# Electrical Technology Grade 12 NSC Supplementary 2015 Memo

> Grade 12 Electrical Technology memorandum from the NSC Supplementary Exam 2015 (English), 14 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/electrical-technology-feb-march-2015-memo-eng/82be59ce-c297-46f5-8f07-e3b597271b57
- Updated: 2026-09-27
- Subject: Electrical Technology
- Grade: Grade 12
- Document type: Memorandum
- Year: 2015
- Exam period: NSC Supplementary Exam
- Curriculum: CAPS
- Pages: 14

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## About this document

Electrical Technology · Grade 12 · NSC Supplementary Exam · 2015. Memorandum, 14 pages. Read online or download the PDF.

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## Document text

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                                   GRAAD 12

                                  NATIONAL
                             SENIOR CERTIFICATE

                                   GRADE 12

                          ELECTRICAL TECHNOLOGY

                            FEBRUARY/MARCH 2015

                                 MEMORANDUM

MARKS: 200

                       This memorandum consists of 14 pages.

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Electrical Technology                             2                          DBE/Feb.–Mar. 2015
                                          NSC – Memorandum

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/Feb.–Mar. 2015
                                      NSC – Memorandum

QUESTION 1: OCCUPATIONAL HEALTH AND SAFETY

1.1       Working on a live system with exposed conductors. 
          Working with portable electric equipment that is not insulated correctly.             (2)

1.2       Do not touch the person with your bare hands. 
          Switch off the supply immediately.
          Use the emergency stop button if the workshop has one. 
          Use an insulated item to free the person.
          Apply the correct medical procedure.
          (Any order)                                                                            (3)

1.3       South Africans should be provided the same work and economic
          opportunities for capable people without any discrimination due to gender.
          No person should be exposed to sexual harassment irrespective of their
          gender.                                                                                (2)

1.4       Do all machines have the required guards?
          Are walkways clearly indicated.
          Does the workshop have sufficient lighting?
          Are signage clearly and correctly put up.
          (Answers must relate to preventing accidents to be considered.)                        (3)
                                                                                                 [10]

QUESTION 2: THREE-PHASE AC GENERATION

2.1       For alternators of similar frame size, three phase machines produce more
          power than single phase machines. 
          The same size of machinery is necessary to drive three phase and single
          phase alternators.
          Three phase alternators can be connected parallel to obtain a combined
          power.
          For the same amount of energy, three phase power is cheaper to generate
          than single phase power.
          The phase of a three phase generator can be connected in star to create a
          neutral point.                                                                         (2)

2.2
               Vmax     ER EY EB

                  0

              V-max
                        120° 120°
                                                                                                 (3)

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Electrical Technology                                  4             DBE/Feb.–Mar. 2015
                                               NSC – Memorandum

2.3        PT = P1 + P2            
                                   
             = 700 + (−290)
             = 410W         
                                                                                          (3)

2.4       2.4.1         I L = 3I Ph        
                                           
                          = 3 X 18
                                           
                          = 31,18 A                                                       (3)

          2.4.2                 VL           
                        VPh =
                                 3
                                380 V        
                              =
                                   3
                              = 219,39 V     
                                                                                          (3)

          2.4.3                    VPh       
                          Z Ph =
                                   I Ph
                            219,39           
                          =
                            31,18 A
                          = 7,04 Ω                                                       (3)

          2.4.4         P = 3 VL I L cos θ                 

                          = 3 x 380 x 31,18 x cos 14°      
                                                           
                          = 19,91 kW                                                      (3)
                                                                                          [20]

QUESTION 3: THREE-PHASE TRANSFORMERS

3.1       Core type ✓and
          Shell type✓                                                                     (2)

3.2       It improves the insulation of the transformer✓ and conducts the heat away
          from the windings and core of the cooling surface✓                              (2)

3.3       Insufficient ventilation✓
          Loose connections✓
          Overloading
          Insulation Breakdown
          Short circuits                                                                  (2)

3.4       Copper losses✓
          Iron losses✓                                                                    (2)

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Electrical Technology                                                5          DBE/Feb.–Mar. 2015
                                                             NSC – Memorandum

3.5       Bucholtz relay✓
          Over voltage relay✓
          Under voltage relay
          Over current relay
          Pressure control device                                                                    (2)

3.6       3.6.1                VL = VPH                 ✓
                           ∴ VPH = 11kV                 ✓                                            (2)

          3.6.2            VL = 3 VPH                   ✓
                                               VL
                           ∴ VPH =
                                                3       ✓
                               380
                           =
                               3
                           = 219,40V                    ✓
                                                                                                     (3)

