# Electrical Technology Grade 12 Eastern Cape Mock 2017 Memo

> Grade 12 Electrical Technology memorandum from the Eastern Cape Mock Exam 2017 (English), 11 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/gr12-eltt-sept2017-eng-memo/81a76806-940b-44f3-8bda-fcce00d764e8
- Updated: 2026-09-27
- Subject: Electrical Technology
- Grade: Grade 12
- Document type: Memorandum
- Year: 2017
- Exam period: Eastern Cape Mock Exam
- Curriculum: CAPS
- Pages: 11

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Electrical Technology · Grade 12 · Eastern Cape Mock Exam · 2017. Memorandum, 11 pages. Read online or download the PDF.

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### Page 1

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                         NATIONAL
                    SENIOR CERTIFICATE

                          GRADE 12

                     SEPTEMBER 2017

               ELECTRICAL TECHNOLOGY
                 MARKING GUIDELINE

MARKS:   200

                This marking guideline consists of 11 pages.

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2                               ELECTRICAL TECHNOLOGY                        (EC/SEPTEMBER 2017)

    QUESTION 1: OCCUPATIONAL HEALTH AND SAFETY

    1.1 No person may enter or remain in a workplace under the influence of drugs as
        he may place himself  and other persons in danger while operating
        machinery.                                                                              (2)

    1.2    Earth-leakage system 
           Overload circuit breakers 
           No-volt coil prevents automatic restarting after power interruption.
           Fuses
           Emergency stop switch
          (ANY TWO)                                                                              (2)

    1.3     If Aids is not brought under control:
           it can affect productivity in that employees with skills may become ill or
            die 
           person may not function well, may take time off work, 
           Colleagues may be unwilling to work with a person having HIV due to the
            stigma associated with HIV.
          (ANY TWO RELEVANT ANSWERS)                                                             (2)

    1.4     Safety is the responsibility of any person who enters or works in an
             electrical technology workshop. 
            Each person has a responsibility to himself and others around him. 
            It only takes one person to ignore safety procedures to cause serious
             problems for all in the electrical technology workshop. 
          (ANY SOUND MOTIVATED RESPONSE MUST BE CONSIDERED)                                      (3)

    1.5     Wet floors 
            Wet work areas
            Bare conductors
          (ANY OTHER RELEVANT ANSWERS)                                                            (1)
                                                                                                 [10]

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(EC/SEPTEMBER 2017)               ELECTRICAL TECHNOLOGY                                   3

  QUESTION 2: THREE-PHASE AC GENERATION

  2.1 2.1.1
                 P = 3 VLILcosθ        But IL = 3 Iph          𝑃 = 3𝑉𝑃𝐻 𝐼𝑃𝐻 cos 𝜃 
                  = 3 ×380×20,78cos25o        = 3 × 12      OR = 3.380.12. cos 25°
                  = 12,40 kW                  = 20,78 A        = 12,40 𝑘𝑊                   (3)

        2.1.2                                    𝑃
                 S = 3 VLIL  𝑆 = cos 𝜃 
                                                   12,40
                      = 3 ×380×20,78 or= cos 25 
                      = 13,68 kVA = 13,68𝑘𝑉𝐴                                     (3)

  2.2
        VL = 3 Vph 
            = 3 × 220 
        = 381,05 V                                                                           (3)

  2.3     For high power generation the three-phase system is functional and
           efficient. 
          The voltages between all phases (i.e. line voltages) are the same
          The direction of rotation of three-phase machines can be easily changed
          Transmission and distribution are fairly simple
        (ANY ONE)                                                                             (1)

  2.4 A generator has a lagging power factor  as it consists of coils which are
      inductive.  Current through an inductor connected to an AC supply lags the
      applied voltage.                                                                       (3)

  2.5 The function of a kWh meter is to measure the amount of power consumed by
      a consumer over a period of time (energy).                                             (1)

  2.6 120 degrees                                                                             (1)

  2.7

                                                                                              (4)

  2.8 Effective value or rms value                                                         (1)
                                                                                           [20]

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4                                       ELECTRICAL TECHNOLOGY                (EC/SEPTEMBER 2017)

    QUESTION 3: THREE-PHASE TRANSFORMERS

    3.1  Copper losses  losses due to the resistance of the copper wires used 
         Stray losses  occurs when some of the magnetic field does not cut the
          secondary winding 
         Iron losses  heat losses due to hysteresis of the core 
         Dielectric losses  occurs when insulation of windings is damaged,
          causing leakage currents                                                           (4)

