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Province of the
EASTERN CAPE
EDUCATION
NATIONAL
SENIOR CERTIFICATE
GRADE 12
SEPTEMBER 2013
ELECTRICAL TECHNOLOGY
MEMORANDUM
MARKS: 200
This memorandum consists of 11 pages.
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Electrical Technology Grade 12 P1 2013 Memo (Eng)
Grade 12 Electrical Technology Paper 1 memorandum from the Eastern Cape Mock Exam 2013 (English), 11 pages. Read online or download the free PDF.
Electrical Technology · Grade 12 · Eastern Cape Mock Exam · 2013. Memorandum, 11 pages. Read online or download the PDF.
- Subject
- Electrical Technology
- Grade
- Grade 12
- Document type
- Memorandum
- Year
- 2013
- Exam period
- Eastern Cape Mock Exam
- Paper
- 1
- Curriculum
- CAPS
- Pages
- 11
- File size
- 684.3 KB
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This exam
Eastern Cape Mock Exam 2013
Questions in this paper
- Question 1Practical Assessment Task (PAT)14 marks
- Question 2Practical Assessment Task (PAT)10 marks
- Question 3Occupational Health and Safety (OHS)10 marks
- Question 4Specialisation: Power Systems [Power Systems]10 marks
- Question 5RLC Circuits and AC/DC Theory30 marks
- Question 6Electrical and Electronic Components25 marks
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2 ELECTRICAL TECHNOLOGY (SEPTEMBER 2013)
QUESTION 1: TECHNOLOGY, SOCIETY AND THE ENVIRONMENT
1.1 Marketing ✓
Communication ✓
Presentation ✓
Financial literacy
Costing of materials, etc. (Any 3) (3)
1.2 Make sure that you do not come into contact with blood. ✓ (1)
1.3 1.3.1 Air
During the use and application of technological processes the
atmosphere can be polluted by factory emissions. ✓✓ (2)
1.3.2 Water
Water in some areas is unsafe to drink. In some situations water is
contaminated with oils or poisonous chemicals. ✓✓ (2)
1.3.3 Land
In some areas, the land is contaminated with hazardous waste, to
an extent that people cannot live on this land or travel over it. ✓✓ (2)
[10]
QUESTION 2: TECHNOLOGICAL PROCESS
2.1 Poor sound quality of the radio ✓✓ (2)
2.2 Design and construct a simple hand controlled lever device that can be
operated by a disabled child who has lost the use his/her legs. ✓✓
NB: If reference is made to the control of current flow, ✓ the learner must
be credited with full marks even if no mention is made of loss of
limbs. ✓✓ (5)
2.3 Observation: this includes analysing the product in order to
understand its design, and jotting down notes about the product. ✓
Interviews: this is about finding out from the target audience, whether
the product satisfy their needs. ✓
Questionnaires survey: to find out how many in the target audience
are satisfied with the product. ✓ (3)
[10]
QUESTION 3: OCCUPATIONAL HEALTH AND SAFETY
3.1 Make sure you are standing firmly. ✓
Make sure you remove the chuck key from the chuck after loosening
or tightening the bit. ✓ (2)
3.2 Make sure that the meter is connected in series in the circuit. ✓
Make sure the leads of the meter are connected into the correct
sockets of the meter. ✓ (2)
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(SEPTEMBER 2013) ELECTRICAL TECHNOLOGY 3
3.3 Poor ventilation may lead to drowsiness, which could cause loss of
concentration which may result in an accident. ✓✓ (2)
3.4 Horseplay in the workshop. ✓
Working on a machine which does not have correct guards. ✓
Working with the mains on (Any 2) (2)
3.5 Be careful not to damage the conductor which may weaken it leading to
poor connections which will result in future problems. ✓✓ (2)
[10]
QUESTION 4: THREE-PHASE AC GENERATION
4.1 Add power factor correcting capacitors in parallel with the load. ✓ (1)
4.2 The power distributed in each phase is the same. ✓✓ (2)
4.3 The motor will draw more current to deliver the same power. ✓
This will result in the motor running hotter than it should and the cost
of running the motor will be high. ✓ (2)
4.4 ✓= ✓ = 2,94 kW ✓ (3)
4.5 = ✓ = 219,4 V ✓
√ √ (2)
[10]
QUESTION 5: R, L AND C CIRCUITS
5.1 5.1.1 ✓= ✓ = 10,37 A ✓ (3)
5.1.2 ✓= ✓ = 5,83 A ✓ (3)
5.1.3 ✓= ✓ = 11 A ✓ (3)
5.1.4 √ ✓= √ ✓= 11,9 A ✓ (3)
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4 ELECTRICAL TECHNOLOGY (SEPTEMBER 2013)
5.1.5
(4)
5.2 5.2.1 ✓= ✓= 43,98 Ω ✓ (3)
5.2.2 ✓ = ✓= 64,96 Ω ✓ (3)
5.2.3 √ ✓ √ ✓= 45,17 Ω ✓ (3)
5.2.4 ✓= ✓ = 4,87 ✓ (3)
5.3 5.3.1 Inductive reactance ✓ (1)
5.3.2 Capacitive reactance ✓ (1)
[30]
QUESTION 6: SWITCHING AND CONTROL CIRCUITS
6.1 Functional operation of a DIAC
If a rising voltage is applied to a DIAC it acts like an open switch. ✓ When
the DIAC's trigger voltage is reached, the internal resistance of the DIAC
breaks down✓ allowing the DIAC to conduct. It operates in both
directions. ✓ When the current falls below the holding current, it switches
off. A DIAC switches on at the same time in both directions. ✓. (4)
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(SEPTEMBER 2013) ELECTRICAL TECHNOLOGY 5
6.2
(5)