          3.6.3                    Np          VPH(P)              ✓
                                         =
                                   Ns          VPH(S)
                                                    11000          ✓
                                         =
                                                    219,40
                                   Np          50
                                         =
                                   Ns          1
                                   N p : N S = 50 : 1              ✓                                 (3)

3.6.4     The voltage ratio shows that the transformer is a step-down transformer
          (50:1) ✓ The primary current will be lower in the primary winding as
          compared to the current in the secondary winding for a step down
          transformer. ✓

          Alternative answer:

          The power of the transformer remains the same, whether it is calculated on
          the primary or the secondary side of the transformer. This means that if the
          voltage decreases from primary to secondary winding (due to the transformer
          ratio) the current drawn for any specific load will have to increase in the
          secondary winding as compared to the current drawn from the supply in the
          primary winding.

           Vph (p )       Np       I ph (s )
                      =        =
           Vph (s )       Ns       I ph (p )                                                         (2)
                                                                                                     [20]

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Electrical Technology                                        6          DBE/Feb.–Mar. 2015
                                                     NSC – Memorandum

QUESTION 4: THREE-PHASE MOTORS AND STARTERS

4.1       Stator 
          Rotor 
          End plates
          Fan
          Frame                                                                              (2)

4.2       They require less maintenance as they do not have as many parts as a single
          phase motor
          For the same size frame as a single phase motor they deliver a higher
          torque.                                                                           (2)

4.3       Power factor 
          Voltage rating 
          Frequency 
          Speed
          Type of connection
          Phase displacement                                                                 (3)

4.4       Vs = 400 V
          F = 60 Hz
          No Poles = 12 poles per phase = 6 pole pairs
          Slip (Per Unit) = 0,04

          4.4.1                    f          
                        ns = 60 ×
                                  p
                                  60          
                           = 60 ×
                                  6
                           = 600 rpm          
                                                                                             (3)

          4.4.2         nr = ns (1 − S PerUnit ) 
                        nr = 600 (1 − 0,04 ) 
                        nr = 576 rpm                                                        (3)

4.5       It is important to carry out a mechanical inspection to determine before
          starting if there are any mechanical faults that my cause damage to the
          motor or operator on starting.                                                    (2)

4.6       Continuity of each winding
          Insulation resistance between each winding 
          Insulation resistance between windings and earth
          Exposed windings                                                                   (2)

4.7       The overload relay unit offers protection to the motor and operator under
          fault conditions.                                                                 (3)

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Electrical Technology                                7                         DBE/Feb.–Mar. 2015
                                             NSC – Memorandum

4.8       4.8.1         The function of a star-delta starter is to reduce the starting current
                        of a motor at start-up.                                                    (1)

          4.8.2         Delta                                                                      (1)

          4.8.3         The retain (hold in) MC 1 is the contact that will keep power on the
                        main contactor when energised.
                        The contactor is activated once the start button has been pressed. 
                        When the start is released power will remain on the main coil through
                        the retain contactor MC 1 . 
                        MC 1 opens automatically (the contactor is spring loaded) when
                        supply to the coil is removed, thus preventing automatic restarting
                        when power returns after a power failure.                                   (3)

          4.8.4         When the timer coil is energised the timer will begin to time
                        through.
                        Contact T (N/C) will only open when the timer has timed through a
                        preset time.                                                               (2)

          4.8.5         Contacts MC 2 (N/C) & MC 3 (N/C) in series with the star and delta
                        contactors are normally closed.
                        When MC 2 is energised – MC 2 (N/C) will disconnect MC 3 and
                        prevent it from being accidentally activated. 
                        While MC 3 is de-energised, MC 3 (N/C) will remain closed, allowing
                        power to be transferred via T(N/C) and MC 3 (N/C) to MC 2 .
                        Timer T (N/C) will only transfer power via MC 3 to MC 2 while T is
                        timing through.
                        Once T times out and activates, T (N/O) closes and T(N/C) opens.
                        MC 2 is deactivated when T(N/C) opens, removing power from it and
                        MC 2 resets  (The contactor is spring loaded).
                        When MC 2 resets, MC 2 (N/C) closes and MC 3 is able to be
                        energised through T(N/O) and MC 2 (N/C) which is now closed.
                        This energises MC 3 which in turn opens MC 3 (N/C) thus preventing
                        MC 2 from being energised.                                                  (6)

4.9       4.9.1         P=    3 × VL × I L × cos θ ×η   

                          = 3 × 380 ×12 × 0,8 × 0,9     
                          = 5,69 kW                     
                                                                                                    (3)

