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T500(E)(A5)TAPRIL EXAMINATIONNATIONAL CERTIFICATEELECTRO-TECHNOLOGY N3(11040343)5 April (X-Paper)09:00–12:00This question paper consists of 7 pages and 1 formula sheet of 3 pages.

(11040343)-2-T500(E)(A5)TDEPARTMENT OF HIGHER EDUCATION AND TRAININGREPUBLIC OF SOUTH AFRICANATIONAL CERTIFICATEELECTRO-TECHNOLOGY N3TIME: 3 HOURSMARKS: 100INSTRUCTIONS AND INFORMATION1.Answer ALL the questions.2.Read ALL the questions carefully.3.Number the answers according to the numbering system used in this questionpaper.4The correct information must be copied from the question paper andsubstituted for the correct symbol.5.Keep the subsections of questions together.6.Rule off after the completion of EACH question.7.Sketches and diagrams must be done in pencil.8.The sketches/diagrams must be neat, reasonably large and fully labelled.9.The answers must be worked to THREE decimal places.10.Use the correct units for answers.11.Write neatly and legibly.

(11040343)-3-T500(E)(A5)TQUESTION 11.11.2Choose the correct word(s) from those given in brackets. Write only theword(s) next to the question number (1.1.1–1.1.2) in the ANSWER BOOK.1.1.1The (field winding, yoke, pole shoes) is that part of the DC machinewhich protects the inner parts.1.1.2(Field poles, pole shoes, carbon brushes) are used to increase theefficiency of the magnetic path.(2)(2)Briefly explain the following methods to minimise the effects of armaturereaction.1.2.1Brush shifting(2)1.2.2Interpoles(2)1.2.3Increasing the field flux(2)[10]QUESTION 22.12.2State FOUR factors which the magnitude of an induced EMF in a conductordepends on.Name TWO generators which are dependent on the excitation process inorder to operate, and support your answer with two relevant sketches.Copyright reservedPlease turn over(4)(6)[10]

(11040343)-4-T500(E)(A5)TQUESTION 33.1State TWO important reasons for the decrease in terminal voltage of aseparately excited generator.(2)3.2Name ONE purpose of the separately excited generator.(1)3.3Briefly state ONE application for each of the following types of generator.3.3.1Shunt generator(1)3.3.2Series generator(1)3.4Name TWO variable factors that the torque of a DC motor depends on.3.5Name THREE applications of the series motors.(2)(3)[10]QUESTION 4A brake test was performed on a DC motor and the following information obtained:The drum radiusDrum speedEffective loadThe supply voltageThe current absorbed by the motor 300 mm420 rev/min425 N.m0,21 kV33 000 mADetermine the following:4.1Input power of the motor in kW(3)4.2Output power of the motor in kW(4)4.3The motor efficiencyCopyright reserved(3)[10]Please turn over

(11040343)-5-T500(E)(A5)TQUESTION 55.1The following ordinate points were read from the full cycle of an alternatingquantity.e 1 20 V; e 2 42 V; e 3 83 V; e 4 120 V; e 5 95 V; e 6 35 V; e 7 18 V.Determine the following from the above data:5.25.1.1What type of alternating quantity is mentioned above?(1)5.1.2Actual value(3)5.1.3Average value(3)5.1.4Form factor(1)5.1.5What type of wave form is deduced from the value of the crestfactor, if crest factor is 1,414 and form factor as calculated inQUESTION 5.1.4?(1)Define maximum value.(1)[10]QUESTION 6An RLC circuit consists of a 400 mH inductor, a resistor of 10 Ω and a 50 mFcapacitor. The circuit is connected in series across a 240 V/60 Hz supply.Determine the following:6.1The impedance of the circuit(5)6.2The circuit current(2)6.3The phase angle and state whether it is leading or laggingCopyright reserved(3)[10]Please turn over

(11040343)-6-T500(E)(A5)TQUESTION 77.1State TWO advantages of a star connection.(2)7.2A 380 V, three-phase, star-connected motor is rated at 25 kW. The full loadpower factor is given as 0,8 and the efficiency as 85%.Determine the following:7.2.1The line voltage for the motor when it runs at full load.(1)7.2.2The phase voltage for the motor when it runs at full load.(2)7.2.3The phase current for the motor when it runs at full load.(5)[10]QUESTION 88.1What is the colour of silica gel after it absorbs moisture?(1)8.2Name TWO sources of losses that occur in a transformer.(2)8.3A single-phase transformer has 42 turns on the secondary winding and isconnected to a 210V AC supply. The output voltage is 70V and the primarycurrent is 218 mA.Determine the following:8.3.1Primary number of turns(2)8.3.2Secondary current in amperes(2)8.3.3Secondary VA if ALL losses are ignored.(1)[8]QUESTION 99.19.2Draw a neat labelled sketch of a dynamometer as an electrical measuringinstrument.Name THREE basic mechanisms which are found in measuring instruments.Copyright reservedPlease turn over(7)(3)[10]

