/60th of the luminous intensity per cm2 of a black body radiator
maintained at the temperature of a solidification of platinum.
7. Illumination (E) the illumination of a surface is measured by the
luminous flux received by a unit area of the surface.
MULTIVECTOR REVIEW AND TRAINING CENTER
If a flux of ф lumens is received normally by an area of A, then
E = ф/A
The unit of E on the unit of A.
i.
ii.
iii.
If A in m2, then E is in lm/ m2. Its alternative names are:
metre – candle (m – Cd) or lux.
If A in cm2, then E is in lm/cm2. Its alternative name is phot.
If A is in ft2, then E is in lm/ft2. Its alternative name is footcandle (fc)
1 phot = 1,000 lux
8. The law of illumination is
𝟏 𝒄𝒐𝒔𝜽
E = 𝟐 lm/m2
𝒓
9. The space/height ratio is given by
horizontal distance between lamps
=
mounting height of lamps
10.
The utilization factor or coefficient of utilization is defined as
lumen actually received on working plane
CU =
lumens emitted by the light source
11.
Depreciation factor or maintenance factor is given by
illumination under actual conditions
MF =
illumination when everything is clean
12.
The total flux required on a working plane can be calculated from
the relation.
𝐄𝐀
Ф=
lumens
𝐂𝐔 𝐱 𝐌𝐅
ILLUMINATION
EXERCISES:
REE – April 2006
1. A lamp gives an average output of 628 lumens. What is its M.S.C.P.?
A. 50 Cd
B. 40 Cd
C. 45 Cd
D. 48 Cd
2. A lamp has luminous intensity of 250 Cd. Calculate the total luminous
flux emitted by it.
MULTIVECTOR REVIEW AND TRAINING CENTER
A. 1,000 lm
2,000 lm
B. 785 lm
C. 3,142 lm
D.
REE – Sept. 2004
3. A light source has an illumination of 100 foot-candles (fc), what is the
illumination in lux?
A. 1,204
B. 1,075
C. 1,496
D. 1,360
REE – Sept. 2004
4. A light source which radiates light in all directions has luminous
intensity of 100 candela. What is the illumination E in foot – candles in a
perpendicular working area 10 ft. distance below the source?
A. 2
B. 3
C. 1
D. 2.5
REE – Sept. 2006
5. A certain light source produces an illumination of 800 lux on a surface.
Find the illumination on that surface if the distance to the light is
doubled.
A. 300 lux
B. 100 lux
C. 400 lux
D. 200 lux
6. A lamp giving 200 candelas uniformly below the horizontal plane is
suspended 3 m above a horizontal plane. Calculate the illumination at a
point on the plane
i.
Directly below the lamp
A. 66.66lm/m2
B. 6.66lm/m2
C. 22.22 lm/m2
D. 2.222lm/m2
REE – Sept. 2005
ii. 3 m away from the vertical axis
A. 7.66 lux
B. 7.55 lux
C. 7.86 lux
D. 7.35 lux
REE – Sept. 2005
7. An intensity of 100 candle power is directed at a point 10 ft. distance
from the source. If the receiving plane is tilted through 30 degrees from
the perpendicular, what is the illumination upon the surface?
MULTIVECTOR REVIEW AND TRAINING CENTER
A. 0.866 fc
B. 0.34 fc
C. 0.5 fc
D. 0.7 fc
REE – April 2005
8. A 150 candlepower lamps gives sufficient light to read a book at a
distance of 6 feet. If the lamp depreciates to 75 candlepower, how much
closer should the book be placed to be read with the same luminance in
ft.?
A. 3.4
B. 2.6
C. 2.2
D. 1.8
9. An incandescent lamp having a luminous flux of 3,000 lumens is
enclosed in a 0.3 meter diameter diffusing sphere and having an
absorption factor of 15%. Calculate the average luminance of the sphere.
A. 450
B. 9,020
C. 2,255
D. 3,000
ILLUMINATION
10.
Two lamp posts are 13 m apart and are fitted with 1,000 Cd lamp
each at a height of 5 m above the ground. Calculate the illumination on
the ground:
i.
Under each lamp
A. 41.85 lux
lux
B. 40 lux
ii. Midway between the lamps
A. 37.18 lux
B. 9.07 lux
27.21 lux
C. 80 lux
C. 18.14 lux
D. 20. 92
D.
11.
What must be the candle power of a point source of light which
gives an illumination of 100 lm/m2 at a point A 2m vertically below the
source?
A. 200 Cd
B. 400 Cd
C. 100 Cd
D. 300 Cd
MULTIVECTOR REVIEW AND TRAINING CENTER
12.
In Prob. #11, if a plane mirror which reflects 75% of the light
falling on it is placed horizontally 30 cm. above the source, what will be
the illumination of A?
A. 100 lux
B. 200 lux
C. 44.4 lux
D. 144.4 lux
13.
Two lamps A and B produce equal illumination on the screen of a
photometer when A is 60 cm and B is 50 cm from the screen. A plane
mirror is placed 6 cm behind B, the plane of the mirror being at right
angles to the line from B to the screen and it is then found that to restore
equality of illumination on the screen, the source A must be moved 10
cm nearer screen. What is the reflecting power of the mirror?
A. 68%
B. 50%
C. 62%
D.
65%
14.
After a balance was obtained between two lamps on a photometric
bench, a sheet of plane glass was interposed on one side of the photo
meter. To restore the balance, one of the lamps was moved back through
10 cm making its distance from the photometer 110 cm. What
percentage of light incident on the glass was transmitted by it?
A. 90.9%
B. 82.6%
C. 86.2%
D. 88.6%
15.
A drawing office containing a number of boards having a total
effective area of 70 m2is lit by a number of 40 Wincandescent lamps
giving 11 lm/W. An illumination of 72 lm/m2 is required on the drawing
boards. Assuming that 60% of the total light emitted by the lamps is
available for illuminating the drawing boards, estimate the number of
lamps required.
A. 19
B. 21
C. 23
D. 25
REE – April 2004
16.
A room with dimensions of 72 ft. x 36 ft. requires 50 foot-candles,
maintenance factor of 0.7, coefficient of utilization of 0.46 and RI of 4.
Calculate the required lumens output of the lamps.
A. 144,000
B. 136,800
C. 402,485
D. 74,690
MULTIVECTOR REVIEW AND TRAINING CENTER
REE – Sept. 2001
17.
A 40 ft. by 20 ft. office is to be illuminated with duplex fluorescent
luminaries at a level of 50 foot-candles. The maintenance factor and
coefficient of utilization are estimated to be 70% and 60%, respectively.
Each fluorescent produces 3,300 lumens. How many luminaries are
required?
A. 18
B. 8
C. 14
D. 28
REE – Sept.2008
18.
Which is not found in a rapid start fluorescent lamp?
A. Ballast
B. starter
C. wire harness
capacitor
ILLUMINATION
D.
19.
A colliery workshop 20m x 10m 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. 8 kW, 34.8 A
B. 9 kW, 39.1 A C. 6 kW, 26.1 kW
D. 7 kW,
30.4 A
20.
A drawing hall in an engineering college is to be provided with a
lighting installation. The hall is 30 n x 20 m x 8 m (high). The mounting
height I 5 m and the illumination required is 135 lm/m2. Using metal
lamps, estimate the size and number of single lamp luminaries. Assume
Utilization coefficient = 0.6; maintenance factor = 0.75
Space/height ratio = 1; lm/W for 300 – W lamp = 13; lm/watt for
500 – W lamp =15
A. 500 W, 24
B. 500W, 48
C. 300W, 46
D. 300W, 23
MULTIVECTOR REVIEW AND TRAINING CENTER
ELECTRICAL EQUIPMENT
MULTIVECTOR REVIEW AND TRAINING CENTER
Electrical Equipment are devices and gadgets used to control, protect,
support, and maintain the efficient and safe operation of electrical machines
like alternators, transformers, motors, electrical lighting system and other
related loads.
Equipment on the Generation System
Governor - used to control the frequency of the generated voltage.
Voltage Regulator – used to adjust the exciting current of the generator,
hence, the generated voltage.
Line Drop Compensators – used to make up for the voltage drop along
the conductor connecting the generator and voltage regulator to ascertain
that the voltage sensed by the regulators is the correct value.
Main Exciter – it is the source of the current that produces the magnetic
field on the poles of the main generator.
Pilot Exciter – it is the one supplying excitation to the main exciter.
Disconnect Switches – these are installed on both sides of the alternator
breakers to attain complete isolation during repairs and maintenance
works.
MULTIVECTOR REVIEW AND TRAINING CENTER
Power circuit Breakers – these can be OCB, GCB or even VCB. They
are installed in conjunction with protective relays to protect the
alternator and its prime mover from the different failures that may occur.
Switchboard/ Switchgears – this is where the different measuring
instruments, control switches, protective relays and other auxiliaries are
housed.
Surge Arresters/ Choke Coils – used to provide protection against
surges like the effect of lighting.
Grounding System – this is provided as path for any nuisance current
towards the ground that may be present as occasioned to the operation of the
main machines.
Battery Bank – this will be the source of power for all control devices.
This is so because during shut downs, the power from the alternators is
not yet ready plus the fact that during faults, the system voltage becomes
abnormal and unreliable, hence, the need for an independent power
supply.
ELECTRICAL EQUIPMENT
Substation Equipment
Substation can be classified info several ways: according to service,
mounting, function, Apparatus
And control. Some of the equipment common to all types according to
function are:
Surge Arresters
MULTIVECTOR REVIEW AND TRAINING CENTER
Potential Transformers and or CVTs
Current Transformers
Bus Bars
Hot Dipped Galvanized Steel Structures
Porcelain Insulators (Pin, Post and Suspension)
Disconnect Switches
Power Circuit Breakers (OCB, GCB, VCB and some MCCBs)
Panel/Switchboards and Switchgears with all the measuring instruments
And protective devices housed on them
Station Service Transformers
Power Transformers
Perimeter Lighting System
Grounding System
Battery Bank
CLRs
Transmission and Distribution Lines
Power Conductors
Power Cables
MULTIVECTOR REVIEW AND TRAINING CENTER
Arcing Horns
Porcelain Insulators (Pin, Suspension and Akimbo)
Over Ground Conductors (Overhead Conductors or Overhead Ground
Wires)
Grounding Rods
ELECTRICAL EQUIPMENT
Station, Intermediate and Distribution Surge Arresters
Disconnect Switches
Poles
Line Towers
Pole Line Materials
Transformer Fuse and Cut Out
Potheads
Pedestals
Loads (Lighting and Motors)
Lighting Receptacles (for series and multiple)
Switches
Motor Starters
Panel Boards
MULTIVECTOR REVIEW AND TRAINING CENTER
Conductors and Cables
These are some of the electrical Equipment commonly encountered by
practicing electrical engineers but there is many more which cannot be
accommodated in this discussion due to the limitation of time.
Electrical Equipment Questionnaire Set 3
1. A circuit for electric lighting connected in series and forming an
extended loop instead of running in two neighboring parallel lines:
A. Cascade ckt.
B. string ckt.
C. belt ckt.
D. flash
ckt.
2. A device for measuring the hysteresis and eddy current losses in
conjunction with an AC wattmeter:
A. Einstein hysteresis tester
C. Steinmetz hysteresis
tester
MULTIVECTOR REVIEW AND TRAINING CENTER
B. Eipstein hysteresis tester
tester
D.Eddy hysteresis
3. In relation to corona, these are ions that are both positively and
negatively charged simultaneously:
A. Zenith Ions
B. Position Ions
C. Swizzle Ions
D. Zwitter Ions
4. A reactor type water turbine with adjustable runner blades and especially
suitable for run – of – river installation with variable head and flow:
A. Koppel
B. Kaplan
C. Kirbey
D. Knell
5. This is the specific gravity of lead – lead acid battery at 80 deg. Fah. At
.100% charged:
A. 1.225
B. 1.285
C. 1.290
D. 1.335
6. The ratio of the resistance to the reactance of the circuit is called:
A. Q – factor
B. power factor
C. reactive factor
D. Dissipation factor
7. The most basic outdoor type of an electric power substation:
A. Distribution
B. Transmission C. Sectionalizing
Litztalor
D.
8. A multi – strand wire having transposition to reduce skin effect:
A. Lithards
B. Litzerdraph
C. Litzerdraht
D. Litztalor
9. It is an electrical instrument for measuring and recording automatically
the rise and fall of water level in a river or dam:
A. Buoy graph
B. Aquagraph
C. Grandegraph
D. Fluviograph
10.
In instrument for registering the number of revolutions made by a
rotating body like motors, alternators and the like:
MULTIVECTOR REVIEW AND TRAINING CENTER
A. Revmeter
D. Odometer
B. Tachometer
C. Cyclometer
11.
On a 480/277 volts 3 phase 4 wire system, the color code for the
three hot conductors are as follows except:
A. Brown
B. Red
C. Orange
D. Yellow
12.
The direction of rotation of this motor can be altered by
dissembling/assembling it:
A. Repulsion
B. Universal
C. capacitor
D. Shaded Pole
13.
Electrical measuring instruments can be classified according to
their principle of operations. Among these are, except:
A. Electrostatic
B. Electromagnetic C. Electro thermal
D.
Electrodynamics
14.
This charging of secondary batteries is to neutralize internal
discharging due to local action/ polarization:
A. Standby
B. Slow rate
C. Trickle
D.
Localize
15.
The following phase sequences are the same, except:
A. BCA
B. ABC
C. CBA
CAB
D.
16.
The electrolyte of nickel – Iron battery is:
A. Sulfuric Acid
C. Potassium Hydroxide
B. B. Potassium Sulfate
D. Potassium Chloride
17.
One source of error on analog electrical measuring instrument is:
A. Dampness
B. Parallax
C. Erratic Movement D.
Effect of Gravity
MULTIVECTOR REVIEW AND TRAINING CENTER
18.
This is an extension tank of power transformer to minimize
sludging on the main tank:
A. Overhead
B. Conservator
C. Salient
back – up
D.
19.
This is the type of transformer that supplies power to runway
lighting system.
A. CVT
B. ICT
C. CCT
D.
CCR
20.
This test is conducted on transformer to make sure it can withstand
voltage surges of high magnitude but of short duration.
A. Induced Voltage
B. DI – electric
C. Impulse
D.
Full Wave
21.
This test is conducted on transformers to establish its BIL.
A. Induced Voltage
B. Full Wave
C. Impulse
Applied Potential
D.
22.
The Buchholz relay of a power transformer will initiate alarm on
any of the following:
A. Over Voltage
C. Loss of oil due to leak
B. B. Line to Line Fault
D. Low Frequency
23.
In the IEC designation of current transformer “5P10”, P means the
core is designed for protection and the 5 means:
A. Percent Tolerance
B. Percent Overload C. Percent Error
D. Percent Distortion
24.
CTs designed for protection purposes are laid out in terms of knee
– point Emf, exciting current at the knee point and secondary winding
resistance. Such current transformer are called class:
A. P
B. I
C. Y
D. X
MULTIVECTOR REVIEW AND TRAINING CENTER
25.
These are low power transformers in which the secondary and
primary voltages are proportional and in phase. The primary and
secondary windings are galvanically separated from each other.
A. IPT
B. IVT
C. CVT
D. CCR
26.
When protecting a delta – wye power transformer with its neutral
floating with over current protection scheme, the relay to use is:
A. 50N/51N
B. 86
C. 50/51
D. 27
27.
To properly protect an electrical system, all zones of protection
should be:
A. Independent from one another
C. Overlapping
B. Under Lapping
D. Separated
28.
The device function number for the relay to protect transmission
line is:
A. 67
B. 74
C. 21
D. 52
29.
When a delta – wye connected transformer is to be protected with
differential protection, its CT must be connected:
A. Delta – wye
B. delta – delta
C. wye – wye
D.
wye – delta
30.
To avoid nuisance tripping on the different protective relays of an
electrical system for upstream to downstream, all of them must be:
A. Tested
B. Operational
C. Overlapping
D.
Coordinated
31.
In designing electrical system, the engineer will have to consider
the BIL of power transformer. The average isokeraunic level in Luzon to
be considered is:
A. 60
B. 50
C. 70
D. 80
32.
Surge Arresters can be classified accordingly, except:
MULTIVECTOR REVIEW AND TRAINING CENTER
A. Station Class
B. Transmission Class C. Intermediate Class D.
Distribution Class
33.
The design of the grounding system is considered safe if the
calculated step voltage ESis less than:
A. GPR
B. Es50
C. Es70
D.
Et50
34.
This is the measurement DC contact resistance of a breaker at
close position using 100A micrometer.
A. TTR
B. Ducter
C. Impulse
D. Chop Wave
35.
Among the method used to measure ground resistance are as
follows, except:
A. Multi – point
C. Fall of potential Method
B. Use of high frequency earth measuring instrument D. Two point
method
36.
In the use of electronic digital electrical measuring instrument like
voltmeters and ammeters, if the upper element is defective, these
numbers will show wrong indication:
A. 4,6
B. 1,4
C. 6,0
D. 9,7
37.
The charges in motion to constitute electrical current are moving in
this speed:
A. Speed of Light
B. Speed of Sound
C. Creeping
D. Mach 5
38.
One standard Voltage of the DC system used on power plants and
substations for control purposes:
A. 24 volts
B. 100 volts
C. 12 volts
D. 12
volts
MULTIVECTOR REVIEW AND TRAINING CENTER
39.
N transmission line insulators, the ratio off the arc – over voltage
of a string of “n” insulators to the “n” of insulators multiplied to the arc
– over voltage of single insulator is called:
A. Insulator eff’cy
B. arc – over eff’cy C. String eff’cy
D. Nota
40.
This is the single diagram indicated on the frontage of
switchboards:
A. Skeleton Bus
B. Miniature Bus
C. Single Bus
Bus System
D.
41.
The type of breaker contact with classification “de – ion contact”
was developed by:
A. ABB
B. Siemens
C. Westinghouse
D.
GE
42.
Among sealed type transformer, the gas used to purge them is:
A. Oxygen
B. Nitrogen
C. Helium
D. Argon
43.
This one method to determine the polarity of a transformer whether
this is additive or subtractive
A. Reactive Kick
B. Mutual Kick
C. Inductive Kick
D. Transformer Kick
44.
This is type of induction motor where its rotor is provided with slip
rings:
A. Squirrel Cage
B. Capacitor Start
C. Capacitor Run
D.
Nota
45.
