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GRAAD 12
NATIONAL
SENIOR CERTIFICATE
GRADE 12
ELECTRICAL TECHNOLOGY
FEBRUARY/MARCH 2016
MEMORANDUM
MARKS: 200
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Electrical Technology Grade 12 NSC Supplementary 2016 Memo
Grade 12 Electrical Technology memorandum from the NSC Supplementary Exam 2016 (English), 15 pages. Read online or download the free PDF.
Electrical Technology · Grade 12 · NSC Supplementary Exam · 2016. Memorandum, 15 pages. Read online or download the PDF.
- Subject
- Electrical Technology
- Grade
- Grade 12
- Document type
- Memorandum
- Year
- 2016
- Exam period
- NSC Supplementary Exam
- Curriculum
- CAPS
- Pages
- 15
- File size
- 567.9 KB
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NSC Supplementary Exam 2016
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Electrical Technology 2 DBE/Feb.–Mar. 2016
NSC – Memorandum
INSTRUCTIONS TO THE MARKERS
1. All questions with multiple answers imply that any relevant, acceptable
answer should be considered.
2. Calculations:
2.1 All calculations must show the formulae.
2.2 Substitution of values must be done correctly.
2.3 All answers MUST contain the correct unit to be considered.
2.4 Alternative methods must be considered, provided that the correct
answer is obtained.
2.5 Where an incorrect answer could be carried over to the next step,
the first answer will be deemed incorrect. However, should the
incorrect answer be carried over correctly, the marker has to re-
calculate the values, using the incorrect answer from the first
calculation. If correctly used, the candidate should receive the full
marks for subsequent calculations.
3. This memorandum is only a guide with model answers. Alternative
interpretations must be considered and marked on merit. However, this
principle should be applied consistently throughout the marking session at
ALL marking centres.
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Electrical Technology 3 DBE/Feb.–Mar. 2016
NSC – Memorandum
QUESTION 1: OCCUPATIONAL HEALTH AND SAFETY
1.1 Unsafe conditions include:
− Inadequate guards around machinery.
− No eye protection goggles in the workshop and around the bench
grinder.
− Loose and dangerous components on moving machinery. (1)
1.2 An unsafe action is an action performed by a person/worker that could
cause:
− Injury to him/herself or his/her colleagues
− Cause damage to tools and equipment
− Render a workplace unsafe
− Lead to loss of income to the employer (2)
1.3 Protocol to follow when discovering a person experiencing electrical shock is:
− Do not touch the victim with bare hands until the supply is turned off.
− Switch off the supply.
− Call a teacher or medical person for help.
− If the electricity cannot be switched off and the victim is still in contact
with it push the wire away with an insulated object. (4)
1.4 Productivity is an important work ethic to assist the growth of South Africa's
economy due to the following reasons:
− Safe working procedures are more productive as less time is lost due to
injuries and downtime related to accidents.
− Efficient working methods lead to less use of raw materials, thus
increasing profits.
− Good ergonomics lead to cleaner and safer work environments and
promotes worker satisfaction and morale.
− Workers with a high morale and a good safety record tend to follow
instructions more precisely and enact safe working procedures.
− Effective monitoring and reporting systems lead to better quality
assurance of products, thus placing a better product in the market within
a shorter time frame, thus promoting productivity. (3)
[10]
QUESTION 2: THREE-PHASE AC GENERATION
2.1 A Kilo-watt hour meter ✓ (1)
2.2 Two advantages of three-phase systems over single-phase systems are:
− 3φ can be operated in star or delta. ✓
− Three-phase systems can supply both three-phase and single-phase
installations of power✓
− In generators and motors, for the same size frames, three-phase
machines delivers more power than single-phase machines (2)
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Electrical Technology 4 DBE/Feb.–Mar. 2016
NSC – Memorandum
2.3
Ph1 or L1 or Ph2 or L2 or Ph3 or L3 or
ER✓ Ey✓ ER✓
+Vmax
Showing ✓
amplitude
0
Showing
displacement of
1200 ✓
-Vmax
1200 1200 (5)
Note to markers:
No additional marks are awarded for the manipulation of the formulae.
Manipulation is shown to illustrate how the formula was derived.
