# Electrical Technology Grade 12 Eastern Cape Mock 2022 Memo

> Grade 12 Electrical Technology memorandum from the Eastern Cape Mock Exam 2022 (English), 10 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/elec-tech-gr12-memo-sept2022-power-english-1/1304f4fd-377a-40b4-8abd-4e3860f4fd0d
- Updated: 2026-09-27
- Subject: Electrical Technology
- Grade: Grade 12
- Language: English
- Document type: Memorandum
- Year: 2022
- Exam period: Eastern Cape Mock Exam
- Curriculum: CAPS
- Pages: 10

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## About this document

Electrical Technology · Grade 12 · Eastern Cape Mock Exam · 2022 · English. Memorandum, 10 pages. Read online or download the PDF.

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## Document text

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### Page 1

NATIONAL
                     SENIOR CERTIFICATE

                               GRADE 12

                      SEPTEMBER 2022

    ELECTRICAL TECHNOLOGY: POWER SYSTEMS
              MARKING GUIDELINE

MARKS:   200

               This marking guideline consists of 10 pages.

### Page 2

2                      ELECTRICAL TECHNOLOGY: POWER SYSTEMS                  (EC/SEPTEMBER 2022)

INSTRUCTIONS TO MARKERS

1.    All calculations 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.

      2.6     Markers should consider that learners answers may deviate slightly from
              the marking guideline depending on how and where in the calculation
              rounding off was used.

3.    These marking guidelines are only a guide with model answers.

4.    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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### Page 3

(EC/SEPTEMBER 2022)    ELECTRICAL TECHNOLOGY: POWER SYSTEMS                                    3

QUESTION 1: MULTIPLE-CHOICE QUESTIONS

1.1   1.1      A✓                                                                            (1)
      1.2      C✓                                                                            (1)
      1.3      B✓                                                                            (1)
      1.4      C✓                                                                            (1)
      1.5      D✓                                                                            (1)
      1.6      B✓                                                                            (1)
      1.7      A✓                                                                            (1)
      1.8      D✓                                                                            (1)
      1.9      C✓                                                                            (1)
      1.10     B✓                                                                            (1)
      1.11     A✓                                                                            (1)
      1.12     D✓                                                                            (1)
      1.13     C✓                                                                            (1)
      1.14     B✓                                                                            (1)
      1.15     A✓                                                                            (1)
                                                                                            [15]

QUESTION 2: OCCUPATIONAL HEALTH AND SAFETY

2.1     An occurrence of catastrophic proportions, ✓ resulting from the use of
        machinery or activities at work. ✓                                                    (2)

2.2     The learner does not know safe practices. ✓
        The learner knows better but intentionally conducts the act.                          (1)

2.3     2.3.1 Running could cause you to trip or collide with another learner. ✓
              This could result in you injuring yourself with nearby equipment or
              machinery. ✓                                                                    (2)

        2.3.2 This could cause the outlet to exceed its rated current ✓and could lead
              to short circuits, fires or damaged appliances. ✓                               (2)

2.4     First, I would define all the various threats to safety in the workshop. ✓
        Secondly, I would determine the extent of all the vulnerabilities in the
        workshop. ✓
        Finally, I would devise countermeasures should a risk occur. ✓                       (3)
                                                                                            [10]

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### Page 4

4                        ELECTRICAL TECHNOLOGY: POWER SYSTEMS           (EC/SEPTEMBER 2022)

 QUESTION 3: RLC CIRCUITS

 3.1     3.1.1 It would remain the same. ✓                                              (1)

         3.1.2 XL = 31,83 Ω ✓                                                           (1)

         3.1.3 (a)      XL = 2πfL ✓
                           = 2 × π × 50 × 0,5 ✓
                           = 157,08 Ω ✓                                                 (3)

                 (b)    Z = √R2 + (XL − XC )2 ✓
                          = √1202 + (157,08 − 31,83)2 ✓
                          = 173,46 A ✓                                                  (3)

                 (c)    Vc = I × XC ✓
                           = 1,38 × 31,83 ✓
                           = 43,93 V ✓                                                  (3)

                 (d)    XC = 2πfC
                                 1

                                 1
                        C = 2πfX ✓
                                     C
                                         1
                           =                  ✓
                               2×π×50×42,44
                          = 0,000075 F = 75 μF ✓                                        (3)