6.3 SCR is switched on and switched off.
Switched on.
• A positive potential must be applied to the anode terminal. The SCR
will now be in a state ready to conduct. ✓
• When the correct positive potential is applied to the gate the SCR will
begin to conduct. ✓
Switched off.
• The current flowing through the SCR must be reduced below the
holding value. ✓
• Remove or reverse the potential across the SCR. ✓ (4)
6.4 6.4.1 TRIAC
(3)
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6 ELECTRICAL TECHNOLOGY (SEPTEMBER 2013)
6.4.2 DIAC
(2)
6.5 When the triggering circuit has no capacitor in it, ✓the triggering signal will not
be delayed by a time constant. ✓When the voltage level at the gate reaches
the triggering level, the SCR will be fired. ✓Because the sinusoidal supply
reaches its maximum at 90°, any value after 90° that could trigger the SCR has
already occurred before 90°.✓ (4)
6.6 The advantage of the TRIAC and SCR in power control is that they
have low power loss for the amount of power that is controlled. ✓
Current control is also smooth, fast and accurate. ✓
They conduct in both directions. ✓ (3)
[25]
QUESTION 7: SWITCHING AND CONTROL CIRCUITS
7.1 Open loop voltage gain AV = infinite ✓
Input impedance ZIN = infinite ✓
Output impedance ZOUT = zero ✓
Bandwidth = infinite ✓
Unconditional stability. ✓
Differential inputs i.e. two inputs ✓
Infinite common rejection ✓ (Any 4) (4)
7.2 A differential amplifier consists of two identical sections each having input
terminals. Two output terminals or a single output terminal common to both
amplifiers is provided. Ground is common to both sections. The differential
amplifier only amplifies the difference in signal between the two input
signals. This amplified signal is represented on the output/outputs. ✓✓ (2)
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(SEPTEMBER 2013) ELECTRICAL TECHNOLOGY 7
7.3 Positive Feedback
When the output of a circuit is fed back to the input of the same circuit in
phase with the input signal, the resultant will be ever increasing output. The
result could be distortion or overloading of the circuit. ✓
Example: When the microphone of an audio amplifier is held close to the
speaker positive feedback causes the amplifier to “whistle” causing great
discomfort to the human ear. ✓
Negative feedback
When the output of a circuit is fed back to the input of the same circuit out of
phase with the input, the result is that the output signal becomes smaller
and may even disappear. ✓
Example: Negative feedback is utilised in amplifiers to obtain volume
control and gain control such as is used in the oscilloscope. ✓ (4)
7.4 7.4.1 Non inverting amplifier mode ✓ (1)
7.4.2 This would not be a good audio amplifier because the output
waveform is not an exact replica of the input waveform. ✓ The
wave has been distorted slightly. ✓ (2)
7.4.3 = ✓ = 11 ✓ (2)
7.4.4 By increasing the value of Rf upward, the amount of gain
achieved by the circuit is increased. This is deduced from the
formula used to calculate the total gain of the circuit. ✓✓
Alternatively the value if Ri could be adjusted, where a small
adjustment in R1 will result in a high change in gain of the
circuit. ✓ (3)
7.5 Negative feedback advantages:
Improved bandwidth ✓
Less distortion ✓
Less noise ✓ (3)
7.6 ✓
=
= ✓
= 28,9 Hz ✓ (3)
7.7 Potential divider base biasing ✓ (1)
[25]
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8 ELECTRICAL TECHNOLOGY (SEPTEMBER 2013)
QUESTION 8: THREE-PHASE TRANSFORMERS
8.1 Making the transformer core long and narrow ✓
Rearranging the primary and secondary windings ✓ (2)
8.2 8.2.1 Secondary phase voltage
✓= ✓ = 219,4 V ✓ (2)
√ √
8.2.2 Primary line voltage
✓= ✓ = 45,58 V ✓ (3)
√ √
8.2.3 Power delivered
√ ✓= √ ✓ = 25,5 kW ✓ (4)
8.3 Stray ✓
Dielectric ✓
Iron ✓
Eddy current ✓ (4)
[15]
QUESTION 9: LOGIC CONCEPTS AND PLCs
9.1 A PLC (i.e. Programmable Logic Controller) is a device that was
invented to replace the necessary sequential relay circuits for machine
control. ✓
The PLC works by looking at its inputs and depending upon their
state, turning on/off its outputs. ✓ The user enters a program, usually
via software, that gives the desired results. (2)
9.2 A hard wired system such as a star delta starter is wired to perform a
certain duty. The system is specialised for that particular use and to
alter it means to replace and rewire the system. ✓✓
