Confidential
SENIOR CERTIFICATE EXAMINATIONS/
NATIONAL SENIOR CERTIFICATE EXAMINATIONS
ELECTRICAL TECHNOLOGY: POWER SYSTEMS
MAY/JUNE 2024
MARKS: 200
TIME: 3 hours
This question paper consists of 18 pages and a 2-page formula sheet.
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Electrical Technology Grade 12 NSC June 2024 Question Paper (Eng)
Grade 12 Electrical Technology question paper from the NSC June Exam 2024 (English), 20 pages. Read online or get the free PDF download, memo linked when available.
Electrical Technology · Grade 12 · NSC June Exam · 2024 · English. Question paper, 20 pages. Read online or download the PDF.
- Subject
- Electrical Technology
- Grade
- Grade 12
- Language
- English
- Document type
- Question paper
- Year
- 2024
- Exam period
- NSC June Exam
- Curriculum
- CAPS
- Pages
- 20
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Electrical Technology: Power Systems 2 DBE/May/June 2024
SC/NSC Confidential
INSTRUCTIONS AND INFORMATION
1. This question paper consists of SEVEN questions.
2. Answer ALL the questions.
3. Sketches and diagrams must be large, neat and FULLY LABELLED.
4. Show ALL calculations and round off answers correctly to TWO decimal
places.
5. Number the answers correctly according to the numbering system used in this
question paper.
6. You may use a non-programmable calculator.
7. Calculations must include:
7.1 Formulae and manipulations where needed
7.2 Correct replacement of values
7.3 Correct answer and relevant units where applicable
8. A formula sheet is attached at the end of this question paper.
9. Write neatly and legibly.
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QUESTION 1: MULTIPLE-CHOICE QUESTIONS
Various options are provided as possible answers to the following questions. Choose
the answer and write only the letter (A–D) next to the question numbers (1.1 to 1.15) in
the ANSWER BOOK, e.g. 1.16 D.
1.1 A burn must be treated by …
A covering the burnt area with butter.
B putting ice on the burn wound.
C puncturing the blisters.
D running cold water over the burnt area until the pain reduces. (1)
1.2 The power in an inductor is …
A apparent power.
B active power.
C reactive power.
D true power. (1)
1.3 A parallel RLC circuit is more capacitive when …
A XC > XL
B R=Z
C IC > I L
D VC > V L (1)
1.4 An increase in resistance of a parallel resonant circuit will cause the total
current to …
A double.
B decrease.
C increase.
D be zero. (1)
1.5 Efficiency of electrical systems in industries with large inductive loads
cause …
A high power factor with current leading the voltage.
B low power factor with current leading the voltage.
C high power factor with current lagging the voltage.
D low power factor with current lagging the voltage. (1)
1.6 The line current of a star connected system is 10 A. The phase current
will be …
A 30 A
B 17,32 A
C 10 A
D 5,77 A (1)
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1.7 The apparent power of a delta-connected three-phase system with a line
current of 5 A and phase voltage of 300 V is …
A 2 598,08 VA
B 1 500 VA
C 2 598,08 VAR
D 1 500 VAR (1)
1.8 The following statement describes a core-type transformer:
A The core hides the major part of the windings
B Double magnetic circuit
C Single magnetic circuit
D The core has five limbs (1)
1.9 The factors that can contribute to excessive heating in transformers is/are …
A constant overloading.
B insufficient ventilation.
C transformer oil may be insufficient.
D All the above-mentioned. (1)
1.10 A continuity test on the windings of a three-phase induction motor is carried
out between … in the terminal box when the connectors between windings
have been removed.
A U1 and V1
B V1 and earth
C U1 and W 1
D U1 and U2 (1)
1.11 The … of a contactor is/are wired into the main/power circuit with reference to
motor starters.
A main contacts
B auxiliary contacts
C zero-volt coil
D A1 and A2 terminals (1)
1.12 The ratio of the output power to the input power of a three-phase motor is
known as …
A power factor.
B efficiency.
C slip.
D apparent power. (1) 2024
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1.13 … proximity sensors are used only to detect the presence of metal objects.
A Capacitive
B Ultrasonic
C Photo-electric
D Inductive (1)
1.14 … are bits in the PLC's virtual storage memory which can be used to hold
data and behave as relays.
A Latching contacts
B Markers or flags
C Timers
D Strain gauges (1)
1.15 The purpose of a braking resistor in the regenerative braking process is to ...
A store the excess energy in its electrostatic field.
B limit the amount of excess current.
C dissipate the excess energy in the form of heat.