2.4 2.4.1 S = 3 VL I L
S
VL = ✓
3 x IL
20000 ✓
=
3 x 25
= 461.88 A ✓ (3)
2.4.2 VL = 3VP
VL ✓
VP =
3
462.43 ✓
=
1.73
= 267.67 V ✓
(3)
2.5 2.5.1 PT = P1 + P2 ✓
= (8 + 4) kW ✓
= 12 kW ✓
(3)
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Electrical Technology 5 DBE/Feb.–Mar. 2016
NSC – Memorandum
2.5.2 P = 3 VL I L Cosθ
P
IL = ✓
3 x VL Cosθ
12000 ✓
=
3 x 380 x 0.8
= 22.79 A ✓ (3)
[20]
QUESTION 3: THREE-PHASE TRANSFORMERS
3.1 By laminating and insulating the laminations of the core of the transformer.✓ (1)
3.2 Two similarities between single phase and three-phase are:
− Both have the same functional operations✓
− Transformers are used to step down or step up the supply voltage✓
− Both can have either the closed core type or shell core type
− No electrical connections between primary and secondary windings, with
the exception of the autotransformer. (2)
3.3 Two factors that may cause excessive heating in a transformer are:
− Insufficient ventilation✓
− Constant overloading✓
− Loose connections
− Ineffective tap-changer contacts
− Impure/carbonized/insufficient transformer oil (2)
3.4 Applications of delta-star connected transformers in power distribution
systems are:
− Single and three-phase supply to Commercial sites ✓
− Three-phase supply to Industrial areas ✓
− Single and three-phase supply to domestic installations
− Single and three-phase supply to agricultural installations
− Three-phase supply to other substations (transformers) in a ring-fed
supply network. (2)
3.5 The function of a transformer is to 'step down' ✓ or to 'step up' ✓
(transform) the voltage ✓in a distribution network to the required value. (3)
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Electrical Technology 6 DBE/Feb.–Mar. 2016
NSC – Memorandum
3.6 3.6.1 V L = VPH ✓
∴ VPH = 6.6 kV ✓
(2)
3.6.2 IL = IP ✓
∴IP = 30 A ✓
(2)
3.6.3 I ×V
IPh( p) = Ph( s) Ph( s) ✓
VPh( p)
30 × 230 ✓
=
6600
= 1.04 A ✓
(3)
3.6.4 NP IPH( s)
= ✓
NS IPH( p)
NP 30 ✓
=
NS 1.04
NP : NS = 30 : 1 ✓ (3)
[20]
QUESTION 4: THREE-PHASE MOTORS AND STARTERS
4.1 Applications of three-phase motors include:
− pumps
− lifts
− cranes (1)
4.2 Advantages of a three-phase over single-phase motors:
− For the same size frame a three-phase motor delivers more power
− Three-phase motors are self-starting
− More robust
− Simple construction (3)
4.3 4.3.1 At start motors draw more than the motors rated current. This may
lead to unnecessary tripping of the motor and the increased
current at start causes wear on switch gear. (3)
4.3.2 The motor is started in the star mode which reduces the voltage
across each phase of the motor as in star V PH = √3 V L . As the
voltage across each phase is reduced, the current in each phase will
also be reduced. (3)
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Electrical Technology 7 DBE/Feb.–Mar. 2016
NSC – Memorandum
4.4 The operation of three-phase squirrel cage induction motor:
− Three-phase supply is such that the peak voltage of each phase is
reached at 120o intervals.
− When a squirrel cage motor is connected across a three phase
alternating supply, alternating current will flow in the respective stator
windings.
− As a result the alternating current will set up a magnetic field in and
around the respective stator windings.
− The stator windings are however spaced 120o apart around the axis of
the motor.
− The result is that the magnetic field has a rotating maximum flux density
as the peak in each winding is reached at a 120o interval, resulting in a
rotating magnetic field.
− A squirrel cage rotor consists of cage like conductors bound electrically
with a short circuit ring at each end.
− The rotating magnetic field will sweep across the rotor conductors of the
squirrel cage rotor.
− As the magnetic field sweeps over the rotor conductors, the magnetic flux
will cut/cross the rotor conductors.
− Faraday's law determines that an EMF will be induced in the rotor
windings due to the sweeping action of the rotating magnetic field.
