N3 ELECTROTECHNOLOGY SPOT ON R80 .00

goaoc000L0 | | | Copyright reserved higher education & training Department: Higher Education and Training REPUBLIC OF SOUTH AFRICA aper consists of 6 pages and 3 formula sheets. Please turn over Downloaded from hlayiso.com

(11040343) -2- TAGO(E)(J23)T DEPARTMENT OF HIGHER EDUCATION AND TRAINING REPUBLIC OF SOUTH AFRICA NATIONAL CERTIFICATE ELECTRO-TECHNOLOGY N3 TIME: 3 HOURS MARKS: 100 INSTRUCTIONS AND INFORMATION Answer ALL the questions. Read ALL the questions carefully. Number the answers according to the n paper. Keep subsections of questions together, for example QUESTION 1.1.1, 1.2, et cetera. 3 & The correct information substituted for the corrects “Use.the correct units for answers. rite.neatly and legibly. Copyright reserved Downloaded from hlayiso.com,..,;: turn over

(11040343) -3- TA60(E)(J23)T QUESTION 1 4.1 State THREE purposes of the yoke as applicable to the DC machine. (3) 4.2 Draw a sketch of a four- pole DC machine. Indicate the following components (7) on the sketch: pole core, flux, field winding and pole shoe. [10] QUESTION 2 2.41 Name the type of brush which is relatively cheap and is uged ) motors. low current density. ; (1) 2.2 2.2.1 Armature reaction (2) 2.2.2 Commutation (2) 2.3 optimal commutation. (2) 24 State THREE methods, e j@ one mentioned in QUESTION 2.3, which fe if the armature reaction. (3) [10] 4) ‘ture during the starting process. (2) 3.3 The torque of an eight-pole lap-wound DC motor is 320 N.m. If the armature current is 160 A and the number of armature conductors is 560, determine the useful flux per pole in mWb. (4) [10} Copyright reserved Please turn over Downloaded from hlayiso.com

(11040343) 4 T46O(E)(J23)T QUESTION 4 4.1 A shunt generator delivers an output voltage of 240 V at a current of 90 A. The armature and field resistances are 0,025 Q and 80 © respectively. If the no-load current is 3,09 A, determine the efficiency of the generator. (7) 4.2 Name TWO methods which are used in a friction test to calculat (2) 43 Give another name for friction test. (1) [10] QUESTION 5 5.1 The following ordinate points were read quantity: e,= 23 V;e,=45V;e, =22V. (1) Actuals (3) 5.1.3 Average (3) ~ Formfactot: ~ (1) - “of wave form is deduced from the value of the crest (1) (1) [10] A series circuit comprises of a coil having a resistance of 16 Q, an inductive reactance of 30 Q and a capacitive reactance of 45 Q. The circuit is connected across a 220 V, 60 Hz supply. Determine the following: 6.1 The impedance of the circuit (2) 6.2 The total current flowing in the circuit (2) Copyightresevea Downloaded from hlayiso.comyesss un over

(11040343) 5. T460(E)(J23)T 63 Current flowing through the capacitor (1) 6.4 Voltage drop across the inductor (2) 6.5 Draw a phasor diagram to show the relationship between the resistor, inductive reactance and capacitive reactance. (3) [10] QUESTION 7 7.1 efficiency of 85%. Calculate the line current of the motor in (4) 72 A460 KW, three-phase AC motor is-60 is 400 V if the power factor is 0,8 Determine the following: 7.2.4 Phase voltag: ‘in (2) 7.2.2 Phase.current (4) (10) QUESTION 8 8.1 \;, three-phase transformer has a delta-connected primary and a ine oc secondary. The input voltage is 3,3 kV and the secondary ine the following: The secondary phase voltage (1) 1.2 The maximum allowable secondary phase current (3) 8.1.3 The maximum power output at a power factor of 0,8 (3) 8.2 State the function of a conservator. (2) 8.3 Give the colour of silica ge! as it absorbs moisture. (4) (10) Copyright reserved ‘Please turn over Downloaded from hlayiso.com