D store the excess energy in its electromagnetic field. (1)
[15]
QUESTION 2: OCCUPATIONAL HEALTH AND SAFETY
2.1 State TWO unsafe acts that would be regarded as dangerous practices by a
user when operating machinery. (2)
2.2 With reference to the Occupational Health and Safety Act, 1993 (Act 85 of
1993), give TWO examples that are considered to be offences when reporting
to the safety inspector. (2)
2.3 Discuss the general duties that manufacturers perform when designing or
manufacturing articles used at work. (2)
2.4 Define a non-critical incident. (2)
2.5 Describe a dangerous effect that a current of 200 mA has on the human body. (2)
[10]
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Electrical Technology: Power Systems 6 DBE/May/June 2024
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QUESTION 3: RLC CIRCUITS
3.1 Explain the term reactance with reference to an alternating current circuit. (2)
3.2 FIGURE 3.2 below shows the circuit diagram, waveforms and a partial phasor
diagram of the voltages in a RLC circuit that is connected to an AC supply.
Study the diagrams below and answer the questions that follow.
VL VC VR
10 V 15 V 18 V
VT
Circuit diagram
VR VT
+V
VC VL
VL
0
t
Rotation
VC
-V
Waveform Phasor diagram
FIGURE 3.2: RLC CIRCUIT, WAVEFORM AND PHASOR
Given:
VR = 18 V
VL = 10 V
VC = 15 V
3.2.1 State whether the circuit is predominantly inductive or capacitive.
Motivate your answer. (2)
3.2.2 Calculate the supply voltage. (3)
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3.2.3 Calculate the phase angle. (3)
3.2.4 Redraw and complete the phasor diagram in the ANSWER BOOK. (3)
3.2.5 Explain why it could be assumed that the supply current is leading
the supply voltage. (2)
3.3 Refer to FIGURE 3.3 below and answer the questions that follow.
IT
IC IR
1,59 A 1,15 A
VT XL R
230 V/50 Hz 62,83 Ω 200 Ω
FIGURE 3.3: RLC PARALLEL CIRCUIT
Given:
XL = 62,83 Ω
R = 200 Ω
IR = 1,15 A
IC = 1,59 A
VT = 230 V
f = 50 Hz
Calculate the:
3.3.1 Current flow through the inductor (3)
3.3.2 Total current flow (3)
3.3.3 Power factor (3)
3.3.4 Value of capacitance that would cause resonance when the
frequency and inductor remain constant (4)
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3.4 FIGURE 3.4 below shows the current vs frequency response curves of a
series resonant circuit with a variable resistor. The inductive reactance of the
circuit is 2 000 Ω at resonance and each response curve is for a different
resistance value.
4A Q1 : R = 25 Ω
Current
(I)
2A Q2 : R = 50 Ω
1,25 A Q3 : R = 80 Ω
f1 fr f2 Frequency
(f)
FIGURE 3.4: FREQUENCY RESPONSE
Given:
XL = 2 000 Ω
3.4.1 State how a decrease in resistance affects the Q-factor of the
circuit. (1)
3.4.2 Calculate the Q factor when R = 50 Ω. (3)
3.4.3 Calculate the resonant frequency when f1 = 1 200 Hz and
f2 = 2 100 Hz. (3)
[35]
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QUESTION 4: THREE-PHASE AC GENERATION
4.1 FIGURE 4.1 below shows a block diagram of the national power grid in
South Africa. Answer the questions that follow.
Industrial
400 V
consumers
Generation Transmission Distribution
A 200/275/400/ 7/11 kV Domestic
230 V
765 kV consumers
FIGURE 4.1: NATIONAL POWER GRID
4.1.1 State the voltage generated at block A. (1)
4.1.2 Name the standard international colour code for a three-phase
system with reference to each phase. (3)
4.1.3 Briefly describe the national grid. (2)
4.1.4 Explain why the transmission of electricity is done at such high
voltages. (2)
4.1.5 Explain how the voltage for an industrial consumer is different from
the voltage for a domestic consumer. (2)
4.2 Give TWO reasons why an industrial consumer would prefer a three-phase
supply over a single-phase supply. (2)
4.3 A 180 kW three-phase delta-connected load is powered by a
200 kVA generator. The line voltage is 400 V. Answer the questions that
follow.
Given:
S = 200 kVA
VL = 400 V
P = 180 kW
Calculate the:
4.3.1 Line current (3)
4.3.2 Phase current (3)
4.3.3 Power factor (3)
4.3.4 Reactive power (5)
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4.4 FIGURE 4.4 below shows the result of power factor correction in a three-
phase system. The supply voltage and load remained the same throughout
the process. S1 and Q1 are original values with S2 and Q2 the values after
power factor correction was done.
