# Electrical Technology Grade 12 NSC June 2018 Memo (English)

> Grade 12 Electrical Technology memorandum from the NSC June Exam 2018 (English), 14 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/electrical-technology-may-june-2018-memo-eng/af9f7f90-5412-4c48-96fa-c3de2254a2f4
- Updated: 2026-09-27
- Subject: Electrical Technology
- Grade: Grade 12
- Document type: Memorandum
- Year: 2018
- Exam period: NSC June Exam
- Curriculum: CAPS
- Pages: 14

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## About this document

Electrical Technology · Grade 12 · NSC June Exam · 2018. Memorandum, 14 pages. Read online or download the PDF.

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## Document text

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                     SENIOR CERTIFICATE EXAMINATIONS

                           ELECTRICAL TECHNOLOGY

                                     JUNE 2018

                              MARKING GUIDELINES

                                                                      MARKS: 200

                      These marking guidelines consist of 14 pages.

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Electrical Technology                             2                                  DBE/2018
                                       SCE – Marking Guidelines

INSTRUCTIONS TO MARKERS

1.        All questions with multiple answers imply that any relevant, acceptable
          answer should be considered.

2.        Calculations:

          2.1           All calculations must show the formula/e

          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 same
                        answer is obtained

          2.5           Where an erroneous 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 learner should receive the full
                        marks for subsequent calculations

3.        The memorandum is only a guide with model answers. 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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Electrical Technology                                   3                          DBE/2018
                                             SCE – Marking Guidelines

QUESTION 1: OCCUPATIONAL HEALTH AND SAFETY

1.1       The type of work done in an electrical technology workshop requires the
          correct intensity of light because good visibility  is crucial in performing
          work, e.g. soldering work, panel wiring, measuring electrical quantities and
          for safety requirements.                                                               (2)

1.2       Dangerous practices are processes or activities that have some form of
          inherent risk or hazard when performed and may lead to serious injuries or
          even death.                                                                             (2)

1.3       Be calm. 
          Assess the situation
          Call the emergency services
          Report the injuries to appropriate authority
          Administer first aid and CPR                                                            (2)

1.4       Workers infected with HIV tend to be absent from work regularly due to
          illness because of their weakened immune system and this impacts
          negatively on productivity.                                                            (2)

1.5       It is ethically correct to report a person who is working under the influence of
          medication that may lead to drowsiness because the person may place
          himself/herself and other persons in danger while operating the machinery. 
          The person may also cause damage to the machinery. therefore no person
          may enter or remain in a workplace under the influence of medications.                  (2)
                                                                                                  [10]

QUESTION 2: THREE-PHASE AC GENERATION

2.1       Single phase generate less power 
          It can supply power to single phase load only 
          Single phase is more expensive to generate power                                        (2)

2.2       Three coils are placed at 1200 relative to each other. When these coils are
          rotated anti-clockwise they cut the magnetic field lines which induce the emfs
          of the same frequency and magnitude in these coils. The induced emfs are
          at 1200 out of phase with each other.                                                  (3)

2.3       2.3.1                      VPH
                        I        =            
                            PH       Z PH
                                   220        
                                 =
                                   16
                                 = 13,75 A                                                       (3)

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Electrical Technology                                 4                                DBE/2018
                                           SCE – Marking Guidelines

          2.3.2                              
                        P = I2Z
                                             
                                2
                          =13,75 × 16
                                             
                          = 3025 W
                          = 3,025 kW
                                                                                                      (3)

2.4       2.4.1         By connecting an appropriate capacitor in parallel with the load 
                        By connecting synchronous motors in parallel with the load                    (1)

          2.4.2         Less current will be drawn from the supply 
                        Cost saving from drawing less current 
                        Less heat will be generated as a results there will be fewer losses           (2)

2.5       2.5.1         Input power to the load
                        P = P1 + P2 + P3              
                          = 8000 + 4000 + (−460)      
                          = 11,54 kW                                                                 (3)