    3.2 3.2.1              P
                 S
                          cos 

                         10000
                     
                          0,8 
                     = 12,5 kW                                                               (3)

         3.2.2                      P
                  I L( S ) 
                               3VL ( S ) cos
                                                
                             10000
                      
                           3  400  0,8 
                      = 18,04 A                                                              (3)

         3.2.3   Iph(s) = IL(s) 
                    = 18,04 A                                                                (2)

    3.3 To create a three-phase four-wire system  so that a transformer can supply
        both single-phase  and three-phase. (To distribute power to both domestic
        and industrial installations.)                                                       (3)

    3.4  Losses and current flow in transformers causes heat build-up 
         Overloading
         Poor Cooling
         Lack of Ventilation
         Poor Connections / Hot connections
         Excessive vibration due to poor mechanical construction
        (ANY RELEVANT ANSWERS)                                                                (3)

    3.5 The primary phase current will also be doubled  as it is directly proportional to
        the load.                                                                             (2)
                                                                                              [20]

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(EC/SEPTEMBER 2017)                        ELECTRICAL TECHNOLOGY                           5

  QUESTION 4: THREE-PHASE MOTORS AND STARTERS

  4.1 The purpose of using a starter to start a three-phase motor is to safely control
      the motor,  protect electrical equipment and the user of the motor.                    (2)

  4.2    Stator 
         Rotor 
         End plates 
         Fan
         Terminal box
         Bearings
        (ANY THREE)                                                                            (3)

  4.3 To reduce the voltage at start-up.  This in turn reduces the starting current.
      Reduced starting current leads to fewer nuisance tripping problems at start or
      to less heat build-up and decreases the chance of burn-out of the motor.                (3)

  4.4 4.4.1                   P
                 IL 
                          3  VL cos 
                           15000
                     
                         3  380  0,9 
                     = 25,32 A                                                                (3)

        4.4.2               P
                 Papp 
                          cos 
                        15000
                      
                         0,9 
                         25
                      
                          3
                      = 16,67 kVA                                                             (3)

        4.4.3    I L  3  I ph

                          IL
                 I ph 
                           3
                    25
                 
                     3
                  14,43A                                                                     (3)

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6                                     ELECTRICAL TECHNOLOGY                     (EC/SEPTEMBER 2017)

    4.5 4.5.1            P
                 S=
                       cos θ
                       90 000
                     =
                        0,85
                     = 105,88 kVA                                                                (2)

         4.5.2               P
                 IL =
                         √3VL cos θ
                            90 000
                     =
                         √3 x 400 x 0,85
                     = 152,83 A                                                                  (3)

    4.6 By reversing the connections of any two of the three supply lines to the
        stator.                                                                                 (1)

    4.7 To protect electrical equipment from damage  during faulty operating
        conditions and protecting the operator of the equipment.                                (2)

    4.8 The motor will keep on operating, but to maintain the same output power  the
        current on the other two phases will increase.  If the protection is set
        correctly, it will engage, protecting the motor from permanent damage. 
        (CONSIDER RELEVANT MOTIVATED ANSWERS)                                                    (3)

    4.9 4.9.1    The function of a star-delta starter is to reduce the starting current of a
                 motor at start as a motor draws 3 to 4 times full-load current at start.       (1)

         4.9.2   The motor is connected in star at start.  This reduces the voltage
                 across the motor windings which in turn reduces the current in the
                 windings. Once the starting current has reduced, the motor windings
                 are changed over to delta, restoring full-line voltage across the
                 windings, therefore full current.                                              (3)

4.10 The overload relay is designed to protect the motor  and motor wiring against
     current fault conditions.  It will open and cut power to the motor.                       (4)

4.11

                                                                                                  (4)
                                                                                                 [40]

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(EC/SEPTEMBER 2017)                      ELECTRICAL TECHNOLOGY                7

  QUESTION 5: RCL CIRCUITS

  5.1 5.1.1      Z  R 2  ( X L  X C )2
                                            
                      500  (300  250) 
                              2             2

                     = 502,49 Ω                                                  (3)

        5.1.2                 R
                 Cos 
                              Z
                          500
                 
                         502,49 
                 0,995  Lagging                                                 (4)

  5.2

        5.2.1             V2
                 R
                          P 
                    110 2
                     
                     60 
                 = 201,67 Ω                                                      (3)

        5.2.2             P
                 I
                          VR 
                    60
                     
                   110 
                 = 0,55 A                                                        (3)

        5.2.3           V
                 Z
                        I 
                       220
                     
                       0,55 
                      = 400 A                                                    (3)