As applied to electrical system protection, MCCB stands for:
A. Motor control center board
C. Magnetic Control
Circuit Breaker
B. Motor Circuit control Breaker
D. Nota
MULTIVECTOR REVIEW AND TRAINING CENTER
46.
This is the timer relay that will shut down the diesel engine prime
mover of the stand – by generator when the normal power is back:
A. TDNE
B. TDEN
C. TDEC
D. TDES
47.
LLD as a factor in determining the maintenance factor in relation
to the correct number of luminaires means:
A. Luminaire lamp dampness
C. Luminaire lamp
depreciation
B. Lamp lumen deactivation
D. Lamp lumen
depreciation
47 . Characteristics of over – current relay can be classified accordingly,
except:
A. Definite
B. Inverse
C. Infinite
D. Very Inverse
48.
1 trega – volt amperes is equal to:
A. 1,000,000,000 VA
B. B. 100,000,000,000 VA
C. 10,000,000,000 VA
D. 1,000,000,000,000 VA
49.
In AC electrical system, another name for impedance is:
A. Virtual Resistance
B. Complex Resistance
C. Apparent
Resistance D. Nota
50.
In the apparel operation of alternators, the giving away its load has
more _________ power factor.
A. Lagging
B. Leading
C. Positive
D. Negative
51.
This connection of three phase transformers is unstable due to the
absence of third harmonic:
A. Delta – delta
B. wye – wye
C. Delta – wye
D.
wye – delta
52.
The broken delta transformer has its greatest application on:
A. Metering
B. Relaying
C. Power supply D.
Balancing
MULTIVECTOR REVIEW AND TRAINING CENTER
53.
A fault liable to occur in the armature of a dynamo due to loose
wire:
A. Flying Fault
B. Flying Arc
C. Flying Cross
D.
Flying Fire
54.
It is used as power supply for hand lamps used on flammable areas
like liquifactions trains.
A. Tripping X’former
C. Isolating X’former
B. Insulating X’former
D. Exothermic X’former
55.
An interrupter for making and breaking an electric current:
A. Rheostat
B. Rheotor
C. Rheotic
D.
Rheotome
56.
This is used to reflect back the surge into the line if it approaches
electrical equipment, if its arrester fails:
A. CLF
B. Overhead Ground C. De – ion gaps D. Choke
Coils
57.
This is a motor similar to the DC series motor and can operate an
unflow and changing flow currents.
A. Repulsion
B. Shaded Pole
C. Universal
D. Nota
58.
A device connected or transmission lines to limit the magnitude of
short circuit currents.
A. CLF
B. CLG
C. CLR
D.
CLB
59.
The material commonly used as grounding Rod:
A. CCS
B. CCC
C. CSS
CSC
D.
MULTIVECTOR REVIEW AND TRAINING CENTER
60.
The average resistance of crushed stone as applied to outdoor
substation yard.
A. 1k – 3k ohm – mt
B. 2k – 5k ohm – mt C. 4k – 8k ohm – mt
D. 7k – 12k ohm –mt
61.
The basic insulation level of high voltage electrical machines and
equipment such as transformer is expressed in:
A. Megohm
B. Begohm
C. Kilovolt
D. Kiloamps
62.
The asymmetrical fault current is used as basis in the determination
of circuit breakers:
A. MR
B. IC
C. AT
D.
AF
63.
A fixture for supporting a cluster of incandescent electric lamps:
A. Tri – pod
B. Electrolier
C. Electropad
D. Electrobench
64.
The number of steps needed to manually parallel an incoming
generator to the bus.
A. 10
B. 11
C. 14
D. 15
65.
Using zonal cavity method to determine the correct number of
luminaries, the following are considered except:
A. Color of the room
B. Shape of the room C. Area of the room
D. Room temperature
66.
This is the instrument used to determine the different turns ratio of
a power transformer:
A. TRT
B. TTR
C. TRR
D.
RTT
67.
This is the common name given to the pressure relief value of the
sealed tank transformer:
MULTIVECTOR REVIEW AND TRAINING CENTER
A. Reliever
Diagraph
B. Diaphram
C. Diaver
D.
68.
Auto – transformer is not recommended as a power supply for
continuous operation due to:
A. Expensive
B. Unstable
C. Erratic
D.
Unsafe
69.
The substation whose function is to subdivide bulf power so as not
to be exceed capacities of equipment:
A. Tap
B. Sectionalizing
C. Anti – Lock
D. Anti – operate
70.
A feature incorporated in the circuit breaker or reclosing scheme
whereby, in the event of a permanent fault, repeated operation of the
C.B. is prevented.
A. Anti – on
B. Anti – pump
C. Anti – lock
D. Anti – operate
71.
This is a system of motor speed control intended to provide either a
constant output horsepower or torque:
A. Ward Leonard
B. Kramer System
C. Stott System
D. Thompson System
72.
The instrument is used to be able to pre – determine the direction
of rotation of three phase motor prior to its termination:
A. MRI
B. PSI
C. PAM
D. PAT
73.
In a given distribution system, the location of pole mounted
transformer is determined by considering the following factors except:
A. V.R.
B. Conductor Size
C. Transformer Cost D.
Transformer Size
74.
As per ANSI, power factor of a certain electrical system is
considered zero if it is:
MULTIVECTOR REVIEW AND TRAINING CENTER
A. 30% lagging
60% lagging
B. 20% lagging
C. 50% lagging
D.
75.
As a rule of the thumb, the transformer size to supply a motor is:
A. 1000 VA/1000 Hp
B. 1 kVA/ 1 kW C. 1 kVA/ 1 Hp
D. 1
Hp/ 1 ton
76.
In using hydrogen gas as coolant for turbo – generators, it must
have this percentage by volume to avoid explosion:
A. 90%
B. 80%
C. 70%
D.
60%
77.
This kind of fuse is to designed to cause the abrupt escape of gas
when its element melts:
A. Cartridge
B. Cassette
C. Expulsion
D.
Fast Acting
78.
The type of the contact for oil circuit breaker popularly known as
explosion chamber saw patented by:
A. G.E
B. Westinghouse
C. ABB
D. Siemens
79.
At this elevation, the voltage rating of power circuit breakers are to
be rated by close to 70%:
A. 15,000 ft.
B. 17,500 ft.
C. 20,000 ft.
D. 14,000 ft.
80.
Tis device is designed to protect motor circuits to include single
phasing:
A. MCCB
B. MCP
C. ACB
D. OLP
81.
The type of circuit breaker that requires no de – rating event when
enclosed:
A. Thermal Compensated
C. Atmosphere
Compensated
MULTIVECTOR REVIEW AND TRAINING CENTER
B. Magnetic Compensated
D. Pressure Compensated
82.
The following are types of lighting protection methods on roofs,
except:
A. Mandrel
B. Flat
C. Mansard
D. Gambrel
83.
This is the device to act as lightning rod and is installed above the
structure it will serve:
A. Air Connector
B. Lighting Bowl
C. Air Terminal
D. Air Gap
84.
On a three phase AC system, the use of the metallic conduct per
conductor is:
A. Redundant
B. Dangerous
C. Tedious
D.
As per Code
85.
This is the type of porcelain insulator bused on end poles:
A. Pin
B. Post
C. Suspension
D. Akimbo
86.
This generator has an almost straight external characteristic from
20% to 100% loading:
A. Amplidyne
B. Diverter Pole
C. Compound
D. Series
87.
Some of the AC bridges being used are the following, except:
A. Parson
B. Anderson
C. Hay’s
D. Owen
88.
Aircraft electrical system is not under the jurisdiction of the
electrical engineering because the frequency used belongs to avionics.
The frequency is equal to;
A. 500 Hz
B. 400 Hz
C. 1000 Hz
D. 200 Hz
89.
Mineralized insulting oil that are oxidized are called:
A. Sludge’s
B. Corroil
C. Askarel
D. Inerteen
MULTIVECTOR REVIEW AND TRAINING CENTER
90.
The transfer of energy can be expressed as the product of the
values of the magnetic field and the component of the electric field,
which is perpendicular to the former. This is:
A. Gauss Law
B. Pettier’s Law
C. Poynting’s Law
D. Einstein Law
91.
The kilowatt rating of the pilot exciter with respect to the main
exciter it serves is:
A. 110% to 20%
B. 1.5% to 5%
C. 5% to 10%
D.
10% to 25%
92.
The _______________
chart is used to determine the voltage
regulation of Transmission circuits containing capacitance:
A. Ronald
B. Dwight
C. Copper
D. Moore
93.
This is a device connected on transmission line to damp vibration
of the conductors:
A. Stone bridge damper
B. Stock bridge damper
C. Stall bridge
damper D. Steel bridge damper
94.
This speed of water wheels is attained when it is reduced
homologously to such a size that it will develop 1 Hp under 1 ft. head at
its best efficiency:
A. Water Speed
B. Specific Acceleration C. Specific speed
D. none of the given
95.
To successfully parallel compound generators, this must be
included:
A. Equalizer Ring
B. Equalizer Bus
C. Equalizer Bar D.
Equalizer Winding
96.
Some methods of ventilating alternators are herein provided
except:
MULTIVECTOR REVIEW AND TRAINING CENTER
A. Simple Radial
Circumferential
B. Complex Radial
C. Multiple Path D.
97.
The following are all voltage regulators for alternators of the
rheostatic direct – acting type, except:
A. Silverstat
B. Golden Rheo
C. Diactor
D.
Rocking Arm
98.
A three- wire direct current system is also known as:
A. Thurry System
B. Selsyn System
C. Gilbert System
D. Edison System
99.
The connection of the reactors in series with synchronous
machines for the purpose of limiting the magnitude of fault current is
called:
A. Scott System
B. Spott System
C. Stott System
D.
Scopp System
100
. Electric power substation can be classified according to the
following criteria, except:
A. Function
B. Service
C. Grounding
D.
Control
Prepared by:
Camilo B. Regacho
PEE Board Exam 1st Placer
NOTE: To ascertain your passing if not top notching the REE board
examinations, supplement your review materials with the 6 reviewers of Engr.
C. B Regacho, available at all National Bookstores nation Wide.
MULTIVECTOR REVIEW AND TRAINING CENTER
POWER SYSTEM
Transmission line. A mechanical Structure designed to transmit a certain
maximum amount of energy economically from one place to another with a
specified limit to the permissible voltage variation at the receiving end and
with the least possible risk of interruption to continuous service.
Equivalent Circuit of a Line. Consists of uniform distribution of resistance,
inductance, conductance and susceptance along the length of the line.
r
jx
ig
r
g
jb
jx
i g ig
r
g
jb
jx
ig
For overhead line conductors, conductance (g) is neglected because ic>>> ig
Resistance. The most important cause of power loss in a line
P
RAC = LOSS
2
31
Where: RAC – effective resistance per wire
PLOSS - total line loss
I2 – RMS magnitude of line current
MULTIVECTOR REVIEW AND TRAINING CENTER
Note: Effective resistance is not equal to DC resistance due to skin effect.
Skin Effect. The internal inductance of the conductor gives rise to a voltage
of self – induction so that the voltage drop in the conductor is the sum of the
voltage of self – induction and the ohmic resistance drop.
R=
𝛒𝟏
𝐀
Ω (for solid conductors)
For stranded wires:
R = (1.01)
R = (1.02)
𝛒𝟏
𝐀
𝛒𝟏
𝐀
Ω (three strands)
Ω ( above three strands)
R = resistance in Ω
ρ = resistivity, Ω - cmil./ ft.
l = length, ft.
A = area, cmil.
𝐑𝟏
𝐑
= 𝟏
(𝐓+𝐭 𝟏 )
(𝐓+𝐭 𝟐 )
T = inferred absolute zero resistance temperature. oC
= 234.5 oC (annealed copper)
= 241 oC (hard drawn copper)
= 228 oC (aluminum)
t1 = initial temperature
t2 = final temperature
R1 = resistance temperature t1
R2 = resistance temperature t2
t
t2
MULTIVECTOR REVIEW AND TRAINING CENTER
t1
R
R1
R2
T
Conductivity
Annealed copper – international standard of conductivity
Hard drawn copper – 97.3% conductivity
Hard drawn aluminum – 61% conductivity
Conductors
Advantages of Aluminum over Copper Conductor
 Lower cost
 Higher weight for the same resistance
 Has larger diameter than copper for the same resistance
Types of Aluminum Conductors
 AAC – All Aluminum Conductor
 AAAC – All Aluminum Alloy Conductor
 ACSR – Aluminum Conductor Steel Reinforced
 ACAR – Aluminum Conductor Alloy Reinforced
Inductance. Flux linkages per unit current.
L=
𝚿
𝐈
=
𝐍𝛟
weber – turns ampere or Henry
𝐈
For Single phase systems using solid conductors.
L = Lint + Lext
𝐃
L = 2 x 10-7 ln 𝟏/𝟒
𝐫
Where:
L = 2 x 10-7 ln
𝐃
𝐫𝐞′
D = separation Distance between Conductors
MULTIVECTOR REVIEW AND TRAINING CENTER
r = radius of the conductor
r’ = re ‘4
re’ = the equivalent radius of a fictitious wire that replaces the actual
wire in order to compute for the inductance L of the wire due to internal
and external flux linkages.
Lin = internal inductance
= ½ x 10-7 H/m (µr = 1)
Lex = external inductance
= (2 x 10-7) ln (D/r) henry/ meter
For the complete single phase circuit consisting of two wires,
LT = 4 x 10-7 ln
𝐃
𝐫′
H/m
Stranded Conductors. For stranded conductor, r is replaced by Ds.
Ds – geometric mean radius (GMR) or self GMD or the geometric mean
of all possible distances of every element in a group to all other elements
in the same group including itself.
DS = 𝟗√𝐃𝟏𝟏 𝐃𝟏𝟐 𝐃𝟏𝟑 𝐃𝟐𝟏 𝐃𝟐𝟐 𝐃𝟐𝟑 𝐃𝟑𝟏 𝐃𝟑𝟐 𝐃𝟑𝟑 (three – strand conductor)
𝟐
DS = 𝐧 √𝐃𝟏𝟏 𝐃𝟏𝟐 … . . 𝐃𝐥𝐧 𝐃𝟐𝟏 … . . 𝐃𝐧𝐧 ( n – strand conductor)
Where:
D11,D22 ….. Dnn = r
D12, D13 ….. Dn(n-1) = center to center distance between strands
Dm = geometric mean distance (GMD) or mutual GMD or the geometric
mean at all possible distances of every element in a group to all other
elements in the other group.
Illustration:
MULTIVECTOR REVIEW AND TRAINING CENTER
a
d
b
e
c
m
n
Dm = 𝐦𝐧√𝐃𝐚𝐝 𝐃𝐚𝐞 𝐃𝐛𝐝 𝐃𝐛𝐞 𝐃𝐜𝐝 𝐃𝐜𝐞
Inductance of stranded conductors
L = 2 x 10-7 ln
𝐃𝐦
𝐃𝐬
Three Phase Lines
Tree Phase lines with equilateral spacing
L = 2 x 10-7 ln
𝐃𝐦
𝐃𝐬
H/m per phase
Transposition. Exchanging the position of the conductors at regular
intervals along the line so that each conductor occupies the original
position of every other conductor over an equal distance.
C
B
A
A
C
B
B
A
MULTIVECTOR REVIEW AND TRAINING CENTER
C
Three Phase Lines with unsymmetrical spacing
L = 2 x 10-7 ln
𝐃𝐞𝐪
𝐃𝐬
H/m per phase
Where: Deq = 𝟑√𝐃𝐚𝐛 𝐃𝐛𝐜 𝐃𝐜𝐚
Bundled Conductors. Two or more conductors per phase in close proximity
compared with the spacing between phases.
𝟒
𝐃𝐛𝐒 = √𝐃𝟐𝐒 𝐃𝟐 = √𝐃𝐒 𝐃(two – bundle conductor)
𝟗
𝐃𝐛𝐒 = √𝐃𝟑𝐒 𝐃𝟔 = √𝐃𝐒 𝐃𝟐 (three – bundle conductor)
𝟏𝟔
𝐃𝐛𝐒 = √𝐃𝟒𝐒 𝐃𝟏𝟐 = √𝐃𝐒 𝐃𝟑 (four – bundle conductor)
Where:
DbS = GMR of bundled conductors
Parallel Circuits
a
d
b
e
c
First Circuit
Second Circuit
MULTIVECTOR REVIEW AND TRAINING CENTER
𝐩
-7
L = 2 x 10 ln
𝐃𝐞𝐪
𝐩
𝐃𝐒
𝟔
𝐩
Where: 𝐃𝐒 = √𝐃𝟑𝐒 𝐃𝐚𝐚 , 𝐃𝐛𝐛 , 𝐃𝐜𝐜 ,
𝐩
𝐃𝐞𝐪 = 𝟔√𝐃𝐚𝐛 , 𝐃𝐚𝐛 , 𝐃𝐛𝐜 , 𝐃𝐛𝐜 , 𝐃𝐜𝐚 , 𝐃𝐜𝐚 ,
Capacitance of Transmission Lines
Potential Difference (or voltage) between two points due to a charge.
V12 =
𝐪
𝟐𝛑𝐤
ln
𝐃𝟐
𝐃𝟏
[ from pt. 1 to pt. 2]
Where: q – charge (coulombs/meter)
k – absolute permittivity of free space
= 8.85 x 10-12 F/m
= (1/ 36π) x 10-9 F/m
kr – relative permittivity (di – electric constant)
Capacitance of two Infinitely Long Parallel Straight Conductors
C=
𝟖.𝟖𝟓𝛑 𝐱 𝟏𝟎−𝟏𝟐
Cn =
𝐃
𝐥𝐧 𝐫
F/m [line to line]
𝟏𝟕.𝟕𝛑 𝐱 𝟏𝟎−𝟏𝟐
𝐃
𝐥𝐧 𝐫
F/m [line to neutral]
For stranded wires, r is replaced by do/ 2 where do is the outside diameter of
the conductor.
MULTIVECTOR REVIEW AND TRAINING CENTER
Charging Current, Ic. The leakage current in the insulting medium.