S1
Q1
S2
θ1
θ2 Q2
P
FIGURE 4.4: POWER FACTOR CORRECTION
4.4.1 State where power factor correcting devices may be installed. (1)
4.4.2 Identify ONE improvement with reference to FIGURE 4.4 above. (1)
4.4.3 Explain how power factor correction affected the current drawn
from the supply by comparing the power triangle before and after
power factor correction. (2)
4.5 Differentiate between wattmeters and energy meters with reference to their
applications. (2)
4.6 Two wattmeters are used to measure the total power of a three-phase
system. If the total power is 3,5 kW and the reading on meter 1 is displayed
as 1 300 W, determine the reading on meter 2. (3)
[35]
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QUESTION 5: THREE-PHASE TRANSFORMERS
5.1 Refer to FIGURE 5.1 below and answer the questions that follow.
Primary
L1 L2 L3
L1 L2 L3
Secondary
FIGURE 5.1: THREE-PHASE TRANSFORMER
5.1.1 State whether this is a core-type or shell-type three-phase (1)
transformer.
5.1.2 State ONE advantage of a shell-type transformer over a core-type
transformer. (1)
5.1.3 State the configuration of the primary and secondary windings in
FIGURE 5.1 above.
(2)
5.1.4 Briefly describe the principle of operation of the three-phase
transformer in FIGURE 5.1 above. (4)
5.2 With reference to transformers, answer the following questions.
5.2.1 Describe the effect that an increase in the load will have on the
primary current of a transformer. (2)
5.2.2 Name ONE function of the dielectric oil used in transformers. (1)
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5.2.3 Name TWO types of losses other than copper losses that occur in
transformers. (2)
5.2.4 Name and explain TWO cooling methods used for dry
transformers. (4)
5.2.5 Compare conductor sizes between a single-phase transformer and
a three-phase transformer for the same power rating. (2)
5.3 FIGURE 5.3 below shows a 300 kW delta-connected load with a power factor of
0,87 that is connected to a delta-star transformer. The primary and secondary
line voltages of the transformer are 8 kV and 600 V respectively. Assume that
the transformer is 100% efficient. Answer the questions that follow.
N1 N2 PT = 300 kW
VL2
VL1 600 V
8 kV
PF = 0,87
FIGURE 5.3: THREE-PHASE DELTA-STAR CONNECTED TRANSFORMER
Given:
VL1 = 8 kV
VL2 = 600 V
PT = 300 kW
pf = 0,87
Calculate the following:
5.3.1 Line current of the load (3)
5.3.2 Phase current of the load (3)
5.3.3 Primary line current (3)
5.4 Protection devices used with three-phase transformers are categorised in two
categories, relay and switches. Name TWO protection relays used in three-
phase transformers (2)
[30]
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Electrical Technology: Power Systems 13 DBE/May/June 2024
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QUESTION 6: THREE-PHASE MOTORS AND STARTERS
6.1 Name TWO rotating parts of a three-phase induction motor. (2)
6.2 FIGURE 6.2 below shows how the coils of a three-phase induction motor are
placed in a stator. Answer the questions that follow.
Stator housing
U1
Stator coils
U2
FIGURE 6.2: THREE-PHASE STATOR
6.2.1 Determine the angle between coil U and coil V. (1)
6.2.2 Explain how a rotating magnetic field is created when a three-
phase supply is connected to the stator windings (4)
6.2.3 State why a squirrel cage is less of a fire hazard (explosion proof). (1)
6.3 A three-phase induction motor has 12 poles. When operating from a
50 Hz supply, the slip is 5%.
Given:
f = 50 Hz
p =2
slip = 5%
Calculate the following:
6.3.1 Synchronous speed (3)
6.3.2 Rotor speed (3)
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6.4 The following information is available on a delta-connected three-phase
induction motor:
Input current = 20 A
Supply voltage = 400 V
Input power = 11,75 kW
Total losses = 1 750 W
Calculate the:
6.4.1 Efficiency (3)
6.4.2 Apparent power (3)
6.4.3 Power factor (3)
6.4.4 Output power of the motor (3)
6.5 Refer to FIGURE 6.5 below and answer the questions that follow.
L
Temporary
connection
STOP
START A START B
MC1
N/O 1
MC2
N/O
OL 2
OL 1
MC1 MC2
(Motor 1) (Motor 2)
N
Control Circuit
Conveyor 1 Motor
1
Conveyor 2 Motor
2
Conveyor system
FIGURE 6.5: MOTOR STARTER CONTROL CIRCUIT AND CONVEYOR SYSTEM
6.5.1 Identify the motor starter control circuit in FIGURE 6.5 above. (1)
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6.5.2 Explain the function of MC1 with reference to the main (power)
circuit. (2)
6.5.3 State the advantage of having two overload relays in the circuit. (1)
6.5.4 Explain the disadvantage of connecting the overloads as shown in
the control circuit in FIGURE 6.5 when MC1 and MC2 are
controlling the two motors on the conveyor belt system. (3)
6.5.5 Currently, MC2 can be energised either before or after energising
MC1. Explain how the control circuit can be altered to prevent MC 2
from being energised without energising MC1 first. (2)
[35]
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Electrical Technology: Power Systems 16 DBE/May/June 2024
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QUESTION 7: PROGRAMMABLE LOGIC CONTROLLERS (PLCs)