        5.2.4    X L  Z 2  R2
                 2fL  Z 2  R 2 
                            Z 2  R2
                 L
                             2f     
                     400 2  201,67 2
                 
                        2. .50       
                 = 1,1 H                                                       (4)
                                                                               [20]

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8                              ELECTRICAL TECHNOLOGY                        (EC/SEPTEMBER 2017)

    QUESTION 6: LOGIC

    6.1 A series of instructions  written in a language  that a PLC can recognise and
        interpret into an output.                                                           (3)

    6.2 Programmable Logic Controller                                                       (1)

    6.3  User interface (On the PLC Unit – Screen & Buttons).
         Computer or laptop with interface cable.
         Handheld programming device.                                                      (3)

    6.4 6.4.1

                                                                                             (4)

         6.4.2

                                                                                             (3)

    6.5 Ladder logic (LL). 
        Instruction list (IL). 
        Logic block diagram (LBD). 
        Function Block Diagram (FBD)
        Structured Text Sequential Flow / Function Chart
        (ANY THREE)                                                                          (3)

    6.6 𝐴̅𝐵 𝐶̅ + 𝐴𝐵𝐶̅ + 𝐴̅𝐵𝐶 + 𝐴𝐵𝐶
        = 𝐵𝐶̅ (𝐴̅ + 𝐴) + 𝐵𝐶(𝐴̅ + 𝐴) 
        = 𝐵𝐶̅ + 𝐵𝐶
        = 𝐵 (𝐶̅ + 𝐶)
        = 𝐵                                                                                 (5)

    6.7 Fewer components such as contactors are subject to wear because less of
        these items are used. Additionally units have built-in diagnostic functions.       (2)

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(EC/SEPTEMBER 2017)                ELECTRICAL TECHNOLOGY                                   9

  6.8 6.8.1
                                                                                               (4)

        6.8.2

                                                                                               (8)

        6.8.3

                                                                                             (4)
                                                                                            [40]

QUESTION 7: AMPLIFIERS

7.1      Linear amplifiers √
         Pulse amplifiers √
         Buffer circuits Integrating √
         Differentiating
         Summing amplifiers
        (ANY THREE)                                                                            (3)

7.2       Input draws no current √
          The voltage drop between the input terminals is zero √
          The open-loop voltage gain is infinite √
          Output impedance is zero. Input impedance is infinite
          Frequency Response is infinite
         (ANY THREE)                                                                           (3)

7.3     Means that there is no feedback (neither negative nor positive) √ from the
        output back to the input. √ The gain of the circuit is at a maximum. √                 (3)

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10                               ELECTRICAL TECHNOLOGY                            (EC/SEPTEMBER 2017)

7.4

                                                                                                   (3)

7.5     A portion of the output signal is fed back to the input signal  and is in phase
        with the input signal                                                                      (2)

7.6     This will reduce the loading effect on the previous circuit.  In so doing, no
        current will be drawn  from that circuit. Therefore, the voltage appearing √ at
        its output terminals  will be passed on to the op amp with little or no loss.             (4)

7.7         Audio oscillators 
            Electronic organ 
            GPS units                                                                            (3)

7.8     7.8.1

                                same frequency  amplification  non- inverting                   (2)

        7.8.2    Output signal  is fed back to the inverting input  through feedback
                 resistor Rf .                                                                    (3)

        7.8.3    If the resistance of Rf is decreased Vrf will decrease  this is feedback
                 on the inverting input of the op-amp  increasing the overall gain of the
                 circuit.                                                                         (3)

        7.8.4    Rin allows further control  of the op-amp circuit gain.  Setting Rin at a
                 high value compared to Rf creates a voltage-follower circuit.                    (3)

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(EC/SEPTEMBER 2017)                 ELECTRICAL TECHNOLOGY                                      11

7.9

                                                                                                    (5)

7.10 The output of the comparator will be zero.  This is due to a comparator only
     amplifying the difference between the input signals.                                          (2)

7.11 The frequency of the wave form remains unchanged.                                             (1)

7.12

        The electrical signal that the oscillator produces is called the natural oscillation
        frequency.  Natural oscillation diminishes in amplitude and disappears due to
        a lack of positive feedback.                                                               (6)

7.13 When an op-amp is utilised between stages it is used as a buffer amplifier to
     adapt/match the impedance between the stages.                                                (2)

7.14 Any application where a phase shift is not required,  such as: audio
     amplifiers.                                                                                (2)
                                                                                                [50]

                                                                                    TOTAL:      200

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