EE 51 POWER SYSTEM 1
I. General Background of an Electric
Power System
(Students enrolling this course are presumed to have comprehensive knowledge in
Mathematics, Circuit theories and theories and principles of operation of different
Electrical Machineries)
An electric power system is a network of electrical components deployed to supply,
transfer, and use electric power. An example of an electric power system is the grid that
provides power to an extended area. An electrical grid power system can be broadly
divided into the generators that supply the power, the transmission system that carries
the power from the generating station to the load, and the distribution system that feeds
the power to nearby homes and industries.
Learning Outcomes:
At the end of this unit, the learner will be able to:
● Identify and determine parameters involve in power system
● Employ related equations in calculating power and relationship between
Voltages and currents
Pre-Test:
Instructions:
Answer the each question/problem quietly and encircle the letter of
the correct answer.
1. At a lower temperature, transmission cables may be coated with ice, what will
happen to its resistance?
(a) increases (b) decreases
(c) the same
(d) increase slightly
2. For short transmission line where length is up to 50 miles, the parameters
considered are usually _____________.
(a) capacitance and inductance, the resistance is neglected.
(b) resistance and capacitance, inductance is neglected.
(c) resistance, inductance and capacitance
(d) Resistance and inductance, capacitance is neglected.
3. In a balance 3-phase star-connected system line voltage is
Compiled and Adapted by: JOY C. IMPERIAL, M.Eng’g, REE, LPT
EE 51 POWER SYSTEM 1
(a)
(b)
(c)
(d)
the phasor difference of the two phase voltages
the phasor sum of the two phase voltages
0.707 times the phase voltages
1.414 times the phase voltages
4. In transmission line conductors, maximum tension occurs in ______.
(a) the lowest point
(b) the supports
(c) the horizontal
(d) the vertical
5. Is the process of alternating the positions of each phase conductor of a line in
equal intervals so that each phase will have the same inductance.
(a) superposition
(b) interpolation
(c) transition
(d) transposition
6. When a unit weight is uniformly distributed along the horizontal or along its
span, a cable assumes the form of?
(a)
hyperbola
(b)
catenary
(c)
parabola
(d)
line
7. The following are reasons why grounded wires are installed on top of the
towers or poles of a line, except one.
(a) to shield the line from direct lightning strokes
(b) to reduce the amount of voltage induced on the line due to electrostatic
fields produced by charged clouds.
(c) to give provisions for neutral
(d) to give provisions for discharging surge current by means of lightning
arrester.
8. A line has GMD of 10.33-ft, GMR of 0.3732-inch and length of 96 km. What is
the admittance?
(a) 3.59 x 10-4 mho
(c) 2889.04 mhos
(b) 9.34 x 10-7 mho
(d) 42.26 mhos
9. In a delta connected system the line current is
(a)
(b)
(c)
(d)
1.414 times the phase current
phasor sum of the two phase currents
equal to the phase current
1.732 times the phase current
10. A (n) _______ stores and returns energy to a circuit while a(n) ________
dissipates energy.
(a) resistance, impedance
(c) inductor, resistor
(b) resistor, inductor
(d) inductor, reactance
11. For an RL circuit, the power factor cannot be less than ______ or greater than
_______.
(a)
1, 0
(b)
0, -1
(c)
0, 1
(d)
-1, 0
Compiled and Adapted by: JOY C. IMPERIAL, M.Eng’g, REE, LPT
EE 51 POWER SYSTEM 1
12. The advantage of using static capacitor to improve the power factor is because
they
(a)
(b)
(c)
(d)
are not variable
