ELECTRICAL TECHNOLOGY
GRADE 12
MEMORANDUM
NATIONAL SENIOR CERTIFICATE
NOVEMBER 2016
INSTRUCTIONS TO THE MARKERS
QUESTION 1: OCCUPATIONAL HEALTH AND SAFETY
1.1
Note: An unsafe action is where an incorrect action may inadvertently lead to an accident or injury. This case relates to a human related intervention. (1)
1.2
Note: Dangerous practice relates to tasks that are inherently dangerous, but if actioned correctly would not lead to an accident or injury. This case relates to the working environment. (1)
1.3
1.4
1.5 Proper ventilation is essential to keep people working efficiently✓ the use of chemicals can make people drowsy and create an unsafe condition.✓ (2)
1.6
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QUESTION 2: THREE-PHASE AC GENERATION
2.1 VPH = VL✓
=380V✓
NOTE:
If the candidate indicates V 380VPH=, award 2 marks. (2)
2.2
NOTE:
1 mark = Drawing phasor diagram
1 mark = Labelling the phases
1 mark = Labelling the angle between the phases
Markers may consider rotation if a learner leaves out one aspect mentioned above.
Alternative Method: Closed Form Phasor Diagram
(3)
2.3
2.3.1 IL = S ✓
√3 x VL
=20 x 103
√3 x 380✓
30,39A✓ (3)
2.3.2
P = √3 VLIL𝐶𝑜𝑠𝜃 ✓
= √3 x 380 x 30,39 x 0.87 ✓
= 17,4kW ✓
Alternative Method:
𝐶𝑜𝑠𝜃 = P
S
P = S x 𝐶𝑜𝑠𝜃
P = 20000 x 0,87
P = 17,4 kW (3)
2.4 The function of a kWh meter is to measure the amount of energy power consumed ✓ by a consumer over a period of time. ✓
Alternative – To measure energy. (2)
2.5 Adding power factor correcting capacitors in parallel with the load.✓
Use synchronous motors. ✓ (2)
2.6
2.6.1 Pin = P1 + P2✓
= 120 + 50✓
= 170 W✓ (3)
2.6.2
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QUESTION 3: THREE-PHASE TRANSFORMERS
3.1
3.2
3.3
3.4 A transformer cannot step up power because all transformers have internal losses✓ that will reduce the output power. ✓
In an ideal transformer input and output power ✓ is the same. ✓ This is due to the current (and Power) of an installation being determined by the load not the transformer. (2)
3.5 If the load of the transformer is increased the transformer will draw more power from the supply✓ to deliver the increased load. ✓As the voltage is constant the primary current will increase. ✓ (3)
3.6
3.6.1 IPL = S ✓
√3VPL
= 20000
√3 x 6600 ✓
= 1,75A ✓ (3)
3.6.2 𝑉𝑃𝐻 = 𝑉𝐿 ✓
√3
𝑉𝑃𝐻=380 ✓
√3
=219,39𝑉✓ (3)
3.6.3 NP = VPH(P)✓
NS VPH(S)
= 6600
219,39 ✓
TR = 30:1 ✓ (3)
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QUESTION 4: THREE-PHASE MOTORS AND STARTERS
4.1
4.1.1
4.1.2
4.1.3 The motor will still rotate ✓but would not develop ✓the correct torque (output power). ✓ (3)
NOTE:
This question may have different interpretations.
The motor may not develop enough torque to turn the motor at all, depending on its efficiency and torque rating.
4.2 They require less maintenance as they do not have as many parts as a single phase motor✓
For the same size frame as a single phase motor they deliver a higher torque.✓ (2)
4.3 To determine that the integrity of the insulation is sound✓ so that the motor may be energised✓without an electrical fault occurring. ✓ (3)
4.4
4.5
4.5.1 nS = 60 xf ✓
p
= 60 x 50 ✓
3
= 1000r/min ✓ (3)
4.5.2 nR = nS (1-S) ✓
nR = 1000(1-0,04)✓
nR = 960r/min✓(3)
NOTE:The answers are rounded off as the answers are given in r/min. Learners must be given credit whether they have or haven't rounded off.
4.6
4.6.1 √3 x VL x IL✓
= √3 x 380 x 8,5 ✓
=5,59 kVA✓(3)
4.6.2 P = √3 x VL x IL x 𝐶𝑜𝑠𝜃 x η ✓
√3 x 380 x 8,5 x 0,8 x0,95 ✓
= 4,25 kW✓ (3)
4.7 The overload unit relay offers protection to the motor and operator ✓under fault conditions.✓ When activated it will remove power from the main contactor disabling the circuit. ✓ (3)
4.8
4.8.1 Conveyor belt. ✓Start the first drive motor and after a pre-determined time start the second drive motor. (1)
NOTE: Any viable explanation must be accepted.