          4.9.2         The active power of the motor will increase.                               (1)

          4.9.3         I L = 3 I PH                                                               (1)

          4.9.4         If the power factor of the motor is improved the motor will draw a
                        lower current as it will still deliver the same output power.             (2)
                                                                                                    [40]

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Electrical Technology                                  8                DBE/Feb.–Mar. 2015
                                               NSC – Memorandum

QUESTION 5: RLC

5.1       Connect an ammeter in the circuit and adjust the frequency of the power
          supply until the reading on the ammeter is at a maximum. 
          OR
          Connect a volt meter across the coil and capacitor and then adjust the
          frequency of the power supply until the readings on the two meters are the
          same.
          OR
          Connect a volt meter across the supply and resistor and then adjust the
          frequency of the power supply until the readings on the two meters are the
          same.                                                                              (3)

5.2       Install a power factor correcting capacitor 
          Make use of synchronous motors                                                     (1)

5.3       When the frequency is decreased, XL will decrease and Xc will increase.
          IXL will then increase and IXC will decrease.
          IZ will increase as the phase angle changes.

          Alternative answer:

          At resonance XL = XC and the resultant current in the circuit IZ=IR and this
          current is in phase with the supplied voltage.
          When the supply frequency is reduced the circuit will move out of resonance.
          Lowering the supply frequency will cause the inductive reactance X L to
          decrease, thus increasing I XL
          Lowering the supply frequency will cause the capacitive reactance X C to
          increase, thus reducing I XC .
          The circuit will now become increasingly inductive and the resultant current
          will be lagging with respect to the supply voltage.
          As I XL and I XC oppose each other the resultant reactance will change thus
          altering the phase angle, resulting in an increase in the current drawn from
          the supply.                                                                        (3)

5.4       5.4.1                     1
                        fr =                                    
                               2π ( LC )
                                           1
                        =
                            2π (0,2 ×10 −3 ) × (160 ×10 − 6 )   
                        = 890,15 Hz                             
                                                                                             (3)

          5.4.2             V                        
                        I =                          At resonance Z=R
                            Z
                          240
                        =                            
                           30
                        = 8A                                                                (3)

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Electrical Technology                                 9                                DBE/Feb.–Mar. 2015
                                              NSC – Memorandum

          5.4.3              XL                             
                        Q =
                              R
                            2πfL
                          =
                              R                                Calculation of X L 
                            2 × π × 889 × 0,2 ×10 −3
                          =
                                      30
                          = 0,04                                                                           (4)

          5.4.4                   1
                        fr =
                           2π LC
                              1                             
                        C=
                           4π f r L
                             2 2

                                         1
                          =                                 
                               4 x π ×1000 2 × 0,2 ×10 −3
                                      2

                          =126,6 µF                         
                                                                                                            (3)
                                                                                                            [20]

QUESTION 6: LOGIC

6.1       6.1.1         The interface is the connection (platform) between real world
                        devices and the CPU (controller).
                        This allows for level conversion of the input information rendering it
                        suitable to be fed into the PLC solid state processor.                              (3)

          6.1.2         Relay contacts
                        Pushbuttons
                        Limit switches
                        Analogue sensors
                        Selector switches
                        (Any TWO)                                                                           (2)

          6.1.3         L.E.Ds 
                        Power Transistors
                        Thyristors
                        (Any TWO)                                                                           (2)

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Electrical Technology                           10                            DBE/Feb.–Mar. 2015
                                         NSC – Memorandum

          6.1.4         STEP 1- check input status – the controller examines the status of
                        the input devices and checks whether it is at a 'HIGH' or 'LOW' , '0'
                        OR '1' 
                        STEP 2- EXECUTE THE PROGRAM- the PLC executes the
                        program one instruction at a time and will store the results as
                        parameters for making further decisions when continuing the scan
                        cycle. 
                        STEP 3- UPDATE OUTPUT STATUS- the PLC then changes the
                        output according to its programmed instruction list, ladder logic
                        program or logic block diagram(LBD)                                       (6)

6.2         6.2.1

                                                                                                   (6)

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Electrical Technology                           11                            DBE/Feb.–Mar. 2015
                                         NSC – Memorandum

            6.2.2

                                                    Grouping                                      (6)

          6.2.3

                                                                                                   (7)

          6.2.4         The Set/Reset function can be used in motor control
                        The reset function takes priority over the set function
                        This could be used to simulate the start and stop function of a
                        direct-on-line-starter (or any other type of starter).
                        (Any acceptable example other than this must be considered)                (3)

