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MARKING GUIDELINE-1ELECTRO-TECHNOLOGY N3MARKING GUIDELINENATIONAL CERTIFICATENOVEMBER EXAMINATIONELECTRO-TECHNOLOGY N323 NOVEMBER 2016This marking guideline consists of 7 pages.Copyright reservedT440(E)(N23)T

MARKING GUIDELINE-2ELECTRO-TECHNOLOGY N3T440(E)(N23)TQUESTION 11.11.1.11.1.2Commutation – is the reversal of the EMF and currentü to theshort-circuited coilü during its transfer from one commutatorsegment to the next.ü(3)Pole pitch – is the angular distance between the centre of twoadjacent poles. ü(1)1.2(4 marks for 4 labels. 2 marks for correct sketch)(6)[10]QUESTION 22.12.22.3 Damaging brushes OR the commutator surface. To short circuit by formation of an arc.(2)Short compound motor – is when a series field coil is connected inseriesüwith a parallel connection of a shunt field coil and armature. ü(2)2.3.12.3.22.4Motor commutation – is to provide a difference in polarity betweenthe armature and the field in order to produce motion.(1)Generator commutation – is to produce a steady voltage whichremains constant.(1)Given: P 4; C 2; Ф 40 mWb 0,04 Wb; N 14 r/sec; E 120 VTotal number of conductors in armature ?2 pFZN60CE 60CZ 2 pFN120V 2 üü2 4 0,04 14 53,571üü or 54 conductorsE Copyright reservedAnswer(4)[10]Please turn over

MARKING GUIDELINE-3ELECTRO-TECHNOLOGY N3T440(E)(N23)TQUESTION 33.13.23.3 Friction loss Wind loss Iron loss(Any 2 x 1)(2) Speed [N] Flux [Ф]3.3.1(2)T c0,318 Z p I a980 2 4 üü0,318 660 4 400Flux [Ф] 0,023 Wb OR 23,346 mWbü3.3.2Answer(3)T c0,318 Z p I a980 2 üü0,318 660 4 400Flux [Ф] 0,00584 Wb OR 5,836 mWbüAnswer(3)[10]QUESTION 44.14.24.1.1When the current reaches overload magnitude, ü the heating coilwill heat the bimetal stripü to the extent that it will bend farenoughü to operate the tripping contacts. ü(4)4.1.2By the time taken for the bi-metal strip to bend.(1)4.2.1Oil4.2.2Overload coil4.2.3Oil dashpot4.2.4Piston4.2.5PlungerCopyright reserved(5)[10]Please turn over

MARKING GUIDELINE-4ELECTRO-TECHNOLOGY N3T440(E)(N23)TQUESTION 55.15.1.15.1.25.1.35.25.2.15.2.2210 V70 W 3 AüCurrent through Resistor [ I R ] (1)AnswerCurrent through Capacitor {I C ] 210 V 2 3,142 60 130 10 -6 ü 10,293 A üAnswer(2)210 Vü2 3,142 60 0,04 13,924AüAnswerCurrent through Inductor [I L ] R 2 ( X C - X L ) 2 70 2 (20.402 - 15.082) 2 üü 70,202 ΩüAnswerZ It (2)210Vü70,202 W 2,991 AüVtZ(3) Answer(2)[10]QUESTION 66.12222i1 i2 i3 .inampsnThe virtual value 10 2 20 2 36 2 38 2 28 2 14 2ü6100 400 1296 1444 784 196 ü6 703,333 ü 26,520 A üAnswer 6.26.3(4)i1 i2 i3 . inn10 A 20 A 36 A 38 A 28 A 14 A ü6146 ü6 24,333 A üAnswerThe average value I rmsI ave26,520 A 24,333A 1,089 ü(3)The form factor Copyright reservedAnswer(1)Please turn over

MARKING GUIDELINE6.46.5-5ELECTRO-TECHNOLOGY N3ImI rms40 26,520 1,508 üT440(E)(N23)TThe crest factor Answer(1)Peak wave or sine wave or sinusoidal wave. ü(1)[10]QUESTION 77.1Power VI70ü140 0,5 AüCurrent 7.2(2)AnswerVoltage IR140ü0,5 280 ΩüResistance 7.37.4VoltageCurrent200 V ü0,5 A 400 Ω ü(2)AnswerImpedance (Z) (2)AnswerThe inductance value of the circuit (L) ?Z R2 X L2X L 4002 - 2802 ü 285,657 Ω üX L 2цfL285,657WL ü2 3,142 60 0,758 H or 758 mHüCopyright reservedAnswerPlease turn over(4)[10]

MARKING GUIDELINE-6ELECTRO-TECHNOLOGY N3T440(E)(N23)TQUESTION 88.18.2 It delivers higher power. It generates a higher torque. A three-phase supply is more versatile when connected in star(V L and V PH ). In the case of alternators, the same size prime mover is required for bothsingle and three-phase.(Any 3 x 1)8.2.18.2.2The apparent power [S} 3 VI 3 x 380 V x 9 A 5 923,613 VAü 5,924 kVAüThe actual power [P] Answer(2)3 V L I L Cos q 3 380 9 0,85 ü 5 035,072 Wü 5,035 kWü8.2.3(3)The phase current of the motor winding [I ph ] (3)AnswerIL3Deltaconnection 9Aü3 5,196 AüAnswer(2)[10]QUESTION 99.1 9.29.2.1Copyright reservedMoving-iron InstrumentUneven scalePointer/NeedleBalance weightSpring Dynamo-meter InstrumentUneven scalePointer/NeedleBalance weightSpringAs a voltmeter:V6VRt t I t 0,006 A 1 000 WAnswerüR series Rt - Rm 1 000 Ω – 120 Ωü 880 ΩAnswerü(4)(3)Please turn over

MARKING GUIDELINE9.2.2-7ELECTRO-TECHNOLOGY N3T440(E)(N23)TAs an ammeter:Voltage across ammeterV m I m Rm 0,006 A 120 W 0,72 VAnswerüCurrent through shuntI s I t - I m 2 A - 0,006 A 1,994 ARs Vs 0,72 V I s 1,994 A 0,361 ΩAnswerü(3)[10]AnswerüQUESTION 1010.110.1.15 valence electrons(2)10.1.23 valence electrons(2)10.2Capacitor(1)10.3Motor-speed control; time-delay circuits; battery charger; heater control;relay circuit; regulated power supplies; static switches; phase controls(Any 5 x 1)TOTAL:Copyright reserved(5)[10]100