IC = j𝝎CabVab(single phase)
IC = j𝝎CanVan(three phase)
Capacitance of Three Phase Lines (Neglecting the Effect of the Earth)
With Equilateral Spacing
Cn =
𝟏𝟕.𝟕𝛑 𝐱 𝟏𝟎−𝟏𝟐
𝐥𝐧
𝐃
𝐫
F/m per phase [line to neutral]
With unsymmetrical
Cn =
𝟏𝟕.𝟕𝛑 𝐱 𝟏𝟎−𝟏𝟐
𝐃𝐞𝐪
𝐥𝐧 𝐫
F/m per phase [line to neutral]
Where: Deq = 𝟑√𝐃𝐚𝐛 𝐃𝐛𝐜 𝐃𝐜𝐚
Bundled Conductors
Cn =
𝟏𝟕.𝟕𝛑 𝐱 𝟏𝟎−𝟏𝟐
𝐃
𝐥𝐧 𝐛
F/m per phase [line to neutral]
𝐃𝐒
Where:
𝐃𝐛𝐒 = √𝐫𝐝( 2 - bundle)
𝐃𝐛𝐒 = √𝐫𝐝𝟐 ( 3 - bundle)
𝟒
𝐃𝐛𝐒 = 𝟏. 𝟎𝟗 √𝐫𝐝𝟑 ( 4 - bundle)
Note: D may be Deq depending on the arrangement of conductors
Parallel Circuits
Cn =
𝟏𝟕.𝟕𝛑 𝐱 𝟏𝟎−𝟏𝟐
𝐩
𝐃𝐞𝐪
𝐈𝐧 𝐩
𝐃
𝐒
F/m
MULTIVECTOR REVIEW AND TRAINING CENTER
𝐩
𝟔
Where: 𝐃𝐒 = √𝑫𝟑𝑺 𝐃𝐚𝐚 , 𝐃𝐛𝐛 . 𝐃𝐜𝐜,
𝐩
𝐃𝐞𝐪 = 𝟔√𝑫𝒂𝒃 𝐃𝐚𝐛 , 𝐃𝐛𝐜 . 𝐃𝐛𝐜, , 𝐃𝐜𝐚 . 𝐃𝐜𝐚,
Representation and performance of Transmission Lines
Levels of Transmission
 Short Transmission Line ( L≤ 80 km)
 Medium Length Transmission Line (80 km < L≤ 240 km)
 Long Transmission Lines ( L < 240 km)
Short Transmission Line. for short lines, only the resistance and inductance
are being considered
VSN = VRN∠ 0o + IZ
Z = R + JX
PR
IR =
∠ ± cos-1pf R
√3VR pfR
Medium Length Transmission Line. analysis involves the use of lumped
parameters
 Nominal 𝝅 Equivalent Circuit
Z
I
VSN
y
I2
2
VSN = [𝟏 +
𝒁𝒀
𝟐
] VRN + IRZ
y
I2`
2
IR
L
O
A VRN
D
MULTIVECTOR REVIEW AND TRAINING CENTER
𝒁𝒀
𝒁𝒀
IS = Y [𝟏 + ] VRN + [𝟏 + ]IR
𝟒
𝟐
pf R = cos𝛉R
and pf S = cos𝛉S
PLOSS =𝟑𝟏𝟐𝐑 R
𝐕
−𝐕
%VR = 𝐑𝐍𝐋 𝐑𝐅𝐋 x 100%
𝐕𝐑𝐍𝐋
 Nominal T Equivalent Circuit
Z/2
Z/2
Is
IR
Vsn
VSN= [1 +
𝑍𝑌
2
V1
] VRN + Z [1 +
IS= YVRN + [1 +
𝑍𝑌
4
PLOSS = 3 [12𝑠 + 12𝑅 ]
𝑍𝑌
4
Y It
VRN
] IR
] IR
𝑅
2
For both nominal PI and nominal T equivalent circuit
𝟐
VRNL = VSN
𝟐+𝐙𝐘
Efficiency =
𝐏𝐑
𝐏𝐒
x 100 % =
𝐏𝐑
𝐏𝐑 + 𝐏𝐋𝐎𝐒𝐒
x 100%
Using the Generalized Transmission Constant
VSN = AVRN + BIR
IS = CVRN + DIR
VRN = DVSN - BIS
IR = - CVSN + AIS
MULTIVECTOR REVIEW AND TRAINING CENTER
For Nominal Pi Circuit
𝐙𝐘
A=D=1+
For Nominal T Circuit
𝐙𝐘
A=D=1+
𝟐
B = Z [𝟏 +
B=Z
C = Y [𝟏 +
VR =
𝐙𝐘
𝟒
]
𝐕𝐑𝐍𝐋 − 𝐕𝐑𝐅𝐋
𝐕𝐑𝐅𝐋
𝟐
𝐙𝐘
𝟒
]
C=Y
𝐕𝐒𝐅𝐋
− 𝐕𝐑𝐅𝐋
𝐀
=
𝐕𝐑𝐅𝐋
Long Transmission Line
r
jx
r
jx
jb
r
jx
jb
i = r + j𝜔l – series impedance per unit length
y = jb = j𝜔c = shunt admittance per unit length (g is neglected)
Voltage and current Relationship
𝐕
VSN = [ 𝐑𝐍
+ 𝐙𝐂 𝐈𝐑
𝟐
𝐕
IS =
Where :
𝐈𝐑 + 𝐙𝐑𝐍
𝐂
𝟐
𝐕
] 𝒆𝜹𝑿 + [ 𝐑𝐍
− 𝐙 𝐂 𝐈𝐑
𝟐
] 𝒆𝜹𝑿
𝐕
𝒆𝜹𝑿 +
𝐈𝐑 + 𝐙𝐑𝐍
𝐂
𝟐
𝒆𝜹𝑿
𝒛
ZC = √ = characteristic impedance
𝒚
𝜹 = √𝒛𝒚 = ∝ + j𝜷 = propagation constant
MULTIVECTOR REVIEW AND TRAINING CENTER
∝ = attenuation constant (neper per unit length)
𝛃 = Phase constant (radians per unit length)
In hyperbolic Form
VSN = VRN Cosh 𝜹x + ZCIR Sinh 𝜹x
𝐕
IS = IR Cosh 𝜹x + 𝐑𝐍Sinh 𝜹x
𝐙𝐂
Velocity of Propagation
V=
𝟐𝝅𝒇
𝜷
unit length per second
Incident and reflected Wave
𝑉
[ 𝑅𝑁
[
+ 𝑍𝐶 𝐼𝑅
2
𝑉𝑅𝑁 − 𝑍𝐶 𝐼𝑅
2
] 𝒆𝜹𝑿 – incident wave (voltage)
] 𝒆𝜹𝑿 – reflected wave (voltage)
Note: at infinite line, reflected wave is zero. If the line is terminated by its
characteristic impedance, ZC, the reflected wave is zero (both voltage and
current).
Sometimes. ZC is called the surge impedance. However, surge impedance is
associated with “load less” line (R and g are zero).
L
Therefore, ZC = √ – surge impedance, resistive impedance)
C
The power transmitted by a line that is terminated by its surge impedance is
called surge impedance loading (SIL)
SIL =
𝟐
𝐕𝐋𝐋
𝐋
√
𝐂
; pf = 1.0 (resistive)
MULTIVECTOR REVIEW AND TRAINING CENTER
Sag – Stress Analysis (Mechanical Design of Transmission Lines)
A uniform conductor suspended in still air from two supports takes the form
of a catenary (sag is more than 10% of span).
𝐓
𝐰𝐱
𝐰
𝐓𝐎
𝐓𝐎
𝐰𝐱
𝐰
𝐓𝐎
S = [ 𝐎] Sinh
y = [ ] Cosh(
− 𝟏)
T =√𝐓𝐎𝟐 + (𝐰𝐬)𝟐 = TO√𝟏 + (𝐬𝐢𝐧𝐡
𝐰𝐱
𝐰𝐱
)
𝐓𝐎
T = TO Cosh
𝐓𝐎
Where:
s = half – span length of the conductor, m
w = “dead weight” of the conductor, kg/m
y = sag
T = maximum tension
TO = Minimum tension (tension at the lowest point)
x = half – span length
Approximate Equations:
Applicable only for parabolic conductors. The sag is less than 10% of the
span. The weight is assumed to be uniformly distributed along the horizontal.
s=x
𝐰𝐱 𝟐
y=
= d (sag)
𝟐𝐓
T = TO
Un stretched length of conductor
𝐰𝐥𝟐
L1 = L 𝐄𝐀𝐝
Where:
L = stretched length
= 2s = 2l +
𝐰 𝟐 𝐥𝟑
𝟑𝐓𝐎𝟐
MULTIVECTOR REVIEW AND TRAINING CENTER
d = sag
E = modulus of elasticity
A = cross – sectional area of the conductor
Supports at Unequal Elevation
h
d1
h2
h1
C
X1
X1 = 1 -
X2
hT
X2 = 1 +
2wl
hT
2wl
Note:
if l< hTO / 2wl, then x1 is negative. A negative value of x1 means
that lowest point of the conductor falls outside the span as in the case of a very
steep mountain side.
Effect Wind Load
WW
C
P
WC
WT
MULTIVECTOR REVIEW AND TRAINING CENTER
2
WT = √WC2 + WW
Where: WC = “dead weight” of the conductor per unit length
P = wind pressure of projected area.
WW = wind load
= P x d per unit axial length of the conductor
d = diameter of the conductor
Note: if wind load is considered. All derived equations are still valid provided
that w is replaced by WT.
Variation of Sag with load and temperature.
Weill’s Equation
𝐖𝟐𝟐 𝐈𝛆𝟐
𝐰𝟏𝟐 𝐈𝛆𝟐
𝟔𝐀
𝟔𝐟𝟏𝟐 𝐀𝟐
𝟐
(
)
𝟐 = 𝐟𝟐 [∝ (𝐭 𝟐 − 𝐭 𝟏 𝛆 + 𝐟𝟐 − 𝐟𝟏 + (
)]
Underground Transmission. Underground transmission system has higher
cost than the overhead transmission system, however, it replaces overhead
transmission for comparatively short distances where the latter system is not
suitable. Commonly used in populous districts or highly urbanized cities
where overhead construction is not advisable permissible.
Advantages of underground system over overhead system
 Not subject to damage by wind, thunderstorms and the like.
 The danger to life is reduced
Submarine Cables – used to transmit electric energy across water that cannot
be spanned by overhead conductors.
Parts of Submarine Cable
 Conductor
 Insulation – impregnated paper or rubber mineral compound
MULTIVECTOR REVIEW AND TRAINING CENTER
 Metal Sheath – protect the conductor from moisture, gases or damaging
liquids in the soil
 Bedding – protect the sheath from galvanic corrosion
 Armoring – layers of galvanic steel wire or steel tapes to protect the
cable from mechanical injury during laying or handling
 Serving – protect armoring from atmospheric conditions.
Insulation Resistance:
R=
𝛒
𝟐𝛑𝐋
𝐫
ln 𝟐 ohms
𝐫𝟏
Where: r1= radius of the conductor
r2 = radius of the cable
= r1+ thickness of insulation
𝜌 = specific resistance of the insulation
Capacitance of Underground Cables
Single Core Cables
C=
C=
𝟐𝛑𝐤 𝐨 𝐤 𝐫 𝐥
𝐫
𝐫𝟏
𝐥𝐧 𝟐
𝟐𝛑𝐤 𝐨 𝐥
𝐫
𝐥𝐧 𝟐
𝐫
∑( 𝟏)
𝐤𝐫
(if the insulation consists of different materials)
Where:
ko = absolute permittivity
kr= relative permittivity
r1 = radius of the conductor
r2 = radius of the cable (r1+ thickness of insulation)
Three – metal Sheathed cable
Cn – CS + 3CC
MULTIVECTOR REVIEW AND TRAINING CENTER
Where:
Cn = capacitance to neutral
CS= capacitance between conductor and sheath of the cable
CC = capacitance between the conductor of the cable
Electric Stress in Cables
g=
𝒒
𝐯𝐨𝐥𝐭𝐬
𝟐𝛑𝐤 𝐨 𝐤 𝐫 𝐱
𝐦
where: g = electric stress at any radius x, volts/meter
q = charge per unit length, coulomb meter
ko = permittivity of free space
kr = relative permittivity of the distance
x = radius where electric stress is considered
Maximum Stress in the Dielectric
𝐄 𝐯𝐨𝐥𝐭𝐬
gm =
𝐫𝟐
𝐫𝟏 𝐥𝐧𝐫
𝟏
𝐦
where: gm = maximum electric stress
E = voltage between the conductor and sheath
r1 = radius of the conductor
r2 = external radius of the insulation
Losses in Cables
 I2R (ohmic loss ) – temperature dependent
 Dielectric loss – voltage and temperature dependent
 Shield losses – current dependent losses due to circulating and eddy
currents
 Pipe losses – current dependent losses from hysteresis and eddy current
losses due to incomplete cancellation of magnetic field from three
conductors.
Insulators. The voltage distribution across an insulator string is non –
uniform due to capacitance between the cap – pin – junction and the ground
MULTIVECTOR REVIEW AND TRAINING CENTER
String efficiency =
𝐯𝐨𝐥𝐭𝐚𝐠𝐞 𝐚𝐜𝐫𝐨𝐬𝐬 𝐭𝐡𝐞 𝐢𝐧𝐬𝐮𝐥𝐚𝐭𝐨𝐫 𝐬𝐭𝐫𝐢𝐧𝐠
𝐧 𝐱 𝐯𝐨𝐥𝐭𝐚𝐠𝐞 𝐚𝐜𝐫𝐨𝐬𝐬 𝐭𝐡𝐞 𝐢𝐧𝐬𝐮𝐥𝐚𝐭𝐨𝐫 𝐧𝐞𝐚𝐫𝐞𝐬𝐭 𝐭𝐡𝐞 𝐜𝐨𝐧𝐝𝐮𝐜𝐭𝐨𝐫
Where: n – number of insulator disks
Methods of Improving String Efficiency
 Grading – using larger insulator disk for one or two units adjacent to the
conductor
 Using insulating material instead of using steel for the cap
 Increasing the length of the cross – arm to reduce the cap – pin junction
to ground capacitance
 Providing high resistance instead of infinite resistance glaze to increase
the resistance leakage current (thermal instability)
 Use of guard ring coupled with arcing horn (the fitting of guard ring to
the insulator string has the effect of introducing two air capacitances –
others are negligible.
Corona. If an alternating potential difference between the conductors is
gradually increased, a point is reached at which the air at the surface of the
conductors become ionized. The cylinder of the ionized air surrounding each
conductor is “corona”.
The potential difference between the conductors during corona formation is
called “disruptive critical voltage”.
Visual Critical Voltage – the potential difference at which faint luminous
glow of violet color is to seen to surround each conductor. The luminous glow
is called visible corona.
Disruptive Critical Voltage
VO= 21.2𝜹 r mO ln
𝐝−𝐫
𝐫
kV rms per voltage
MULTIVECTOR REVIEW AND TRAINING CENTER
Where:
𝛿 = ratio between density of air at P (cm Hg) and T (0C) and
density of air at 76 cm of Hg and 250C
d = distance between centers of conductors (cm)
r = radius of the conductors (cm)
mO = irregularity factor
= 1 for smooth, polished, cylindrical conductor
= 0.93 to 0.98 for weathered cylindrical conductor
= 0.87 to less than 0.93 for weathered, more than seven strands
= 0.80 to less than 0.87 for weathered having up to strands.
Visual Critical Voltage
VV = 21.2𝜹 r mV (𝟏 +
𝟎.𝟑
√𝜹𝒓
)ln
𝐝−𝐫 𝐤𝐕 𝐫𝐦𝐬
𝐫
𝐩𝐡𝐚𝐬𝐞
Corona Power Loss
PO = (
𝟐𝟒𝟐
𝜹
𝐫
𝐤𝐖
𝐝
𝐤𝐦
) (f + 25) √ (Vф - VO)2 x 10-5
per phase
Factors Affecting Corona
 The breakdown strength of air which varies with atmospheric conditions
 The irregularity of conductor surfaces
 The conductor radius
 The distance between the conductors
Effects of Corona
 Interference with communications
 Non – sinusoidal charging current which causes non – sinusoidal voltage
drop
 Corona power loss
 Corona helps to attenuate high voltage surges caused by lighting or
switching
DC Distribution System. in DC systems, power may be fed and distributed
by either:
MULTIVECTOR REVIEW AND TRAINING CENTER
 Two – wire system
 Three – wire system
Methods of Feeding
 Feeding at one end
 Feeding at both ends and equal voltages
 Feeding at both ends with unequal voltages
 Feeding at some intermediate zone
Types of Loads
 Concentrated loading
 Uniform loading
 Combination of a and b
DC Distributor Fed at Both Ends
A
C
IAC
D
ICD
E
IDE
B
IEB
I1
I2
I3
𝚺VD = IACRAC + ICDRCD + IDERDE + IEBREB
I4
Distributor Fed at both Ends
A
C
IAC
I1
With Unequal Voltages
D
ICD
E
IDE
I2
B
IEB
I3
MULTIVECTOR REVIEW AND TRAINING CENTER
𝚺VD = VA - VB
𝚺VD = IACRAC + ICDRCD + IDERDE + IEBREB
With Equal Voltages
𝚺VD = VA - VB
𝟎 = IACRAC + ICDRCD + IDERDE + IEBREB
Where:
ICD = IAC – I1
IDE = ICD – I2
IEB = IDE – I3
Uniformly Loaded Distributor
A
B
I
I
I
I
I
I
𝚺VD(AB) = ½ IR
When fed at both ends voltages, the point of minimum potential is the middle
point.
𝟏
VD(MIDDLE POINT) = IR
𝟖
Point of Minimum Potential. The point in the distributor where there is a
reversal of current.
Per unit Calculations
MULTIVECTOR REVIEW AND TRAINING CENTER
Per Unit Value. A value by which voltage, current, power, impedance and
admittance are represented to simplify the calculations in power system
problems such as load flow studies, fault analysis, etc.