          6.2.5         An external Stop Button provides the operator with a physical
                        means of disconnecting supply to the No Volt Coil. 
                        On a PLC system the Start and Stop functions can be programmed
                        into the memory as Set Reset functions. 
                        This programming function would then in turn enable starting and
                        stopping of the motor. 
                        If the output contactor developed a fault and remained in an ON
                        position the operator would lose control of the system,
                        An external Stop button will serve as additional safety and will stop
                        the motor to prevent an accident.                                          (3)

6.3       Contacts activate after a pre-set time.                                                (2)
                                                                                                   [40]

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Electrical Technology                            12                            DBE/Feb.–Mar. 2015
                                          NSC – Memorandum

QUESTION 7: AMPLIFIERS

7.1       A 741 op-amp is an integrated circuit which is a circuit that consists of
          many electronic components situated on a silicon wafer chip. 
          The chip is placed in dual in line package which allows for connection in
          different modus.
          OR
          An op-amp is an integrated circuit that consists of many electronic
          components with a high voltage gain when connected to a DC-voltage source.                (3)

7.2       Infinite bandwidth means that the op-amp can amplify an input signal over a
          range of infinite frequencies.                                                           (2)

7.3       Open loop voltage gain is infinite 
          Input impedance is infinite
          Output impedance is zero
          Unconditional stability                                                                   (2)

7.4       7.4.1         Negative feedback is when a portion of the output signal is fed
                        back to the input signal.
                        It is 180º out of phase with the input signal. 
                        It is therefore subtracted from the input signal.
                        If not controlled negative feedback could attenuate the output to
                        zero.                                                                       (3)

          7.4.2         Positive feedback is when a portion of the output signal is fed back
                        to the input signal 
                        The feedback is in phase with the input signal
                        It is therefore added to the input signal thus increasing the output
                        signal value. 
                        if not controlled positive feedback could lead to run away, making the
                        circuit unstable.                                                           (3)

7.5       The bandwidth is increased. 
          The level of noise (hiss) is decreased 
          The gain is decreased
          The deformation of the input signal is reduced (Any TWO)                                  (2)

7.6       Output must show               OR
          Amplification 
          Same frequency                If open loop gain = ∞
          Correct phase 
          The input and output frequency
          must relate/be the same

                     V1
                                         Vout

                     V2
                                                                                                    (3)
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Electrical Technology                              13                              DBE/Feb.–Mar. 2015
                                            NSC – Memorandum

7.7       7.7.1
                            +V
                                                          Output must show
                            0V                            No inversion 
                                                          Square wave 
                            -V                            At saturation points 
                                                                                                        (3)

          7.7.2         Zero level detector 
                        Threshold detector                                                              (1)

7.8       7.8.1         Non-inverting op amp                                                           (1)

          7.8.2
                                                             Non inverted output must show
                                                             Amplification 
                                 Vn                Vout
                                                             Same frequency 

                                                                                                        (2)

          7.8.3                    Rf
                        AV = 1 +            
                                   Rin
                                   12 000
                           = 1+
                                    3 000   
                                            
                           = 5
                                                                                                        (3)

          7.8.4                Vout     
                          Av =
                                Vin
                        Vout = Av x Vin 
                             =5x 3
                                        
                             = 15 V
                                                                                                        (3)

          7.8.5         The value of Rf determines how much feedback is obtained. If the
                        value of Rf was decreased the amount of negative feedback
                        obtained is increased. 
                        The result is that the voltage gain of the circuit will be decreased.          (2)

          7.8.6         The inverting terminal is connected via R in to the 0 V rail  therefore
                        the value of R in sets the voltage at that input.                              (2)

7.9       As a buffer between stages to provide impedance matching from one stage
          to the other of an amplifier circuit.                                                        (2)

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Electrical Technology                               14                        DBE/Feb.–Mar. 2015
                                             NSC – Memorandum

7.10      7.10.1
                           +V
                                                           Output must show
                                                           Triangular waveform
                           0V                              +V
                                                           -V 

                           -V
                                                                                                   (3)

          7.10.2        R in and C form a time constant which controls the rate at which the
                        output voltage increases which determines the time the voltage
                        reaches saturation.                                                       (3)

7.11      7.11.1        Hartley oscillator                                                        (1)

          7.11.2        L = L1 + L2                   
                           = 0,035 + 0.035            
                           = 0,07 H                   

                                 1
                        Fr =                          
                               2π LC
                                        1
                           =
                               2π 0,07 x 0,47 x10−6   
                           = 877,89 Hz                
                                                                                                   (6)
                                                                                                   [50]

                                                                                    TOTAL:         200

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