                 (e)       V
                        I = RR ✓
                           240
                         = 120 ✓
                         = 2A✓                                                          (3)

 3.2     0A✓                                                                            (1)

 3.3     The small band of frequencies ✓ centred around the resonant frequency. ✓
         The frequencies between the upper cut-off frequency and the lower cut-off
         frequency, centred around the resonant frequency.                              (2)

 3.4     Z is maximum ✓
         I is minimum ✓
         XL = XC                                                                        (2)

 3.5     3.5.1
                 IT = √IR 2 + (IC − IL )2 ✓
                     = √0,832 + (0,63 − 0,5)2 ✓
                     = 0,84 A ✓                                                         (3)

         3.5.2 I = V
                   R
               V = I. R ✓
                 = 0,83 × 120,487✓
                 = 100 V ✓                                                              (3)

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### Page 5

(EC/SEPTEMBER 2022)          ELECTRICAL TECHNOLOGY: POWER SYSTEMS                           5

         3.5.3 Z = V ✓
                   I     T
                         100
                      = 0,84 ✓
                      = 119,05 Ω ✓                                                         (3)

         3.5.4 cosθ = IR
                      I        T
                                       I
                 θ       = cos−1 IR ✓
                                           T
                                 0,83
                         = cos−1 0,84 ✓
                     = 8,85° ✓
                 Leading ✓                                                                (4)
                                                                                         [35]

 QUESTION 4: THREE-PHASE AC GENERATION

 4.1     Three coils are placed 120 electrical degrees apart and then rotated in a
         uniform magnetic field. ✓ This produces three single-phase voltages, with
         each voltage separated by 120°. ✓                                                 (2)

 4.2     4.2.1   Star connected three-phase system. ✓                                      (1)

         4.2.2   A – neutral ✓
                 B – yellow phase ✓                                                        (2)

         4.2.3   By connecting one side of each coil to a common point. ✓                  (1)

         4.2.4   VPH =
                             VL
                                   ✓
                             √3
                             440
                         =         ✓
                             √3
                         = 254,03 V ✓                                                      (3)

 4.3     4.3.1 It is defined as the ratio of the real power flowing ✓ to the apparent
               power in a circuit. ✓                                                       (2)

         4.3.2 The power factor is directly related to the current level. ✓                (1)
             𝑜𝑢𝑡𝑝𝑢𝑡
 4.4     ƞ = 𝑖𝑛𝑝𝑢𝑡 × 100% ✓
             220 × 6,5
          = 380 × 4,5 × 100% ✓
          = 83,63% ✓                                                                       (3)

 4.5     4.5.1 VL = VPH ✓
                  = 380 V ✓                                                                (2)

         4.5.2 Power factor = cosθ
               θ = cos −1 power factor ✓
                = cos−1 (0,89) ✓
                = 27,13° ✓                                                                 (3)

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### Page 6

6                      ELECTRICAL TECHNOLOGY: POWER SYSTEMS                (EC/SEPTEMBER 2022)

         4.5.3 S = √3VL IL ✓
                 = √3 × 380 × 27 ✓
                 = 17 770,84 VA = 17,77 kVA ✓                                              (3)

         4.5.4 P = √3VL IL cosθ ✓
                = √3 × 380 × 27 × 0,89 ✓
                = 15 816,05 W = 15,82 kW ✓                                                 (3)

 4.6     PTOT = W1 + W2 + W3
         W3 = PTOT − (W1 + W2 ) ✓
              = 28,75 − (12 + 7,5) ✓
              = 9,25 kW ✓                                                                  (3)

 4.7     A lagging power factor draws more current from the load. ✓
         The higher currents increase the energy lost and requires larger conductors
         and equipment. ✓
         This results in the electrical utilities charging a higher cost to commercial
         consumers. ✓                                                                      (3)

 4.8                 𝑊 −𝑊
         𝑡𝑎𝑛𝜃 = √3 [𝑊1 + 𝑊2 ]
                       1   2
                                16,6 − 5,5
         𝜃     = tan−1[√3 × (16,6 + 5,5)]
             = 41,02° ✓
         𝑃𝑜𝑤𝑒𝑟 𝑓𝑎𝑐𝑡𝑜𝑟 = 𝑐𝑜𝑠𝜃
                      = cos (41,02) ✓
                      = 0,75 ✓                                                            (3)
                                                                                         [35]

