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NATIONAL
SENIOR CERTIFICATE
GRADE 12
SEPTEMBER 2022
ELECTRICAL TECHNOLOGY: ELECTRONICS
MARKING GUIDELINE
MARKS: 200
This marking guideline consists of 11 pages.
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Electrical Technology Grade 12 Eastern Cape Mock 2022 Memo
Grade 12 Electrical Technology memorandum from the Eastern Cape Mock Exam 2022 (English), 11 pages. Read online or download the free PDF.
Electrical Technology · Grade 12 · Eastern Cape Mock Exam · 2022 · English. Memorandum, 11 pages. Read online or download the PDF.
- Subject
- Electrical Technology
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- Grade 12
- Language
- English
- Document type
- Memorandum
- Year
- 2022
- Exam period
- Eastern Cape Mock Exam
- Curriculum
- CAPS
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- 11
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2 ELECTRICAL TECHNOLOGY: ELECTRONICS (EC/SEPTEMBER 2022)
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.
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(EC/SEPTEMBER 2022) ELECTRICAL TECHNOLOGY: ELECTRONICS 3
QUESTION 1: MULTIPLE CHOICE-QUESTIONS
1.1 A✓ (1)
1.2 C✓ (1)
1.3 B✓ (1)
1.4 C✓ (1)
1.5 B✓ (1)
1.6 B✓ (1)
1.7 C✓ (1)
1.8 D✓ (1)
1.9 B✓ (1)
1.10 A✓ (1)
1.11 C✓ (1)
1.12 A✓ (1)
1.13 A✓ (1)
1.14 A✓ (1)
1.15 A✓ (1)
[15]
QUESTION 2: OCCUPATIONAL HEALTH AND SAFETY
2.1 An occurrence of catastrophic proportions, ✓ resulting from the use of
machinery, or activities at work. ✓ (2)
2.2 The learner does not know safe practices. ✓
The learner knows better but intentionally conducts the act. (1)
2.3 2.3.1 Running could cause you to trip or collide with another learner. ✓
This could result in you injuring yourself with nearby equipment or
machinery. ✓ (2)
2.3.2 This could cause the outlet to exceed its rated current ✓and could
lead to short circuits, fires or damaged appliances. ✓ (2)
2.4 First, I would define all the various threats to safety in the workshop. ✓
Secondly, I would determine the extent of all the vulnerabilities in the
workshop. ✓
Finally, I would devise countermeasures should a risk occur. ✓ (3)
[10]
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4 ELECTRICAL TECHNOLOGY: ELECTRONICS (EC/SEPTEMBER 2022)
QUESTION 3: RLC CIRCUITS
3.1 3.1.1 It would remain the same. ✓ (1)
3.1.2 XL = 31,83 Ω ✓ (1)
3.1.3 (a) XL = 2πfL ✓
= 2 × π × 50 × 0,5 ✓
= 157,08 Ω ✓ (3)
(b) Z = √R2 + (XL − XC )2 ✓
= √1202 + (157,08 − 31,83)2 ✓
= 173,46 A ✓ (3)
(c) Vc = I × XC ✓
= 1,38 × 31,83 ✓
= 43,93 V ✓ (3)
(d) 1
XC = 2πfC
1
C = 2πfX ✓
C
1
= 2 × π × 50 × 42,44 ✓
= 0,000075 F = 75 μF ✓ (3)
(e) V
I = RR ✓
240
= 120 ✓
=2A ✓ (3)
3.2 0A✓ (1)
3.3 The small band of frequencies ✓ centred around the resonant
frequency. ✓
The frequencies between the upper cut-off frequency and the lower cut-off (2)
frequency, centred around the resonant frequency.