(11040343)-7-T500(E)(A5)TQUESTION 1010.110.210.3Draw and label the following gates by its IEC symbols.10.1.1AND gate(2)10.1.2NOR gate(2)Change the following decimal numbers to binary and show ALL necessarysteps.10.2.110,5 10 .(2)10.2.214,25 10 .(2)10.2.3Subtract the answer of QUESTION 10.2.1 from QUESTION 10.2.2and leave the answer in binary number.(2)Briefly explain with the aid of a neat sketch the concept of forward bias.TOTAL:Copyright reserved(2)[12]100

(11040343)-1-T500(E)(A5)TELECTRO-TECHNOLOGY N3FORMULA SHEETAny applicable formula may also be used1.E V - I a Ra2.E V I a Ra3.E 2pФ4.N VK 5.T 0,318I a Zp C6.Efficiency/Rendement VI 100%VI I a Ra I sV C7.Efficiency/Rendement VI ( I a Ra I sV C ) 100%VI8.Efficiency/Rendement 2 N (W S )r 100%60VI9.Efficiency/Rendement 10.E Blv11.e E m Sin2цft12.i I m Sin2цft13.e ave / gem or/of i ave / gem 0,637 E m or/of I m14.e rms / wgk or/of i rms / wgk 0,707 E m or/of I mZN60c2215.I1 100%I1 I 2e1 e2 e3 e4 . enni1 i2 i3 . inOr/of I ave / gem nE ave / gem

(11040343)-2-e1 e2 e3 . enn216.E rms / wgk 222i i2 i3 . inOr/of I rms / wgk 1n22217.Form factor / Vormfaktor 18.Crest factor/Kruinfaktor 19.I 20.X L 2цfL;i VXL21.X C 2цfC;i VXC22.Z 23.Tan 24.V R I x R;25.Z 26.Tan θ 27.P V x I;28.P VICosθ29.Cos θ 30.I active / aktief ICos ;31.P VI Cos θT500(E)(A5)T2E rms / wgkor/ofE ave / gemEmor/ofE rms / wgkI RMS / WGKi AVE / GEMImI rms / wgkVRR2 X L ;2Z R2 X C ;2I VZXXL; Tan θ CRRV L I XL; VC I XCR2 ( X L X C )2 ;Z R 2 ( X C X L ) 2XL XCX XL; Tan θ CRRP I2R;P V2RRWor / ofkW; Cos θ ZVAor / ofkVAI reactive / reaktief ISin Q VI Sin θ

(11040343)-3-T500(E)(A5)T132.fr 33.I IR IL ;Tan θ ILIR34.I I R I C ; Tan θ ICIR35.I I R (I L I C ) 2 ;Tan θ I L ICIR36.I I R (I C I L ) 2 ;Tan θ IC I LIR37.Cos θ 38.V L V p ; I L 3I p39.V L 3V p ;40.W 41.V1 N1 I 2 V2 N 2 I 142.kVA 43.V shunt / sjunt Vmeter ; I s Rs I m Rm44.IT Im Is45.It 2 LC222222VtRtCopyright reservedIRIIL Ip3VL I L Cos 3VL I L1 000

MARKING GUIDELINENATIONAL CERTIFICATEAPRIL EXAMINATIONELECTRO-TECHNOLOGY N35 APRIL This marking guideline consists of 11 pages.

MARKING GUIDELINE-2ELECTRO-TECHNOLOGY N3T500(E)(A5)TQUESTION 11.11.1.11.1.2Yoke Pole shoes (2)(2)1.21.2.11.2.21.2.3 By moving brushes backwards in the motor. By moving brushes forwards in the generator. (2) Interpoles are smaller poles placed between the main poles. Connected in series with the armature and must have the samepolarity as the main poles - passed in the motor – to ensuresparkless commutation. (2) By making use of series winding on the main field poles. Varying the main field to the load condition. (2)[10]QUESTION 22.12.2 The number of pairs of poles used. The strength of the magnetic field or flux. The rate at which the magnetic flux is cut by the moving conductor. The number of active conductors connected in series. Separately excited generator. Self-excited generator(4)(1) For correct labelled sketch(2) (1) For correct labelled sketchCopyright reserved(2)[10]Please turn over

MARKING GUIDELINE-3ELECTRO-TECHNOLOGY N3T500(E)(A5)TQUESTION 3 Effective field flux is reduced due to armature reaction as the loadincreases. Voltage drop due to the armature circuit resistance. (2)3.2Used as the generator in the Ward-Leonard motor generator system. (1)3.33.3.13.13.3.23.43.5Shunt generator – it is used where a constant voltage is required. (1)Series generator – as a booster on DC transmission line. (1) Flux (Ф) Armature current (Ia) Driving crane Train Hoists LiftsTrolley busesElectric vehicleCopyright reserved (2)(Any 3 x 1 )(3)[10]Please turn over