are almost loss free
provide continuous charge of power factor
none of these
13. The advantage of the star-connections over the delta connections for the same
phase voltage is that it gives
(a) step-down current
(b) extra step-up voltage
(c) extra step-up current
(d) extra step-up power
14. The inductance is responsible for the current to _____ the applied voltage.
(a)
lag
(b)
be in phase with
(c)
lead
(d)
be opposite
15. A line has GMD of 10.33-ft, GMR of 0.3732-inch and length of 96 km. What is
the line reactance?
(a) 96 Ohms
(c) 42 Ohms
(b) 26 Ohms
(d) 69 Ohms
Compiled and Adapted by: JOY C. IMPERIAL, M.Eng’g, REE, LPT
EE 51 POWER SYSTEM 1
Content:
The power system engineer
- the power system engineer should know the methods of load making studies,
fault analysis and stability studies and principles of economic load dispatch
because such studies affect the design and operation of the system and the
selection of the apparatus for its control.
The general principal divisions of an electric power system:
1. Generating Stations
2. Transmission Lines
3. Distribution Systems
4. Utilization
Figure 1.1 Structure of a power system and its equivalent one-line-diagram.
Compiled and Adapted by: JOY C. IMPERIAL, M.Eng’g, REE, LPT
EE 51 POWER SYSTEM 1
System modeling
SYSTEM MODEL
1. Circuit Model for synchronous machine
d. Fuse
Ra
Xs
e. Current Transformer
E
Vt
f. Power Circuit Breaker
2. Equivalent Circuit for Transformer
R1
g. Air Circuit Breaker
X1
h. Three-Phase Three-Wire
Delta-Connection
V1
a V2
i. Three-Phase Three-Wire
Y – neutral ungrounded
i. Three-Phase Three-Wire
Y – Neutral-Grounded
3. The One-Line Diagram
a. Machine or Rotating Armature
j. Ammeter and Voltmeter
A
b. Two-Winding Transformer
k. Potential Transformer
or
V
c. Three-Winding Transformer
Transmission lines are the connecting links between the generating stations and the
distribution systems and back to the other power systems over interconnections.
A distribution system connects all the individual loads to the transmission lines at
substations which perform voltage transformation and switching functions.
Voltage of large modern generators usually is in the range of 18 to 24 kV.
- Generator voltage is stepped-up to transmission levels in the range of 115 to
765 kV.
Standard
HV
EHV
UHV
: 115, 138 and 230 kV
: 345, 500 and 756 kV
: levels of 1000 to 1500 kV
Load study - is the determination of the voltage, current, power, and power factor or
reactive power at various points in an electric network under existing or
contemplated conditions of normal operation. It is essential in planning the
development of the system.
Economic dispatch – is the name given to the process of apportioning the total load on
a system between the various generating plants to achieve the greatest economy
of operation.
Compiled and Adapted by: JOY C. IMPERIAL, M.Eng’g, REE, LPT
EE 51 POWER SYSTEM 1
Transmission and Distribution system
Line Parameters:
A.
Resistance (Series)
- The ac resistance is equal to Ridc resistance of the conductor only if the
distribution of current throughout the conductor is uniform. At frequencies
of 60 Hz and below, the difference between ac resistance and the dc
resistance is less than 1% for conductor less than 350 MCM in cross
section. An increase in frequency causes non-uniform current density.
This phenomenon is called skin effect.
Skin Effect – the tendency of current to move outward the surface of
conductor.
Rac = K Rdc
K 
Eq. 1.1
1
1  F2
2
For:: Copper
Aluminum
F = 0.0106 d²f
F = 0.0063 d²f
where;
Rac = ac resistance; Rdc = dc resistance of each of the line
f = frequency in Hz ; d = diameter of conductor in inch
B.
Inductance (L)
 GMD  H/mi
L  0.7411x 10 3 log 