7.1 State THREE advantages of hard wiring. (3)
7.2 Refer to FIGURE 7.2 below and answer the questions that follow.
A
Output
B
A
& Output
B
FIGURE 7.2: LOGIC GATE
7.2.1 Draw the ladder logic diagram. (3)
7.2.2 Complete the truth table for the logic gate by writing down only the
state of the output for the following TWO input conditions.
A B OUTPUT
0 1 (X)
1 1 (Y)
TABLE 7.2.2 (2)
7.3 Refer to analogue and digital inputs on a PLC and answer the questions that
follow.
7.3.1 Give THREE examples of analogue input devices. (3)
7.3.2 Explain why an analogue input should be converted to a digital
input in a computing device. (4)
7.4 Explain the concept latching. (3)
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7.5 Refer to FIGURE 7.5 below and answer the questions that follow.
L1
Stop MC1
N/O2
O/L1 O/L2
N/C N/C
Start MC1 T
N/O1 N/O
MC1 MC2
T
N
FIGURE 7.5: AUTOMATIC SEQUENCE STARTER
7.5.1 Describe the function of T. (2)
7.5.2 State TWO conditions that must be met for MC2 to energise. (2)
7.5.3 Differentiate between ON-delay and OFF-delay timers. (3)
7.5.4 In the ANSWER BOOK, redraw the PLC ladder logic diagram in
FIGURE 7.5.4 below and complete it so as to execute the same
function as the circuit in FIGURE 7.5. above.
N
L stop
FIGURE 7.5.4 (9)
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7.6 Refer to a variable speed drive as a programmable motor controller. Describe
the basic principle of operation of a variable speed drive when it controls the
speed of an AC induction motor. (3)
7.7 FIGURE 7.7 below shows the characteristic curve of speed versus torque
when a VSD is used to control the speed of a three-phase induction motor.
Study the characteristic curve and describe the start-up and run profile of an
induction motor.
400%
400 % Breakdown
speed
300%
%
Torque 200
200%%
100 %
100%
0
Speed
20%
20% 40 % 60
40% 60%% 80 % 100
80% %
100%
FIGURE 7.7: TORQUE VERSUS SPEED CURVE (3)
[40]
TOTAL: 200
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FORMULA SHEET
RLC CIRCUITS THREE-PHASE AC GENERATION
=V Cos θ STAR
XL = 2 fL VL = √3 V
1 V =
XC =
2 fC L=
1 f 1 f2
fr = OR fr = DELTA
2 √LC 2
fr
W= OR W = f2 f1 VL = V
Q
SERIES V =
VR = R L = √3
VL = XL POWER
VC = XC S( app ) = √3 VL L
VT Q( r ) = √3 VL L Sin θ
T= OR T= R= C= L
= √3 VL L Cos θ
= √R2 (XL XC )2 Cos θ =
S
2
VT = √VR 2 (VL - VC ) OR VT = EFFICIENCY
R VR UT
Cos θ = OR Cos θ = = 100
VT
XL XC VL VC 1 L
Q= = = = = √
R R VT VT R C TWO-WATTMETER METHOD
PARALLEL
T= 1 2
VT = VR = VC = VL
VT 1 2
R= tan θ = √3 ( )
R 1 2
VT
C= THREE-WATTMETER METHOD
XC
VT
L= T= 1 2 3
XL
2
T=√R (L C)
2
VT
=
T
R
Cos θ =
T
R R
Q= =
XL XC
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THREE-PHASE TRANSFORMERS THREE-PHASE MOTORS AND STARTERS
STAR STAR
VL = √3 V VL = √3 V
L= L=
DELTA DELTA
VL = V VL = V
L = √3 L = √3
POWER POWER
S( app ) = √3 VL L S( app ) = √3 VL L
Q( r ) = √3 VL L Sin θ Q( r ) = √3 VL L Sin θ
= √3 VL L Cos θ = √3 VL L Cos θ
Cos θ = = √3 VL L Cos θ
S
V 1 1 2
= = Cos θ =
V 2 2 1 S
1
Turns ratio TR = EFFICIENCY
2
UT
losses
= 100 = 100
UT losses
UT
= 100
SPEED
0 f
ns =
p
ns nr
er Unit Slip =
ns
ns nr
Slip = 100
ns
nr =ns (1 slip)
Slip = ns nr
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