NATIONAL
SENIOR CERTIFICATE
GRADE 12
SEPTEMBER 2023
ELECTRICAL TECHNOLOGY: ELECTRONICS
MARKING GUIDELINE
MARKS: 200
This marking guideline consists of 13 pages.
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Electrical Technology Grade 12 Eastern Cape Mock 2023 Memo
Grade 12 Electrical Technology memorandum from the Eastern Cape Mock Exam 2023 (English), 13 pages. Read online or download the free PDF.
Electrical Technology · Grade 12 · Eastern Cape Mock Exam · 2023 · English. Memorandum, 13 pages. Read online or download the PDF.
- Subject
- Electrical Technology
- Grade
- Grade 12
- Language
- English
- Document type
- Memorandum
- Year
- 2023
- Exam period
- Eastern Cape Mock Exam
- Curriculum
- CAPS
- Pages
- 13
- File size
- 539.6 KB
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2 ELECTRICAL TECHNOLOGY: ELECTRONICS (EC/SEPTEMBER 2023)
INSTRUCTIONS TO MARKERS
1. All calculations 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.
2.6 Markers should consider that learners answers may deviate slightly from
the marking guideline depending on how and where in the calculation
rounding off was used.
3. These marking guidelines are only a guide with model answers.
4. Alternative interpretations must be considered and marked on merit. However,
this principle should be applied consistently throughout the marking session at
ALL marking centres.
Copyright reserved Please turn over
(EC/SEPTEMBER 2023) ELECTRICAL TECHNOLOGY: ELECTRONICS 3
QUESTION 1: MULTIPLE-CHOICE QUESTIONS
1.1 B✓ (1)
1.2 A✓ (1)
1.3 C✓ (1)
1.4 C✓ (1)
1.5 D✓ (1)
1.6 A✓ (1)
1.7 D✓ (1)
1.8 B✓ (1)
1.9 C✓ (1)
1.10 C✓ (1)
1.11 A✓ (1)
1.12 C✓ (1)
1.13 B✓ (1)
1.14 A✓ (1)
1.15 A✓ (1)
[15]
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4 ELECTRICAL TECHNOLOGY: ELECTRONICS (EC/SEPTEMBER 2023)
QUESTION 2: OCCUPATIONAL HEALTH AND SAFETY
2.1 • Integrity ✓
• Sense of responsibility ✓
• Emphasis on quality
• Discipline
• Sense of teamwork (Any 2 x 1) (2)
2.2 An event that causes grave or severe physical injury to a person, ✓ threatening
their health and safety. ✓ (2)
2.3 • Avoid direct contact with any chemical ✓
• Always wear personal protective clothing
• Always read and be aware of the warning symbols on containers
(Any 1 x 1) (1)
2.4 An unsafe act is the wilful performance of a task or activity in a manner that
may threaten the health and safety of everyone. ✓
A calculated risk is the probability that injury or damage may occur while
using dangerous equipment. ✓ (2)
2.5 • To prevent oneself from being electrocuted. ✓
• To prevent injury (Any 1 x 1) (1)
2.6 This could damage the equipment and render the equipment unsafe ✓ and
compromise the safety of the user. ✓ (2)
[10]
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(EC/SEPTEMBER 2023) ELECTRICAL TECHNOLOGY: ELECTRONICS 5
QUESTION 3: RLC CIRCUITS
3.1 It is the opposition offered to current flow by the reactive components of an
inductor ✓ when it is connected to an AC supply. ✓ (2)
3.2 VL
I
Y-Axis
Y-Axis
0
90º 180º 270º 360º
Y-Axis
• Full cycles drawn correctly ✓
• Phase difference correct ✓
• Waveforms labelled correctly ✓ (3)
3.3 3.3.1 𝑋𝐶 = 1 ✓
2𝜋𝑓𝐶
1
= 2 × 𝜋 × 60 × 200 × 10–6 ✓
= 13,26 Ω ✓ (3)
3.3.2 𝐼 = 𝑉 ✓
𝑍
110
= 101,65 ✓
= 1,08 𝐴 ✓ (3)
3.3.3 Z = √R2 + (XL − XC )2 ✓
R = √Z 2 + (XL − XC )2
= √101,652 − (31,55 − 13,26)2 ✓
= 100 Ω ✓ (3)
3.3.4 𝑋𝐿 = 2𝜋𝑓𝐿 ✓
𝑋𝐿
𝐿 = 2𝜋𝑓
31,55
= 2×𝜋×60 ✓
= 0,084 𝐻 = 84 𝑚𝐻 ✓ (3)
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6 ELECTRICAL TECHNOLOGY: ELECTRONICS (EC/SEPTEMBER 2023)
3.4 1
fr = 2π√LC ✓
1
= ✓
2 × π × √50 × 10–3 × 60 × 10–6
= 91,89 Hz ✓ (3)