− Lenz's Law will determine that the induced current will be such that its
force will oppose the inducing action (from the stator), thus establishing a
repelling force between the stator and rotor (due to the opposing
magnetic fields), thus forcing the rotor into rotation.
− When the speed of the rotor approaches the speed of the rotating
magnetic field (synchronic speed), the inducing action is reduced and the
rotor tends to slow down. The result is that the rotor rotates at a speed
lower than that of the rotating magnetic field (synchronic speed) (7)
4.5 4.5.1
✓ L1 L2 L3
N✓
U1 V1 W1
✓
✓ ✓
W2 U2 V2 E
(5)
Coil 1:U1 – U2 ✓
Coil 2: V1 – V2✓
Coil 3: W1 – W2✓
L1 , L2, L3 – 3 Phase specified✓
N: Star is specified✓
E – Not requested in the question
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Electrical Technology 8 DBE/Feb.–Mar. 2016
NSC – Memorandum
4.5.2 The reading on the meter should be of a high resistive value, more
than 1 MΩ, this will indicate that there is no electrical connection
between the windings showing sound/good insulation. (3)
4.5.3 A low resistance reading will indicate that there is electrical connection
between the winding and earth. This will indicate any electrical fault
that may lead to an operator being shocked in the motor is
energised. (3)
4.6 60 × f ✓
Pole pairs =
NS
✓
60 × 50
=
600
✓
=5 (3)
4.7 4.7.1 S = 3 VL I L ✓
= 3 x 380 x 16 ✓
= 10.53 kVA ✓
(3)
4.7.2 η=100%
P = S x Cos θ ✓
= 10.53 x 0.85 ✓
✓
= 8.95 kW (3)
4.7.3 Po
η=
Pi
Po = Pi x η
✓
= 8.95 x 0.9 ✓
= 8.06 kW ✓ (3)
[40]
QUESTION 5: RLC
5.1 Factors influencing the capacitive reactance are:
− Value of capacitance✓
− Frequency of the supply✓ (2)
5.2 Reactance is the opposition of the specific reactive component✓ to the flow
of current in AC circuits
Impedance is the total opposition✓ offered to the flow of current in an AC
circuit which contains resistive and reactive components (2)
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Electrical Technology 9 DBE/Feb.–Mar. 2016
NSC – Memorandum
5.3
Z
[Ω]
✓
✓ ✓
X >X XL>XC
R C L
✓
fr F[Hz]
(4)
5.4 XL
Q=
Z ✓
4000
= ✓
50
= 80 ✓
(3)
5.5 5.5.1 1
XC =
2πfC ✓
1
= ✓
2π x 50 x 47 x 10− 6
✓
= 67,73 Ω (3)
5.5.2 X L = 2πfL ✓
= 2π x 50 x 400 x 10− 3 ✓
= 125,66 Ω ✓
(3)
5.5.3 1
fr =
2π LC ✓
1
= ✓
2π 400 ×10− 3 × 47 ×10− 6
= 36.71 Hz ✓ (3)
[20]
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Electrical Technology 10 DBE/Feb.–Mar. 2016
NSC – Memorandum
QUESTION 6: LOGIC
6.1 Advantages of a PLC over a hardwired systems are:
− Reduced space. PLCs are solid state devices and are compact and small
compared to hard wire logic.
− Less maintenance PLCs have error diagnostic units making it easy to
trouble shoot.
− Faster response time it can process thousands of items a second. (2)
6.2 Cycle Repeats
Input ✓ Process✓ Output✓
Check Inputs Execute Update
via input Instructions Outputs
interface✓ according to according to
the program✓ the program
and inputs
and starts
again✓
(6)
6.3 Output devices that can be used with a PLC are:
− Motor contactor
− indicator lamps
− relays
− switches (2)
6.4 F = AB C + A B C + A B C + A B C
= A C (B + B) + AC (B + B)
= AC + AC
= C (A + A)
=C (5)
6.5 PLCs could be programmed using the following methods:
− Ladder diagrams
− Function blocks
− Sequential functions
− Charts
− Instructional list
− Structured text (3)
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Electrical Technology 11 DBE/Feb.–Mar. 2016
NSC – Memorandum
6.6 6.6.1
AB AB AB AB Correct
CD 0 0 0 0 groupings
CD 1✓ 1✓ 1✓ 0
CD 1✓ 0 0 1✓
CD 0 0 0 0
(6)
6.6.2 Y = B C D✓+ A C D✓+ B C D✓ (3)
6.7 6.7.1 Sequence starter with a time delay circuit
Note to markers:
The circuit cannot be star delta starter as there is no lockout function
included in the circuit. (2)
6.7.2
OUTPUT LAMP
O/L STOP START 1
✓
LAMP
TIMER
✓ ✓
✓
O/L STOP TIMER 01 MARKER/FLAG (MO1)
✓ ✓
✓ OUTPUT MOTOR
MO1 START 2
✓
MOTOR
(8)
Note to markers:
− The candidate may use inverse functions (simulating real world NO
and NC switches) in the ladder circuit, which must be considered.