(11040343) 6- T460(E)(J23)T QUESTION 9 9.1 An ammeter has an internal resistance of 20 © and a full-scale deflection when 30 mA flows through it. The instrument is used to measure 15 VDC. Show the necessary calculations to do the conversion from an ammeter to a voltmeter. 9.2 Explain mutual inductance as applicable to a single-phase sketch is not necessary as a part of the explanation. 9.3 Name TWO control mechanisms used in electrical instrumén QUESTION 10 10.1 What is the main purpose of horizo! al and veftical plates in a cathode ray? 10.2 Draw a fully labelled sketch. with THREE inputs using mechanical switches. . 10.3 State TWO applications of 10.4 Subtract the following bin: number: ; Copyright reserved 1111,0010.5-. TOTAL: Downloaded from hlayiso.com (4) [10] 100°

(11040343) a1 TAGO(E)(J23)T ELECTRO-TECHNOLOGY N3 FORMULA SHEET Any applicable formula may also be used. 1. B=V¥ -1yRq 2. BaV + [Ry 3. B= 2p0 2 60C 4 Net K® 5. 72 O18 Ig EPP Cc 6. Efficiency/Rendement = 3,“ * 10 V+ 14, Crmsnegh OPP dens wen = 707 Em or/of hn e+e, $e, ta te, 15. Egg = n i tip ti, +o.ti or/of I Foes hie ea nome ave/gem n Copyright reserved Please turn over Downloaded from hlayiso.com

(11040343) 2 T460(E)(J23)T 2 2 2 2 16 E Cp +es test. te rnsArgk n 2) 3? , 32 92 tp ti thy be ae or/of 7 ot} igh = raw wg 1” 17, Form factor/Vormfaktor = —8 orfof 2st ‘ave-gent aveigent Ln 18, Crest factor/Kruinfaktor = En orfof ros wk rms sgh =e. tay 9 = Xoo R y? 27.0 PaV¥xIl; P=PR,; P=— R 28. P=V¥ICos @ 2. Cos o-3 » Cos g = otot KW ~ VA orfof KVA 30. f, =I cos® auine-aktef =Isin@ | pactivernetie e Downloaded from hlayiso.com goo suis over Copyright reserved

-3- (11040343) T460(E)(J23)T 34, P=VICos 0 Q=VISin 0 i 32. em A 2nVLC 33. f= Vi+l Tan @ = 1 R 34, r= fh+h ; Tan ont R 35, PafR +, -he) : Tang = tte 36. 1=fi+d.-L,) ; Tand= mn oe =Vresee 2 TR. = Tn, nm + i, 45. [= ve R Copyright reserved Downloaded from hlayiso.com

higher education & training Department: Higher Education and Training REPUBLIC OF SOUTH AFRICA MARKING GUIDELINE NATIONAL CERTIFICATE AUGUST EXAMINATION ELECTRO-TECHNOLOGY N3 23 JULY 2014 This marking guideline consists of 9 pages. Copyright reserved Piease turn over Downloaded from hlayiso.com

MARKING GUIDELINE -2- T500(E)(J23)T ELECTRO-TECHNOLOGY N3 QUESTION 1 14 ¢ It forms the outer casing of the machine. e It supports the main field system. “ * It forms the magnetic circuit of the main poles. “ (3) 1.2 (3 marks for correct sketch) Ss Flux : Field winding ——-Pola shoe Pole core (4 marks for indicating parts and labelling) A FOUR-POLE DC MACHINEY (7) [10] QUESTION 2 2.1 Carbon brush “ (1) v 2.2 2.2.1 Armature reaction — is the distortion of the main flux entering and leaving the armature caused by the flux_set_up by armature conductors when they carry current, “ OR It is the effect of the armature ampere-turns upon the value and distribution of the magnetic flux entering and leaving the armature core. (2) 2.2.2 Commutation — is the reversal of the EME and current in the short” circuited coil during its transfer from one commutator segment to the next. (2) 2.3 Disadvantage of brush shifting: The brushes have to be moved each time the current changes. v~ (2) 2.4 Increase the field flux OR Increase the brush contact resistance. “ Interpoles ¥ Compensating winding (3) ° [10] Downloaded from hlayiso.com, .., 4:0. Copyright reserved