3.4 Z is maximum ✓
I is minimum ✓
XL = XC (2)
3.5 3.5.1
IT = √IR 2 + (IC − IL )2 ✓
= √0,832 + (0,63 − 0,5)2 ✓
= 0,84 A ✓ (3)
3.5.2 I = V
R
V = I. R ✓
= 0,83 × 120,487 ✓
= 100 V ✓ (3)
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(EC/SEPTEMBER 2022) ELECTRICAL TECHNOLOGY: ELECTRONICS 5
3.5.3 Z = V ✓
I T
100
= 0,84 ✓
= 119,05 Ω ✓ (3)
3.5.4 cosθ = IR
I T
I
θ = cos−1 IR ✓
T
0,83
= cos−1 0,84 ✓
= 8,85° ✓
leading ✓ (4)
[35]
QUESTION 4: SEMICONDUCTOR DEVICES
4.1 Junction field effect transistor ✓ (1)
4.2 N-channel JFET or NFET ✓ (2)
P-Channel JFET or PFET ✓
4.3 To overcome leakage current between the gate terminal and drain-source
channel, the gate terminal was electrically isolated from the channel ✓ by
means of an extremely narrow layer of metal-oxide-silicon. ✓ (2)
4.4 4.4.1 Enhancement-mode ✓ N-channel ✓ MOSFET (2)
4.4.2 The lamp will switch ON as soon as the gate voltage VGS is raised
✓ to a sufficient level which will forward bias the internal channel of
the MOSFET. ✓ (2)
4.4.3 If RGS is short circuited, it will cause the internal conductive channel
of the MOSFET to disperse, ✓ cutting the current flow ✓ and
switching the lamp OFF. ✓ (3)
4.5 4.5.1 A – Cut off region ✓
D – Saturation region ✓ (2)
4.5.2 At point C the UJT triggers ON. As the UJT is triggered, its internal
resistance and voltage ✓ will decrease ✓ while current increases. ✓
This is contrary to Ohm's law and is called negative resistance. At
point D if the emitter is supplied with sufficient current the UJT
operating point will continue falling until a valley point is reached (D),
UJT goes into its permanent ON condition called saturation region. (3)
4.6 4.6.1 Darlington pair ✓ (1)
4.6.2 Very high current gain ✓
Improved input impedance ✓
When used in common collector pair it develops a very low output. (2)
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6 ELECTRICAL TECHNOLOGY: ELECTRONICS (EC/SEPTEMBER 2022)
4.7 • Infinite gain ✓
• Infinite input impedance ✓
• Zero output impedance ✓
• Infinite bandwidth ✓
• Infinite common mode rejection ratio
• Unconditional stability (4)
4.8 4.8.1 𝑅𝐹
𝐴𝑉 = 1 + ✓
𝑅𝐼𝑁
50 000
𝐴𝑉 = 1 + ✓
10 000
𝐴𝑉 = 6 ✓ (3)
4.8.2 𝑅𝐹
𝑉𝑂𝑈𝑇 = 𝑉𝐼𝑁 × (1 + ) 𝑉✓
𝑅𝐼𝑁
50 000
𝑉𝑂𝑈𝑇 = 1,5 × (1 + ) 𝑉✓
10 000
𝑉𝑂𝑈𝑇 = 9 𝑉 ✓ (3)
4.8.3 If the value of the feedback resistor is decreased the gain of the
amplifier will decrease ✓ causing the output voltage to decrease. ✓ (2)
4.9 4.9.1 This pin sets the voltage at which the 555 IC will trigger. It is used
to maintain ✓ the voltage across the timing capacitor ✓ which is
discharged through pin 7. ✓ (3)
4.9.2 The 555 IC can operate from power supply voltages of between
+5 V ✓ and +18 V. ✓ (2)
4.9.3 In this mode the 555 timer is astable ('free running'), therefore its
output will continuously toggle between high and low ✓ thus
generating a continuous train of square-wave pulses. ✓ (2)
4.9.4 The pin closest to the dot. ✓
The pin to the left of the notch. (1)
4.10 Open loop gain is the gain of an op amp without any from the output to the
input. ✓
