ELECTRICAL TECHNOLOGY: ELECTRONICS
GRADE 12
NOVEMBER 2020
MEMORANDUM
NATIONAL SENIOR CERTIFICATE
INSTRUCTIONS TO THE MARKERS
QUESTION 1: OCCUPATIONAL HEALTH AND SAFETY
1.1 Any article or part thereof which is manufactured, provided or installed ✓in the interest of the health or safety of any person.✓✓(2)
1.2 Your right to fair labour practices.✓✓
Your right to work reasonable hours.
Your right to belong to a trade union.
Your right to earn a living wage.
Your right not to be discriminated against. (1)
1.3
1.4
1.5 In an emergency it can be pushed and it would immediately cut all electric power ✓✓to all the equipment, stopping them,✓✓thus making the workshop safe.✓ (2)
[10]
QUESTION 2: RLC CIRCUITS
2.1
2.1.1 If VX lags VR by 35°
(2)
2.1.2 The voltages represent an RC circuit ✓ because, VR is always in phase with IT and VX lags VR by 35°. ✓
Because VR is in phase with IT and VX is lagging VR, thus proving that the circuit is predominantly capacitive as IT leads VX. (2)
2.2
2.2.1
XL = 2 x π x f x L
= 2 x π x 60 x 20 x 10-3
= 7,54 Ω (3)
OR
XL = VL
IT
XL = 49
6,5
XL = 7,54 Ω
2.2.2
VC = I x Xc
= 6.5 x 25
= 162,5 V (3)
2.2.3 The voltage is lagging, ✓ because the capacitive reactance is greater than the inductive reactance (VC is greater than VL). ✓(2)
2.2.4
(5)
NOTE: 5 marks, 1 mark for each correct label of which VL, VC and VR are priority marks and thereafter any other two correct labels.
2.3 2.3.1 Phasor diagram of a parallel RLC✓ circuit at resonance. ✓ (2)
2.3.2 The voltage drop across the components in a parallel circuit is the same, ✓ hence the voltage is used as the reference. (1)
OR
The applied voltage is common across all components.
2.3.3 In a parallel resonant circuit impedance is at maximum, ✓ and the total current is at minimum. ✓ (2)
OR
The relationship between impedance and current in a parallel RLC circuit is inversely proportional.
2.4
2.4.1
fr = 1
2π√LC
= 1
2 x 3,142√300 x 10-3 x 150 x 10-6
= 23,73 Hz (3)
2.4.2
Q = 1/R x √L/C
= 1 x √300 x 10-3
20 150 x 10-6
= 2,24
(3)
OR
If candidates calculate XL or Xc they can use the following formulae:
XL = 2πfL
XL = 2π(23.73)(300 x 10-3)
XL = 44.73Ω
Q = XL/R
Q = 44.73
20
Q = 2,24
Q = XC/R
Q = 44.73
20
Q = 2,24
2.4.3
Z = 20Ω
Z = R at resonance (2)
2.4.4
C = 1
4 x π2 x L x fr2
= 1
4 x 9,87 x 300 x 10-3 x 4 x 106
= 2,111 x 10-8 F
= 21,11 nF (3)
OR
C = 1
(2πfr)2 x L
= 1
(2 x 3,14 x 2000)2 x 300 x 10-3
= 2,113 x 10-8 F
= 21,13 nF
For resonance. Calculating XL first and then since XL= XC
XL = 2πfL
XL = 2π(2000)(300 x 10-3)
= 3769.91Ω
C = 1
2πfXC
= 1
2π(2000)(3769.91)
= 21.11nF
2.5.
2.5.1 Q1 (1)
2.5.2 Reading the values from the graph where f1 = 30000 Hz and f2 = 35000 Hz the following can be deduced:
fr = f1 + f2
2
= 30000 + 35000
2
= 32500Hz
= 32,5 kHz (3)
NOTE: If the candidate deduced the value 32,5 kHz directly from the graph, full marks will be awarded.
Because the incorrect formula was provided on the formula sheet, the following calculation will be accepted.
fr = f1 + f2
2
= 35000 - 30000
2
= 2500Hz
2.5.3 Deducing the indicated bandwidth of Q1 from the graph as 35 kHz - 30 kHz the candidate will be able to calculate as follows:
BW = f - f
= 35000 - 30000
= 5000Hz
BW = fr
Q
Q = fr
BW
= 32500
5000
= 6,5
OR
BW = fr and BW=(f2 - f1)
Q
(f2 - f1) = fr
Q
Q = fr
(f2 - f1)
= 32500
(35000- 30000)
= 6,5 (3)
[40]
QUESTION 3: THREE-PHASE AC GENERATION
3.1 3.1.1 Reactive power is the power in an AC circuit when the current is out of phase with the voltage ✓ as a result of the capacitive and/or the inductive load. ✓
Reactive power is the power that is wasted and not used to do work on the load. (2)
NOTE: If only the formula is explained or written only 1 mark will be awarded.
3.1.2 Efficiency is the percentage ratio ✓ of the output power to the input power. ✓ (2)
NOTE: If only the formula is explained or written only 1 mark will be awarded.