MARKING GUIDELINE -3- T500(E)(J23)T ELECTRO-TECHNOLOGY N3 QUESTION 3 3.1 Given: V = 230 V5 1, =45 A; R, = 250 mO = 0,250; © = 25 mWb = 0,025 Wb. K = 33,2.:N=? EMF =V-I,R, = 230 V— (45 x 0,25) Vv = 218,75 V v (One mark for correct answer and unit) Speed N = -£ kp = 218,757 33,2 x 0,025Wb = 263,564 r/min “ 4) 3.2 In order to limit a very high starting current from the supply; ¥ which can damage the armature windings, commutator and brushgear. ” (2) 3.3 Given: C = 2p; T = 320N.m; |, = 160A; Z = 560; p =4 = 0.3181, Zp¢ Cc = 320x2x4 0,318 x 160x 560« 4 = 0,0225 Wb ¥ : = 22,5 mWh (4) [10] T vv Copyright reserved Please turn over Downloaded from hlayiso.com

MARKING GUIDELINE | A. T500(E)(J23)T ELECTRO-TECHNOLOGY N3 QUESTION 4 44 4.2 43 Copyright reserved Given: Output voltage = 240 V; [= 90 A; R, = 0,0250;R,=800; ‘ | v= 3,09 A no No load losses = VI,,, = 240 x 3,09 =741,0W _» ¥ Shunt field loss = VI, But | ,,,= = =3A =240x3 =72QW v Full load armature losses = 1,” &, but (j=, +2, = (93) 0,025 = 90 +3 =216,225W v =O3A a Total losses = 7 41,6 + 216,225 + 720 = 1677,825W Output power = 240 x 90 W = 21600 W_» v Input power = 1 677,825 W +21 600 W =23277,825W ¥ Outpnt 199% OR Efficiency = ———!? __-x 100% Input Vel OR, +10 +L VY Ho y 21 600% “100% - 21600 W 23 277,825 21 600 + 216,225 + 741,6 + 720 = 92,79% v = 92,79 % —> ——b Rope brake “ Prony brake (2) Efficiency = (7) - Direct method ” (1) [10] Downloaded from hlayiso.com,..¢ 1s o.¢;

MARKING GUIDELINE 5- T500(E)(J23)T ELECTRO-TECHNOLOGY N3 QUESTION 5 5.1 Given: e,= 23 V; e,= 45 V; e,= 86 V; e, = 123 V; e,= 98 V; e,= 38 V, e, =22V. 5.4.4 Alternating voltage “ (1) 2 2 2 2 2 2 2 e: te, +e, te, te, +e, +e, 61.2 Actual [E ais] = | n = {23° +45 +867 +123? +98? +38? +22? v 7 36611, 7 = 5230143 = 72,320V ¥ —> (3) 5.1.3 Average value [E 4), ] = = (234454864123 +98 + 38+22)/ VA v = 435, 4 v = 62,143 V ¥ ee (3) €) Fe, +, +e, tes +e ef 7 = Ems / = 72320¥, 5.1.4 Form factor = “24s Ep "Aanasy v =1,164Vv (1) ie 5.1.5 Sine wave / peak wave / sinusoidal wave “ (1) 5.2 Maximum value — is the maximum or peak value of an alternating voltage or current. ” (1) [10] Copyright reserved Please turn over Downloaded from hlayiso.com

MARKING GUIDELINE T500(E)(J23)T QUESTION 6 -6- ELECTRO-TECHNOLOGY N3 6.1 Given: R= 16.0; X, =30 Q;X, = 40 Q;V, = 220 V/60 Hz. z= JR +X, -X,P = ¥16? +(45-30)' ¥ (2) 6.2 l= Ve, 220 " Z 21,9320 9 = 10,031A v (2) 63 |, = 1,. [connected in series] 1, =10031A ¥ (1) 6.4 Voltage drop across the inductor V, =1, x, but |,=J, =J, connected in series = 10,031Ax302V = 300,930 V ¥ (@) 6.5 Xi % R (Ko-X,) ¥ xX. 4 (3) [10] QUESTION 7 7A Given: Power = 76 kW; —_-V, = 380 V; Power factor = 0,84; Efficiency = 85% Power (P) = ¥3V,1,Cos0 x1 I= 76 000 Vv ' $3 «380 V x 0,84 x 0,85 = 161,723 AV = 0,162 kA v (4) Downloaded from hlayiso.com,... 1:9 over Copyright reserved

