# Electrical Technology Grade 12 NSC Supplementary 2016 Memo

> Grade 12 Electrical Technology memorandum from the NSC Supplementary Exam 2016 (English), 15 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/electrical-technology-feb-march-2016-memo-eng/9c539077-0a97-4106-8266-0047ddd1abab
- Updated: 2026-09-27
- Subject: Electrical Technology
- Grade: Grade 12
- Document type: Memorandum
- Year: 2016
- Exam period: NSC Supplementary Exam
- Curriculum: CAPS
- Pages: 15

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Electrical Technology · Grade 12 · NSC Supplementary Exam · 2016. Memorandum, 15 pages. Read online or download the PDF.

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                                   GRAAD 12

                                  NATIONAL
                             SENIOR CERTIFICATE

                                   GRADE 12

                          ELECTRICAL TECHNOLOGY

                            FEBRUARY/MARCH 2016

                                 MEMORANDUM

MARKS: 200

                       This memorandum consists of 15 pages.

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Electrical Technology                             2                           DBE/Feb.–Mar. 2016
                                          NSC – Memorandum

INSTRUCTIONS TO THE 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 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.

3.        This 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/Feb.–Mar. 2016
                                     NSC – Memorandum

QUESTION 1: OCCUPATIONAL HEALTH AND SAFETY

1.1       Unsafe conditions include:
          − Inadequate guards around machinery. 
          − No eye protection goggles in the workshop and around the bench
             grinder.
          − Loose and dangerous components on moving machinery.                                 (1)

1.2       An unsafe action is an action performed by a person/worker that could
          cause:
          − Injury to him/herself or his/her colleagues
          − Cause damage to tools and equipment
          − Render a workplace unsafe
          − Lead to loss of income to the employer                                              (2)

1.3       Protocol to follow when discovering a person experiencing electrical shock is:
          − Do not touch the victim with bare hands until the supply is turned off. 
          − Switch off the supply. 
          − Call a teacher or medical person for help. 
          − If the electricity cannot be switched off and the victim is still in contact
              with it push the wire away with an insulated object.                             (4)

1.4       Productivity is an important work ethic to assist the growth of South Africa's
          economy due to the following reasons:
          − Safe working procedures are more productive as less time is lost due to
              injuries and downtime related to accidents. 
          − Efficient working methods lead to less use of raw materials, thus
              increasing profits. 
          − Good ergonomics lead to cleaner and safer work environments and
              promotes worker satisfaction and morale. 
          − Workers with a high morale and a good safety record tend to follow
              instructions more precisely and enact safe working procedures.
          − Effective monitoring and reporting systems lead to better quality
              assurance of products, thus placing a better product in the market within
              a shorter time frame, thus promoting productivity.                                (3)
                                                                                                [10]

QUESTION 2: THREE-PHASE AC GENERATION

2.1       A Kilo-watt hour meter ✓                                                              (1)

2.2       Two advantages of three-phase systems over single-phase systems are:

          −    3φ can be operated in star or delta. ✓
          −    Three-phase systems can supply both three-phase and single-phase
               installations of power✓
          −    In generators and motors, for the same size frames, three-phase
               machines delivers more power than single-phase machines                          (2)

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Electrical Technology                                     4                          DBE/Feb.–Mar. 2016
                                                  NSC – Memorandum

2.3
                                            Ph1 or L1 or       Ph2 or L2 or   Ph3 or L3 or
                                            ER✓                Ey✓            ER✓
                     +Vmax

                                                                                   Showing ✓
                                                                                   amplitude

                               0

                                                                                   Showing
                                                                                   displacement of
                                                                                   1200 ✓

                        -Vmax

                                                        1200         1200                                   (5)

          Note to markers:
          No additional marks are awarded for the manipulation of the formulae.
          Manipulation is shown to illustrate how the formula was derived.