 GMR  phase
(For English Units)
[Henry per mile per conductor (per
phase)]
Eq. 1.2a
 GMD  H/m
L  2 x 10 7 ln 

 GMR  phase
(For S.I. Units)
[Henry per meter per conductor (per
phase)
Eq. 1.2b
Where: GMD = Geometric mean distance; GMR = Geometer mean radius;
L
= Series inductance
Transposition – is the alternating of position of each phase of the line in
equal intervals so that each phase has the same inductance.
C.
Capacitance (Shunt)
Cn 
0.0388 x 10  6
 GMD 
log 

 GMR 
F/mi
phase
(For English Units)
[Farad per mile per phase (neutral)]
Eq. 1.3a
Compiled and Adapted by: JOY C. IMPERIAL, M.Eng’g, REE, LPT
EE 51 POWER SYSTEM 1
(For S.I. Units)
2 π x 10 12
F/m
[Farad
per
meter
per
phase
(neutral)]
phase
 GMD 
ln 

Eq. 1.3b
 GMR 
Where: Cn – shunt capacitance of the line
Cn 
Note: GMD and GMR and r must have the same units.
2
d12
d23
1
GMD 
Figure 1.2
D.
d31
3
3
d12  d23  d31
Eq. 1.4
Conductors of the three phases spaced d1, d2 and d3
between centers respectively of a power system. For solid
conductors, GMR = r (radius)
Conductance – (shunt) negligible
- It is mainly due to corona and leakage to insulators. Corona is not likely to
occur when the diameter of the conductor is small compared with the
distance between conductors. High voltage small wires and close spacing
contribute to a high voltage gradient which may induce corona. A rough
and dirty surface on a conductor increases the probability of the
occurrence of the corona, and damp weather increases the loss.
- In general it is due to the leakage current at the insulators of overhead
lines and through the insulations of cables and is taken to be negligible,
thereby conductance between conductors of an overhead line is assumed
to be zero.
Corona –A luminous discharge caused by ionization of the gasses around
the high voltage conductor usually accompanied by a hissing sound and
characterized by the presence of ozone.
Reasons why grounded wires are installed on top of the transmission line towers or
poles
1. To shield the transmission line from direct lightning strokes.
2. To reduce the amount of voltage induced upon the transmission line
conductor due to the electrostatic field produced by a charged cloud.
3. High voltage will occur between wires during lightning, so it must be
discharged to the ground by means of lightning arrester.
Compiled and Adapted by: JOY C. IMPERIAL, M.Eng’g, REE, LPT
EE 51 POWER SYSTEM 1
Effect of temperature (surrounding temperature) on the resistance of conductor
Recall:
R2  R 1 [ 1  α1 (t 2  t1 )]
Where:
- R1 and R2 are resistances respectively at temperatures t1 and t2.
(R1 = resistance of the conductors at ambient temperature)
α1 
1
/Co
T  t1
=
Ohms
temperature coefficient of resistance at
temperature t1
T = inferred absolute zero temperature of the conductor
(for copper T = - 234.5 ºC)
Voltage and Current Relations
-
(For 3-phase and 1-phase AC transmission and distribution systems)
1.
All loads connected at the receiving end (load side) are considered to be
balanced. Likewise, it follows that the voltage and currents are also
balanced.
2.
All voltage equations must be done in per phase basis.
3.
All vector / phasor diagrams must be done in a per phase basis.
4.
Circuit representation (representation of a transmission lines) must be
done in per neutral basis, or per phase basis, or per wire basis, or per
conductor basis.
5.
Line parameters (resistance, inductance, and capacitance – impedance)
must be treated per wire, or per phase, or per conductor, per line basis
(even if it is not specified to be).
6.
For transmission (short) lines: up to 50 miles in length (use only
resistance and inductance, neglect capacitance effect).
7.
For medium length transmission lines: from 51 to 150 miles in length (use
all the three parameters).
8.
For long transmission lines– 151 miles up (use all the three parameters).
Conventional symbols:
Receiving end data: Load Side
ERL
ER
=
=
Receiving end line to line voltage
Receiving end line to neutral (per phase) voltage
E
- - - for 3-phase T.L.
ER  RL
3
Compiled and Adapted by: JOY C. IMPERIAL, M.Eng’g, REE, LPT
EE 51 POWER SYSTEM 1
P.F.R
=
ERL
- - - for 1-phase T.L.
2
Receiving end power factor
ER 
Eq. 1.5a
P.F.R  Cos θR
Eq. 1.5b
Where: θR
R.F.R
IR
PR
QR
SR
=
=
=
=
=
= Receiving end p.f. angle
Receiving end reactive factor
P.F.R  Sin θR
Eq. 1.5c
Receiving end line current
Receiving end real (true) power in kW, MW,GW
Receiving end reactive power in kVARs, MVARs
Receiving end apparent power in kVA, MVA
Sending end data: Source
ESL
ES
=
=
P.F.S
=
Sending end line to line voltage
Sending end line to neutral (per phase) voltage
E
- - - for 3-phase T.L.
ES  SL
3
E
- - - for 1-phase T.L. .
ES  SL
2
Sending end power factor
P.F.S  Cos θS
Where: θ S
R.F.S
IS
PS
QS
SS
=
=
=
=
=
Eq. 1.6a
Eq. 1.6b
= Sending end p.f. angle
Sending end reactive factor
P.F.S  Sin θS
Eq. 1.6c
Sending end line current
Sending end real (true) power in kW, MW,GW
Sending end reactive power in kVARs, MVARs
Sending end apparent power in kVA, MVA
Power:
Power triangle
P
S
θ
S
-jQ
jQ
θ
P
Compiled and Adapted by: JOY C. IMPERIAL, M.Eng’g, REE, LPT
EE 51 POWER SYSTEM 1
Apparent Power (S):
S  P  jQ
in VA, kVA, or MVA
-
Where:
(+) j Q - - - Capacitive VARs
( -) j Q - - - Inductive VARs
Three-Phase:
/S / 
3 EL IL
1000
/S/ 
3E I
1000
Eq. 1.7a
Single-Phase:
EL IL
/S / 
1000
- - kVA
/S/ 
- - kVA
- - kVA
2E I
- - kVA
1000
Eq. 1.7b
True (Real) Power (P):
P  S  Cos θ
-
Three-Phase:
P 
in W, kW, or MW
3 EL IL Cos θ
3 E I Cos θ