MARKING GUIDELINE-4ELECTRO-TECHNOLOGY N3T500(E)(A5)TQUESTION 4Given: R 300 mm 0,3 m;V 0,21 kV 210 V;4.14.24.3N 420 r/min; effective load 425 N.mI 33 000 m A 33AInput Power [P] IV 33 A x 210 V 6 930 W 6,93 kW Answer(3)2 x NWr602 3,142 420 425 0,3 60 5608,47 W 5,609 kW AnswerOutput Power [P] Output 100%Input5,609 kW 100% 6,93 kW 0, 80938 x 100% 80,938 % Answer(4)Efficiency Copyright reserved(3)[10]Please turn over

MARKING GUIDELINE-5ELECTRO-TECHNOLOGY N3T500(E)(A5)TQUESTION 5Given: :- e 1 20 V; e 2 42 V; e 3 83 V; e 4 120 V; e 5 95 V; e 6 35 V; e 7 18 V.5.15.1.1Alternating voltage. 5.1.2(1)e 1 e 2 e 3 e 4 e5 e 6 e 7n2Actual [E RMS ] 222222202 422 832 1202 952 352 182 734 027 7 4 861 69.721 V Answer 5.1.3Average Value [E AVE ] (3)e1 e2 e3 e4 e5 e6 e77 20 42 83 120 95 35 18 7 413 7 59 V Answer5.1.4Form factor E RMSE AVE 69,721V 59V 1,1825.1.55.2 Sine wave Peak wave Sinusoidal wave(3)Answer (1)Maximum value – is the maximum or peak value of an alternating voltage orcurrent. Copyright reserved(1)(Any 1 x 1)(1)[10]Please turn over

MARKING GUIDELINE-6ELECTRO-TECHNOLOGY N3T500(E)(A5)TQUESTION 66.1XC X L 2 fL12 fC12 3,142 60 50 10 6 53,045 Ω Answer 2 x 3,142 x 60 Hz x 400 x 10 3 150,816 Ω AnswerImpedance of the circuit [Z] R2 ( X L X C )2 102 (150,816 53,045) 2 100 9559.168 98,281 Ω Answer6.26.3VtZ240V 98,281 2,442 A (5)Circuit current [I t ] Phase angle: Tan θ Answer(2)X L XCR150,816 53,045 10θ Tan 1 9,777 84,160º lagging Copyright reservedAnswer(3)[10]Please turn over

MARKING GUIDELINE-7ELECTRO-TECHNOLOGY N3T500(E)(A5)TQUESTION 77.1 With a star connection two voltages are available, namely V L and V ph . By earthing the neutral, earth leakage protection is simplified. 7.27.2.1V L 380 V .7.2.2VL (Given)(1)3VPH380 V 3 219,393 V V ph 7.2.3(2)Input power Output Answer(2) 10025 000 100 85 29 411,765 W Pin 3 VL I L Cos 29 411,765 WIL 3 380 0.8 55,858 AAnswer Therefore I C I ph (In star connection) I ph 55.858 A Copyright reservedAnswerAnswer(5)[10]Please turn over

MARKING GUIDELINE-8ELECTRO-TECHNOLOGY N3T500(E)(A5)TQUESTION 88.1Pink 8.2 Winding Core8.38.3.18.3.28.3.3(1) (2)210V 70V 126 turns Primary number of turns[N 1 ] 42 V1 I 2 V2I1210V 0,218AI2 70V 0,654 A Answer(2)Answer(2)Secondary volt-ampere VI 70 Volt x 0,654 A 45,78 VA Answer(1)[8]

MARKING GUIDELINE-9ELECTRO-TECHNOLOGY N3T500(E)(A5)TQUESTION 99.1FOUR marks for ANY RELEVANT correct labellingTHREE marks for correct sketch.9.2 A deflecting device A controlling device A damping device Copyright reserved(7)(3)[10]Please turn over

MARKING GUIDELINE-10ELECTRO-TECHNOLOGY N3T500(E)(A5)TQUESTION 1010.110.1.1 for correct diagram and labelling(2)10.1.2(2) for correct diagram and labelling10.210.2.12 102 52 22 110,5 1010.2.2Copyright reserved01010,5 x 2 1 (2)Answer2 142 72 32 110.2.3remainderremainderremainderremainder 1010,1 2 10011,11 2 00,25 x 2 010,5 x 2 111 14,25 10 1110,01 2 (2)Answer(2)AnswerPlease turn over

MARKING GUIDELINE-11ELECTRO-TECHNOLOGY N3T500(E)(A5)T10.3 1 for correct sketch and 1 for labellingTOTAL:Copyright reserved(2)[12]100