          2.5.2         Power dissipated
                        P = 3VL I L Cos θ         
                                11540             
                        IL =
                             3 × 380 × 0,8
                          = 21,92 A                                                                  (3)
                                                                                                      [20]

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Electrical Technology                                 5                                        DBE/2018
                                           SCE – Marking Guidelines

QUESTION 3:THREE-PHASE TRANSFORMERS

3.1       3.1.1      Core type transformer 
                     Shell type transformer                                                                 (2)

          3.1.2      The iron core of transformers is laminated to reduce circulating eddy
                     currents in the core which generate losses in the form of heat.                        (1)

3.2       When an alternating voltage is connected to the primary windings, a
          magnetising current will flow, and an alternating magnetic field is set up in
          the core.  This alternating magnetic field links with the secondary windings,
          and an emf is induced in the secondary windings                                                   (3)

3.3       Same voltage and turns Ratio (both primary and secondary voltage rating is
          same) ✓
          Same KVA ratings✓
          Same frequency rating
          Same polarity
          Similar impedance                                                                                  (2)

3.4       It improves the insulation qualities of the windings 
          It cool-down the temperature rise due to heat generated in the transformer
          during its circulation process.                                                                    (1)

3.5       3.5.1      VL = 3 × VPH
                                                  
                          = 3 × 220               
                          = 381,05 V                                                                        (3)

          3.5.2         S = 3 × VL × I L
                                                            I PH = I L           
                               S                                                  
                     IL =                                    ∴ = 454,55 A
                          3 × VL
                                  300000
                            =                    
                                 3 × 381,05
                                = 454,55 A       
                                                                                                             (5)

          3.5.3      P = S Cos θ                              P = 3VL I L Cos θ
                                                
                       = 300000 × 0,8                  or       = 3 × 381,05 × 454,55 × 0,8
                       = 240 kW                                 = 240 kW                                    (3)
                                                                                                             [20]

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Electrical Technology                                  6                                DBE/2018
                                            SCE – Marking Guidelines

QUESTION 4: THREE-PHASE MOTORS AND STARTERS

4.1       4.1.1      It houses the bearings that hold the rotor axis                                  (1)

          4.1.2      The purpose of the cooling fins is to dissipate generated heat to the
                     outside of the motor                                                             (1)

          4.1.3      It houses the end wires of the three coils and earth                             (1)

                     It is used to connect the supply to the motor

4.2       High starting torque 
          Can be connected in star or delta 
          Wide range of application 
          Higher efficiency
          They are self-starting                                                                       (3)

4.3       4.3.1         Synchronous speed of the motor.

                             60 × f                
                        ns =
                               p
                             60 × 50               
                           =
                                2
                           = 1500 RPM                                                                 (3)

          4.3.2         Rotor speed of the motor.

                        n r = n s (1 - S)          
                           = 1500(1 - 0,04)        
                           = 1440 RPM              
                                                                                                       (3)

4.4       4.4.1           P = 3 × V × I × Cos θ × η          
                                     L L
                            = 3×380×15×0,8×0,9               
                                                             
                            = 7,108 kW                                                                 (3)

          4.4.2         The supply current will be less✓ for the same output power. ✓                  (2)

4.5       4.5.1         The start button normally is open and spring loaded, when it is
                        pressed and released, the main contact (MC 1 ) will energise, 
                        closing MC 1 (N/O) and current will flow through it which will enable
                        the circuit to operate until the normally closed stop button is
                        pressed.                                                                       (4)

          4.5.2         The two motors are running, because the MC 1 (N/O), MC 2 (N/O) 
                        and T (N/O) are closed.                                                       (3)

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Electrical Technology                              7                                     DBE/2018
                                        SCE – Marking Guidelines

          4.5.3         The overload relays are connected to each motor so that each motor
                        can be disconnected from the supply the moment the current in
                        that motor exceed the value that the overload is specifically
                        designed for without affecting the running of the other motor.                (3)