2.4       2.4.1         S = 3 VL I L
                                        S
                        VL =                        ✓
                                   3 x IL
                                      20000         ✓
                               =
                                     3 x 25
                                   = 461.88 A       ✓                                                       (3)

          2.4.2         VL = 3VP
                             VL               ✓
                        VP =
                              3
                             462.43           ✓
                           =
                              1.73
                           = 267.67 V         ✓
                                                                                                            (3)

2.5       2.5.1         PT = P1 + P2    ✓
                           = (8 + 4) kW ✓
                           = 12 kW      ✓
                                                                                                            (3)

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Electrical Technology                                5                    DBE/Feb.–Mar. 2016
                                             NSC – Memorandum

          2.5.2         P = 3 VL I L Cosθ
                                   P
                        IL =                     ✓
                               3 x VL Cosθ
                                12000            ✓
                          =
                             3 x 380 x 0.8
                          = 22.79 A              ✓                                                (3)
                                                                                                  [20]

QUESTION 3: THREE-PHASE TRANSFORMERS

3.1       By laminating and insulating the laminations of the core of the transformer.✓           (1)

3.2       Two similarities between single phase and three-phase are:
          − Both have the same functional operations✓
          − Transformers are used to step down or step up the supply voltage✓
          − Both can have either the closed core type or shell core type
          − No electrical connections between primary and secondary windings, with
             the exception of the autotransformer.                                                (2)

3.3       Two factors that may cause excessive heating in a transformer are:

          −    Insufficient ventilation✓
          −    Constant overloading✓
          −    Loose connections
          −    Ineffective tap-changer contacts
          −    Impure/carbonized/insufficient transformer oil                                     (2)

3.4       Applications of delta-star connected transformers in power distribution
          systems are:
          − Single and three-phase supply to Commercial sites ✓
          − Three-phase supply to Industrial areas ✓
          − Single and three-phase supply to domestic installations
          − Single and three-phase supply to agricultural installations
          − Three-phase supply to other substations (transformers) in a ring-fed
              supply network.                                                                     (2)

3.5       The function of a transformer is to 'step down' ✓ or to 'step up' ✓
          (transform) the voltage ✓in a distribution network to the required value.               (3)

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Electrical Technology                                  6              DBE/Feb.–Mar. 2016
                                               NSC – Memorandum

3.6       3.6.1         V L = VPH              ✓
                        ∴ VPH = 6.6 kV         ✓
                                                                                             (2)

          3.6.2         IL = IP          ✓
                        ∴IP = 30 A       ✓
                                                                                             (2)

          3.6.3                   I     ×V
                        IPh( p) = Ph( s) Ph( s) ✓
                                      VPh( p)
                                    30 × 230       ✓
                                  =
                                      6600
                                  = 1.04 A         ✓
                                                                                             (3)

          3.6.4         NP IPH( s)
                          =                    ✓
                        NS IPH( p)
                        NP   30                ✓
                           =
                        NS 1.04
                        NP : NS = 30 : 1 ✓                                                   (3)
                                                                                             [20]

QUESTION 4: THREE-PHASE MOTORS AND STARTERS

4.1       Applications of three-phase motors include:
          − pumps
          − lifts
          − cranes                                                                           (1)

4.2       Advantages of a three-phase over single-phase motors:
          −   For the same size frame a three-phase motor delivers more power 
          −   Three-phase motors are self-starting 
          −   More robust
          −   Simple construction                                                            (3)

4.3       4.3.1 At start motors draw more than the motors rated current. This may
                lead to unnecessary tripping of the motor  and the increased
                current at start causes wear on switch gear.                                (3)

          4.3.2 The motor is started in the star mode which reduces the voltage
                across each phase of the motor as in star V PH = √3 V L .  As the
                voltage across each phase is reduced, the current in each phase will
                also be reduced.                                                            (3)

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Electrical Technology                           7                           DBE/Feb.–Mar. 2016
                                        NSC – Memorandum