3.2
VPH = VL
√3
=380
√3
= 219, 39 V (3)
3.3
3.3.1
3.3.2
3.4 Kilo-watt-hour meter/Energy meter. ✓ (1)
3.5
3.5.1
IL = √3 x IPH
= √3 x 30
= 51.96A
= 52A (3)
3.5.2
VL = VPH
= 380 V
ZPH = VPH
IPH
=380
30
= 12,67 Ω (5)
When candidates substitute Vph correctly as 380 V without calculating it, two additional marks should be awarded to the 3 mark calculation.
3.5.3
P= √3 x VL x IL x Cos θ
= √3 x 380 x 52 x 0,9
= 30,8 kW (3)
3.6
PT = P1 + P2
= 10000 + 3000
= 13000W
= 13kW (3)
3.7 A wattmeter is used to measure the input power/output power ✓ in an AC system.
Any acceptable application that uses a wattmeter to indicate power usage. (1)
[30]
QUESTION 4: THREE-PHASE TRANSFORMERS
4.1
4.2
4.3
4.4
4.5 If the load is increased, the secondary current will increase✓ resulting in an increase✓ in the primary current. This increase is due to the increased magnetic force that results due to higher current flow. (2)
4.6 Delta – star ✓
Star – delta ✓
Star – star
Delta – delta
Step-up
Step-down
Isolation transformer (2)
4.7
4.8 (3)
CORE-TYPE THREE-PHASE TRANSFORMER CONSTRUCTION
4.9
4.10
4.10.1
POUT = 200000 x 0.85
= 170000W
η = POUT x 100
PIN
= 170000 x 100
171800
= 98.95% (5)
OR
If the learner calculated the input power by using the given line current, the following answer will be accepted.
PIN = √3VLIL x cosθ
= √3(11000)(30) x 0.85
= 485840.25W
η = POUT x 100
PIN
= 170000 x 100
485840.25
=34.99%
4.10.2
TR = Vph(1)
Vph(2)
= 11000
219,39
= 50 :1 (3)
4.10.3
IL(2) = S
√3 x VL(2)
=200000
√3 x 380
= 303,86 A
OR
IL(2) = P
√3Vcosθ
= 170000
√3 380 0.85
= 303.86A (3)
[30]
QUESTION 5: THREE-PHASE MOTORS AND STARTERS
5.1
5.2
5.3.
5.3.1 Sequence motor starter ✓ without a timer
NOTE: This cannot be a forward / reverse motor due to no lock-out. (1)
5.3.2
5.3.3
5.4
5.4.1
η =60 x f
p
= 60 x 50
6
= 500rpm (3)
5.4.2
nr = ns(1 - S)
= 500 x (1 - 0,05)
= 475 rpm (3)
NOTE: If the candidate calculated 5% of the synchronous speed
0,05 x 500 = 25 rpm the following can apply:
S = ns – nr
Nr = Ns – slip(revolutions)
Nr = 500 – 25
Nr = 475 r.p.m
5.5
5.5.1
IL = P
3 x VL x Cos θ
= 18000
√3 x 380 x 0,8
= 34,18 A (3)
5.5.2
Papp = √3 x VL x IL
= √3 x 380 x 34,18
= 22,5 kVA
S = P
Cos θ
= 18000
0.8
= 22.5kVA (3)
[30]
QUESTION 6: PROGRAMMABLE LOGIC CONTROLLERS (PLCs)
6.1 6.1.1 Hardware refers to all the physical parts✓ (e.g. CPU) of the device that can be seen and touched. ✓ (2)
6.1.2 Software refers to machine/graphical language (e.g. windows, software programmes) that is installed on a computer or written into the PLC ✓ that instructs it to interact ✓ with its input and output hardware. ✓ (3)
6.1.3 An opto-coupler is a semiconductor device that uses light ✓ (e.g. Light Emitting Diode) to transmit a signal between two circuits ✓ that are electrically isolated. ✓ (3)
6.2 Input module✓
Output module ✓
Power supply✓
Central processing unit (Microprocessor) (3)
6.3
6.4
6.4.1
NOTE: One mark for the inputs
One mark for the logic function (2)
NOTE: See examination instruction
6.4.2
6.5 The inductive proximity sensor is applied in:
6.6
6.6.1 An ON-delay timer contact✓
Edge triggered wiping relay (Siemens) (1)
6.6.2
6.7 The circuit will not latch ✓ when the start button is released. (1)
6.8
6.8.1 The "Red Lamp" will be ON ✓
The "Green Lamp will be OFF ✓ (2)
6.8.2 The function of the MC (contactor coil) is to close ✓ and open✓the contacts the moment it is energised. (i.e. MC/NO1 and MC/NO2 will close and MC/NC1 will open)
OR
The purpose of MC (contactor coil) is to magnetise the core which is connected to the contacts. When the core moves the contacts are closed.
When the coil is no longer magnetised, a coil spring opens the contacts again. (2)
6.8.3
(9)
6.9
NOTE: If the following response is given, 2 marks will be awarded.
When a three-phase or single-phase supply is connected to a VSD, its circuitry will increase or decrease the frequency of the output AC power signal supplied to the motor. (5)
6.10 Synchronous motor with a permanent magnet stator. ✓
Synchronous motor with brushes or brushless.✓ (2)
6.11
6.11.1 Pulse A ✓ has shorter on time and Pulse B ✓ has longer on time. (2)
6.11.2 The greater the width of the pulses ✓ the higher ✓ the average voltage. (2)
6.12
6.13
6.13.1 Braking Unit ✓ (1)
6.13.2 Braking resistor ✓ (1)
6.14
[60]
TOTAL: 200