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### Page 7

(EC/SEPTEMBER 2022)             ELECTRICAL TECHNOLOGY: POWER SYSTEMS                         7

QUESTION 5: THREE-PHASE TRANSFORMERS

5.1   •   Iron losses ✓
      •   Copper losses ✓                                                                   (2)

5.2   The Bucholtz relay monitors the flow of gas ✓
      A severe fault will cause a lot of gas and the relay will operate immediately ✓
      For slow faults, the gradual build-up of gas will trigger the relay as soon as a
      certain concentration is reached ✓                                                    (3)

5.3   5.3.1    B✓                                                                           (1)
      5.3.2    D✓                                                                           (1)
      5.3.3    A✓                                                                           (1)
      5.3.4    C✓                                                                           (1)

5.4   VLP = VPP = 6 600 V ✓
            𝑉
      VPS = 𝐿𝑆
               √3
              381,05
          =             ✓
               √3
         = 220 𝑉 ✓
      𝑡𝑢𝑟𝑛𝑠 𝑟𝑎𝑡𝑖𝑜 = 𝑉𝑃𝑃 ∶ 𝑉𝑃𝑆 ✓
      = 6 600 ∶ 220 ✓
      = 30 ∶ 1 ✓                                                                            (6)

5.5   5.5.1 𝑉𝐿𝑆 = √3𝑉𝑃𝑆
                  𝑉
            𝑉𝑃𝑆 = 𝐿𝑆 ✓
                           √3
                          11 000
                     =              ✓
                             √3
                     = 6 350,85 V = 6,35 kV ✓                                               (3)

      5.5.2 𝑆 = √3𝑉𝐿𝑃 𝐼𝐿𝑃 ✓
             = √3 × 33 000 × 9,72 ✓
             = 555 572,62 VA = 555,57 kVA ✓                                                 (3)

      5.5.3 𝑃 = √3𝑉𝐿𝑆 𝐼𝐿𝑆 𝑐𝑜𝑠𝜃
                   𝑃
            𝐼𝐿𝑆 =          ✓
                     𝑐𝑜𝑠𝜃√3𝑉𝐿𝑆
                            500 000
                    =                        ✓
                         √3 × 11 000 × 0,9
                    = 29,16 𝐴 ✓                                                             (3)

      5.5.4 𝐼𝐿𝑃 = √3𝐼𝑃𝑃
                  𝐼
            𝐼𝑃𝑃 = 𝐿𝑃 ✓
                          √3
                         9,72
                     =          ✓
                          √3
                     = 5,61 𝐴 ✓                                                             (3)

      5.5.5 ƞ =             𝑃𝑂𝑈𝑇
                                        × 100% ✓
                𝑃        𝑂𝑈𝑇 +𝑙𝑜𝑠𝑠𝑒𝑠
                            500 000
                 = 500 000+1 800 × 100% ✓
                 = 99,64 % ✓                                                               (3)
                                                                                          [30]
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### Page 8

8                               ELECTRICAL TECHNOLOGY: POWER SYSTEMS        (EC/SEPTEMBER 2022)

QUESTION 6: THREE-PHASE MOTORS AND STARTERS

6.1   •   Robust ✓
      •   Dependable ✓
      •   Economical ✓
      •   Requires less maintenance
      •   More efficient                                                                    (3)

6.2   •   Lathes ✓
      •   Drill presses ✓
      •   Bench grinders                                                                    (2)

6.3   6.3.1   Continuity test ✓
              Insulation resistance between windings and earth ✓                            (2)

      6.3.2   No. ✓ There is no continuity between V1 and V2 ✓ and V2 of the coil is
              short circuited to earth. ✓                                                   (3)

6.4   6.4.1   Yes, it is suitable. ✓                                                        (1)

      6.4.2   𝑃 = √3VL IL cosθ ✓
                = √3 × 380 × 24 × 0,8 ✓
                = 12 637,04 W = 12,64 kW ✓                                                  (3)