4.4       The operation of three-phase squirrel cage induction motor:

          −    Three-phase supply is such that the peak voltage of each phase is
               reached at 120o intervals.
          −    When a squirrel cage motor is connected across a three phase
               alternating supply, alternating current will flow in the respective stator
               windings.
          −    As a result the alternating current will set up a magnetic field in and
               around the respective stator windings.
          −    The stator windings are however spaced 120o apart around the axis of
               the motor.
          −    The result is that the magnetic field has a rotating maximum flux density
               as the peak in each winding is reached at a 120o interval, resulting in a
               rotating magnetic field.
          −    A squirrel cage rotor consists of cage like conductors bound electrically
               with a short circuit ring at each end.
          −    The rotating magnetic field will sweep across the rotor conductors of the
               squirrel cage rotor.
          −    As the magnetic field sweeps over the rotor conductors, the magnetic flux
               will cut/cross the rotor conductors.
          −    Faraday's law determines that an EMF will be induced in the rotor
               windings due to the sweeping action of the rotating magnetic field.
          −    Lenz's Law will determine that the induced current will be such that its
               force will oppose the inducing action (from the stator), thus establishing a
               repelling force between the stator and rotor (due to the opposing
               magnetic fields), thus forcing the rotor into rotation.
          −    When the speed of the rotor approaches the speed of the rotating
               magnetic field (synchronic speed), the inducing action is reduced and the
               rotor tends to slow down. The result is that the rotor rotates at a speed
               lower than that of the rotating magnetic field (synchronic speed)                  (7)

4.5       4.5.1
                                          ✓ L1       L2        L3
                                                                      N✓

                                           U1        V1        W1
                                                 ✓
                                                         ✓ ✓

                                           W2        U2        V2               E
                                                                                                   (5)

                     Coil 1:U1 – U2 ✓
                     Coil 2: V1 – V2✓
                     Coil 3: W1 – W2✓
                     L1 , L2, L3 – 3 Phase specified✓
                     N: Star is specified✓
                     E – Not requested in the question

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Electrical Technology                                8                     DBE/Feb.–Mar. 2016
                                             NSC – Memorandum

          4.5.2 The reading on the meter should be of a high resistive value,  more
                than 1 MΩ, this will indicate that there is no electrical connection
                between the windings  showing sound/good insulation.                            (3)

          4.5.3 A low resistance reading will indicate that there is electrical connection
                between the winding and earth. This will indicate any electrical fault
                that may lead to an operator being shocked in the motor is
                energised.                                                                       (3)

4.6                         60 × f     ✓
          Pole pairs =
                             NS
                                       ✓
                         60 × 50
                        =
                           600
                                       ✓
                        =5                                                                        (3)

4.7       4.7.1         S = 3 VL I L             ✓
                            = 3 x 380 x 16       ✓
                            = 10.53 kVA          ✓
                                                                                                  (3)

          4.7.2         η=100%

                        P = S x Cos θ            ✓
                            = 10.53 x 0.85       ✓
                                                 ✓
                            = 8.95 kW                                                             (3)

          4.7.3                Po
                        η=
                               Pi
                        Po = Pi x η
                                             ✓
                            = 8.95 x 0.9     ✓
                            = 8.06 kW        ✓                                                    (3)
                                                                                                  [40]

QUESTION 5: RLC

5.1       Factors influencing the capacitive reactance are:
          − Value of capacitance✓
          − Frequency of the supply✓                                                              (2)

5.2       Reactance is the opposition of the specific reactive component✓ to the flow
          of current in AC circuits

          Impedance is the total opposition✓ offered to the flow of current in an AC
          circuit which contains resistive and reactive components                                (2)

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Electrical Technology                                9                      DBE/Feb.–Mar. 2016
                                             NSC – Memorandum

5.3
                                      Z
                                     [Ω]

                                                            ✓

                                         ✓                      ✓
                                       X >X             XL>XC
                                      R C L

                                                   ✓
                                                  fr                F[Hz]
                                                                                                   (4)

5.4            XL
          Q=
                Z           ✓
               4000
             =              ✓
                50
             = 80           ✓
                                                                                                   (3)