Per unit value =
𝐚𝐜𝐭𝐮𝐚𝐥 𝐯𝐚𝐥𝐮𝐞
𝐛𝐚𝐬𝐞 𝐯𝐚𝐥𝐮𝐞
Bases:
SB – base power
ZB – base impedance
YB – base admittance
IB = base current
VB – base voltage
With VB and SB known,
Single Phase
Three Phase
𝐒
IB = 𝐁
IB =
𝐕𝐁
𝐒
SB =
YB = 𝐁𝟐
𝐕𝐁
SB =
𝐙𝐁
Conversion to new base
ZPU(NEW) = ZPU(OLD)[
𝐒𝐁(𝐍𝐄𝐖)
𝐒𝐁(𝐎𝐋𝐃)
][
𝐕𝐁(𝐎𝐋𝐃)
𝐕𝐁(𝐍𝐄𝐖)
Short Circuit Calculations
SSC =
I F=
Where:
𝐒𝐁
𝐗 𝐏𝐔
𝐒𝐒𝐂
√𝟑𝐕𝐁
√𝟑𝐕𝐁
𝐒
YB = 𝐁𝟐
𝐕𝐁𝟐
𝐒𝐁
(MVA)
(amperes)
SSC – short circuit power
]
𝐕𝐁
𝐕𝐁𝟐
𝐙𝐁
MULTIVECTOR REVIEW AND TRAINING CENTER
XPU – per unit value of positive sequence reactance
- Thevein’s equivalent impedance or reactance as viewed at the
point of the fault.
Three Phase Fault
IF = IF(PU) x IB=
𝐒𝐁
√𝟑𝐕𝐁 𝐗 𝟏𝐏𝐔
Line to Line fault
IF= IF(PU) x IB=
𝐒𝐁
𝐕𝐁 (𝐗 𝟏𝐏𝐔 + 𝐗 𝟐𝐏𝐔 )
Single Line to Ground Fault
IF= IF(PU) x IB=
√𝟑𝐒𝐁
𝐕𝐁 (𝐗 𝟏𝐏𝐔 + 𝐗 𝟐𝐏𝐔 + 𝐗 𝟎𝐏𝐔 )
Where:
SB = rated apparent power
VB = rated apparent power
X1PU = per unit value of the positive sequence reactance
X2PU = per unit value of the negative sequence reactance
X0PU = per unit value of the zero sequence reactance
NOTE: If the base values are not given, the rated values may be used (rated)
values are normally selected as base values).
MULTIVECTOR REVIEW AND TRAINING CENTER
POWER SYSTEM
EXERCISES:
REE – Oct. 2000
1. Which of the following is not a standard distribution voltage?
A. 6.24 kV
B. 16 kV
C. 13.8 kV
D. 11 kV
2. The fact that the outer layer of the conductor carries more current as
compared to core is known as _____.
A. Corona
B. permeability
C. fault
D.
skin effect
3. How many stands are there for a three – layer stranded conductor?
A. 19
B. 37
C. 53
D. 27
4. An ACSR conductor having seven steel strands surrounded by 25
aluminum conductor is specified as __.
A. 7/25
B. 25/7
C. 14/50
D. 50/14
REE – May 2008
5. A single phase, 20 km line has a total inductance of 35 mH. The distance
between the two conductors is 59 inches. Find the GMR of the
conductor.
A. 1.75 cm
B. 1.89 cm
C. 1.65 cm
D. 1.99 cm
REE – Sept. 2008
6. A three- phase, 60 Hz, transmission line has its conductors arranged in a
triangular configuration so that the two distances between conductors are
7.62 m and the third is 12.8 m. The conductors are Dove with GMR of
0.0314 ft. If the line is 75 km long, what is the inductive reactance of the
line?
A. 39.89Ω
B. 37.57Ω
C. 37.18Ω
D. 38.73Ω
REE – April 2007
MULTIVECTOR REVIEW AND TRAINING CENTER
7. A three – phase transposed distribution line is designed with equilateral
spacing of 12ft. It is decided to build the line with horizontal spading
(D13 = 2D12 = 2D23). The conductors are transposed. What should be the
spacing between adjacent conductors in order to obtain the same
inductance as in the original design?
A. 9.5246
B. 9.5056
C. 9.6437
D. 3.6190
REE – April 2007
8. A 3 – phase circuit 60 Hz, 230 kV transposed transmission line is
composed of two 1,272 MCM 54/19 conductors per phase with
horizontal configuration. The bundle conductors are 50 cm in distance
while phase spacing between the centers of the bundle is 10 meters. If
the GMR of the ACSR conductors is 0.0466 ft., find the inductive
reactance in ohm per km per phase of the transmission line.
A. 0.3775
B. 0.3398
C. 0.3628
D. 0.3964
9. A double circuit line consists of 300,000 cmil 26/6 ACSR Ostrich
conductor arranged vertically with distance between phases of 15 ft., 15
ft. and 30 ft. The horizontal distance between the circuits is 23 ft.
Determine the inductive reactance of the line in ohms per mile per phase
if the GMR of each conductor is 0.0229 ft.
A. 0.28
B. 0.46
C. 0.56
D.
0.39
10.
The conductors are bundled primarily to _________________.
A. Increase reactance
C. reduce reactance
B. Reduce radio interference
D. reduce resistance
POWER SYSTEM
REE – April 2001
11.
A 115 kV line has a vertical configuration with 9 ft. spacing. The
diameter of the 336.4 MCM ACSR conductor used is 0.721 inch. What
is the total capacitance if the length of the line is 20 km.
A. 0.282 µF
B. 0.187 µF
C. 2.2 µF
D.
5.25 µF
MULTIVECTOR REVIEW AND TRAINING CENTER
REE – Sept.2002
12.
A three – phase transmission line is 30 km long. The outside
diameter of the conductors is 0.721 inch and has an equilateral spacing
of 4 ft. Determine the capacitance per phase of the line.
A. 4.57 µF
B. 1.023 µF
C. 0.341 µF
D. 6
µF
13.
Calculate the capacitive reactance in kohm – km of a bundled 60
Hz, three – phase line having three conductors per bundle with 45 cm
between conductors of the bundle. The outside diameter is 0.175 in and
the spacing between bundle centers 10, 10 and 20 m.
A. 243
B. 765
C. 486
D. 382
REE = Sept. 2008
14.
A single circuit 745 kV 60 Hz three – phase transposed
transmission line is composed of four ACSR 1,272,000 cmils, 54/19,
pheasant conductors per phase with horizontal configuration. The phase
spacing between the center of the bundles is 14 m and the bundle
spacing is 45 cm. The conductors have a diameter of 1.382 inch and a
GMR of 0.5592 inch. Determine the capacitance per phase per km of the
transmission line in micro farad per kilometer.
A. 0.01424
B. 0.01266
C. 0.01583
D. 0.01234
15.
Shunt capacitance in neglected in _______________ transmission
lines.
A. Short
B. medium length
C. long
D. all
of these
REE – April 2004
16.
A 30 – mile 3 – phase transmission line is to deliver 20,000 kW at
69 kV at 85% power factor. The line resistance is 0.324 ohm per mile
and inductive reactance is 0.737 ohm per mile. What is the line loss?
A. 1,050 kW
B. 376.7 kW
C. 997 kW
D.
1,130.3 kW
MULTIVECTOR REVIEW AND TRAINING CENTER
17.
Estimate the distance over which a load of 15 MW at 80% power
factor can be delivered by a three – phase transmission line having
ACSR conductors each of resistance 0.61 ohm/km. The voltage at the
receiving end is to be 138 kV and the loss of the line is to be limited to
7.5% of the load.
A. 100 km
B. 33km
C. 76 km
D. 7836 km
REE – Sept. 2001
18.
A 250 km transmission line has the following parameters:
resistance per kilometer 0.05 ohm, capacitive reactance per kilometer
625,000 ohms, and inductive reactance per kilometer 0.2 ohm. What is
the series impedance?
A. 37.5 + j150 Ω
B. 12.5 – j2,450 Ω
C. 12.5 + j50 Ω
D.
75.5 – j2,450 Ω
REE – Sept. 2005
19.
A three – phase, 50 Hz transmission line with a length of 10 km
delivers 2,500 KVA to a load at 6,000 volts and 0.8 p.f. lagging. The
line resistance is 0.3 Ω/km and inductance is 1.28 mH/km. Determine
the sending end voltage.
A. 7.24 KV
B. 7.88 KV
C. 8 KV
D.
8.024 KV
POWER SYSTEM
REE – Sept.2006
20.
A three – phase short transmission line having per phase
impedance of 2 + j4 ohms has an equal line to line receiving end and
sending end voltages of 115 KV, while supplying a load of 0.8 p.f.
leading. Find the power supplied by the line?
A. 872.8 MW
B. 860.2 MW
C. 839.2 MW
D. 822.4 MW
REE – Sept. 2005
MULTIVECTOR REVIEW AND TRAINING CENTER
21.
A three – phase 60 Hz transmission line delivers 750 KW at 95
percent leading power factor and at delta voltage of 6,900 volts. The
resistance per conductor is 4.3 ohms and the reactance per conductor is
2.6 ohms. Assuming both source and load are wye connected, what is
the load impedance per phase?
A. 69Ω
B. 64.5Ω
C. 72.2Ω
D. 60.27Ω
REE – April 2005
22.
A three – phase 60 HZ transmission line delivers 750 KW at 95
percent leading power factor and at delta voltage of 6,900 volts are
measured at the load end. The resistance per conductor is 4.3 ohms and
the reactance per conductor is 2.6 ohms. Assuming both source and load
are wye connected, determine the voltage regulation of the transmission
line in percent?
A. 3.9
B. 4.3
C. 4.6
D. 5.6
REE – April 2005
23.
A three – phase 60 HZ, 20 – mile transmission line delivers 20,000
KW at 66,000 V and 75% power factor lagging. The conductors are 3/0
AWG copper spaced 72 inches. The resistance is 0.326 ohm per mile
and reactance is 0.742 ohm per mile. Capacitance is nil. What is the
transmission efficiency in percent?
A. 93
B. 95
C. 96
D. 97
REE – Sept. 2007
24.
A69 kV, three – phase short transmission line is 16 km long. The
line has a per phase series impedance of 0.125 +j0.4375 ohm per km.
Determine the transmission efficiency when the line delivers 70 MVA,
0.80 lagging power factor at 64 kV.
A. 98.75%
B. 96.36%
C. 94.67%
D. 95.90%
25.
When the load at the receiving end of a long transmission line is
removed or the line is lightly loaded, the end voltage is less than the
receiving end voltage. This phenomenon is called ______________
MULTIVECTOR REVIEW AND TRAINING CENTER
A. Ferranti effect
skin Effect
B. Proximity effect
C. Kelvin Effect D.
REE – April 2005
26.
The capacitive reactance of a 40 km 34.5 KV line is 90,000 per
kilometer. What is the total capacitive reactance of the line?
A. 2,250Ω
B. 1.08 x 107 Ω
C. 6,750Ω
D. 3.6 x
6
10 Ω
REE – Oct.2006
27.
A 60 Hz, three phase transmission line delivers 20 MVA to a load
at 66 kV and 80% power factor lagging. The total series impedance of
each line is 15 + j75 ohms. If nominal “pi” circuit is used, what would
be the transmission efficiency if the admittance is j6 x 10 -4mhos?
A. 90.8%
B. 91.7%
C. 94.67%
D. 95.90%
POWER SYSTEM
28.
A single circuit, 60 Hz, three – phase transmission line is 300 mile
long and has the following parameters: R = 0.30Ω/mi, L = 2.10 mH/mi,
C = 0.014µF/mi. What is the surge impedance loading of the line if the
receiving end voltage is 132 kV?
A. 45 MW
B. 54 MW
C. 15 MW
D. 38 MW
REE – Sept. 2007
29.
A three – phase, 115 kV, 60 Hz transmission line has a per phase
series impedance of Z = 0.05 + j0.45 ohm/km and a per phase shunt
admittance of Y = j3.4 x 10 raised to negative 6 Siemens/km . The line
is 120 km long. Using the nominal Pi line model, determine the
transmission line D constant.
A. 0.999 + j0.001248
C. 0.989 + j0.001224
B. 0.969 + j0.001199
D. 0.001212
MULTIVECTOR REVIEW AND TRAINING CENTER
30.
For a line terminated by its characteristics impedance, the reflected
wave is equal to ____________.
A. Zero
C. half of the incident wave
B. B. equal to incident wave
D. twice of the
incident wave
31.
A transmission line delivers 1200 MW at 500 kV and 85% power
factor. The series impedance of the line is 0.42 + j 0.65 ohms/mi and the
shunt impedance to neutral is – j0.12 + 106 ohms – mi. What is the
velocity of propagation of the line?
A. 560 x 106 km/hr
B. 890 x 106 km/hr C. 750 x 106 km/hr
D. 510 x 106 km/hr
32.
Which of the following protects a cable against mechanical injury
during handling and laying?
A. Serving
B. armoring
C. sheath
D. bedding
33.
A three – core sheathed cable is being tested at 15 kV. The
capacitance measured between any two conductors is 0.2 µF/km.
Determine the charging current per km of the cable at 60 Hz.
A. 0.65A
B. 1.3 A
C. 1.8 A
D. 2.6 A
REE – Oct. 1994
34.
A lead sheath cable for underground service has a copper
conductor (diameter = 0.35 inch) surrounded by 0.2 inch wall of rubber
insulation. Assuming a dielectric constant at 4.3 for rubber, calculate the
capacitance of the cable.
A. 1.01 µF/mile
B. 0.504 µF/mile
C. 0.76 µF/mile
D.
0.252 µF/mile
35.
The total resistance of a 500 m two – core distributor is 17 x 10-5
Ω/m Feeding points A and B are supplied at 255 V and 250 V,
respectively. Load currents of 120 A, 160 A and 100 A are taken from
points C, D and E which are 150 m 250 m and 400 m from feeding point
A. what is the point of minimum potential?
MULTIVECTOR REVIEW AND TRAINING CENTER
A. B
B. C
C. D
D. E
36.
An overhead transmission line has a span of 300 m. It is supported
by two towers of heights 20.5 m and 30.5 m, respectively. If the required
clearance between the conductor and the ground midway between
towers is 19 m and the weight of conductor is 0.80 kg/m, what is the
tension in the conductor?
A. 1524 kg
B. 1458 kg
C. 1232 kg
D. 1385 kg
37.
The conductor of the transmission line has a diameter of 19.6 mm
and weighs 0.865 kg/m. Its ultimate strength is 8070 kg. If the
permissible slant sag with a horizontal wind pressure of 3.91 g/cm2 is
6.28 m, calculate the maximum span between two consecutive level
supports considering a factor of safety of 3
A. 171 m
B. 296 m
C. 342 m
D. 592 m
POWER SYSTEM
38.
A string efficiency of 100% means that _____________.
A. One of the insulator discs is shorted
C. potential across
each disc is the same
B. The potential across each disc is zero
D. potential across the
large disc is very large
39.
A string of three insulators is used to suspend one conductor of a
33 kV, three – phase overhead line. the air capacitance between each
cap/pin junction and the tower is 1/10 of the capacitance of each unit.
Determine the string efficiency.
A. 86.8%
B. 88.6%
C. 68.8%
D. 78.6%
REE – Oct.1996
40.
The causes of nearly all high voltage flashovers in transmission
lines are due to one of the following. Which one is the following?
A. High humidity
C. corona
B. dust and dirt
D. lightning and discharge
MULTIVECTOR REVIEW AND TRAINING CENTER
41.
a 500 kV line has a total corona loss of 280 kW. When energized
at 230 kV, the corona loss is 42 kW. What will be the corona loss if
used to transmit power a voltage of 385 kV?
A. 117 kW
B. 153 kW
C. 207 kW
D. 183 kW
REE – April 2001
42.
The percent impedance of a line is 6% at 34.5 kV and 100 MVA
base. What is the ohmic impedance?
A. 2.32
B. 3
C. 0.72
D. 1.2
REE – Sept. 2006
43.
A single phase transmission line supplies a reactive load at lagging
p.f. The load draws 1.2 per unit current at 0.6 per unit voltage while
drawing 0.5 per unit power. If the base voltage is 20 kV and the base
current is 160 A, what is the p.f. of the load?
A. 0.80
B. 0.75
C. 0.85
D.
0.69
REE – April 2002
44.
At a certain point in an electric network, the available fault MVA
is 400 A, 15 MVA, 34.5 kV, 2.5% impedance, wye – grounded
transformer is installed at that location. Determine the short circuit MVA
at the secondary side of the transformer.
A. 600
B. 625
C. 240
D. 500
REE – April 2001
45.
At a 34.5 kV substation, the available fault current is 10 pu. What
is the available fault MVA if the base is 50 MVA?
A. 50 MVA
B. 100 MVA
C. 250 MVA
D. 500 MVA
REE – Sept. 2002
46.
At a certain point in a 69 kV transmission line, the positive
sequence impedance is j0.15 p.u., and the zero sequence impedance is
MULTIVECTOR REVIEW AND TRAINING CENTER
j0.55 p.u. Calculate the fault current if a line fault occurs. The base is 50
MVA.
A. 3,511 A
B. 1,890 A
C. 420 A
D. 2,414 A
47.
At a certain point in a system, the thevenin’s equivalent of the
network is 0.2 pu at 100 MVA base. A 115/34.5 kV, 10 MVA
transformer with 5% impedance is tapped at this point. If a three – phase
fault occurs at the secondary, find the fault current at the primary.
A. 2390 A
B. 718 A
C. 1380 A
D. 1240 A
POWER SYSTEM
REE – April 2001
48.
The transformer used to serve a customer is rated 5 MVA,
13.8/0.48 kV, 5% impedance. The cable connecting the breaker to the
transformer has an impedance of 0.032 ohm per phase. What is the fault
current if a three – phase fault occurs at the breaker?
A. 8,000 A
B. 5,000 A
C. 6,000 A
D. 1,200 A
49.
NPC supplies energy to Clark Development Corporation at 69 kV
from a 50 MVA transformer whose impedance is 4%. The short circuit
MVA at the primary of the transformer is 600 MVA. To limit the fault
current, a three – phase limiting reactor is connected on the secondary of
the transformer. What is the reactance of the reactor required to limit the
short circuit MVA at the secondary to 200 MVA.
A. 0.127 pu
B. 0.167 pu
C. 0.173 pu
D. 0.139 pu
50.
Surge arrester is needed in transmission lines for the following
purpose. Which one is important?
A. Regulate the voltage
C. Protect the system from
high transients
B. Prevent the lightning from striking the line
D. Protect the
line from short circuit currents
REE – Oct. 2000
MULTIVECTOR REVIEW AND TRAINING CENTER
51.
What arrester rating shall be used to protect an 11 kV ungrounded
system?
A. 18 kV
B. 30 kV
C. 12 kV
D. 9 kV
REE – April 2002
52.
The distribution system is 34.5 kV, grounded. Which arrester shall
be installed to protect a distribution transformer on the system?
A. 27 kV
B. 30 kV
C. 34.5 kV
D. 38 kV
53.
Which of the following is not one of the classes of arresters?