Closed loop gain is the of an op amp with feedback from the output to the
input. ✓ (2)
4.11 Basic timing functions like turning a light on or off ✓
Pulse, oscillation and waveforms generation ✓
Digital logic probes ✓
Creating a warning light that flashes on and off
Produce musical notes of a particular frequency (3)
[45]
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(EC/SEPTEMBER 2022) ELECTRICAL TECHNOLOGY: ELECTRONICS 7
QUESTION 5: SWITCHING CIRCUITS
5.1 5.1.1 Two external inputs ✓
Two stable states ✓ (2)
5.1.2 The LED will be destroyed. ✓ (1)
5.1.3 The current flowing through the LED will not be limited. ✓ This will
cause the LED to draw more current than what it is able to
handle.✓ (2)
5.1.4 Pressing the SET button will pull Pin 2 “low” ✓ and cause the IC
output to switch to “high”. ✓ As Pin 6 is deliberately held “low” the
IC cannot reset itself thus staying “high”. ✓ (3)
5.1.5 These resistors are known as “pull up” resistors. ✓ When both SET
and RESET buttons are open, the pull up resistors keep the voltage
on the input high. ✓ (2)
5.2
RF ✓
+15 V
-
✓
+
-15 V R1✓
C✓ V OUT ✓
R2 ✓
(6)
5.3 The feedback resistor is connected from the output ✓ of the op amp to the
inverting input. ✓ This allows a part of the output to flow back to the inverting
input. ✓ (3)
5.4 It is used to eliminate switch bounce. ✓ (1)
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8 ELECTRICAL TECHNOLOGY: ELECTRONICS (EC/SEPTEMBER 2022)
5.5 • A light dependent resistor (LDR) and a100 kΩ resistor is connected in
series. This forms a voltage divider. ✓
• The voltage divider feeds the non-inverting input of the op amp. ✓
• The inverting input is fed by a 100 kΩ variable resistor. ✓
• Less light on the LDR, the resistance rises and in turn the voltage on the
non-inverting also rises. ✓
• When the voltage level increases to a level higher than the level set by the
variable resistor, the op amp output will go high immediately. ✓
• This will switch the transistor on, and the alarm will be energized. ✓ (6)
5.6
VOUT✓
+10 V✓
-1 V✓ VIN✓
+1 V✓
-10 V✓
(6)
5.7
✓ ✓
✓ ✓ (4)
5.8 5.8.1 A summing amplifier is used to add two or more different input
signals ✓ to create one amplified output signal. ✓ (2)
5.8.2 𝑉𝑂𝑈𝑇 = − (V1 + V2 + V3 ) ✓
= − (0,5 + 1,2 + 0,9) ✓
= − 2,6 V ✓
OR
𝑅 𝑅 𝑅
𝑉𝑂𝑈𝑇 = − (𝑉1 𝑅𝐹 + 𝑉2 𝑅𝐹 + 𝑉3 𝑅𝐹) 𝑉 ✓
1 2 3
20 000 20 000 20 000
𝑉𝑂𝑈𝑇 = − (0,5 20 000 + 1,2 20000 + 0,9 20 000 𝑉 ✓
𝑉𝑂𝑈𝑇 = − 2,6 𝑉 ✓ (3)
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(EC/SEPTEMBER 2022) ELECTRICAL TECHNOLOGY: ELECTRONICS 9
5.8.3 𝑅 𝑅 𝑅
𝑉𝑂𝑈𝑇 = − (𝑉1 𝑅𝐹 + 𝑉2 𝑅𝐹 + 𝑉3 𝑅𝐹) 𝑉 ✓
1 2 3
40 000 40 000 40 000
𝑉𝑂𝑈𝑇 = − (0,5 5 000 + 1,2 10 000 + 0,9 20 000 𝑉 ✓
𝑉𝑂𝑈𝑇 = − 10,6 𝑉 ✓ (3)
5.8.4 𝑅 𝑅 𝑅
𝑉𝑂𝑈𝑇 = − (𝑉1 𝑅𝐹 + 𝑉2 𝑅𝐹 + 𝑉3 𝑅𝐹) 𝑉 ✓
1 2 3
−𝑉𝑂𝑈𝑇
𝑅𝐹 = 𝑉1 𝑉 𝑉
+ 2+ 3
𝑅1 𝑅2 𝑅3
−10,4
𝑅𝐹 = 0,5 1,2 0,9 ✓
+ +
20 000 20 000 20 000
𝑅𝐹 = 80 𝑘Ω ✓
OR
𝑅𝐹
𝐴𝑉 = −
𝑅𝐼𝑁
𝑅𝐹 = 𝐴𝑉 × 𝑅𝐼𝑁 Ω
𝑅𝐹 = 4 × 20 000 Ω
𝑅𝐹 = 80 000 Ω
𝑅𝐹 = 80 𝑘Ω (3)
5.8.5 V
A V = − OUT
VIN
VOUT
= − ✓
1
V + V 2 + V3
5,2
= − ✓
0,5 + 1,2 + 0,9
=− 2 ✓ (3)
[50]