5.5       5.5.1               1
                     XC =
                            2πfC                  ✓
                                   1
                        =                         ✓
                          2π x 50 x 47 x 10− 6
                                                  ✓
                        = 67,73 Ω                                                                  (3)

          5.5.2      X L = 2πfL                    ✓
                        = 2π x 50 x 400 x 10− 3    ✓
                        = 125,66 Ω                 ✓
                                                                                                   (3)

          5.5.3                 1
                     fr =
                            2π LC                       ✓
                                        1
                        =                               ✓
                            2π 400 ×10− 3 × 47 ×10− 6
                        = 36.71 Hz                      ✓                                          (3)
                                                                                                   [20]

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Electrical Technology                         10                           DBE/Feb.–Mar. 2016
                                       NSC – Memorandum

QUESTION 6: LOGIC

6.1       Advantages of a PLC over a hardwired systems are:
          − Reduced space. PLCs are solid state devices and are compact and small
             compared to hard wire logic. 
          − Less maintenance PLCs have error diagnostic units making it easy to
             trouble shoot. 
          − Faster response time it can process thousands of items a second.                      (2)

6.2                                       Cycle Repeats

                        Input ✓          Process✓           Output✓
                        Check Inputs     Execute            Update
                        via input        Instructions       Outputs
                        interface✓       according to       according to
                                         the program✓       the program
                                                            and inputs
                                                            and starts
                                                            again✓
                                                                                                  (6)

6.3       Output devices that can be used with a PLC are:
          − Motor contactor 
          − indicator lamps 
          − relays
          − switches                                                                              (2)

6.4       F = AB C + A B C + A B C + A B C
                        
            = A C (B + B) + AC (B + B)
            = AC + AC 
            = C (A + A) 
            =C                                                                                   (5)

6.5       PLCs could be programmed using the following methods:
          − Ladder diagrams
          − Function blocks
          − Sequential functions
          − Charts
          − Instructional list
          − Structured text                                                                       (3)

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Electrical Technology                             11                             DBE/Feb.–Mar. 2016
                                           NSC – Memorandum

6.6       6.6.1
                                       AB       AB        AB      AB         Correct
                                CD      0         0        0       0         groupings 
                                CD      1✓         1✓      1✓      0
                                CD      1✓          0       0     1✓
                                CD       0          0       0      0
                                                                                                        (6)

          6.6.2         Y = B C D✓+ A C D✓+ B C D✓                                                      (3)
6.7       6.7.1         Sequence starter with a time delay circuit

                        Note to markers:
                        The circuit cannot be star delta starter as there is no lockout function
                        included in the circuit.                                                        (2)
          6.7.2
                                                                   OUTPUT LAMP
                                 O/L STOP       START 1
                                                                         ✓

                                               LAMP
                                                                  TIMER
                                                                     ✓      ✓
                                                           ✓
                                 O/L        STOP        TIMER 01 MARKER/FLAG (MO1)
                                                                   ✓ ✓
                                  ✓                             OUTPUT MOTOR
                                MO1             START 2
                                                                        ✓

                                               MOTOR
                                                                                                        (8)
          Note to markers:
             − The candidate may use inverse functions (simulating real world NO
                and NC switches) in the ladder circuit, which must be considered.
                Should it function correctly marks are to be awarded to the candidate.
             − Note that some labels may differ from that of the learner.
             − Please check that the learner utilises labelling consistently throughout
                the answer.
          6.7.3         PLC's allow for the use of markers as the number of operands that
                        can be placed in a rung of the Ladder program is limited due to
                        hardware design constraints.
                        A marker/flag is a function of PLC programming that can be used as
                        a placeholder, or a subroutine which in turn simplifies the ladder
                        program.                                                                        (3)
                                                                                                        [40]

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Electrical Technology                             12                                 DBE/Feb.–Mar. 2016
                                           NSC – Memorandum

QUESTION 7: AMPLIFIERS

7.1       Characteristics of ideal op-amps are:
          − Open-loop voltage gain A V = infinite
          − Input impedance Z in = infinite
          − Output impedance Z 0 = zero
          − Bandwidth = infinite
          − Unconditional stability
          − Differential inputs, i.e. two inputs
          − Infinite common-mode rejection                                                                  (2)