1000
1000
EL IL Cosθ
2 EI Cosθ
P 

1000
1000
Single-Phase:
kW
kW
Eq. 1.7c
Reactive (Imaginary) Power (Q):
Q  S  Sin 
-
Three-Phase:
Q 
3 EL IL Sin θ
1000
Single-Phase:
Q 
EL IL Sin θ
1000
in VAR, kVAR, or MVAR


3 E I Sin θ
1000
2 EI Sin θ
1000
kVARs
kVARs
Eq. 1.7d
Compiled and Adapted by: JOY C. IMPERIAL, M.Eng’g, REE, LPT
EE 51 POWER SYSTEM 1
Learning Activities
Activity 1
Understanding Basic Concepts and Nomenclature of Power
System
Direction.
Do research, study and understand the basic concept, operation,
characteristics and technical features of the following.
1.
2.
3.
4.
5.
Single Line Diagram of Power Supply System
Classification of Substations
Difference Between Transmission and Distribution Line
Difference Between Power Transformer and Distribution Transformer
Different Layouts for Substation
Activity 2
Application of Related Equations Employed in Power System
Direction.
Study and understand the following illustrative problems below.
Illustrative Problem 1.1
A three-phase 60-Hz line has flat horizontal spacing. The conductors have a
GMR of 0.0133m with 10m between adjacent conductors. Determine the
inductive reactance per phase in ohms per kilometer.
Solution:
Given Data:
3-phase transmission line @ f=60 Hz
Spacing:
Flat horizontal (10 meters between adjacent conductors)
d12 = 10m; d23 = 10m; d31 = 20m
Conductor:
GMR = 0.0133 inch = 3.38 x10-4 m
Required:
Inductive reactance (XL)per phase in ohms per kilometer.
Recall: XL  2π f L Ohms
Inductance (L) per conductor using Eq. 1.2b (For S.I. Units)
 GMD  H/m 1000m
L  2 x 10 7 ln 
x

km
 GMR  phase
Where:
GMD 
3
101020  12.60m
L  2.11 mH per km
hence

XL  2π 60 2.11x103
 0.794 Ohms per km

Compiled and Adapted by: JOY C. IMPERIAL, M.Eng’g, REE, LPT
EE 51 POWER SYSTEM 1
Illustrative Problem 1.2
A three-phase 60-Hz line spaced triangularly with distances 12ft, 12ft and 16ft
respectively between the three phases. The conductors have an outside
diameter of 3.28 cm. Determine the capacitive reactance to neutral in ohmmeters and the capacitive reactance of the line in ohms if its length is 125 mi.
Solution:
Given Data:
3-phase transmission line @ f=60 Hz
Spacing:
Triangular
d12 = 12 ft ; d23 = 12 ft; d31 = 16 ft
Conductor:
D = 3.28 cm
GMR (Assume solid conductor) = r = D/2
= 0.0164 m
Length = 125 mi = 201,137.5m
Required:
Capacitive reactance (Cn) to neutral in ohm-meters and the capacitive
reactance (XCn) of the line in ohms
Capacitive reactance to neutral (XCn):
1
Recall: XCn 
Ohms
2ππXCn
Capacitance per conductor to neutral using Eq. 1.3b for
S.I. Units
2 π x 10 12
F/m
x length
GMD
phase


ln 

 GMR 
2 π x 10 12
F/m
Cn 
phase
3
 1 
 121216  

 3.28  
ln 


0.0164




F/m
12
Cn  1.126 x 10
phase
Cn 
Hence,
XCn 
1
 2,354,782,955 Ω  meter
2π 60 1.126 x 1012
XCn 
2,354,782,955 Ω  meter
 11,707.330 Ohms
201,137.50m