A. Transmission class
B. Station class
C. distribution class
D. intermediate class
REE – April 1997
54.
In transmission lines, the most effective protection against
lightning strikes is one of the following. Which of one is this?
A. Lightning rods
C. Peterson coils
B. Lightning arresters
D. Overhead wires
REE – Sept. 2001
55.
A combination of switch and fuse.
A. Fuse cut – out
B. relay
D. circuit breaker
C. safety switch
REE – April 1997
56.
The transformer bus scheme has the following characteristics
except one. Which one is this?
A. It allows the disconnection of circuit breaker for maintenance without
interrupting the service
B. It is more flexible
C. It allows better continuity of service as compared with the single bus
D. It is more costly than that of the single bus system
MULTIVECTOR REVIEW AND TRAINING CENTER
Power Plant engineering
It is the art of designing and installing generating plant that will result in
maximum return on investment over the expected life of the equipment. And
also operating this equipment to achieve reliable, continuous and cheap power
service.
Power Plant. A station or establishment which houses the prime mover,
electric generators and auxiliaries for conversion of mechanical, chemical and
or nuclear energy into electrical energy.
Types of power Plant (As to Source of Energy)
1. Thermal Power Plant – an electric generating station using heat as a
source of energy.
a) Oil Fired Steam – makes used of heavy fuel oil, light fuel oil or
bunker oil for production of energy.
b) Coil Fired – makes used of pulverized coal as a fuel.
c) Dendro – makes used of wood such as ipil – ipil as a fuel.
d) Nuclear Steam – makes used of steam generated in a reactor by
heat from the fission process of nuclear fuel (Uranium 235,
Uranium 238)
e) Gas Turbine – makes used of combustible gases as a fuel from a
gas turbine engine prime mover.
f) Geothermal – makes used of generated heat from the inherent
steam from the earth’s magma fuel.
g) Solar Steam – makes used of steam generated from solar radiation.
2. Hydro Electric Power Plant – an electric generating station using
flowing water to drive the prime mover
(Hydraulic Turbine)
MULTIVECTOR REVIEW AND TRAINING CENTER
Hydraulic Turbine – sometimes referred to as water wheel, is the
most important element in hydraulic power plant.
Types of Hydraulic Turbine
a) Reaction Turbine – the water under pressure is partly converted into
velocity before it enters the turbine runner.
1. Francis Type – The water enters the spiral case from the penstock,
passes through the stay ring guided by the stationary stay ring
vanes, then through which it flows into a tailrace or tail water
reservoir.
2. Propeller – It is the same as Francis Type but it has an
unshrounded blades.
3. Axial Flow Turbine – is a propeller type runner with either fixed or
adjustable blades.
b) Impulse Turbine – The water under pressure is entirely converted into
velocity before it enters the turbine runner. It consists of one or more
free jets of water discharging into an aerated space and impinging on
a set of buckets attached around the periphery of a disk or wheel.
Turbine Components:
1. By pass valve
2. Scroll case
3. Guided Vanes
tube
10. Turbine Sleeve ring
11. Head and Button Covers
12. Discharge Ring and Draft
MULTIVECTOR REVIEW AND TRAINING CENTER
4. Wicket Gates
5. Wicket Gates Gland
6. Wicket Gate Bushing
7. Operating Ring
8. Operating Servomotor
9. Governor
13. Runner and Cone
14. Tail water and Tailrace
15. Surge Chamber
16. Vacuum Valves
17. Diffuser Tripod
Types of Hydroelectric Plant
a) Run – of – river- using pondage or stream flows as it occurs, more
power can be generated in a rainy season than in dry season.
b) Plant with storage capacity – associated with large reservoir. This
permits regulated supply of water so that the power output is constant
through the year
c) Pump Storage Plant – where energy is generated during the periods of
high system demand using water, which has been pumped, into reservoir
usually during periods of relatively low system demands.
3. Diesel Power Plant – a plant of internal combustion engine (ICE) prime
mover using diesel as fuel in producing energy.
4. Windmill Plant – using a series of windmills as prime mover
5. Sea Wave Ocean / Ocean Tides Plant - makes used of natural rising of
tide to simulate flowing water.
6. Solar or Photo Voltanic Plant – chemical conversion of radiant energy of
the sun to electric (DC) energy.
Types of power Plant (As to Use)
MULTIVECTOR REVIEW AND TRAINING CENTER
1. Base Load Plant – Plant that assumes load requirement under normal
condition.
2. Peaking plant – Plant that is normally operated to provide power only
during peak load periods.
3. Regulating Plant – plant capable of carrying load for the interval either
during off – peak or peak period and usually respond to the change in
system frequency.
4. Reserve or Stand by Plant – For peak or system deficiencies.
Types of Plant Reserves
a) Cold Reserve – portion of the installed reserve kept in operable
condition and available for service but not for immediate loading.
b) Operating Reserve – refers to the capacity in actual service in excess of
peak load.
c) Hot Reserve – refers to the unit available, maintained at operating
temperature and ready for service although not in actual operation.
d) Spinning Reserve – generating capacity connected to the bus ready to
take load.
Definition of Terms
1. Demand – the load at the receiving terminals averaged over a specified
interval of time.
2. Capacity – the rated load – carrying ability expressed in Kilovolt –
Ampere or Kilowatts of generating equipment or other electric
apparatus.
3. Capability - the maximum load carrying ability expressed in Kilovolt –
Ampere or Kilowatts of generating equipment or other electric apparatus
under specified condition for a given interval of time.
MULTIVECTOR REVIEW AND TRAINING CENTER
4. Net Capability – the maximum generation expressed in Kilowatt – hour
per hour which a generating unit; station, power source, or system can
be expected to supply under optimum operating conditions.
5. Maximum Demand or Peak Load (MD) – it is the largest demand of
load on the power station during a period of time.
6. Minimum Demand – the smallest of a particular type of demand
occurring within a specified period.
7. Connected Load (CL) – it is the sum of all continuous rating of all
utilization equipment in the consumer premises connected to the supply
system.
8. Average Load / Average Demand (AL) –the average loads occurring
on the power station in a given period of time.
9. Load Diversity – the difference between the sum of the maxima of two
or more individual loads and the coincident or combined maximum load,
usually measured in kilowatts over a specified period of time.
10.
Coincident Demand –any demand that occurs simultaneously
with any other demand; also the sum of any set of coincident demand.
11.
Non coincident Demand – the sum of the individual maximum
demand regardless of time of occurrence within a specified period.
12.
Ratchet Demand – the maximum past or present demand, which
is taken into account to establishing billings for the previous or
subsequent period.
MULTIVECTOR REVIEW AND TRAINING CENTER
13.
Seasonal Diversity – load diversity between two (or more) electric
systems which occurs when their peak loads are in different seasons of
the year.
14.
Off peak Energy – energy supplied during designed periods of
relatively low system demand.
15.
Off peak Energy – energy supplied during designed periods of
relatively high system demand.
16.
Demand Factor (DF) - It is the ratio of maximum demand on the
power station to its connected load
MD
DF = ≤ 1
CL
17.
Diversity Factor – it is the ratio of the sum of the individual
maximum demands to the simultaneous maximum demand on the power
station.
=
Σ MD
Simulataneous MD
≥1
18.
Coincidence Factor – the ratio of the maximum coincident total
demand of a group of costumers to the sum of the maximum power
demand of the individual consumers comprising the group both taken at
the same point of supply for the same time.
19.
Load Factor (LF) – it is the ratio of the average load (AL) to the
maximum demand during a given period.
=
AL
MD
=
TEP
MD x Sh
operating hours
Sh= service hours or
MULTIVECTOR REVIEW AND TRAINING CENTER
20.
Capacity Factor – it is the ratio of the actual energy produced to
the maximum possible energy that could have been produced during a
given period.
TEP
=
IC = installed capacity or rated
IC x Pr
capacity as an individual
Pr = Period
Capacity factor – It is the product of the Utilization factor (UF) and
local factor (LF)
21.
Plant Use (Output) Factor – it is the ratio of the energy generated
to the product of plant capacity and the number of hours for which the
plant was in operation.
=
TEP
IC x Sh
22.
Utilization factor (UF) – it is the ratio of the maximum generator
demand to the generator capacity.
=
MD
IC
23.
Operating Plant Factor – it is the ratio of the average load for a
given period to the operating capacity in actual service only (for multi –
set plant)
=
TEP
OPC x Sh
OPC = Operating Plant Capacity
24.
Operating Load Factor – it is the ratio of the average load for a
given period of time to the average of daily maximum demand during
the same period.
MULTIVECTOR REVIEW AND TRAINING CENTER
25.
Operating (Service) Factor – it is the ratio between the capacity
factors (CF) to the plant use factor (PUF) or it is the ratio of service hour
the period hour.
S
= h
Pr
26.
Plant Ratio – it is the per annum measure of the plant’s capacity
factor of the ratio of Net k Wh output for the year to the nameplate
capacity of the plant times hours of the year.
=
𝐓𝐄𝐏 𝐩𝐞𝐫 𝐲𝐞𝐚𝐫
𝐈𝐂 𝐱 𝐏𝐫 (𝐨𝐫 𝟖𝟕𝟔𝟎)
27.
Connected Load Factor – it is the product of the load factor (LF)
and the demand factor (DF).
Economics of Power Generation
Characteristics of an Electric Rate (or Tariff)
The following elements enter into the cost of the electric energy to consumers.
1. Fixed Element – to start the enterprise includes the following
a. Power plant, lads, building structure, etc.
b. Primary distribution lines and substation including supports
c. Management, administrative staff
d. Depreciation
e. General maintenance required whether the plant is in operation or
not.
2. Energy Element – sometimes called running cost, it is the operating
expense to generate power/ energy which includes the following:
a. Fuels
b. Labor of Operation
c. Facilities, water, oil supply
d. Maintenance required when the plant is in operation only
MULTIVECTOR REVIEW AND TRAINING CENTER
3. Customer Element – cost to bring power to the users/ consumers which
includes the following
a. Secondary distribution system, including service drop and meter
b. Personnel’s for lines, meter readers, collection bills
c. Advertisement
d. Franchise (50 yrs. period)
4. Investors profit/ Return on Investments - controlled by the
government
Note:
Fixed Element is dependent on the size of enterprise
Energy Element is dependent on the volume/ level of the generated
KW or Kwh
Customer Element is dependent on the number of customer
Types of Tariff
1. Uniform Rate Tariff – there is a fixed rate per kwh consumed.
2. Flat Rate Tariff – when different types of consumers are changed at
different uniform rates. It is generally used for domestic purposes.
3. Block Rate Tariff – when a block of energy is charged at a specified
rate and the succeeding blocks of energy are changed at progressively
reduced rates. This tariff has the advantage of encouraging energy
consumption. (This is also called “room rates”).
Ex.
P 1.75 per unit for the first 100 units
P 1.5 per unit for the next 200 units
P 1.00 per unit for the rest
4. Two Part Tariff – there is a fixed charge on the basis of their maximum
demand and running charge based on the energy consumed. For the
fixed charge, either a maximum demand indicator is installed, or
otherwise the charge is based on the installed capacity (it is also called
“two charge tariff”).
MULTIVECTOR REVIEW AND TRAINING CENTER
5. Maximum Demand Tariff – I is similar to two part tariff with the only
difference is that maximum demand is actually measured by installing a
maximum demand meter in the premises of the consumer. It is mostly
applied to the large users.
6. Power Factor Tariff – power factor of consumers load is taken into
consideration.
Billing (KWH) = meter reading x Fk
Fk = 1.0 if pf is from 80.1% to 85.0%
Fk> 1.0 if pf is lower than 80.1% (Fk is from 1.01 to 1.09)
Fk< 1.0 if pf is higher than 85.0% (Fk is from 0.95 to 0.99)
Most Economical Power Factor
∅2 = corrected power factor angle
From Calculus:
Sin ∅2= Y/X
∅2 ∅1
X – P per KVA of maximum demand (fixed
charge in tariff)
Y – P per annum per improvement plant cost
∅1 = original power factor angle
Leading kVAr supplied by power factor correction equipment
= kW (tan ∅1±∅2)
∅1= original power factor angle
power factor
∅2 = corrected
MULTIVECTOR REVIEW AND TRAINING CENTER
a. Maximum KVA demand Tariff – in this case instead of Kw, KVA
consumption is recorded and the charges are estimated partly or fully on
this demand.
b. Sliding scale – in this case the average power factor is fixed say 0.8
lagging
Power Factor Improvement Equipment
1. Static Capacitor
Three
phase
Three
phase
2. Synchronous Condenser
Three
phase
3. Phase Advancers.
MULTIVECTOR REVIEW AND TRAINING CENTER
7. Three Phase Tariff – also called “Three charge rate” od “Doherty rate”;
total charge is split into three.
Rate Making
1. Domestic Consumer
2. Industrial Consumer
3. Public Utilities
4. Public Institution
5. Commercial Consumer
MULTIVECTOR REVIEW AND TRAINING CENTER
POWER PLANT
REE – Oct. 1998
1. To produce 1 kW – hr a power plant burns 0.9 lb. of coal with a heating
value of 13,000 BTU per lb. What is the heat rate of the plant?
A. 6,250 BTU/ kW – hr
C. 9,500 BTU/ kW hr
B. 11,700 BTU kW/hr
D. 8,700 BTU/ kW - hr
REE – Oct.1998
2. A 100 MW power plant has a heat of 2.88 x 106 calories per kW – hr. It
is a base load plant and runs at full load 24 hours a day. How Many tons
of coal is needed per day?
A . 250
B. 625
C. 830
D. 960
REE – April 1997
3. An effective remover of dust, carbon particles and other from the flue
gas of the power plant is called ________.
A . Electrostatic Precipitator
C. Mechanical
Collector
B . School Blower and collector
D. Dust Scrubber
REE – March 1998
4. In fission , energy is produced when
A . Chemical Compound splits to its constituent
B . nuclear particle splits
C. Nuclear particles combine
D. Molecules combine
REE – march 1998
5. In the list below which one is not a type of present day nuclear power
plant.
A . boiling water
C. low pressure reactor
B . Fusion – Fission reactor
D. fast breeder reactor
MULTIVECTOR REVIEW AND TRAINING CENTER
REE – April 1990
6. A thermal power plant using dry organic material such as wood and
other wastes for its fuel.
A . Geothermal power plant
C. Dendro power plant
B . Diesel power plant
D. Hydro power plant
REE – Oct. 1994
7. Type of water wheels used in hydroelectric plants.
A . propeller
B. impulse
C. reaction
these
D. all of
REE – April 2001
8. A hydroelectric plant generates 100 MW at an available head of 200 m
and an overall efficiency of 75%. What quantity of water in cubic meter
per second in required.
A . 1,200
B. 122
C. 26
D. 68
REE – Oct. 1998
9. A ten – year investigation of rivers potential gave an average water flow
of 25 cm per second at the bottom and 90 cm per second at the surface.
The average cross section at the same location is 80 sq. m. What is the
average flow rate in cubic meter per hour?
A . 165,600
B. 57,500
C. 46,500
D. 5,760
POWER PLANT
REE – March 1998
10.
A power plant gets water from the dam from a height of 122.45
meters at the rate of 1,000 cubic meters per minute. If the output of the
plant is 15,000 kW. What is the efficiency?
A . 80%
B. 70%
C. 75%
D.
65%
REE – Sept. 2002
MULTIVECTOR REVIEW AND TRAINING CENTER
11.
A power plant is driven by a turbine at a rate of 10 cubic meters
per second and with a head of 235 m. If the plant efficiency is 0.65,
calculate the power plant output.
A . 10,000 KW
B. 15,000 KW
C. 35,000 KW
D. 25,000 KW
12.
A hydroelectric station has a turbine efficiency of 86% and a
generator efficiency of 92%. The effective head of water is 150 m.
Calculate the volume of water used when delivering a load of 40 MW
for 6 hours.
A . 74.18 x 104m3
B. 4.158 x 104m3
C. 82.54 x 103m3
D. 94.48 x 104m3
13.
An industry having a maximum demand of 1000 KW works for
320 days in a year, 8 hours per day. Assuming that it works at full
capacity all the time during working days, calculate the daily load factor.
A . 56.2%
B. 67%
C. 33.33%
D. 29.5%
14.
What is the total energy generated by a station in KWHR in a year
if its maximum demand and annual load factor is 150 – KW and 45%
respectively?
A . 591,300
B. 1,314,000
C. 24,637
D.
67, 500
15.
A 100 MW power station delivers 100 MW for 2 hrs., 50 MW for
6 hrs. and is shut for the rest of each day. It is also shut down for
maintenance for 45 days each year. Calculate the annual load factor.
A . 18.3%
B. 56.9%
C. 20.83%
D. 29.5%
16.
A power station has a load cycle as follows:
260 MW for 6 hrs.
200 MW for 8 hrs.
160 MW for 4 hrs.
100 MW for 6 hrs.
A . 61.1%
B. 34.5%
C. 23.6%
D. 76.5%
17.
The current loads of four circuits are as follows:
MULTIVECTOR REVIEW AND TRAINING CENTER
Circuit no.1 = 25 amperes
Circuit no. 3 = 18
amperes
Circuit no.2 = 38 amperes
Circuit no. 4 = 45
amperes
If the diversity factor is 1.5. What is the minimum ampacity of the
feeder conductor?
A . 126 amperes
B. 189 amperes
C. 83 amperes
D.
152 amperes
18.
The metering of power customer was read and gave the following
data:
KW – hr = 200,000 KVAR – HR = 180, 000
Maximum demand =
380 kW
What is the load factor?
A . 68%
B. 80%
C. 73%
D.
62%
REE – Sept. 2001
19.
The power customer is applying for source and listed the following
loads:
5 – 15 hp motors
lighting loads – 8 kw
3 – 5 hp motors
miscellaneous loads 3
– 5 kw
The demand factor is 65%. Load factor is 46% and the power factor is
80%. What is the maximum KW demand of the customer?
A . 78
B. 64
C. 69
D. 57
POWER PLANT
20.
The power customer has four circuits of 220 – volt three phase.
The circuits have the maximum demands as follows:
Circuit no.1 = 35 amperes
Circuit no. 3 = 72
amperes
Circuit no.2 = 46 amperes
Circuit no. 4 = 57
amperes
MULTIVECTOR REVIEW AND TRAINING CENTER
The diversity factor is 1.5, the load factor is 50% and the power factor is
80%. What is the approximate KW demand of the customer?