MARKING GUIDELINE ia T500(E)(J23)T ELECTRO-TECHNOLOGY N3 . 72 Given: Power = 160 kW; V , = 400 V; Power factor = 0,86 7.2.1 In delta: V,=¥, V,=400V V,=0,4 kV v , —_ (2) 7.2.2 Power (P) = V3 V,1,Cos@ | = 160x 10° 1 NB x 400 ¥ x 0,86 = 268,535A vv but], = VBI, in delta connection 268,535A4 | =e v ' v3 = 155,039A = 0,155 kA vw (4) [10] QUESTION 8 8.1 Given: Rating = 400 kVA; Delta-connected primary; Star-connected secondary; Input voltage = 3,3 kV; Output voltage = 380 V _ 380 8.1.1 Secondary phase voltage (V ,.] = B = 219,393 V v (1) 8.1.2 Maximum allowable secondary phase current: /,, = Fir L 400 000 se V3 x 380 = 607,737A ¥ —> (3) 8.1.3 Maximum power output at a power factor: P= 3 V,,L,,Cos? ORSxCos@ ¥ = ¥3x380x607,737x0,8 OR 400000x0,8 v = 319 999,93 W OR 320000 W v =320kW Answer OR 320 kW (3) 8.2 A conservator makes allowances for the expansion and contraction when temperature changes. v ” (2) 8.3 Silica gel turns pink as it absorbs moisture. : (1) {10] Copyright reserved Please turn over Downloaded from hlayiso.com

MARKING GUIDELINE -8- T500(E)(J23)T ELECTRO-TECHNOLOGY N3 QUESTION 9 9.1 9.2 9.3 Copyright reserved As a voltmeter: Rt Therefore: R,=R, +R Given: R,, = 20 Q;1,, =30mA=0,03 A; V, = 15V = I 0,034 =5002% series (500 - 20) QV =480QV__, (4) R series Two coils are brought together namely primary (input) and secondary windings (output). The supply is connected to input windings. The voltage forms a magnetic field between the primary and secondary winging. ¥ The variation of magnetic field induces an EMF of mutual inductance into the secondary winding. ~ There is no electrical connection between the two coils, only magnetic connection. ¥ (4) Gravily control “ Control spring (2) [10] Downloaded from hlayiso.com,,,. ,,,, 00,

MARKING GUIDELINE -9- T500(E)(J23}T Regulated power supplies Static switches Phase controls (Any 2 x dmark) (2) ELECTRO-TECHNOLOGY N3 QUESTION 10 10.1 Horizontal and vertical plates are used to control the path of the electrons ” (1) 10.2 ° A 8 o— —— c . ; Swi tches t “oe (3 marks for correct diagram) (3) OR gate with THREE inputs 10.3 e Motor speed control e Time delay circuits ° Battery charger e Heater control e Relay circult e o 10.4 1114,0010 41014,0114 0011,1011,—» v 0014,4011, =2'+2°+274+04+27% 4274 Vv =2+1+0,5 +) +0,125 + 0,0625 v = 3,6875,) “ (4) [10] TOTAL: 100 Copyright reserved Downloaded from hlayiso.com

higher education & training Department: Higher Education and Training REPUBLIC OF SOUTH AFRICA | peonoc000 ELECTRO-TEGHNOLOGY N3 (14040343) Me 2014 (Y-Paper) 13:00-16:00 r consists of 6 pages and 1 formula sheet of 3 pages. Copyright reserved Please turn over Downloaded from hlayiso.com