4.8.2
4.8.3
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QUESTION 5: RLC
5.1 Reactance is the opposition offered to the flow of current in an AC circuit✓ by either a capacitor or an inductor or both✓. Impedance is the opposition offered to the flow of current in an AC circuit✓ by the combination of the resistance and the reactance of the circuit✓. (4)
5.2 The phase angle indicates the angle between the supply voltage✓ and the current ✓this indicates the power factor of the circuit. (2)
5.3
5.3.1
5.3.2
(3)
5.4
5.4.1 (3)
5.4.2 θ = cos-1 R ✓
Z
=cos-1 30 ✓
36,05
=33,68º ✓ (3)
5.5 VC = ICXC
f = Ic ✓
Vc x 2 x π x C
= 10 x 10-3 ✓
20 x 2 x π 1,47 x 10-6
= 54,13Hz ✓(3)
NOTE: This is a complex calculation and is rated as a higher order, difficult question.
[20]
QUESTION 6: LOGIC
6.1
6.2
6.3
6.4 A low current device can be activated by connecting it via the PLC output relay✓ to a source✓
Transistor output PLC's can be connected directly as it delivers an output voltage via a transistor. (2)
6.5
NOTE:
Please take note of the mark allocation indicated for grouping.
Be aware of the alternative method of populating the Karnaugh map when labelling is different.
(11)
6.6 (5)
6.7
6.7.1 (1)
6.7.2 X = A + C (1)
6.7.3 Y = A.B (1)
6.7.4 (2)
6.8
6.8.1 Normally open contact / switch ✓ Input
Input (1)
6.8.2 Normally closed contact✓ Input
Inverted input (1)
6.8.3 (Coil) output✓
Output (1)
6.9
6.9.1 Alternative circuit using NON Inverted inputs are acceptable.
(5)
Note: One mark may be awarded for a correct drawing with NO labels.
6.9.2 Motor starter✓ (1)
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QUESTION 7: AMPLIFIERS
7.1
7.2 Unconditional stability means that the operation of the amplifier is not✓ influenced by temperature changes. ✓ (2)
7.3 Positive feedback occurs in an amplifier circuit when a portion of the output✓ signal is fed back into the input.✓ The portion of the wave fed back will be in phase with the input therefore added to the input waveform. (3)
7.4 Oscillator circuit. ✓ (1)
7.5
7.6 (3)
7.7
7.7.1 Inverting op-amp ✓ (1)
7.7.2 Must show amplification inversion and same frequency.
NOTE: It has been found that some PDF’s and printouts do not show all the graphics.
Be lenient when considering drawings on separate axis. All three conditions however should be met. (3)
7.7.3𝐴 𝑉=−𝑅𝐹 ✓
𝑅𝐼𝑁
𝐴𝑉=−12000✓
2200
𝐴𝑉=−5,45✓ (3)
7.7.4 𝐴𝑉=𝑉𝑂𝑢𝑡 ✓
𝑉𝐼𝑁
𝑉𝑂𝑢𝑡=𝐴𝑉×𝑉𝐼𝑁 ✓
𝑉𝑂𝑢𝑡=−5,45×5 ✓
𝑉𝑂𝑢𝑡=−27,25𝑉 (3)
7.7.5 If RF was decreased the voltage gain of the op-amp will decrease✓
as it is directly proportional the value of the feedback resistor. ✓ (2)
7.8
7.8.1 This circuit allows for various input signal voltages to be fed into the circuit ✓thus producing a single output signal ✓that is the sum of the input signals. ✓(3)
7.8.2
𝑉𝑂𝑢𝑡=−(𝑉1+𝑉2+𝑉3)
𝑉𝑂𝑢𝑡=−(2−10+5)
𝑉𝑂𝑢𝑡=−(−3)
𝑉𝑂𝑢𝑡=3𝑉 (3)
7.9
7.9.1 Timing circuit ✓ (1)
7.9.2
(7)
Note: Labelling of T1 and T2 is not critical. If only T1 or is labelled, award 2 marks.
7.9.3 t = 5RC ✓
= 5 x 12000 x 47 x 10-6✓
= 2,82 s ✓ (3)
7.10
(3)
7.11 The differentiator is one type of op-amp where the magnitude of the output is determined by the rate ✓ at which the voltage applied to its input changes.✓
The differentiator is one type of op-amp where it changes a triangular wave into a square wave. (2)
7.12 Op-amps are packaged as an integrated circuit in a hard plastic body✓ with external pins for connections into circuits. ✓
Op-amps may also be packed in an SMD package. (2)
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TOTAL: 200