A . 15 KW
B. 21 KW
C. 43 KW
D. 53 KW
21.
The power customer has four circuits of 220 – volt three phase.
The circuits have maximum demand as follows:
Circuit no.1 = 35 amperes
Circuit no. 3 = 72
amperes
Circuit no.2 = 46 amperes
Circuit no. 4 = 57
amperes
The diversity factor is 1.5, the load factor is 50% and the power factor is
80%. Determine the maximum KVA of the customer?
A . 46
B. 80
C. 26.5
D. 53
22.
A diesel station supplies the following loads to various customers:
Industrial consumer = 1500 kw
Commercial
establishment = 750 kw
Domestic power = 100 kw
Domestic light = 50
kw
If the maximum demand on the station is 2500 kw and the number of
kwhr generated per year is 45 x 105, determine the diversity factor and
the load factor.
A . 1.12, 20.5%
B. 0.97, 31.2%
C. 1.53, 65.7%
D. 21.1, 50.2%
REE – Sept. 2004
23.
A distribution transformer has a connected load of 300 HP motor
operating at 70% load factor and 50% demand factor. The energy cost is
4 pesos per KWHR. What is the average KW load if the efficiency is
80%?
A . 280
B. 98
C. 140
D.
196
MULTIVECTOR REVIEW AND TRAINING CENTER
24.
A 100 MW coal fired power plant has an average heat rate of 9,500
BTU/kwhr. The plant load factor is 75% and the heating value of coal is
12,000 BTU/lb. Calculate the amount of coal usage for one day.
A . 1.425 x 106lbs.
B. 2.235 x 105 lbs.
C. 1.425 x 108 lbs.
D. 1.826 x 106lbs.
25.
A power plant has a maximum demand of 15 MW. The annual
load factor and the capacity factor are 50% and 40%, respectively.
Determine the reserved capacity of the plant.
A . 7350 kW
B. 5370 kW
C. 3750 kW
D. 3075 kW
REE – Oct. 1997
26.
A coal thermal power plant has a boiler, turbine and the alternator
efficiencies of 35%, 86% and 93%, respectively. Coal with heating value
of 12,000 BTU per lb. cost P1.5 per lb. What is the fuel cost in
producing 1 kWhr?
A . P 1.9 per kWHR
B. P 1.52 per kWHR C. P 1.75 per kWHR
D. P 1.62 per kWHR
REE – Oct. 1998
27.
A certain coal fired power plant has a heat rate of 2.88 x 106
calories per kW – hr. coal costs P 2,500 per ton. How much is the fuel
cost component of producing 1 kW – hr?
A . P 2.50
B. P 1.00
C. P 1.75
D. P 1.25
POWER PLANT
28.
A power station has a load cycle as under
260 MW for 6 hrs.
200 MW for 8 hrs.
160 MW for 4 hrs.
100 MW for 6 hrs.
If the power station is equipped with 4 sets of 75 MW each, calculate the
daily fuel requirement if the calorific value of the oil used were 10,000
Kcal/kg and the average heat rate of the station were 2,860 Kcal/
kWHR.
MULTIVECTOR REVIEW AND TRAINING CENTER
A . 1890 tons
D. 1258.4 tons
B. 3452 tons
C. 7641 tons
REE – Oct.1998
29.
A diesel generator set burns diesel with a heating value of 18,000
BTU per lb. The diesel engine has an efficiency of 30% and the
alternator has an efficiency of 95%. Determine the fuel cost component
of producing one kW – hr if diesel cost P2.8 per lb.
A . P0.15
B. P 2.1
C. 3.28
D.
1.86
REE – Oct. 2000
30.
A diesel power plant has an overall efficiency of 30% . If the
calorific value of diesel is 18,000 BTU/ lb. and cost P 6.00 per liter, how
much is the fuel cost of production per kWhr of energy? Take the
density of diesel as 900 grams per liter.
A . P1.91 / kW – hr
B. P1.85 / kW – hr C. P2.12 / kW – hr
D. P2.45 / kW – hr
REE – Sept. 2007
31.
The power generated by some steam power plants drops in the
summer because ___________.
A . there is no rain
C. many people go
swimming in the lake
B . the water in the lake decreases
D. the lake
temperature increases
REE – Sept. 2006
32.
Diesel power plant is best suited as
A . Peak load
B . Standby
REE – Sept. 2008
C. Base load
D. General purpose
MULTIVECTOR REVIEW AND TRAINING CENTER
33.
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 . 5.48
B. 48.75
C. 54.87
D. 4.87
34.
One advantage of computer control of generating unit is that
A . Var. outputs of the units are minimized
B . All units under the control of the computer will be loaded to the
same load
C. Loading of the units will be frequently adjusted to maintain them at
equal incremental cost
D. It can control all equipment
35.
Inadvertent accumulations between power systems are normally
balanced by scheduling compensating deliveries of power.
A . During off – peak load
B. During periods corresponding to those in which the inadvertent
energy was accumulated
C. On weekends
D. On base load plant
POWER PLANT
36.
Online load flow program of a SCADA system
A . Used the actual operating data
B . Used calculated valued of loading of lines and equipment
C . Cannot Provide transmission loss factor
D . Are used only on system designed function
37.
Spinning reserve is generating capacity that
A . Is available by starting gas – Turbine generating units
B. Is synchronized and online, with capacity in excess of existing load
C. Has the lowest fuel cost
D. A policy matter
MULTIVECTOR REVIEW AND TRAINING CENTER
38.
Heating value of coal largely depend on
A . ash content
B. Moisture content
D. size of coal particle
C. volatile matter
39.
Power plant using coal work closely on which of the following
cycle
A . Rankine Cycle
C. Otto Cycle
B . Binary Vapour cycle
D. size of coal
particle
40.
In steam turbine cycle, the lowest pressure occurs in
A . Turbine Inlet
B. Boiler
C. Condenser
heater
D. Super
41.
For low head and high discharge the hydraulic turbine used is.
A . Kaplan
B. Francis Type
C. Pelton Wheel D. Jonvas
42.
Overall efficiency of Gas Turbine is
A . Equal to Carnot cycle efficiency
cycle efficiency
B . Equal to Rankine cycle efficiency
diesel cycle efficiency
C. Less than Diesel
D. More than otto or
43.
The function of moderator in Nuclear Reaction is.
A . To accelerate reaction
C. To absorb excessive
neutrons
B . To control the reaction
D. To liberate excessive
neutrons
44.
During Load shedding
A . system voltage is reduced
factor is changed
B . some loads are switched off
is reduced
C. system power
D. system frequency
MULTIVECTOR REVIEW AND TRAINING CENTER
45.
A low utilization factor for a plant indicates that
A . plant is used for based loads only
C. plant is under
maintenance
B . plant is used for standby purpose only
D. plant is used for
peak load
46.
Which of the following relations is correct?
A . Capacity factor = Utilization factor x Load Factor
B . Load factor x Maximum load = average load
C . Demand factor x Connected load = maximum demand
E. None of the above
47.
A diversity factor of 2.5 gives a saving of ________ % in
generating equipment .
A . 60
B. 50
C. 40
D. 25
CONTROL SYSTEM
System – an arrangement, set or collection of things connected or related in
such a manners as to form an entirely or whole.
Control system- an arrangement of physical components connected or related
in such a manner as to command, direct, or regulate itself or another system.
Input – is the stimulus or excitation applied to a control system from an
external source, usually in order to produce a specified response from the
control system.
Output – is the actual response obtained from a control system. it may or may
not be equal to the specified response implied by the input.
MULTIVECTOR REVIEW AND TRAINING CENTER
Feedback – is that property of a closed – loop system which permits the
output (or some other controlled variable of the system) to be compared with
input to the system (or an input to some other internally situated component or
subsystem of the system) so that the appropriate action may be formed as
some function of the output and input.
Block Diagram – is a shorthand, pictorial representation of the cause and
effect relationship between the input and output of the physical system.
Input; stimulus CONTROL
system
Desired Response
Output; Response
Actual Response
Input
Output
system
r(t)
c (t)
(a)
Input Subsystem
Subsystem
r(t)
Output
c(t)
(b)
Note:
Subsystem
The input, r(t), stands for reference input
The output, c(t), stands for controlled variable
CONTROL SYSTEM
MULTIVECTOR REVIEW AND TRAINING CENTER
Canonical Form of a Feedback Control System
R
+
E
C
G
±
B
H
G = direct transfer function
function = control ratio
= forward transfer function
C/R = closed – loop transfer
H = feedback transfer function
error ratio
E/R = actuating signal =
GH = loop transfer function
= open loop transfer function
B/R = primary feedback ratio
In the following equation, the (-) sign refers to a positive feedback system,
and the (+) sign refers to a negative feedback system.
C/R = G / (1 ± GH)
B/R = GH / (1 ± GH)
E/R = 1/ (1 ± GH)
Item No.
f(t)
F(s)
1
𝛿(𝑡)
1
MULTIVECTOR REVIEW AND TRAINING CENTER
Item No.
1
2
u(t)
3
tu(t)
4
tn u(t)
5
-at
e u(t)
6
Sin 𝜔 tu(t)
7
cos 𝜔 tu(t)
4
5
6
7
sn+1
1
s+a
ω
s 2 + ω2
s
s 2 + ω2
CONTROL SYSTEM
Theorem
Name
∝
L[ƒ(t) ] = F(s) = ∫ ƒ (𝑡)e – Definition
0
st
2
3
1
s
1
S2
n!
dt
L[kƒ (t) ] = kF (s)
L[ƒ1(t) + ƒ2 (t) ] = F1(s) +
F2(s)
L[e-at ƒ(t) = F (s+ a)
L[f(t – T)] = e–sT F(s)
1
𝑠
𝑎
𝑎
L[ƒ(at)] = F ( )
𝑑ƒ
Linearity Theorem
Linearity Theorem
Frequency Shift Theorem
Time Shift Theorem
Scaling Theorem
L[ ]= sF(s) – ƒ(0-)
Differentiation Theorem
8
𝑑𝑡
𝑑2 ƒ
L[ 2 ]= s2 F(s) – sƒ(0-) –
𝑑𝑡
ƒ(0-)
Differentiation Theorem
9
L[ 𝑛 ]= s2 F(s) –
𝑑𝑡
𝑛
∑𝑘−1 𝑠 𝑛−𝑘 ƒ k-1 (0-)
𝑑𝑛 ƒ
Differentiation Theorem
10
L [∫0− ƒ (𝜏)𝑑𝜏]=
Integration Theorem
11
1
ƒ(∝) = lim sF (s)
𝐹(𝑠)
𝑠
Final Value Theorem1
MULTIVECTOR REVIEW AND TRAINING CENTER
s
12
→0
ƒ(0 +) = lim sF (s)
s
→∝
Initial Value Theorem2
CONTROL SYSTEM
MULTIVECTOR REVIEW AND TRAINING CENTER
Voltage – current, voltage – charges and impedance relationships for
capacitors, resistors and inductors
𝟏
𝟏
𝒅𝒗(𝒕)
v(t)= ∫𝟎 𝒇(𝝉)𝒅𝝉
𝐂
i(t) = C
v(t) = Ri (t)
i(t) = v(t)
𝟏
𝟏
𝐂
𝐂𝐬
v(t) = q(t)
𝒅𝒕
Cs
capacitor
𝟏
𝐑
𝟏
v(t) = R
𝒅𝒒(𝒕)
R
𝒅𝒕
=𝑮
resistor
𝑹
v(t) = L
Ls
Inductor
𝟏
𝒅𝒊(𝒕)
𝟏
𝟏
v(t)= ∫𝟎 𝒇(𝝉)𝒅𝝉
𝐋
𝒅𝒕
v(t) =
𝒅𝟐 𝒒(𝒔)
𝒅𝒕𝟐
𝑳𝒔
Note: The following set of symbols and units is used throughout this book:
v(t) = V (volts). i(t) = A (amps.), q(t) = Q (coulombs), C = F (farads), R = Ω
(ohms), G = Ʊ (mhos.), L = H (henries).
Component
Impedance
Force-
Force -
Velocity
Displacement
ZM(s) = F(s)
X(s)
Spring
v(𝝉)
f(t)
Viscous Damper
v(t)
𝟏
f(s) = K ∫𝟎 𝒗(𝒕)𝒅𝒕
f(s) = Kx(t)
K
MULTIVECTOR REVIEW AND TRAINING CENTER
f(s)
f(s) = fv(t)
f(t) = fv
𝒅𝒙(𝒕)
fv
𝒅𝒕
fN
mass
v(t)
M
f(s) = M
𝒅𝒗(𝒕)
𝒅𝒕
f(t) =M
𝒅𝟐 𝒙(𝒔)
𝒅𝒕𝟐
Ms2
f(t)
Note: The following set of symbols and units is used throughout this book.
F(t) = N (newtons), v(𝜏) = m (meters), v(t) = m/s(meters/ second), K = N/m
(newtons/meters) fN= Nms m (newtons* seconds/meter), M = ky 1kilograms =
newton* seconds2/meters
CONTROL SYSTEM
Component
Torque –
Angular
Torque –
Angular
Velocity
Displacement
M(𝜹)
Spring
𝟏
T(t) = K ∫𝟎 𝝎(𝝉)𝒅𝝉
K
K
Viscous
T(s) = K𝜽 (t)
Impedance
ZM (𝜹) = T(𝜹)
MULTIVECTOR REVIEW AND TRAINING CENTER
damper
T(t) = D𝝎(𝒕)
T(t) = D
𝒅𝜽(𝒕)
𝒅𝒔
Ds
D
inertia
T(s)= j
ds
J
2
𝒅𝝎(𝒕)
𝒅𝒕
T(s) = d
𝒅𝟐 𝜽(𝒕)
𝒅𝒕𝟐
Note: The following set of symbols and units is used throughout this book:
T(s) = N-m(newton – meters), D(t) = rad(radians), 𝜔(t) rad/s (radian /second,
K = N –m rad (newton – meters/radian, D = N – m-s/rad(newton- metersseconds/ radian), J = kg-m2 (kilogram- meters2 = newton – meters – second
MULTIVECTOR REVIEW AND TRAINING CENTER
CONTROL SYSTEM
1. Find the inverse Laplace transform of
2
F(s) = (s+1)(s+2)
A. (2 e-t - 2 e-2t) u (t)
B. (2 e-t - 2 e-2t)
C. (2 e-2t - 2 e-t) u (t)
D. 2 e-t
2. Given the following differential equation, solve for y(t) if all initial
conditions are zero
d2 y
dt2
+
12dy
dt
+ 32 y = 32 u (t)
A. (1 – 3e-4t + e -7t )u(t)
)u(t)
B. (1 – 5e-2t + 6e -8t )u(t)
C. (1 – 2e-4t + e -8t
D. (1 – 2e-4t)u(t)
3. Find the transfer function represented by
A.
1
(s+1)
B.
1
(s+2)
dc(t)
dt
+2c(t) = r(t)
C.
1
s
D. 1
4. In an RL circuit, it is given that L = 3H, R = 6Ω, voltage source is 24V
dc and i (o) = 1A. Determine the current at t >0
A. 4 - 3e-2t
B. 3 - 4e-2t
C. 4 - 3e-t
D. 4 - 3e-2t- 5e-5t
5. In a RC circuit, it is given that R = 4Ω, C = 1/8F, voltage source is 14V
dc, and voltage across capacitor is 2V at t = 0. Determine the equation of
the current at t >0
A. 2e-2tA
B. 3e-2tA
C. 5e-2tA
D. 6e-tA
MULTIVECTOR REVIEW AND TRAINING CENTER
6. In a RL circuit, it is given that L is 1H, R = 2Ω and the voltage source is
an impulse function δ(t). Determine the i(t) at t >0
A. 2e -2t u(t)A
B. 3e -t u(t)A
C. e -4t u(t)A
D. e -2t u(t)A
7. In a RLC circuit, determine the transfer function wherein the capacitor
voltage is the output and the voltage source is the input. All initial
conditions are zero.
1/LC
1/ LsC
A. 2 Rs
s + L +1/LC
B. 2 Rs
C. 2
1
s +RC/S+1/LC
s + L +1/LC
1 /LC
D. 2 Rs
s + L +1/LC
8. In a transitional mechanical system, the spring constant is K, the
coefficient of viscous friction is fv and the mass M. Determine its series
analog.
A. Inductor =M
C. Inductor =M
Resistor = fv
Capacitor = 1/K
Voltage source = applied force
force
Mesh current = velocity
Resistor = fv
Capacitor = K
Voltage source = applied
Mesh current = velocity
CONTROL SYSTEM
B. Inductor =fv
D. Inductor =M
Resistor = M
Resistor = fv
MULTIVECTOR REVIEW AND TRAINING CENTER
Capacitor = 1/K
Voltage source = applied force
force
Mesh current = velocity
Capacitor = K
Voltage source = applied
Mesh current = velocity
9. In a transitional mechanical system, the spring constant is K, the
coefficient of viscous friction is fv and the mass M. Determine its series
analog.
A. Capacitor = 1/M
C. Capacitor = M
Resistor =1/ fv
Inductor = 1/K
Current source = velocity
velocity
Node voltage= applied force
applied force
B. Capacitor = M
Resistor =1/ fv
Inductor = 1/K
Current source = velocity
applied force
Node voltage= applied force
velocity
Resistor =1/ fv
Inductor = 1/K
Current source =
Node voltage=
C. Capacitor = M
Resistor = fv
Inductor = K
Current source =
Node voltage=
10.
In the canonical form of the feedback control system, it is given
that C = input signal, R = output signal, G = forward transfer function
and H = feedback transfer function. Determine the closed – loop transfer
function of control ratio.
MULTIVECTOR REVIEW AND TRAINING CENTER
A.
G
B.
1±GH
1
C.
1±GH
GH
1±GH
D.1 ± GH
11.
In a canonical feedback control system w/a forward transfer
function of G and a feedback transfer function H, determine the open
loop transfer function and the error ratio (or actuating signal ratio).
A. GH,
1
B. G,
1±G
G
C. GH,
1±GH
1
D. G,
1±GH
GH
1±GH
12.
In a canonical feedback control system w/a forward transfer
function of G and a feedback transfer function of H, determine the
primary feedback ratio.
A.
GH
B.
1±GH
G±H
.
G
C.
1±GH
1±GH
D.
G
G
13.
In a canonical feedback control system the forward transfer
K
function is 1 and the feedback transfer function is K2S. Determine
s(s+p)
the error ratio.
s+p
A.
s+p− K1 K2
s+p+ K1 K2
D.