      6.4.3           𝑃
              Ƞ = 𝑃𝑂𝑈𝑇 × 100% ✓
                          𝐼𝑁
                       10 000
                 =                    × 100 ✓
                      12 637,04
                 = 79,13 % ✓                                                                (3)
                               ns −nr
      6.4.4   %slip =            ns
                          nr
              ns = 1−slip ✓
                          3000
                 = 1− 0,04 ✓
                 = 3125 rpm ✓                                                               (3)

6.5   6.5.1 The timer ensures that motor 2 can only be started after a preset time. ✓       (1)

      6.5.2 380 V ✓                                                                         (1)

      6.5.3 MC2 would de-energise ✓ as soon as the start motor 2 button is
            released. ✓                                                                     (2)

      6.5.4 Motor 1 must start first. ✓
            Motor 2 can only be started after a preset time has elapsed. ✓
            Both motors must have independent overload protection. ✓
            The stop button switches the whole circuit off. ✓                               (4)

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### Page 9

(EC/SEPTEMBER 2022)         ELECTRICAL TECHNOLOGY: POWER SYSTEMS                                  9

6.6    The star contactor ensures that the motor is connected in star during
       start-up, ✓ thus reducing the starting current by √3 or 58% of full load
       current. ✓
       The delta contactor ensures that the motor is connected in delta after a preset
       time. ✓ This delivers maximum power to be delivered when the motor is at full
       speed. ✓                                                                                   (4)

6.7    P = √3VL IL cosθ
             P
       IL =         ✓
              cosθ
              √3VL
                 6 000
         =                   ✓
              √3×380×0,78
         = 11,69 A ✓                                                                             (3)
                                                                                                [35]

QUESTION 7: PROGRAMMABLE LOGIC CONTROLLERS (PLS’s)

7.1   • Input scan ✓
      • Process scan ✓
      • Output scan ✓                                                                            (3)

7.2   7.2.1
                     A   0       >=1                              A
                                  ≥1      0
                                                OUTPUT                          OUTPUT
                                                           OR     B
                     B   0

                                 1 mark for symbol correct ✓
                                 2 marks for correct labels ✓✓                                    (3)

      7.2.2     a–0✓
                b–0✓                                                                              (2)

7.3   They are bits in the storage memory which can be used to hold data ✓ and
      behave as relays, ✓ being able to be switched on or off ✓ and to switch other
      devices on or off. ✓                                                                        (4)

7.4              ✓                                                        ✓              ✓
                                    ✓                     ✓

                                 DIODE
            AC                   BRIDGE                                INVERTER     MOTOR
           LINE                                          FILTER
                                 RECTIFIER

         SPEEED
                                                                              X-
         REFERENCE                              ✓        VOLTAGE AND         Axis
                                                         FREQUENCY
                                                         CONTROL         ✓
                                       X-Axis                                                     (7)

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### Page 10

10                          ELECTRICAL TECHNOLOGY: POWER SYSTEMS            (EC/SEPTEMBER 2022)

7.5    PART 1 – converts the AC supply voltage to a DC voltage by rectification. ✓ It
       consists of three pairs of rectifying diodes, each pair rectifying one of the
       phases. ✓
       PART 2 – smooths the AC ripple out with filter capacitors. ✓ This ensures a
       true DC voltage on the DC rail. ✓                                                       (4)

7.6    •   Variable air volume air conditioning systems ✓
       •   Water pumping systems
       •   Exhaust air systems
       •   Speed control of fan systems
       •   Heating systems for air and liquid applications                                     (1)

7.7    A switch changes the state of a device from 1 to 0 or vice-versa. ✓
       A sensor converts a physical condition into an electrical signal that can be read
       by a PLC. ✓                                                                             (2)

7.8    7.8.1
                                                               ✓
                                                                            ✓
                             stop ✓        Start 1 ✓         O/L1
                                                                           MC1

                                           N/O
                                           MC1 ✓
                                 N/O
                     stop        MC1                         O/L2               ✓
                                       ✓   start 2 ✓
                            ✓
                                                                           MC2
                                                                ✓
                                           N/O✓
                                           MC2

                                                                                              (11)

       7.8.2    The motors may have different rated current values, ✓ so the method
                offers independent overcurrent prptection to each motor. ✓                     (2)

       7.8.3    The N/O MC1 contact is connected to ensure that motor 1 must be
                energised, before motor 2 can be energised. ✓                                  (1)
                                                                                              [40]

                                                                                    TOTAL:    200

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