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NATIONAL
SENIOR CERTIFICATE
GRADE 12
SEPTEMBER 2016
ELECTRICAL TECHNOLOGY
MEMORANDUM
MARKS: 200
This memorandum consists of 9 pages.
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Electrical Technology Grade 12 Eastern Cape Mock 2016 Memo
Grade 12 Electrical Technology memorandum from the Eastern Cape Mock Exam 2016 (English), 9 pages. Read online or download the free PDF.
Electrical Technology · Grade 12 · Eastern Cape Mock Exam · 2016. Memorandum, 9 pages. Read online or download the PDF.
- Subject
- Electrical Technology
- Grade
- Grade 12
- Document type
- Memorandum
- Year
- 2016
- Exam period
- Eastern Cape Mock Exam
- Curriculum
- CAPS
- Pages
- 9
- File size
- 449.1 KB
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Eastern Cape Mock Exam 2016
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2 ELECTRICAL TECHNOLOGY (EC/SEPTEMBER 2016)
QUESTION 1: OCCUPATIONAL HEALTH AND SAFETY
1.1 Remove the person from the heat source. √
If you are helping someone with a serious burn, keep the burn areas
elevated to reduce swelling. √
(ANY TWO RELEVANT ANSWERS) (2)
1.2 Attendance √
Character √
Team work √ (3)
1.3 The type of work that will be done in an electrical technology workshop
requires the correct lighting level because good visibility is crucial in
soldering work, making electrical connections and other electrical
work. √√
OR
Proper lighting improves safety as well as better quality
workmanship. √√ (1 x 2) (2)
1.4 Risk analysis is a process that will help people to adopt a policy of safe
practice as an on-going process. √
As projects in a workshop change according to people’s needs the
manufacturing processes also has to change. √
Safety practices must be included in all stages of planning. √ (3)
[10]
QUESTION 2: THREE-PHASE AC GENERATION
2.1 Power factor meter enables the consumer √ to determine the power factor in
the commercial installations. √ (2)
2.2 Solar √
Wind turbines √ (2)
A star connected alternator generator produces 1,1 kW per phase. Each
phase has an impedance of 44 Ω.
2.3 2.3.1 VF =√𝑃𝑅 √
=√1100 × 44 √
= 220 V √ (3)
2.3.2 VL = √3𝑉𝐹 √
= √3 x 220 √
= 381,05 V √ (3)
2.4 To be able to supply consumers with both single-phase √ and three-phase
power. √ (2)
2.5 Measure power √ (1)
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(EC/SEPTEMBER 2016) ELECTRICAL TECHNOLOGY 3
2.6 Apparent power is the power without considering the efficiency, losses and
power factor of the circuit. S = √3VLIL √√ (2)
2.7 Three-phase systems are more versatile and they can operate in both star
and delta. √ (1)
2.8 Pact = Papp x cos θ √
= 300 000 x 0,8 √
= 240 kW √ (3)
2.9 Less current drawn from the supply.
Cost saving from drawing less current.
Less maintenance.
(ANY CORRECT ANSWER) √ (1)
[20]
QUESTION 3: THREE-PHASE TRANSFORMERS
3.1 Yes √
The primary circuit is complete √ and will have a supply connected across it. √
Therefore current will flow and it will be dependent upon the impedance of the
primary circuit and the supply voltage. √ (4)
3.2 3.2.1 𝑁𝑆 ×𝑉𝑃
𝑉𝑆 = √
𝑁𝑃
1 ×11000
= √
46
= 239, 13 𝑉 √ (3)
3.2.2 S = √3VLIL √
= √3x 11 000 x 6 √
= 114,32 kVA √ (3)
3.3 3.3.1 The current will also double. √ (1)
3.3.2 The voltage will stay the same. √ (1)
3.3.3 The input power will also double. √ (1)
3.4 To reduce eddy currents circulating in the core. √ This causes heat to be
generated in the core of the transformer. √ (2)
3.5 One mark for primary
connection
One mark for secondary
connection
One mark for single-phase
transformers (5)
[20]
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4 ELECTRICAL TECHNOLOGY (EC/SEPTEMBER 2016)
QUESTION 4: THREE-PHASE MOTORS AND STARTERS
4.1 4.1.1 The current drawn by the motor will decrease. √ Pout =√3VLILcos θ
Therefore IL must decrease. √ (2)
4.1.2 The output power remains constant. √
The line voltage remains constant. √ (2)
4.2 4.2.1 𝑃𝐼𝑁
𝐼𝐿 = √
√3×𝑉𝐿 ×cos 𝜃
12 000
= √
√3×400×0,8
= 21,65 𝐴 √ (3)
4.2.2 Papp = √3VLIL √
= √3x 400 x 21,65 √
= 15 kVA √ (3)
4.3 4.3.1 Three-phase induction motor. √ (1)
4.3.2 1 Lifting eye √
2 Shaft √
3 Laminated steel rotor √ (3)
4.4 The same size frame a three-phase motor delivers more power. √
High starting torque. √
(ANY RELEVANT ANSWERS) (2)
4.5 60 ×𝑓
𝑃𝑜𝑜𝑙𝑝𝑎𝑖𝑟𝑠 𝑝𝑒𝑟 𝑝ℎ𝑎𝑠𝑒 = √
𝑁𝑆
60 ×50
= √
500
= 6√√
This mean 6 x 2 = 12 poles per phase
The total number of poles will be 12 x 3 = 36 √ (4)