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10 ELECTRICAL TECHNOLOGY: ELECTRONICS (EC/SEPTEMBER 2022)
QUESTION 6: AMPLIFIERS
6.1 Class A – the transistor is biased with the Q-point on the midpoint ✓ of the
load line, allowing for the full signal (360°) to be amplified. ✓ (2)
6.2 6.2.1 C2 serves as the AC coupling component between the two stages. ✓
C2 also blocks or decouples the DC component of the signal. ✓ (2)
6.2.2 • When an AC voltage is applied to the input of the first
amplifier stage, ✓an alternating current will flow in the
collector circuit of transistor (Q1). ✓
• An alternating voltage will develop across the collector resistor
(RC1). ✓
• The developed alternating voltage across the RC1 will be
transferred through capacitor C2 ✓ to the base of the transistor
(Q2) in the amplifier's second stage (stage 2). ✓
• The process will be repeated and the amplified output will
measured between C3 and 0 V. ✓ (6)
6.2.3 • Impedance matching ✓
• DC isolation ✓
• Correct frequency response (Any 2 x 1) (2)
6.3 6.3.1 A – Low frequencies ✓
B – Middle frequencies ✓
C – High frequencies ✓
D – Voltage gain✓ (4)
6.3.2 Frequency response is the ability of the circuit ✓ to respond to a
range of frequencies applied to the transistor. ✓ (2)
6.3.3 At lower frequencies the reactances of the decoupling capacitors
across the emitter resistors rises. ✓
These reactances each combine with the resistance of their emitter
resistors, causing the total impedance to rise ✓ limiting the stage
gain. ✓ (3)
6.4 6.4.1 Impedance matching can be achieved by selecting a transformer ✓
with the required number of turns ✓ that will coincide with the
impedances of the respective stages. (2)
6.4.2 The reason for using a transformer is that the relatively high output
impedance of the second stage ✓ is connected to the relatively low
impedance of the speaker ✓ thus matching the output impedance ✓
of the amplifier to the load. (3)
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(EC/SEPTEMBER 2022) ELECTRICAL TECHNOLOGY: ELECTRONICS 11
6.4.3 Poor frequency response – gain is constant over only a small
range of frequencies ✓
Coupling transformers are bulky and expensive at audio
frequencies✓
Low frequencies receive less amplification than high frequencies
Tends to introduce hum in the output
Increased cost (2)
6.5 6.5.1 An oscillator is a device which generates an AC output signal ✓
without any externally applied input signal. ✓
(2)
6.5.2 The RF coil resistance against the change in the collector current
✓and causes the collector voltage Vc to decrease. ✓ (2)
6.5.3 The resistors R1 and R2 form a voltage divider ✓ to bias the base of
the transistor. ✓ (2)
6.5.4 The Colpitts oscillator uses TWO capacitors and an inductor in the
tank circuit. ✓
The Hartley oscillator uses TWO inductors and a capacitor in the
tank circuit. ✓ (2)
6.5.5 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)
6.5.6 1
Fo = 2𝜋√𝐿×𝐶 ✓
LT = L1+L2 ✓ (2)
[45]
TOTAL: 200
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