Compiled and Adapted by: JOY C. IMPERIAL, M.Eng’g, REE, LPT
EE 51 POWER SYSTEM 1
Illustrative Problem 1.3
Given the following data for 3-phase transmission line; distance, 25 mile;
power, 6,000 kW; power factor, 0.80 ,lagging current; voltage at load, 33,000
volts, 3-phase, 60-cycles; size wire, No. 00 AWG solid copper (diameter, 365
mils, ρ = 11 ohms per cir. mil-ft. @ 600C); spacing of wires, 48 in., triangular.
Determine (a) AC resistance per conductor at 600C; (b) inductive reactance per
conductor and capacitive reactance to neutral; (c) receiving end current.
Solution:
Given Data:
3-phase transmission line
Load:
PR =6,000kW
P.F.R = 0.80 lagging
ERL = 33,000 Volts at 60Hz
Conductor:
No. 00 AWG solid copper
d = 365 mils
ρ = 11 ohms per cir. mil-ft
Length = 25 miles
Spacing:
Triangular
d1 = d2 = d3 = 48 inches
Required:
(a) AC resistance per conductor at 600C
Using Eq. 1.1
Rac = K Rdc
Where:
K 
FCopper
1
1  F2
2
 0.0106d2 f
Rdc  ρ 
Rac
Rac
(b)
L
A
2

2
1 1 0.0106   365  60

 1000 


 
2




 10.92 Ohms


   25 x 5280 
 11 

   3652 



inductive reactance per conductor and capacitive reactance to neutral
Inductive reactance per conductor (XL)
Recall: XL  2π f L Ohms
Compiled and Adapted by: JOY C. IMPERIAL, M.Eng’g, REE, LPT
EE 51 POWER SYSTEM 1
Inductance per conductor (L) using Eq. 1.2a for English Units
 GMD  H/mi
L  0.7411x 10 3 log 
x length

 GMR  phase
Where: From Eq. 1.4
GMD 
3
48  48  48  48 inches
GMR  r 
365
 0.1825 inch
1000 2
 48  H/mi
L  0.7411x 10 3 log 
x 25 mi

 0.1825  phase
L  44.84 mH
Hence,
 44.84 
XL  2π 60
  16.90 Ohms
 1000 
Capacitive reactance to neutral (XCn):
1
Recall: XCn 
Ohms
2ππXCn
Capacitance per conductor to neutral using Eq. 1.3a for
English Units
0.0388 x 10  6
 48 
log 

 0.1825 
 40.10x108 F
F/mi
x 25 mi
phase
Cn 
Cn
Hence,
XCn 
(c)
1
 6.62 kΩ
2π 60 40.10 x 108


Receiving end current (IR)
For three-phase system from Eq. 1.7c
3 ER IR CosθR
PR 
1000
Hence,
PR x 1000
6,000x1000
IR 