Should it function correctly marks are to be awarded to the candidate.
− Note that some labels may differ from that of the learner.
− Please check that the learner utilises labelling consistently throughout
the answer.
6.7.3 PLC's allow for the use of markers as the number of operands that
can be placed in a rung of the Ladder program is limited due to
hardware design constraints.
A marker/flag is a function of PLC programming that can be used as
a placeholder, or a subroutine which in turn simplifies the ladder
program. (3)
[40]
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Electrical Technology 12 DBE/Feb.–Mar. 2016
NSC – Memorandum
QUESTION 7: AMPLIFIERS
7.1 Characteristics of ideal op-amps are:
− Open-loop voltage gain A V = infinite
− Input impedance Z in = infinite
− Output impedance Z 0 = zero
− Bandwidth = infinite
− Unconditional stability
− Differential inputs, i.e. two inputs
− Infinite common-mode rejection (2)
7.2 The bandwidth is the range of frequencies within which an amplifier can
amplify without distorting the output signal (retain linearity) or losing
gain. (3)
7.3 Positive feedback is obtained when the output signal of an amplifier circuit is
fed back to the input of the circuit. When the signal is fed back to the
input, it is in phase with the input signal, thus added to the input signal. (3)
7.4
+Vsupply (+Vcc)
Inverting input -
Output
Non-inverting input +
-Vsupply (-Vcc)
(6)
7.5 7.5.1
Output
Input
Amplification
Cycles coincide/
same frequency
Inversion
(3)
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Electrical Technology 13 DBE/Feb.–Mar. 2016
NSC – Memorandum
7.5.2
Output
Amplification
V1 Cycles coincide/
V2
same frequency
Inversion due to V1 on
inverting input and V2 on
non-inverting input
Common mode
(3)
7.6 Advantages of using negative feedback are:
− Bandwidth is increased.
− Level of noise (hiss) is decreased.
− Gain is decreased.
− Deformation of the input signal is reduced. (2)
7.7 7.7.1 Non inverting op-amp (1)
7.7.2
Output
Amplification
In phase
Non inversion
Input
(3)
7.7.3 Vout R ✓
AV = =1+ f
Vin R in
45x103
=1+ ✓
15x103
=4 ✓ (3)
7.7.4 Vout = A v × Vin ✓
= 4 x 2.5 ✓
= 10 V ✓ (3)
7.8 The a-stable multivibrator can be used as a:
− Tone generator
− Clock pulse generator
− Square wave generator
− Fake alarm circuit if used with two LED's (2)
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Electrical Technology 14 DBE/Feb.–Mar. 2016
NSC – Memorandum
7.9
+V
0 Trigger
-V pulse
+V
Vout
0
-V
(4)
7.10 7.10.1 Positive feedback (1)
7.10.2 Sine wave generator
Tone Oscillator
Buzzer driver (1)
7.10.3 1 ✓
f RC =
2π 6RC
1
= ✓
2π 6 × 15 × 150 × 10− 9
= 43.31 Hz ✓
(3)
7.11 7.11.1 Timing circuit (1)
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Electrical Technology 15 DBE/Feb.–Mar. 2016
NSC – Memorandum
7.11.2
+V
Vin=
0
Square wave
2 Marks
-V
+V
Vout
0 if T=smaller
-V
Note:
Higher order answers will show the wave starting from
zero, thus receiving additional 2 marks
Other answers will receive 4 marks maximum.
Vout
Alternative
if T=larger
(6)
50
TOTAAL: 200
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