In a PLC system or soft wired system, relays are wired to the PLC. All
the control aspects such as timers, lockouts, etc. are controlled by the
PLC. Changing the system now to accommodate a change in
manufacturing needs or process adaptations is as easy as changing
the programming that controls the relay sets. ✓✓ (4)
9.3 9.3.1 A PLC works by continually scanning a program. ✓ We can think
of this scan cycle as consisting of 3 important steps. There are
typically more than 3 but we can focus on the important parts and
not worry about the others. Typically the others are checking the
system and updating the current internal counter and timer
values. ✓ (2)
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(SEPTEMBER 2013) ELECTRICAL TECHNOLOGY 9
9.3.2 Step 1: CHECK INPUT STATUS ✓ – First the PLC takes a look at
each input to determine if it is on or off. In other words, is the
sensor connected to the first input on? How about the second
input? How about the third ... It records this data into its memory to
be used during the next step. ✓
Step 2: EXECUTE PROGRAM ✓ – Next the PLC executes your
program one instruction at a time. Maybe your program said that if
the first input was on then it should turn on the first output. Since it
already knows which inputs are on/off from the previous step it will
be able to decide whether the first output should be turned on
based on the state of the first input. It will store the execution
results for use later during the next step. ✓
Step 3: UPDATE OUTPUT STATUS✓ – Finally the PLC updates
the status of the outputs. It updates the outputs based on which
inputs were on during the first step and the results of executing
your program during the second step. Based on the example in
step 2 it would now turn on the first output because the first input
was on and your program said to turn on the first output when this
condition is true. After the third step the PLC goes back to step
one and repeats the steps continuously. One scan time is defined
as the time it takes to execute the 3 steps listed above. ✓ (6)
9.4 9.4.1
(4)
9.5 9.5.1 A.B + AB = X will only partly satisfy ✓ the problem. An added
memory component is needed to fully satisfy the problem. ✓
A = High Level Sensor
B = Low level Sensor
X = Motor (2)
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10 ELECTRICAL TECHNOLOGY (SEPTEMBER 2013)
9.5.2
0000 – Low Level Sensor ✓
0001 – High Level Sensor ✓
1000 – Retaining Relay ✓
0500 – Motor ✓ (7)
9.5.3 Addresses are assigned to inputs and outputs, for the PLC device
to correctly identify the assigned inputs and to its programmed
function. ✓✓ (2)
9.6 9.6.1
(6)
[35]
QUESTION 10: THREE-PHASE MOTORS AND CONTROL
10.1 10.1.1 Line current and phase current
√ ✓
✓= ✓ = 15,19 A ✓
√ √
= ✓ = 8,77 A ✓ (5)
√ √
10.1.2 Input in kVA
√ ✓=√ ✓ = 9,997 kVA✓ (3)
10.2 To reduce the voltage at start-up. ✓ This in turn reduces the starting
current. ✓ Reduced starting current leads to less heat build-up and
decreased chance of burn out. ✓ (3)
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(SEPTEMBER 2013) ELECTRICAL TECHNOLOGY 11
10.3 No-volt protection prevents a motor from restarting after a power failure. ✓
The operator needs to re-engage the start switch in order to restart the
motor. ✓ This protects both the operator and the equipment. ✓ (3)
10.4 A normally closed contact is closed in a de-energised state and opens in an
energised state. ✓✓ (2)
10.5 Earthing the enclosure of the motor will keep it at earth potential, ✓ thus
preventing it from becoming live / hot and placing the life of the operator in
danger. ✓ This precautionary measure ensures the safety of the worker. ✓ (3)
10.6 By swapping the connections to any two phases of the three supply lines to
the stator. ✓✓ (2)
10.7 The I2R ✓ heat losses in the stator and rotor of the motor due to the
resistance of the winding. ✓ Iron losses, eddy current and hysteresis losses
in iron core ✓ due to magnetic fields. ✓ Mechanical losses ✓ like friction
between moving parts such as bearings and wind resistance of the cooling
fan. ✓ (6)
10.8 Mechanical (Any 1 correct answer)
Is the cooling fan intact and turning freely but mounted securely on
the motor shaft? ✓
Electrical (Any 2 correct answers)
Is the frame of the motor earthed? ✓
Has the insulation test been performed on the stator and are the
readings above 1 MΩ? ✓ (3)
[30]
TOTAL: 200
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