4.6 1 Overload → Automatically disconnect the motor in the event of an
overload. √√
2 Interlocks/holding out → prevent the timer from being energised before
contactor 1. √√
3 Normally open contact of a timer. → Connect the second motor after the
time has lapsed. √√
4 Normally open contact of the main contactor → Latches after the start
button has been released. (Holding contact) √√ (8)
4.7 Check for obvious open/exposed wires, any bare wire carrying electricity
is just bad news. √√
Check the electrical connection in the terminal box. It is vital important
that the nuts where the supply leads/cables are joined to the terminal
box are tight. √√ (4)
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(EC/SEPTEMBER 2016) ELECTRICAL TECHNOLOGY 5
4.8
(6)
4.9 Copper losses √
Due to resistance of the copper wire in the rotor. √
(ANY RELEVANT ANSWER) (2)
[40]
QUESTION 5: RLC CIRCUITS
5.1 Frequency is inversely proportional to the capacitive reactance √ so the
increase in capacitive reactance causes the impedance to increase, √ the
current will decrease √ and will cause the brightness of the incandescent lamp
to decrease. √ (4)
5.2 Radio tuning circuit √
Filtering circuit √
Timing circuit √
(ANY RELEVANT ANSWER) (3)
5.3 Current is maximum √
Z=R√
XL = XC √
(ANY RELEVANT ANSWER) (3)
5.4 5.4.1
𝐼𝑇 = √𝐼𝑅 2 + (𝐼𝐿 − 𝐼𝐶 )2√
= √42 + (9 − 6)2√
=5𝐴√ (3)
5.4.2 𝐼𝑅
Cos θ = √
𝐼𝑇
4
= √
5
= 0,8 √ lagging √ (4)
5.4.3 𝑉
Z=𝐼 √
𝑇
230
= √
5
= 46 Ω √ (3)
[20]
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6 ELECTRICAL TECHNOLOGY (EC/SEPTEMBER 2016)
QUESTION 6: LOGIC
6.1 The internal memory of the PLC store the instruction sets √ /programming for
the CPU to access when operating. √ (2)
6.2 Economical √ For a control panel with more than ten relays, a PLC is cheaper. √
(ANY RELEVANT ANSWER) (2)
6.3 PLCs are used to automate machinery in assembly lines and were developed
as a substitute for large relay panel √√
No rewiring is needed when the sequence is changed. √√
(ANY RELEVANT ANSWER) (4)
6.4
(6)
6.5 6.5.1
CARRY = AB + AC + BC √√ (5)
6.5.2
(4)
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(EC/SEPTEMBER 2016) ELECTRICAL TECHNOLOGY 7
6.5.3
(3)
6.6
Full marks if answer is obtained using Karnaugh map (4)
6.7 Sequencing of robot √
Control of machinery in factories √
Automation of machinery in assembly lines √
(ANY RELEVANT ANSWERS) (3)
6.8 6.8.1 Normally closed contact √ (1)
6.8.2 Normally open contact √ (1)
6.8.3 Coil √ (1)
6.9
TWO MARKS FOR GROUPING
F = B √√ (4)
[40]
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8 ELECTRICAL TECHNOLOGY (EC/SEPTEMBER 2016)
QUESTION 7: AMPLIFIERS
7.1 Negative feedback is 180o output of phase √ with the input signal.
When the input goes positive the feedback signal will go negative
diminishing the gain of the amplifier due to a smaller resulting input
signal. √√ (3)
7.2 The op-amp amplifies the difference between the input signals.
It operates as a differential amplifier. √√ (2)
7.3
(5)
7.4 7.4.1 Inverting amplifier √ (1)
7.4.2 𝑅𝑓
𝑉𝑂𝑈𝑇 = −[ × 𝑉𝐼𝑁 ] √
𝑅𝐼𝑁
100 000
= −[ × 5] √
10 000
= -50 V √ (3)
7.4.3
(2)
7.4.4 The gain is proportional to the value of the feedback resistor √ so die
and the gain will also increase. √ (2)
7.5 Voltage sensitive switch. √
Supplies a digital output. √
Output frequency is the same as input frequency. √
Have two fixed trigger voltages. √ (4)
7.6 A differential amplifier will only amplify the difference between two input
signals. √ If they are the same no amplification will take place. √ (2)
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(EC/SEPTEMBER 2016) ELECTRICAL TECHNOLOGY 9
7.7
(8)
7.8 Audio mixer √
Digital-to-analogue converter √
Bipolar to unipolar converter √ (Any 2 x 1) (2)
7.9 To integrate a signal representing water flow. √
It can also be used as a ramp generator. √ (2)
7.10
It can be used for generating RF signal. √
It is widely used as local oscillator in radio receivers. √ (6)
7.11 7.11.1 1
𝑓= √
2𝜋√6𝑅𝐶
1
= √
2𝜋√6×103 ×250×10−12
= 41,09 𝐻𝑧 √ (3)
7.11.2 It is used at audio frequency as audio oscillator √
Electronic organ √ (2)
7.12 Day night switches √
The result is that the output of the Schmidt trigger is changed to switch
a light on √
Wave Shaping Circuits √
Schmitt trigger swings into an upper or lower saturation point. √
(ANY RELEVANT ANSWERS) (2)
[50]
TOTAL: 200
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