DEPARTMENT OF HIGHER EDUCATION AND TRAINING REPUBLIC OF SOUTH AFRICA NATIONAL CERTIFICATE ELECTRO-TECHNOLOGY N3 TIME: 3 HOURS | (11040343) -2- T460(E)(M26)T MARKS: 100 INSTRUCTIONS AND INFORMATION 1. Answer ALL the questions. 2. Read ALL the questions carefully. 3. Number the answers according to th num system used in this question paper. 4. Keep subsections of questions together, for example QUESTION 1. 1.4, 1.2, et cetera.. é ; 6. The correct inforttation must "be copied from the question paper and substituted for'the correct symbol. 7. Sketches and diagrams must be done in pencil and must be neat, reasonably large ahd fully labelled. ts to THREE decimal places after the comma. Copyright oseves Downloaded from hlayiso.comyess tin ove:

(11040343) -3- T460{E}(M26)T QUESTION 1 Choose ONE correct option from those given in brackets. Write only the correct option next to the question number (1.1-1.5) in the ANSWER BOOK, 1.4 The voltage drop in the load characteristics of a separately excited generator is due to (residual magnetism, armature reaction, hysteresis). 41.2 When the ampere turns of the two windings oppose each otter, the machine is said to be (cumulatively compounded, compound-connected, differs compounded). ue ° 1.3 Pole pitch is the (metres, distance, displacement) bé poles. : 1.4 1.6 (10] QUESTION 2 24 ' are dependent on the excitation process in order to operate. «Support the answer with TWO relevant sketches. (6) 2.2 State FOUR factors (4) [10] , labelled schematic diagrams to explain how the direction of the ing DC motors can be reversed. Shunt motor Compound motor (2 x 2) (4) 3.2 What is the purpose of a starter resistor which is connected in series with the armature? (2) 3.3 ADC motor has a full-load current of 45 A and an armature resistance of 0,65 ©. The supply voltage is 180 V. Determine the back EMF in kV. (4) [10] Copyright reserved Please turn over Downloaded from hlayiso.com

(11040343) ; 4- T460(E)(M26)T QUESTION 4 A brake test was performed on a DC motor and the following information was obtained: The effective radius = 300 mm Drum speed : = 420 r/min Effective load = 0,425 kN.m Supply voltage = 220V Current absorbed by the motor = 33 000 mA Determine the following: 41 Input power of the motor in kW 4.2 Output power of the motor in kW 43 The motor efficiency QUESTION 5 6.1 A resistor, an inductor and a capacitor are each in turn connected to a 220 V, 60 Hz supply (thatis not in parallel or series). Calculate the current flow in each of the following: 130{IF capacitor 51.3% 70,0 resistor he components in QUESTION 5.1 are connected in parallel, determine The total current flowing in the circuit The circuit impedance (3) (4) 3) (10) | (2) (2) (1) (2) [10] Copyightresevea Downloaded from hlayiso.comesse tun ove

(11040343) 5 QUESTION 6 7460(E)(M26)T An inductor is connected in series to an 80 W, 100 V, incandescent lamp so that it can operate from a 240 V, 60 Hz supply. Determine the following: 6.1 The current flow in the circuit 6.2 The resistance of the lamp 6.3 The impedance of the circuit 64 The inductance value of the circuit QUESTION 7 (2) (2) (2) (4) [10] 7.4 An alternating voltage of 230 V af frequen y'@f,90 Hz is applied to a circuit consisting of a resistor of 12 Q-anin étor of 0,02°H and a capacitor 70 pF connected in series. : : Determine the resonange frequéncy of the'circuit. 7.2 ‘Acsin nsformer has 350 windings on the primary coil spectively. Calculate the following: 6.1.1 The number of windings on the secondary side 8.1.2 The primary current 8.1.3 The rating of the transformer in VA 8.2 State the dual purpose of the oil-immersed transformer. 8.3 Name TWO iron losses that occur in the transformer. Copyright reserved Downloaded from hlayiso.com of (6) (2) [10] which is actéd to. a 210 V supply. The secondary voltage and current is 42 V and (3x2) 6) (2) @) [10] Please turn over