B.
s+p
s+p+ K1 K2
C.
s+p
K1 K2
s+p
14.
Two blocks are in series. One block has a transfer function of P1
and the other one has transfer function P2. The equivalent transfer
function of the two blocks would be
P P
A. P1/P2
B. P1P2
C. P1 + P2
D. 1 2
P1 +P2
CONTROL SYSTEM
15.
In an undamped second – order system, the peak time is defined as
the time
MULTIVECTOR REVIEW AND TRAINING CENTER
A. Required to reach the maximum value
B. Required to reach the minimum value
C. Required to reach either the minimum or maximum value
D. Required to go to 0.9 of the final value
16.
In an undamped second – order system the time required for the
transients damped oscillations to reach and stay within ± 2% of the
steady – state value.
A. Setting time
B. rise time
C. fall time
D. over time
17.
In the transfer function G (s) = 2
frequency and the damping ratio.
A. 8, 0.35
B. 7, 0.35
36
, determine the natural
S +4.2 s+36
C. 6, 0.23
D. 6, 0.35
MULTIVECTOR REVIEW AND TRAINING CENTER
Electronic Communication System
The conveyance of information from a source to a destination through
the use of electronic and / or electrical devices.
Antenna
Wireless
Information
Transmitter
information
Receiver
Wired
Information makes take the form of any of the following:
1. Voice
2. Music
within the audio
3. Picture
range = 20 Hz to 20 kHz
4. Code
5. Data
Division of Frequency Spectrum
Band
Frequency
ELF (Extremely Low Frequency)----------------------------------------------Below
300 Hz
ILF (Infra Low Frequency) -----------------------------------------------------300 to
3000 Hz
MULTIVECTOR REVIEW AND TRAINING CENTER
VLF (Very low Frequency)-----------------------------------------------------3 to 30
kHz
LF (Low Frequency) ------------------------------------------------------------ 30 to
300 kHz
MF (Medium Frequency)------------------------------------------------------- 300
kHz to 300 kHz
HF (High Frequency)------------------------------------------------------------ 3 to
30 MHz
VHF ( Very High Frequency) -------------------------------------------------- 30
MHz to 300 MHz
UHF (Ultra High Frequency) -------------------------------------------------- 300
MHz to 3000 MHz
SHF (Super high Frequency) -------------------------------------------------- 3 GHz
to 30 GHz
EHF ( Extremely High Frequency) ------------------------------------------- 30
GHz to 300 GHz
THF (Tremendously High Frequency) --------------------------------------- 300
GHz to 3000 GHz
*Infrared Region ----------------------------------------------------------------- 300
GHz to 300 THz
*Light Region--------------------------------------------------------------------- 300
THz to 3,000 THz
*Voice Frequency ---------------------------------------------------------------- 300
to 3400 Hz
*Velocity of Electromagnetic Waves (RF) in free space = 3 x 108 m/sec
= 186,000 miles/second
= 982,000,000 ft./sec
*Velocity of Sound in air = 1130 ft./sec
344.4 m/sec
Fundamental and Harmonics
MULTIVECTOR REVIEW AND TRAINING CENTER
Fundamental frequency – the lowest frequency component that can be
produced by an instrument or device
Harmonics – multiples of the fundamental
Example: If 300 Hz is the fundamental, its 2nd , 3rd,4th,….. harmonics are 600
Hz, 900 Hz, and 1200 Hz and so on, respectively.
*Octave – range of frequencies wherein the ratio of the highest to the lowest
frequency is equal to two.
Transmission Modes:
a. Simplex – information is sent in only one direction
b. Half Duplex – information is sent in both directions but not
simultaneously
c. Full – duplex – information is sent in both directions simultaneously.
Tx
Rx
Tx
Rx
Rx
Tx
b. half – duplex
a. Simplex
Tx
Rx
Rx
Tx
c . Full - duplex
Modulation – the process of putting in or superimposing an information
(modulating signal) inside an RF (radio frequency) “carrier” wave.
- The process of varying some parameters or characteristics of the
“carrier” (RF) wave (e.g., frequency or amplitude) in accordance
to the information wave variation.
MULTIVECTOR REVIEW AND TRAINING CENTER
Demodulation – getting or retrieving information contained or superimposed
from an RF carrier wave.
- The opposite or reverse process of modulation.
Types of modulation
1. Amplitude Modulation (AM)
2. Angle modulation
a. Frequency Modulation (FM)
b. Phase Modulation (PM)
Advantages/ Need(s) for modulation:
1. To allow the use of practicable size antenna
2. To separate differing transmissions for proper transmission and
reception
3. To overcome difficulties in transmission at very low frequencies.
Amplitude Modulation (AM)
The amplitude of the RF carrier is varied by the amplitude of the
modulating signal (information).
Em
Emax
Ec
Emin
Modulation Index (m)
=
=
Emax − Emin
Emax + Emin
Em
Ec
MULTIVECTOR REVIEW AND TRAINING CENTER
Instantaneous AM Voltage
CC(t) = ECsin𝜔Ct +
Carrier
mEc
2
cos (𝜔c– 𝜔m ) t
mEc
2
cos (𝜔c– 𝜔m )t
low side band (LSB) upper side band (USB)
Where:
𝜔c = carrier radian frequency
𝜔m= modulating (radian) frequency
EC = carrier voltage amplitude
 AM Broadcast Band = 525 to 1705 kHz
Channel bandwidth = 9 kHz + 1 kHz Guard Band = 10 kHz
Power in AM
Pt = PC + PLSB + PUSB
= PC +
m2
4
Pt = PC (1 +
m2
PC +
m2
2
4
)
It = IC√1 + m2 /2
where: Pt = total transmitted power
PC
PC = carrier power
PLSB = power in the lower side band
PUSB = upper side band power
m = modulation index
Frequency Modulation
The frequency of the carrier is varied by the amplitude of the modulating
signal whereas its amplitude is held constant.
Modulating
Wave
Output FM
waveform
MULTIVECTOR REVIEW AND TRAINING CENTER
*FM broadcast Band = 88 to 108 HHz
*Channel Bandwidth = 200 KHz
FM Equation, CC (t):
CC(t) = ECsin(𝝎c t– mfsin𝝎mt)
Where: EC = carrier voltage
𝜔c= carrier radian frequency
𝜔m = modulating (radian) frequency
mf = FM modulating index
Forms of AM Transmission
1. Double sideband Full Carrier (Standard AM ) – A3E
2. Single sideband Suppressed Carrier – J3E
3. Single Sideband Reduced Carrier – R3E
4. Single Sideband Full Carrier – H3E
5. Independent sideband Emission (ISB) – B8E
6. Vestigial Sideband – C3F
Total Modulation Index by several Waves (AM)
mt2 = m12+ m22+ m32 + ………+ mn2
Where:
mt= resultant modulation index
m1, m2, m3,…..mn = individual modulation indices
Bandwidth of AM Transmission (BW) = 2fm
Where:
fm= highest modulating frequency
Modulation Index of FM Transmission
MULTIVECTOR REVIEW AND TRAINING CENTER
mt=
δ
fm
where:
𝛿 = maximum frequency
fm = modulating frequency
FM Bandwidth
1. Approximate bandwidth (Carson’s Formula)
BW = 2 (𝛿 +fm )
2. Exact bandwidth
BW = 2 x fm x no. of significant sidebands pairs.
PRINCIPLES OF COMMUNICATIONS
Television System and Standards
American standard – National Television Standard Committee (NTSC)
European standard – Phase Alternation by line (PAL)
France – Sequential Technique and Memory Storage
MULTIVECTOR REVIEW AND TRAINING CENTER
Fiber Communication System
- A communication system that uses light as the carrier of
information
Analog or
Digital
Interface
Voltage to
Current
Converter
Light Source
Source to
Fiber
Interface
Fiber to
Light
Detector
Interface
Light
Detector
Current to
Current
Converter
Analog or
Digital
Interface
Fiber Types
1. Plastic Core and Cladding
2. Glass Core with plastic Cladding (PCS)
3. Glass Core and Glass Cladding (SCS)
Refraction – bending of signal due to different densities of the medium
Modes of Propagation
1. Single Mode – one path
2. Multimode – more than one path
Index Profile – value of reactive index across the fiber
Index Types:
MULTIVECTOR REVIEW AND TRAINING CENTER
1. Step Index – central core uniform refractive index, cladding uniform
also
2. Graded Index – refractive index is highest at the center and decreases
gradually until it reaches the index of the cladding.
Configurations:
1. Single mode step index
2. Multimode step index
3. Multimode graded index
Antenna – a length of wire or metal tubing designed to radiate and receive
electromagnetic waves through/from free space.
Basic Types:
1. Ungrounded (half wave) = Hertz
2. Grounded (Quarter Wave) = Marconi
Antenna Gain & Effective Radiated Power (ERP)
1. Hertzian dipole – 1.5:1 = 1.76 dB
2. Half Wave Dipole – 1.64:1 = 2.15 dB
ERP = Pfed *Ap
Networking
Types of Network:
1. LAN – Local Area Network
- Limited within a building or cluster of building
2. MAN – Metropolitan Area Network
- Formed by connecting 2 LAN, communication spreads over a city
or state.
3. WAN – Wide Area Network
- Communications spreads over countries and Continents.
Network Topology
1. BUS
MULTIVECTOR REVIEW AND TRAINING CENTER
2. STAR
3. RING
Transmission Lines
a. Parallel two – wire line: (balanced line)
Space
s
d
MULTIVECTOR REVIEW AND TRAINING CENTER
ZO =
276
2S
Ω
where:
d = wire diameter
s = separation distance
k = permittivity constant
ZO = characteristic impedance
b. Coaxial Cable: (Unbalanced Line)
√k
log
d
inner conductor
jacket
d
outer conductor
ZO =
138
√k
log
D
d
Ω
D
Where: d = outer diameter of the inner conductor
D = inner diameter of the outer conductor
ZO = characteristic impedance
k = permittivity constant
Equivalent Circuit:
Characteristic Impedance, ZO:
MULTIVECTOR REVIEW AND TRAINING CENTER
ZO = √
R+jωL
G+jωC
≈√
L
C
Where:
R = resistance per unit
G = conductance per unit length
L = inductance per unit
C = capacitance per unit
PRINCIPLES OF COMMUNICATION
1. The process of varying some characteristics of a high frequency sine
wave in accordance with information signal is called
A. Communication
B. multiplexing
C. propagation
D. modulation
2. Which of the following characteristics of a sine wave can be modulated?
A. Amplitude
B. angular phase C. angular velocity D. all
of these
3. The following are advantages of modulation except
A. Allow the use of practical – size antenna
B. Separate differing transmission
C. Circuit simplicity
D. Overcomes difficulty in transmission at low frequencies
4. In AM, carrier amplitude variation depends on modulating signal’s
A. Amplitude
C. amplitude and frequency
B. Frequency
D. none of these
5. A 400 watts carrier is modulated to a depth of 75 percent. Calculate the
total power in the modulated wave?
MULTIVECTOR REVIEW AND TRAINING CENTER
A. 512.5w
B. 8.47w
C. 530w
D. 515.5w
6. A broadcast radio transmitter radiates 10 kw when the modulation
percentage is 60. How much of this is carrier?
A. 8.2kw
B. 8.47 kw
C. 5.3 kw
D. 6.8 kw
7. The difference frequency of the carrier and modulating signal is also
known as
A. Lower side frequency
C. middle frequency
B. Upper side frequency
D. bass frequency
8. Assume a frequency range of 405.0125 MHz to 405.0.875 MHz and 25
kHz channeling plan, determine the center frequency of the second
channel from the lower limit.
A. 405.030 MHZ
B. 405.025 MHZ
C. 405.050 MHZ
D. 405.075 MHZ
9. A certain transmitter radiates 9 kw with the carrier un modulated and
10.125 kw when the carrier is sinusoidally modulated. Calculate the
modulation index.
A. 1
B. 0.50
C. 0.60
D.
0.70
10.
From Prob. No. 9. If another sine wave corresponding to 40
percent modulation is transmitted simultaneously determine the total
radiated power?
A. 10.84 kw
B. 6.56 kw
C. 6.19 kw
D. 5.19 kw
PRINCIPLES OF COMMUNICATION
MULTIVECTOR REVIEW AND TRAINING CENTER
11.
A single sideband suppressed carrier transmission in AM uses the
code
A. A3E
B. R3E
C. J3E
D.
H3E
12.
The instantaneous voltage of a modulated FM carrier is given by v
= 10 sin(750 x 106t t + 10 sin 1000 t). The carrier frequency is
A. 750 MHz
B. 1000 MHz
C. 119.37 MHz
D. 1000 Hz
13.
Which of the following is not true about AM
A. The carrier amplitude varies
B. The carrier frequency remain constant
C. The carrier frequency changes
D. The information signal amplitude changes the carrier amplitude
14.
The circuit used to “superimpose” the intelligence on the carrier
wave is also as the
A. Modulator
C. variable gain amplifier
B. Demodulator
D. Multiplexer
1. For ideal AM, which of the following is true?
A. M = 0
B. m = 1
C. m < 1
D. m >1
2. Find the carrier and modulating frequencies, modulation index and the
maximum deviation of the FM wave represented by the voltage equation
v = 12 sin (6x 108 + 5 sin 1250t).
A. 199 Hz, 5995 Hz
B. 199 Hz, 990 Hz
C. 199 Hz, 6,980 Hz
D. none of these
MULTIVECTOR REVIEW AND TRAINING CENTER
3. From (16) what power will this FM wave dissipate in a 10 Ω resistor.
A. 7.2 w
B. 7.9 w
C. 6 kw
D. 7.2 kw
4. Most of the power in a “standard” AM signal is in the
A. Carrier
B. upper sideband
C. lower sideband
modulating signal
D.
REE – April 2003
5. Which of the following carries the most number of channels?
A. 50 multiplexer
C. 100 chips
B. 100 pairs of cables
D. 1 fiber optic
REE – April 2005
6. What is the optical source for the transmitting end of an optic fiber
cable?
A. Hologen
C. injection Laser Diode
B. LED
D. Photocell
REE – Sept. 2004
7. What is the most commonly used fiber optic cable?
A. Single mode step index
C. multi-mode step index
B. Single mode graded index
D. multi-mode graded
index
PRINCIPLES OF COMMUNICATION
8. The ratio of the maximum deviation to the modulating frequency being
used is referred to as
A. Modulation index
C. deviation ratio
B. Percent modulation
D. carrier swing
REE – Sept. 2003
9. What is the coon unit of the sound absorption coefficient?
MULTIVECTOR REVIEW AND TRAINING CENTER
A. Watt
B. erg
C. Sabine
D. decibel
10.
Speaker is a device that ____________.
A. Converts current variations into sound waves
B. Converts electrical energy to mechanical energy
C. Converts sound waves into current and voltage
11.
Tendency of sound energy to spread.
A. Reflection
B. diffraction
D. refraction
C. rarefaction
12.
___________ is used to measure speech volume.
A. Volume meter
C. Volume Unit Meter
B. Audio frequency meter
D. Speech meter
13.
A large speaker having a large diameter (15 cm and above)
A. Coaxial speaker
B. woofer
C. tweeter
D. triaxial
speaker
14.
Given the modulation index is 5 and the maximum variation in
carrier is 75 kHz, the maximum modulating frequency is
A. 10 kHz
B. 11 kHz
C. 12 kHz
D. 15 kHz
15.
The wavelength of a sound of a 20 kHz frequency is
A. 16.5 m
B. 16.5 cm
C. 16.5 mm
mm
D. 165
16.
A technique that enables more than one data source to share the
use of a common facility
A. Companding
B. multiplexing
C. sharing
D. all of
these
17.
A hardware and software rules and procedures for making sure that
any transmission errors are detected.
MULTIVECTOR REVIEW AND TRAINING CENTER
A. Standard
protocol
B. parity
C. hamming code
D.
18.
The input is 0.1 W and the network gain is 13 dB, the output is
A. 2.0 W
B. 2.5 W
C. 1.5 W
D. 1. W
19.
A ground wave is
A. Surface wave
B. Tropospheric wave
and sky waves
C. sky wave
D. combination of space
REE – Sept. 2005
20.
What replaces coaxial cable for terrestrial transmission?
A. Copper cable
C. aluminum cable
B. Point to point wireless transmission
D. fiber optics
PRINCIPLES OF COMMUNICATION
21.
The farthest distance from the earth a satellite obit reaches is called
A. Perigee
B. line of apsides
C. altitude
D. apogee
22.
Satellite engine use
A. Liquid fuel
B. Jet propulsion
C. ion propulsion system
D. solar jet
23.
Station located on an object which is beyond and is intended to go
beyond the major portion of the earth’s atmosphere.
A. Terrestrial station
C. satellite station
B. Space station
D. mobile satellite station
24.
Interconnects devices within a building or campus
A. WAN
B. LAN
C. MAN
D. internet
25.
What is the optimum length of a Marconi antenna for the
transmission of a 300 MHz signal? The antenna velocity factor is 0.85.
MULTIVECTOR REVIEW AND TRAINING CENTER
A. 0.22 m
B. 0.23 m
C. 0.21
D. 0.24 m
26.
When testing transmitter to prevent interfering with other stations,
which type of antenna must be used?
A. Dummy antenna
B. Hertzian antenna C. void antenna
D. none of these
27.
What mode of transmission is being used when the communication
is strictly one way?
A. Semi simplex
B. real – time
C. simplex
D. duplex
28.
One octave above 600 Hz is
A. 601 Hz
B. 800 Hz
Hz
C. 1400 Hz
D. 1200
29.
If the receive signal is 600 kHz in an AM broadcast, what is the
local oscillator frequency?
A. 600 KHZ
B. 145 KHZ
C. 1055 KHZ
D. 455 KHZ
30.
What will be the percent modulation of an FM signal, which is
being broadcast in 88 – 108 MHz band, having a carrier swing of 125
kHz?
A. 82.3%
B. 83.3%
C. 85%
D. 90%
31.
The gain of a Hertzian dipole with respect to an isotropic antenna
is
A. 1.76 dB
B. 2.15 dB
C. 4.15 dB
D. 5.15 dB
32.
Between any two adjacent super groups there is a void band of
frequencies called
A. Empty band
B. null band
C. guard band
D. clearance band
REE – Sept. 2003
MULTIVECTOR REVIEW AND TRAINING CENTER
33.
