Electrical Technology Grade 12 Eastern Cape Mock 2017 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 12 SEPTEMBER 2017 ELECTRICAL TECHNOLOGY MARKING GUIDELINE MARKS: 200 This marking guideline consists of 11 pages.

Downloaded from hlayiso.com 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] Copyright reserved Please turn over

Downloaded from hlayiso.com (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] Copyright reserved Please turn over

Downloaded from hlayiso.com 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] Copyright reserved Please turn over

Downloaded from hlayiso.com (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) Copyright reserved Please turn over

Downloaded from hlayiso.com 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] Copyright reserved Please turn over

Downloaded from hlayiso.com (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] Copyright reserved Please turn over

Downloaded from hlayiso.com 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) Copyright reserved Please turn over

Downloaded from hlayiso.com (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) Copyright reserved Please turn over

Downloaded from hlayiso.com 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) Copyright reserved Please turn over

Downloaded from hlayiso.com (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 Copyright reserved Please turn over