(11040343) -6- T460(E)(M26)T QUESTION 9 9.1 State ONE function of the following devices as applicable to electrical measuring instruments: 9.4.4 Deflecting torque 9.1.2 . Controlling torque 9.1.3 Damping torque 9.2 A moving-coil instrument gives a full-scale deflection if 28 ra ws through it. The resistance of the coil is 50 Q. a Determine the following: 9.2.1 Voltage across the meter 9.2.2 The resistance required'to use: 2 the instrument as an ammeter for a 1 A current flow ; 9.2.3 it as a voltmeter up to 12 V QUESTION 10 10.1 10.2 10.3 ‘the main advantage of a cathode-ray oscilloscope as comoared to other lectrical measuring instruments. Sta ONE component which is used in the filtering circuit. Draw a truth table of an AND gate with TWO inputs logic. TOTAL: Copyright reseved Downloaded from hlayiso.com (3) (3) [10] (1) (2) (2) (2) [10] 100

(411040343) -1- T460(E)(M26)T ELECTRO-TECHNOLOGY N3 FORMULA SHEET Any applicable formula may also be used. 1, E=V—-IgRa 2. EB=V+IgRg 3. E=2p® 2N 60C 4, = K® 5. SLEEL ZpP Cc 6. Efficiency/Rendement = offor i. = 0,637 Em orfof Im ave gens 14, Cronengh LF bens ge = 0,707 Em orlof im 8 Fete; tute, l 5. EF aveigem ~ n ip tip ti, +..ti i, — 1 2 3 or/of Vavergem = * n Copyright reserved Please turn over Downloaded from hlayiso.com

(11040343) 2 T460(E)(M26)T 2 2 _ fef tes tey tutes ‘raswgh = ris 3? al 42 32 +t tip tu.ti = a 2 3 - A OL/OF Tpsagt = rs E Z, 17. Form factor/Vormfaktor = —°""" or/of mst ‘ave gem ave geur 18. Crest factor/Kruinfaktor = En or/of fn rms. gh mason gk a Xe : Tan@= Kem ¥ R x, y 27.0 P= V¥KI;P=PR; Pam 28. P=VICos @ 7 " a 29, cos b= ® ; Cosg = VP KH Z VA or/of KVA 30. L ‘acne abver = 1 COS O if =[sin@ reactivelreaktie Copyright reserved Downloaded from hlayiso.comesss tun over

(11040343) 3 T460(E}(M26)T 31. P=VICos 0 Q=VISin @ 32. fra , 2nV LC 3. g-V+h : Tanga 43, =F, meter? LR, =1,R, any 44. dp =H, +1, 43. pate R Copyright reserved Downloaded from hlayiso.com

higher education & training Department: Higher Education and Training REPUBLIC OF SOUTH AFRICA MARKING GUIDELINE NATIONAL CERTIFICATE APRIL EXAMINATION ELECTRO-TECHNOLOGY N3 26MARCH 2014 This marking guideline consists of 8 pages. Copyright reserved Please turn over Downloaded from hlayiso.com

MARKING GUIDELINE -2- T460(E)(M26)T ELECTRO-TECHNOLOGY N3 QUESTION 1 14 Armature reaction 4.2 Differentially compounded 1.3 Distance 14 Lap-wound 1.5 Series-connected (5x2) [10] QUESTION 2 2.1 « Separately excited generator” e Self-éxcited generator” (2) x ae e SEPARATELY EXCITED GENERATOR (2) 4 req 1 raat ") field windings SELF-EXCITED GENERATOR (2) . Copyright sewed Downloaded from hlayiso.comagss. ius ove

MARKING GUIDELINE -3- * T460(E)(M26)T ELECTRO-TECHNOLOGY N3. 2.2 ° The number of active conductors connected in series. ¥ e The number of pairs of poles used.” e The strength of the magnetic field or flux.” ° The rate at which the magnetic flux is cut by moving conductors.“ ito} QUESTION 3 3.1 3.1.1 v eK I ~ ¥ SHUNT MOTOR (2) 3.1.2 v '. [= | (i | lab ; iv COMPOUND MOTOR (2) v v 3.2 It is to prevent/minimise/protect excessively high current from entering the starter. (2) 3.3 Given: |, =45A R, = 0,370 Vv = 180V EMF =V-1,R,% = 180 V=(45 A x 0,85 0) = 180 V~ 29,25 VY = 150,75 W¥ = 0,15 075 kV ORO, 151 KV Answer ; (4) [10] Copyright reserved Please turn over Downloaded from hlayiso.com