 131.216 Amp
3 ER CosθR
 33,000 
3
0.80
3 

Compiled and Adapted by: JOY C. IMPERIAL, M.Eng’g, REE, LPT
EE 51 POWER SYSTEM 1
Activity 3
Self-Assessment: Practice Problems
Direction.
Analyze and identify data given on the following problems and employ
related equations to determine the corresponding required data for
each problem. (See Answer-Key to verify your answers)
Practice Problem 1.1
A three-phase 60-Hz line spaced triangularly with distances 5.2m, 5.2m and
8.6m respectively between the three phases. The conductors have an outside
diameter of 1.824-inch. Determine the capacitive reactance to neutral in gigaOhm-meters and the capacitive reactance of the line in Mega-Ohm s if its
length is 160 km.
Practice Problem 1.2
A three-phase 60-Hz line has flat horizontal spacing. The conductors have a
GMR of 1.58cm with 32ft between adjacent conductors. Determine the
inductive reactance per phase in ohms per mile.
Practice Problem 1.3
Given the following data for 3-phase transmission line; distance, 120km;
power, 16,000 kW; power factor, 0.84 ,lagging current; voltage at load, 69kV,
3-phase, 60-cycles; size wire, No. 000 AWG solid copper (diameter, 410 mils,
ρ = 11 ohms per cir. mil-ft. @ 600C); spacing of wires, 4.8m., triangular.
Determine (a) AC resistance per conductor at 600C; (b) inductive reactance per
conductor and capacitive reactance to neutral; (c) receiving end current.
Assessment
Instructions:
Answer the each question/problem quietly and encircle the letter of
the correct answer.
1. In a balance 3-phase star-connected system line voltage is
(a) the phasor sum of the two phase voltages
(b) 1.414 times the phase voltages
(c) 0.707 times the phase voltages
(d) the phasor difference of the two phase voltages
2. In transmission line conductors, maximum tension occurs in ______.
(a) the lowest point
(b) the vertical
(c) the horizontal
(d) the supports
3. Is the process of alternating the positions of each phase conductor of a line in
equal intervals so that each phase will have the same inductance.
(a) superposition
(b) transposition
(c) transition
(d) interpolation
Compiled and Adapted by: JOY C. IMPERIAL, M.Eng’g, REE, LPT
EE 51 POWER SYSTEM 1
4. When a unit weight is uniformly distributed along the horizontal or along its
span, a cable assumes the form of?
(a)
parabola
(b)
catenary
(c)
hyperbola
(d)
line
5. The following are reasons why grounded wires are installed on top of the
towers or poles of a line, except one.
(a) to shield the line from direct lightning strokes
(b) to give provisions for neutral
(c) to reduce the amount of voltage induced on the line due to electrostatic
fields produced by charged clouds.
(d) to give provisions for discharging surge current by means of lightning
arrester.
6. A (n) _______ stores and returns energy to a circuit while a(n) ________
dissipates energy.
(a) resistance, impedance
(c) resistor, inductor
(b) inductor, resistor
(d) inductor, reactance
7. For an RL circuit, the power factor cannot be less than ______ or greater than
_______.
(a)
0, 1
(b)
0, -1
(c)
1, 0
(d)
-1, 0
8. The advantage of the star-connections over the delta connections for the same
phase voltage is that it gives
(a) step-down current
(b) extra step-up voltage
(c) extra step-up current
(d) extra step-up power
9. The inductance is responsible for the current to _____ the applied voltage.
(a)
lag
(b)
be in phase with
(c)
lead
(d)
be opposite
10. A three-phase 60-Hz line spaced triangularly with distances 18-ft, 18-ft and 24ft respectively between the three phases. The conductors have an outside
diameter of 765mils. Determine the capacitive reactance to neutral in gigaOhm-meters and the capacitive reactance of the line in Mega-Ohm s if its
length is 100 mi.
(a) 23.56 GΩ-meter; 14.727 MΩ
(c) 2.716 GΩ-meter; 18.876 kΩ
(b)
(d)
2.716 GΩ-meter; 16.876 kΩ
23.56 GΩ-meter; 147.27 MΩ
11. A three-phase 60-Hz line has flat horizontal spacing. The conductors have a
GMR of 0.840-inch with 9m between adjacent conductors. Determine the
inductive reactance per phase in ohms per mile.
(a) 0.808 Ohms per mile
(b) 8.08 Ohms per mile
(c) 0.761 Ohms per mile
(d) 7.61 Ohms per mile
12. Given the following data for 3-phase transmission line; distance, 120mi; power,
20,000 kVA; power factor, 0.80 ,lagging current; voltage at load, 66kV, 3-
Compiled and Adapted by: JOY C. IMPERIAL, M.Eng’g, REE, LPT
EE 51 POWER SYSTEM 1
phase, 60-cycles; size wire, No. 0000 AWG solid copper (diameter, 460 mils, ρ
= 11 ohms per cir. mil-ft. @ 600C); spacing of wires, 20-ft., triangular.
Determine (a) AC resistance per conductor at 600C; (b) inductive reactance per
conductor and capacitive reactance to neutral; (c) receiving end current.
(a) RAC = 34.011 Ω; XL = 101.23 Ω; XCn = 19.150 kΩ; IR = 139.968 A
(b) RAC = 62.427 Ω; XL = 61.764 Ω; XCn = 24.015 kΩ; IR = 133.882 A
(c) RAC = 34.011 Ω; XL = 101.23 Ω; XCn = 15.190 kΩ; IR = 139.968 A
(d) RAC = 26.427 Ω; XL = 61.764 Ω; XCn = 24.015 kΩ; IR = 133.882 A
Answer Key
Pre-Test
1.
2.
3.
4.
5.
b
d
a
b
d
6.
7.
8.
9.
10.
c
c
c
d
c
11.
12.
13.
14.
15.
c
b
b
a
c
Activity 3
1. 2.356 GΩ-meter;
14.727 kΩ
2. 0.808 Ohms per mile
3. (a)
(b)
(c)
RAC = 26.427 Ohms;
XL = 61.764 Ohms ;
XCn = 24.015 kΩ
IR = 133.882 Amp.
Compiled and Adapted by: JOY C. IMPERIAL, M.Eng’g, REE, LPT