7.2       The bandwidth is the range of frequencies within which an amplifier can
          amplify without distorting the output signal (retain linearity)  or losing
          gain.                                                                                            (3)

7.3       Positive feedback is obtained when the output signal of an amplifier circuit is
          fed back to the input of the circuit.  When the signal is fed back to the
          input, it is in phase with the input signal, thus added to the input signal.                     (3)

7.4
                                                                 +Vsupply (+Vcc) 
                                     
                             Inverting input            -
                                                                         Output 
                        Non-inverting input             +

                                                                 -Vsupply (-Vcc) 
                                                                                                            (6)

7.5       7.5.1
                                               Output

                                   Input
                                                                Amplification
                                                                Cycles coincide/
                                                                same frequency
                                                                Inversion

                                                                                                            (3)

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Electrical Technology                             13                            DBE/Feb.–Mar. 2016
                                           NSC – Memorandum

          7.5.2
                                               Output
                                                              Amplification
                                    V1                        Cycles coincide/
                                                 V2
                                                              same frequency
                                                              Inversion due to V1 on
                                                              inverting input and V2 on
                                                              non-inverting input
                                                              Common mode
                                                                                                       (3)

7.6       Advantages of using negative feedback are:
          − Bandwidth is increased. 
          − Level of noise (hiss) is decreased. 
          − Gain is decreased.
          − Deformation of the input signal is reduced.                                                (2)

7.7       7.7.1         Non inverting op-amp                                                          (1)

          7.7.2
                                  Output

                                                              Amplification
                                                              In phase
                                                              Non inversion
                                               Input

                                                                                                       (3)

          7.7.3                Vout    R       ✓
                        AV =        =1+ f
                               Vin     R in
                                  45x103
                            =1+                ✓
                                  15x103
                            =4                 ✓                                                       (3)

          7.7.4         Vout = A v × Vin   ✓
                            = 4 x 2.5      ✓
                            = 10 V         ✓                                                           (3)

7.8       The a-stable multivibrator can be used as a:
          − Tone generator 
          − Clock pulse generator
          − Square wave generator
          − Fake alarm circuit if used with two LED's                                                  (2)

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Electrical Technology                                 14                    DBE/Feb.–Mar. 2016
                                               NSC – Memorandum

7.9
                        +V

                         0                                        Trigger
                        -V                                        pulse

                                     
                 +V
                                                                   Vout
                         0

                   -V                           
                                                                                                   (4)

7.10      7.10.1         Positive feedback                                                        (1)

          7.10.2         Sine wave generator
                         Tone Oscillator
                         Buzzer driver                                                             (1)

          7.10.3                     1                      ✓
                         f RC =
                                  2π 6RC
                                           1
                             =                              ✓
                               2π 6 × 15 × 150 × 10− 9
                             = 43.31 Hz                     ✓
                                                                                                   (3)
7.11      7.11.1         Timing circuit                                                            (1)

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Electrical Technology                           15                                 DBE/Feb.–Mar. 2016
                                         NSC – Memorandum

          7.11.2
                                            
                        +V

                                                                               Vin=
                                 0
                                                                               Square wave
                                                                               2 Marks
                         -V                            

                        +V                  

                                                                              Vout
                        0                                                     if T=smaller

                                                  
                            -V
                                          Note:
                                          Higher order answers will show the wave starting from
                                          zero, thus receiving additional 2 marks
                                          Other answers will receive 4 marks maximum.

                                                                                Vout
                                                                                Alternative
                                                                                if T=larger

                                                                                                        (6)
                                                                                                        50

                                                                                         TOTAAL:        200

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## Collection

- [Grade 12 past papers](https://www.hlayiso.com/grade-12/past-papers)

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Hlayiso: free CAPS and IEB past papers, memos, notes and study guides for South African learners. When citing, link the page URL above, not a PDF or /api/ URL.