          4.5.4         When the start button is pressed, the MC 1 will be energised, 
                        closing MC 1 (N/O 1 ) and MC 1 (N/O 2 ) and Motor 1 will run.  Timer
                        contact (T) will also be energised and after pre-determined time T
                        (N/O) will close and Motor 2 will run. Both motors can be
                        stopped by pressing the stop button.                                            (4)

          4.5.5         Conveyer belt                                                                   (1)

4.6       Check the:
          • shaft bearings for smooth movement 
          • motor housing for cracks 
          • end plates for proper even fastening
          • cooling fan blades
          • protection cover over the fan                                                               (3)

4.7       The motor cannot be used because there is continuity between winding U
          and W due to damaged insulation of the two windings  which will result in
          a short circuit and further burning of wires.                                                 (3)

4.8       The function of the zero volt coil is to switch off the supply to the circuit in
          the event of power failure.  and when the power is restored, the system will
          remain off until it is re-started.                                                            (2)
                                                                                                        [40]

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Electrical Technology                                  8                              DBE/2018
                                            SCE – Marking Guidelines

QUESTION 5: RLC

5.1      Frequency 
         Inductance 
         Capacitance                                                                            (3)

5.2      5.2.1          The power factor is lagging                                             (1)

         5.2.2                                   R                  
                                     Cos θ =
                                                 Z
                                                                    
                                      but Z = R 2 + (XL − XC )2
                                            = 20 2 + (60 − 35)2
                                        = 32 ,02 Ω                  
                                           20                       
                        therefore Cos θ =
                                          32,02
                                        = 0,62                                                  (5)

         5.2.3          The Q factor will be halved because the Q factor is inversely
                        proportional to the resistance of the circuit. The doubling of the
                        inductance and capacitance as a ratio will have no effect on the Q-
                        factor.                                                                 (3)

5.3      At resonance, the impedance of the circuit is at a minimum and hence the
         current is at a maximum. When there is a change in frequency from
         resonance, the impedance will increase  causing the current to decrease.              (2)

5.4       5.4.1                  1
                        fr =                                           
                               2π LC
                                                 1
                          =
                               2π (0,5 × 10 ) × (0,0147 × 10 -6 )
                                            −3
                                                                       
                          = 58,705 Hz                                  
                                                                                                 (3)

          5.4.2             E                  
                        Ir =
                            R
                            240                
                          =
                             20
                          = 12 A                                                                (3)
                                                                                                 [20]

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Electrical Technology                            9                                     DBE/2018
                                      SCE – Marking Guidelines

QUESTION 6: LOGIC

6.1       6.1.1                
                                     or
                          X1                   X1                                                     (1)

          6.1.2

                                                                                                      (1)

6.2       Relay logic                                                                                (1)

6.3       Laptop
          Handheld device
          Desktop computer
          PLC interface                                                                               (1)

6.4       - High power SCRs or Triacs
          - Solid state relays                                                                        (1)

6.5       A series of instructions written in ladder logic/function block or instruction list
          that is used to control the operation of a PLC.                                            (2)

6.6       Function block diagram
          Instruction list                                                                           (2)

6.7       6.7.1      A timer is a function in PLC programs that waits for a predetermined
                     time before executing a certain programmed function.                           (2)

          6.7.2      Markers are registers in the memory used to store information until
                     needed by the program.          Markers are also used to retain
                     information from the end one rung to the beginning of the next rung of           (2)
                     a program. 