In 1885, who invented antenna for a long distance transmission of
electromagnetic waves?
A. Benjamin Franklin
B. Guglielmo Marconi
C. Isaac Newton
D. Carl Siemen
REE – April 2004
34.
An antenna that transmits and receives equally well in all direction
is said to be _________.
A. Omnidirectional
B. Bidirectional
C. Unidirectional
D. Quasidirectional
PRINCIPLES OF COMMUNICATION
35.
A device used to enable two separate transmitter to operate on a
single antenna is called
A. Duplexer
B. switch
C. combiner
D.
mixer
36.
The __________ angle measured the satellite position clockwise
from the direction of true north
A. Elevation
B. azimuth
C. depression
D.
critical
37.
The carrier swing necessary to provide 80% modulation in the FM
broadcast band is ___________.
A. 150 kHz
B. 120 kHz
C. 60 kHz
D. 75
kHz
38.
Radiated power per unit area
A. Air density
B. sound density C. power density
signal density
D.
REE – April 2004
39.
It refers to a device that merely boosts a signal and passes it on
A. Amplifier
B. booster
C. op – amp
D.
repeater
MULTIVECTOR REVIEW AND TRAINING CENTER
40.
If the transmitter output is 700 watts, transmission lines loss is 100
watts and the antenna gain is 10, what is affective radiated power?
A. 610 kW
B. 6 kW
C. 811 kW
D. 7 kW
41.
The earth’s area that the satellite can receive from or transmit to
A. Primary area
B. footprint
C. secondary area
D.
coverage area
REE – April 2004
42.
A type of communication that uses light rather than electricity to
communicate is called ___________.
A. Laser carrier
B. Microwave
C. Power line carrier D.
fiber optic
REE – Sept. 2007
43.
What contains in each cell site of cellular telephone system?
A. Repeater
C. direct – link to branch
B. Touch tone processor
D. control computer
MULTIVECTOR REVIEW AND TRAINING CENTER
P
In an n – type semiconductor the three
electrons are the majority carries, while
the holes are the minority carriers. In a p
– type semiconductor, the holes are the
majority carriers while the free electrons
are the minority carriers.
n
Fig.1
Schematic symbol for a rectifier diode
R
VS
VD
Fig. 2
Forward – biased diode
When an external voltage opposes the
barrier potential, the diode is forward –
biased. If the applied voltage is greater
than the barrier potential, the current is
large. In other words, current flows
easily in a forward – biased diode.
R
VS
VD
When an external voltage aids the
barrier potential, the diode is reverse –
biased. The width of the depletion layer
increases when the reverse voltage
increases. The current is approximately
zero. In other words, a reverse –biased
diode acts approximately like a switch.
MULTIVECTOR REVIEW AND TRAINING CENTER
Fig. 3
reverse – biased diode
Diode approximations:
First approximation
IDEAL
I
V
reverse = biased
Fig. 4
Ideal diode curve
forward biased
Common – emitter
configuration
RC
Rb
VCC
The emitter is grounded or common is
in a CE circuit. The base – emitter part
of the transistor acts like an ordinary
diode. The base – collector parts acts
like a current source that is equal to 𝛽dc
times that the current.
Vbb
Common - base
configuration
In the CB connection, the
common or ground side of each
supply is connected to the base.
MULTIVECTOR REVIEW AND TRAINING CENTER
Re
Rc
Vee
Vcc
Other Semiconductor devices
Anode
Silicon controlled rectifier
(SCR) – this thyristor has an
extra lead called a gate. By
applying a trigger to the gate, we
can turn on the SCR.
At high frequencies,
the varactor acts the
same as a variable
capacitance.
Triac – acts like two SCR’s in
parallel. On account of this, the
triac can control current in either
direction.
Opt isolator – a
package that contains
an infrared LED and a
photo detector. The
wavelength response of
each is tailored to be as
identical as possible to
permit the highest
measure of coupling
possible.
gate
cathode
anode
gate
cathode
drain
gate
Mosfet (metal Oxide
semiconductor field effect
transistor) – off when the gate
source voltage is zero. Fully on
when gate – source voltage is
around 15V.
Second Approximation
MULTIVECTOR REVIEW AND TRAINING CENTER
2ND Approximation
I
0.7 V
V
Reverse- biased
Forward - biased
Third Approximation
I
3RD Approximation
0.7 V
V
Reverse- biased
Forward - biased
Rectifiers
Half wave
A rectifier diode conducts
current in one direction and
opposes direction. On account of
this, it is utilized to convert
alternating current to direct
current.
Full Wave
MULTIVECTOR REVIEW AND TRAINING CENTER
Center – tapped
ELECTRONICS
1. Electrons at the outermost shell are called
A. Free electrons
B. outer electrons
D. active electrons
Bridge Type
C. valence electrons
2. Which has the least number of valence electrons?
A. Conductor
C. insulator
B. Elemental semiconductor
D. compound
semiconductor
3. Energy required by a valence electron before it can move towards the
conduction band.
A. Energy gap
B. conduction band C. work function
D. valence band
4. A semiconductor that is free from impurities
A. Intrinsic semiconductor
C. extrinsic semiconductor
B. Compensated semiconductor
D. elemental semiconductor
5. The process of adding impurities in a semiconductor material
MULTIVECTOR REVIEW AND TRAINING CENTER
A. Growing
depleting
B. diffusion
6. If the substance used in doping has less
`
A. acceptor
B. donor
pentavalent
7. P – type semiconductor has an excess
A. Electron
B. hole
neutron
8. Donor – doped semiconductor becomes a
A. N – type semiconductor
B. P-n type semiconductor
C. doping
C. trivalent
C. proton
D.
D.
D.
C. good conductor
D. N – type semiconductor
REE – Oct. 1997
9. A semiconductor frequently used to rectify AC current to DC is called
A. Zinc oxide
B. transistor
C. tube
D. diode
REE – April 1997
10.
A germanium diode which has an offset voltage of 0.2 V and an
incremental resistance of 20 ohms is connected in series with a 10,000
ohm resistance across a 100 V DC source. What is the current?
A. 9.98 A
B. 10.0 A
C. 9.78 A
D. 9.96 A
REE – April 1997
11.
The resistor is connected across an AC supply of 220 V. The
power drawn is 1000 W. If a diode were connected in series with the
resistor, what would be the power absorbed by the resistor?
A. 850 W
B. 250 W
C. 200 W
D. 500 W
REE – April 1997
MULTIVECTOR REVIEW AND TRAINING CENTER
12.
A simple half – wave rectifier consists of a diode and a load. If the
internal resistance of the diode is 1 Ω and the load is 5 Ω, what is the
DC load current if the supply voltage is 12 V AC?
A. 2.7 A
B. 2.4 A
C. 2.0 A
D. 3.6 A
13.
The forward voltage across a 1N4001 diode is 0.93 V when the
current is 1A. What is the power dissipation for these values?
A. 0.93 W
B. 0.1 W
C. 0.0 W
D. 1.0 W
ELECTRONICS
REE – Oct. 1996
14.
A circuit consists of a resistor which is connected in series with a
perfect diode. When connected across a 20 V battery, then current
measured 10 A. However, when connected across a 30 V battery but
with reverse polarity, what would be the new current?
A. 15 A
B. zero current
C. 20 A
D.
none of these
15.
A half wave rectifier, utilizing a silicon diode with a bulk
resistance of 0.25Ω, has a secondary terminal voltage 12 V (rms) and a
resistance of 0.1 Ω. Assuming a load resistance of 20Ω, determine the
average load voltage and average load current.
A. 4.983V, 0.455 A
B. 5.021 V, 0.355 A C. 5.093 V, 0.255 A
D. 6.013 V, 0.155 A
16.
A full –wave rectifier has a ripple voltage which is approximately
_______ that of a half – wave rectifier
A. One – fourth
B. one half
C. twice
D. thrice
17.
The pulsating DC voltage output of a full waver rectifier without a
filter capacitor can be used in a
A. Battery charger
B. radio
C. computer
D. stereo system
MULTIVECTOR REVIEW AND TRAINING CENTER
18.
In order to measure the voltage signals from a filtered rectifier
circuit, DC and AC voltmeters were used. The readings were 30 V DC
and 2.5 V rms respectively. Determine the ripple of the output voltage.
A. 9%
B. 8.33%
C. 7%
D.
7.33%
19.
In a full – wave bridge rectifier with a capacitor filter, the output
voltage is 30 V with a ripple of 9%. Assuming an input frequency of 60
Hz, determine the ripple voltage (rms) and its output frequency
A. 1V. 120 Hz
B. 30 V, 100 Hz C. 0 V, 60 Hz
D.
2.7 V,120 Hz
20.
A center – tapped transformer full – wave rectifier has a 0.5
kilohm load. The secondary voltage of the transformer is 24 V (rms).
The type of silicon diode used has a resistance of 0.30 ohm. Find the
average output voltage and rectifier efficiency.
A. 10.50 V, 81.06%
B. 9.3 V, 83.5%
C. 11.35 V, 98.35%
D. 24 V, 100.0%
REE – Sept. 2004
21.
An ideal full wave rectifier with a non-inductive load has zero
transformation losses. What I the efficiency?
A. 100
B. 90
C. 80
D. 70
22.
Suppose s bridge rectifier has a DC load current of 10 mA and a
filter capacitance of 470 µF. What is the peak to peak ripple out of a
capacitor – input filter?
A. 0.155 V
B. 0.024 V
C. 0.177 V
D. 0.24 V
REE – Oct. 1996
23.
An unregulated power supply which consists of a transformer, a
rectifier and a filter has some characteristics which are as follows
EXCEPT one. Which one is this?
A. It has good voltage regulation
B. It is relatively expensive
MULTIVECTOR REVIEW AND TRAINING CENTER
C. The output voltage varies with the load
D. The output voltage depends on the input voltage
ELECTRONICS
24.
This is special diode optimized for operation in the breakdown
region.
A. Light – emitting diode B. zener diode
C. rectifier diode
D. schottky diode
25.
The resistance of a semiconductor is known as
A. Bulk resistance
C. extrinsic resistance
B. Intrinsic resistance
D. dynamic resistance
26.
A device is formed when an N – type and P – type semiconductors
are brought together.
A. Pn junction
C. depletion region
B. Semiconductor junction
D. junction diode
27.
The minimum voltage required before the diode can totally
conduct in a forward direction
A. Triggering voltage
B. saturation voltage C. breakdown voltage
D. threshold voltage
28.
Half wave rectifiers has a maximum efficiency of
A. 40.6%
B. 78.5%
C. 81.2%
D. 50%
29.
Diode circuit that is used to cut a portion of the input signal.
A. Clipper
B. clamper
C. peal detector
D. level
shifter
30.
Ripple factor of a full wave rectifier.
A. 0.24
B. 2.42
0.48
C. 1.21
D.
MULTIVECTOR REVIEW AND TRAINING CENTER
31.
The output of a full wave rectifier with an input frequency of 60
Hz.
A. 60 Hz
B. 120 Hz
C. 200 Hz
D. 100 Hz
32.
The voltage output of a bridged rectifier is
A. Half wave signal
C. sine wave
B. Bridged rectified
D. full wave signal
33.
Half wave rectifiers has a DC voltage of 0.138 Vmax while a
bridged type has
A. 0.385 Vmax
B. 0.305 Vmax
C. 0.636 Vmax
D. 2
Vmax
34.
The ripple of a filter output voltage with the following output
signal measurement of Vdc = 30 volts and vrms = 1.5 V is
A. 9.5%
B. 95%
C. 5%
D.
50%
35.
Zener diode normally requires _________ biasing.
A. Forward
B. reverse
C. maximum
impedance
36.
Zener diodes are used as
A. Rectifiers
B. switches
D. regulators
D.
C. amplifiers
37.
Two resistors R1 (with a resistance 1K) and R2 (with resistance of
3K) are connected in series. The resistors draw power from a 16 V DC
power supply. A zener diode with a breakdown voltage of 10 V is then
connected across R2. Determine the power dissipation across the zener
diode
A. 0.05 W
B. 0.04 W
C. 0.03 W
D. 0.02 W
ELECTRONICS
MULTIVECTOR REVIEW AND TRAINING CENTER
38.
A parallel combination of a germanium diode and a silicon diode is
connected is series with a 1K resistor. A 10 V DC source is then
connected to the circuit in such a way that its positive terminal is
connected to the anodes of the diodes and the negative terminal to the
ground. Determine the voltage across the 1K resistor
A. 0 V
B. 0.3 V
C. 9.7 V
D. 10 V
39.
A transistor has __________ doped regions
A. 1
B. 2
C. 3
D. 4
40.
What is one important thing that transistors do?
A. Amplify weak signals
C. emit sound
B. Rectify line voltage
D. emit light
41.
An advantage of a transistor is
A. Small size
B. lack of heater C. it lasts indefinitely D. all
of these
42.
A transistor has a collector current of 2 mA. If the current gain
135, what is the base current?
A. 14.8 µA
B. 15.8 µA
C. 16.8 µA
D. 17.8 µA
43.
In a CE connection, the base supply voltage is 10 V and the base
current – limiting resistance is 100 K. the collector – emitter voltage is
6.4 V, collector current is 1 mA and collector – current limiting
resistance is 3.6 K. Determine the required collector supply voltage and
the transistor power consumption.
A. 10 V, 6.4 mW
B. 6.4V, 6.4 mW
C. 12 V, 4.6 mW
D. 5.6 V, 7.8 mW
44.
In a CE connection, the base and collector supply voltages are both
15 V. the base and collector current limiting resistances are 470K and
3.6 K respectively. Determine the collector emitter voltage.
A. 8.03 V
B. 2.30 V
C. 3.56 V
D. 4.06 V
MULTIVECTOR REVIEW AND TRAINING CENTER
45.
The reason, why a metal tab of a power transistor is fasten to the
chassis of an electronic equipment is
A. To hold the transistor firmly
B. For aesthetic value purposes
C. For heat to escape easily
D. To contribute to the rigidly of the equipment
46.
The relationship between DC alpha and DC beta is
A. βdc = αdc / (1 - αdc)
C. βdc = αdc
B. βdc = (1 - αdc)/ αdc
D. βdc = αdc/ (1 - αdc)
REE – April 1998
47.
it is very powerful small component of a computer. Which one?
A. Diode
B. triode
C. chip
D. internet
REE – Sept. 2004
48.
What is the equivalent of the collector terminal of transistor in
FET?
A. Grid
B. gate
C. source
D. drain
49.
A JFET
A. Voltage – controlled device
B. Current controlled device
controlled device
ELECTRONICS
C. both a and b
C. frequency –
50.
The gate – source voltage across a JFET is normally
A. Positive
B. negative
C. both a and b
D. 0
51.
The depletion – mode mosfet normally ________ when the gate –
source voltage is zero.
A. Is off
C. does not depend on gate
voltage
B. Is on
D. fails
MULTIVECTOR REVIEW AND TRAINING CENTER
52.
An enhancement mode mosfet normally ________ when the gate –
source voltage is positive.
A. Is off
C. does not depend on gate
voltage
B. Is on
D. fails
53.
A semiconductor device that can couple an input signal to the
output circuit is
A. Opt isolator
B. pulse transformer C. coupler
D.
diac
54.
The ____ diode is a diode with almost zero reverse recovery time.
A. Varactor
B. SCR
C. schottky
D. light emitting
55.
The _____ can be used for turning radio and television sets.
A. Varactor
B. opt isolator
C. triac
D. Schmitt trigger
56.
The _____ has been designed to control powers as high as 10 MW
with individual ratings as high as 2000 A at 1800 V.
A. Power mosfet
B. power transistor C. SCR
D. op
– amp
57.
A SCR has a gate and anode current – limiting resistance of 1K
and 100 ohms respectively. The anode supply voltage is at 15 V. What is
the output voltage when the SCR is off? What is the input voltage that
triggers the SCR? If the anode supply voltage is decreased until the SCR
opens, what is the corresponding value of the anode supply voltage?
Assume a trigger voltage of 0.7, an on voltage of 0.7V, a trigger current
of 7mA, and a holding current of 6mA.
A. 15 V, 7.75 V, 1.3 V
C. 12 V, 6 V, 0 V
B. 14 V, 9.75 V, 2.4 V
D. 10 V, 5 V, 0 V
MULTIVECTOR REVIEW AND TRAINING CENTER
58.
A 2N4441 SCR (having a trigger voltage at 0.75) is utilized
together with zener diode (having a breakdown voltage of 5.6 V) and a
resistance of 68 ohm to form crowbar. Calculate the supply voltage that
runs on the SCR.
A. 6.65 V
B. 6.55 V
C. 6.45 V
D. 6.35 V
59.
In a circuit which contains an inductor and a transistor, a free –
wheeling diode is often included in order to
A. Protect the inductor
C. both a and b
B. Protect the transistor
D. rectify the AC waveform
60.
Ideally. For fast switching, the base emitter voltage should start at
zero volts and the base – current should be at large at switch on, but
rapidly settle down to the minimum required for saturation. Also, switch
– off should be accomplished by a large reverse bias voltage which
quickly returns to zero. This problem can easily be solved by means of a
A. Schottky diode
C. SCR
B. Commutating capacitor
D. zener diode
ELECTRONICS
61.
A proto transistor can be used for
A. High isolation
B. lighting control
of these
C. relays
D. all
62.
An inverting op – amp with a feedback resistance of R1 and
another resistance of magnitude R1 has a voltage gain of
A. –Rf/R1
B. Rf/R1
C. 1 + Rf/R1
D. 1
63.
An non - inverting op – amp with a feedback resistance of R1 and
another resistance R1 has a voltage gain of
B. –Rf/R1
B. Rf/R1
C. 1 + Rf/R1
D. 1 Rf/R1
REE – Sept. 2005
MULTIVECTOR REVIEW AND TRAINING CENTER
64.
A negative feedback amplifier has 0.5% of output voltage feedback
and anti-phase with it. The gain of the amplifier then is 150. Calculate
the amplifier gain without feedback.
A. 600
B. 450
C. 800
D. 700
REE – April 2006
65.
A triode has an anode – slope resistance of 12 kΩ and an
amplification factor of 15. Determine the stage gain if it is used as an
amplifier with an anode load consisting of a choke having a resistance of
20 kΩ and an inductance of 7,500 µH when the input voltage has a
frequency of 600 KHZ
A. – 12.17
B. – 15.65
C. 14.66
D. – 29.21
MULTIVECTOR REVIEW AND TRAINING CENTER