3.5 3.5.1 V
IC = X ✓
C
100
= 11,83 ✓
= 8,45 A ✓ (3)
3.5.2
IT = √IR 2 + (IC − IL )2 ✓
= √9,092 + (8,45 − 4,54)2 ✓
= 9,9 A ✓ (3)
3.5.3 cosθ = IR ✓
I
T
9,09
= 9,9 ✓
= 0,918 ✓ (3)
3.5.4 Leading, ✓ because Ic is larger than IL ✓ (2)
3.6 Selectivity is a measure of how well a resonant circuit responds to a range of
frequencies. ✓ and separates other frequencies. ✓ (2)
3.7 • The value of the series resistor ✓
• The LC ratio ✓ (2)
[35]
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(EC/SEPTEMBER 2023) ELECTRICAL TECHNOLOGY: ELECTRONICS 7
QUESTION 4: SEMICONDUCTOR DEVICES
4.1 ✓
DRAIN (D)
✓
GATE ( G)
✓
SOURCE (S) (3)
4.2 4.2.1 1 – Source ✓
2 – Gain ✓
3 – Drain ✓
4 – Enhanced channel ✓ (4)
4.2.2 N-type material ✓ (1)
4.3 Cut off: with zero on the emitter the pn junction is reverse biased and a small
reverse leakage current flows. ✓ By increasing the emitter voltage, the reverse
leakage current falls until a point is reached where the emitter voltage equals the
voltage in the bar point. ✓ No current flows either direction. ✓ If the emitter voltage
rises until the pn junction is on the point of becoming forward biased, forward
current begins to flow at the cut off region. ✓ (4)
4.4 + VCC. ✓
R1 RC
Output
C
Input. T1 . ✓
✓ C1 B
T.2 ✓
E
R2 RE CE . ✓
0V (5)
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8 ELECTRICAL TECHNOLOGY: ELECTRONICS (EC/SEPTEMBER 2023)
4.5 4.5.1 A = Non inverting op-amp ✓
B = Inverting op-amp ✓
C = Output ✓ (3)
4.5.2 The Op-amp is ideal for amplifying AC voltages because of its dual
voltage supply ✓ which allows the output terminal to rise and fall above
and below zero volts. ✓ (2)
4.6 4.6.1 Inverting amplifier. ✓ (1)
4.6.2 V RF
OUT = VIN (− R ) ✓
IN
VOUT × R IN
RF =
VIN
8 × 1,8 × 103
RF = ( )✓
0,4
= 36 000 Ω ✓ (3)
4.7 4.7.1 The signal will be 10 × greater than the input signal / ✓the output will be
1 V ✓ and the output signal will be 180° out of phase with the input
signal. ✓ (3)
4.7.2 The signal will be 11 × greater than the input signal / ✓the output will
be 1,1 V ✓ and the output signal will be in phase with the input signal. (3)
✓
4.8 • The amplifier circuits gain will be reduced. ✓
• The amplifier becomes much more stabilised. ✓
• Distortion of the output signal is reduced. ✓
• The range of frequencies which can be amplified is increased. ✓ (4)
4.9 4.9.1 2 – Trigger ✓
3 – Output ✓
5 – Control voltage ✓
8 – Power supply ✓ (4)
4.9.2 It provides the discharge path ✓ for the timing capacitor ✓ and timing
resistor ✓ (3)
4.9.3 The 555 IC can only operate at power supply voltages between
+5 V ✓ to +18 V. ✓ (2)
[45]
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(EC/SEPTEMBER 2023) ELECTRICAL TECHNOLOGY: ELECTRONICS 9
QUESTION 5: SWITCHING CIRCUITS
5.1 5.1.1 Monostable multivibrator ✓ (1)
5.1.2 Bistable multivibrator ✓ (1)
5.2 RF
✓
+ 12 V. ✓
A
-
B +
- 12 V. ✓ R1
C1 . ✓ VOUT. ✓
VC
R2
t
(5)
5.3 5.3.1 Bistable ✓multivibrator ✓ (2)
5.3.2 Resistor 𝑅1 and 𝑅2 are pull-up ✓ resistors ✓ (2)
5.3.3 When set is pressed, it pulls pin 2 ‘low’ ✓ (0 V) and cause the output to
go 'high'. ✓ turning the LED on. ✓ (3)
5.3.4 Threshold pin 6 is purposefully held at 0 V ✓causing the IC not to
reset, ✓ keeping the output high ✓ when S1 is pressed. (3)
5.4 5.4.1 The output signal represents an inverting Schmitt trigger ✓ because the
output signal is inverted ✓ with reference to the input signal. ✓ (3)
5.4.2 ✓
✓ + VCC
VIN ✓
✓ VOUT
✓ RF ✓
-VCC
R1 ✓
0V
(7)
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10 ELECTRICAL TECHNOLOGY: ELECTRONICS (EC/SEPTEMBER 2023)
5.5 5.5.1 The gain of the amplifier is determined by the ratio ✓ of the feedback