6.8       It would reduce the complexity of the ladder program and the reaction and
          execution time would be quicker with a smaller program.                                    (2)

6.9                                 
                                                 
               A                                                 
               B                      &         ≥1                                
                                                                 ≥1                     X
                                                                             ≥1
               C                                
                                                                         
               D                                    &
                                                                                                      (6)

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Electrical Technology                              10                          DBE/2018
                                         SCE – Marking Guidelines

6.10       X= DEF+DEF+DEF+DEF
                                 
           = D F( E + E) + D F( E + E)
           = DF + DF
                  
           = F( D + D)
                                                                                              (6)
           =F

6.11                         

                         

             
                                        
                                 

                                                                                            (8)

6.12

                                                                  
                 O/L             Stop          Start                MC 1

                 X1              X2             X3                  Q0

                                               MC 1
                                               N/O (Q0)

                                                                                              (5)

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Electrical Technology                              11                             DBE/2018
                                         SCE – Marking Guidelines

QUESTION 7: AMPLIFIERS
7.1       7.1.1      Negative feedback is when a portion of the output is fed back to the
                     input and is 180º out of phase with the input                             (2)

          7.1.2      The bandwidth refers to the range of frequencies that the op amp
                     can operate without distortion (linearly)                                  (2)

7.2       7.2.1
                                                          V REF = 2 V
                             0V

                            +V CC                   

                        VOUT 0 V

                            -VCC              
                                                                                                 (4)

          7.2.2      Threshold detector
                     Voltage level detector                                                      (1)
7.3       Oscillator circuits                                                                   (1)

7.4

                                                                                                 (2)

7.5       7.5.1      Summing amplifier circuit                                                  (1)

          7.5.2      Audio mixer
                     Temperature measurement                                                     (1)

          7.5.3                    V V V 
                     Vout = − R f  1 + 2 + 3  
                                    R1 R2 R3 
                                0,5 0,2 0,2 
                          = − 2       +   +    
                                2       2   2  
                          = − 0,9 V                                                             (3)
                     (All values in kilo ohm)

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Electrical Technology                               12                               DBE/2018
                                          SCE – Marking Guidelines

7.6         7.6.1       By replacing the feedback capacitor with a feedback resistor              (2)

            7.6.2

                                                                     
                                                

                                                            

                                                                                                    (4)

7.7       7.7.1         Audio oscillators
                        Electronic organs                                                           (1)
          7.7.2                1
                        f =π                              
                            2 6RC
                                        1
                          =                               
                           π2 6 × 8000 × 275 × 10 −12
                         = 43,805 Hz                      
                                                                                                    (3)

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Electrical Technology                               13                                         DBE/2018
                                          SCE – Marking Guidelines

          7.7.3

                                               •                                 output

                                                               

                                                   •60°                          Input a

                                                                   

                                                     •                           Input b
                                                     120°

                                                               

                                                          •180°                  Input c

                                                                                                              (3)

7.8
                                                               Upper trigger
                                                                  level

                                                                          Input signal

                              .   .

                                       
                                                              Lower trigger
                                                                  level

                                  
                                                                                                              (4)

7.9                       1
               fr =                                        
                  2π LC
          but L = L1 + L2
                  = 30 + 30
                  = 60 mH
                                  1
               fr =                                        
                      2π 0,06 × 0,35 × 10 −6
                  = 1098,27 Hz                                                                               (3)

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Electrical Technology                                    14                         DBE/2018
                                               SCE – Marking Guidelines

7.10                               1
                        fr =                              
                            2π LC
                             CC                           
                     but C = 1 2
                            C1 + C2
                            30 × 10 −9 × 60 × 10 −9
                          =
                            30 × 10 −9 + 60 × 10 −9       
                          = 2 × 10 −8
                                                          
                                           1
            therefore f r =
                               2π 2,6 × 2 × 10 −8         
                          = 697 Hz                                                             (5)
7.11
                 +V

                                                        Input pulses

                 -V

                                           
            +V                

                                                        Output

           -V
                                                                                              (5)

7.12      7.12.1        used as a buffer                                                      (1)

          7.12.2        Used for isolation  because of its high input impedance and low
                        output impedance                                                      (2)
                                                                                               [50]

                                                                                   TOTAL:      200

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