resistance ✓ to the input resistance of each. ✓ (3)
5.5.2 VOUT = - (𝑉1 𝑅1 + 𝑉2 𝑅2 + 𝑉3 𝑅3)
𝑅𝑓 𝑅𝑓 𝑅𝑓
✓
100 𝑘Ω 100 𝑘Ω 100 𝑘Ω
= − (200 𝑚𝑉 20 𝐾Ω + 300 𝑚𝑉 10 𝐾Ω + 400 𝑚𝑉 25 𝑘Ω ) ✓
= −(200 𝑚𝑉 × 5) + (300 𝑚𝑉 × 10) + (400 𝑚𝑉 × 4) ✓
= −5.6 𝑉 ✓ (4)
5.5.3 VOUT = VIN × Gain
𝑉
Gain AV = 𝑉𝑂𝑈𝑇 ✓
𝐼𝑁
5,6
= 0,2 + 0,3 + 0,4 ✓
= 6,2 𝑉 ✓ (3)
5.5.4 With a variable resistor in the feedback loop, the gain ✓ of the amplifier
can be varied / controlled. ✓ (2)
5.5.5 If 𝑅2 is changed to 5 kΩ, the gain for 𝑉2 will increase ✓ causing the total
output voltage to increase. ✓ (2)
5.6 +V C
F ✓
Input ✓
R +V ✓
0V VIN IN ✓
t -
0V V
OUT ✓
+
Output ✓
-V
- VSAT 0V
(6)
5.7 • The inputs draw zero current. ✓
• The two inputs will always have the same voltage. ✓
• The capacitor will charge at a constant rate when a constant current is
supplied. ✓ (3)
[50]
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(EC/SEPTEMBER 2023) ELECTRICAL TECHNOLOGY: ELECTRONICS 11
QUESTION 6: AMPLIFIERS
6.1 6.1.1 Vce = Vcc✓ (1)
6.1.2 𝑉𝑐𝑐
𝐼𝐶𝑚𝑎𝑥 = 𝑅𝑐 ✓
12 𝑉
= 600 Ω ✓
= 2 𝑚𝐴 ✓ (3)
6.2 6.2.1 • RC coupled amplifier. ✓
• Common emitter amplifier. (Any 1 x 1) (1)
6.2.2 The emitter-base junction (EB) must be forward biased above the cut-
off region, ✓ and below the saturation region. ✓ The collector-base
junction (CB) must be reversed biased ✓ for the transistor to operate in
the active region for amplification to take place. ✓ (4)
6.2.3 VRC = VCC – VCE ✓
= 8 – 2,38 ✓
= 5,62 V ✓ (3)
6.3 6.3.1 Push-Pull amplifier ✓ (1)
6.3.2 NPN transistor ✓ (1)
6.4 6.4.1 𝐴𝐼 = 20𝑙𝑜𝑔10 𝑂𝑈𝑇 ✓
𝐼
𝐼 𝐼𝑁
15,3 𝑚𝐴
= 2𝑂𝑙𝑜𝑔10 3,6 𝑚𝐴 ✓
= 20 × 4,25
= 85 𝑑𝐵 ✓ (3)
6.4.2 𝐴𝑃 = 10𝑙𝑜𝑔10 𝑃 𝐼𝑁 ✓
𝑃
𝑂𝑈𝑇
750 × 10−3 𝐴
=10 𝑙𝑜𝑔10 ✓
28
= 10 × 0,026
= 0,268 𝑑𝐵 ✓ (3)
6.4.3 𝑉 ✓
A V = 20𝑙𝑜𝑔10 𝑉𝑂𝑈𝑇
𝐼𝑁
219 ✓
= 20𝑙𝑜𝑔10 230
= 19 𝑑𝐵 ✓ (3)
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12 ELECTRICAL TECHNOLOGY: ELECTRONICS (EC/SEPTEMBER 2023)
6.5
+ Vcc ✓
R1 ✓
Q1
NPN ✓
R2
C
R3
AC . ✓ Q2
PNP. ✓
R4
(5)
6.6
✓
✓
✓
Pass
band
Fr ✓
(4)
6.7 6.7.1 An oscillator is a device which generates an AC output signal, ✓without
any externally applied input signal. ✓ (2)
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(EC/SEPTEMBER 2023) ELECTRICAL TECHNOLOGY: ELECTRONICS 13
6.7.2 • The RC-phase-shift oscillator uses three sets of RC combinations ✓
to create a phase-shift of 180° of the output waveform from the
amplifier ✓ and attenuates it.
• Together with the 180º phase-shift between the base voltage and
collector voltage of the amplifier, ✓ the output signal with a phase-
shift of 360º is achieved and amplified before supplying it to the
RC-network. ✓
• The cycle repeats itself. (4)
6.8 6.8.1 Hartley Oscillator ✓ (1)
6.8.2 • When first switched ON, the collector voltage rises and allows the
capacitor in the tank circuit to charge. ✓
• The voltage drop across the inductors is in an inverted form, driving
the transistor's base in the opposite direction thereby switching it
OFF. ✓
• The capacitor will discharge through the inductors and push the tank
circuit into oscillation. ✓
• During oscillation the voltages at each end of the tank circuit are
180° out of phase with each other, relative to their 0 V common
centre tap point. ✓
• This ensures that the collector voltage is 180° out of phase with the
base voltage. ✓
• The freewheeling effect of the tank circuit’s operation then begins to
drive the transistor alternately ON and OFF which in turn continually
re-charges the tank circuit keeping it oscillating at a constant
amplitude. ✓ (6)
[45]
TOTAL: 200
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