MARKING GUIDELINE 4. ELECTRO-TECHNOLOGY N3 QUESTION 4 Given: R = 300 mm = 0,3 m; N = 420 rimin = 7 r/s ; Effective load = 0,425 kN.m =425N.m; V=220V; 1=33A 41 Input power (P] = IV = 33A x 220VY = 7260 WY =7,26 kW ——+ Answerv ax “ 4.2 Output power [P] = — - 2x 3,142 x 7x 425x0,3 vv | 5608,47 WY 5,609 kW ——> Answer” Output 4.3 Efficiency = x 100% Input = 5,609 kW nom 7,26 kW = 0,7 726 x 100%Y = 77,26 % » Answer” QUESTION & 5.1 Given: V = 220/60 Hz 5.14 X_L=2x 3,142 x 60 x 40 x 10° = 15,082” = 220 15,082 = 14,587A__» AnswerY Xoa—e anfC = 4/2 x 3,142 x 60 x 130 x 10° = 20 401A 220 20,401 = 10,783 A> A ry e coprist exoney DOWNlGaded trom hlayiso. T46O(E)(M26)T (3) (4) (3) [10] (2) COMpase turn over

MARKING GUIDELINE ~5e ELECTRO-TECHNOLOGY N3 5.1.3 V, 220 =3,143A ___, Answer” 52 824 y LP eU, Ie)? = 314? + 04,587 — 10,783) ¥ = 24330 =4,933A ___» Answer” 822 7K. MY Ty 4,933 = 44,597Q——» Answer” QUESTION 6 Given: Incandescent lamp = 80 WW/100 V. Voltage supply = 240 V/60 Hz 61 Power = VI i= 227 100 =0,8 A ——» AnswerY = 300 Q ——» Answer” Copyright reserved Downloaded from hlayiso.com T460(E}(M26)T (3) (2) [10] (2) (2) Please turn over

MARKING GUIDELINE -6- T460(E)(M26)T ELECTRO-TECHNOLOGY N3 6.4 The inductance value of the circuit (L) =? Ze YR4X/ X= ¥300? -125? ¥ = 272.7170 —+> v¥ X,=2nfL = XxX Ls“. ev 2x 3,142 x60 272,117 sito 2x 3,142 x 60 = 0,723 H or 723mH __»AnswerY (4) [10] QUESTION 7 it 74 Resonance frequency [F ,] = ———— yIFiI 2TIViC = ly 2x 3.142 x ¥0,02 x 70x 10° _ 1 0.007435 = 134,493 Hz _____» Answery” (6) 7.2 e The power factor decreases when KVA increases. e Losses in the transmission lines increase (I? R).”” ° Larger cables are needed to carry a larger current. “ ° Switches, switchgear, transformer and instruments must be capable of : operation.” (Any 2 x 1) (2) 7.3 Capacitors” Synchronous machines” (2) : [10] QUESTION 8 8.1 Given: N, = 350 windings Input voltage = 210 V 8.144 yn. = 8h eZ = 350 x 42 y 240 = 70 windings ——» Answer’ (2) ° Downloaded from hlayiso.com, Copyright reserved lease lurn over

MARKING GUIDELINE 7- T460(E)(M26)T ELECTRO-TECHNOLOGY N3 8.1.2 Woh Vv, Z, = 42x8 v 210 =1,6A_, Answer’ (2) 8.1.3 Rating = V,x/, = 210x 1,6“ . = 336 VA __, Answer” (2) 8.2 e |t improves the insulation qualities of the insulation. “ * Itconducts the heat away from windings and core to the cooling surface.” (2) 8.3 e Hysteresis loss¥” e Eddy-current loss¥ : (2) {10] QUESTION 9 9.1 9.4.4 Deflecting torque — causes a deflection of the pointer from its zero position. ~ 9.1.2 Controlling torque — opposes the deflecting torque and increases with the deflection.“ 9.1.3 Damping torque — prevents oscillation of the moving system — enables its final position quickly.” (3x 1) (3) 9.2 Given:-1,, = 25 mA;R,, =509; |= 1A @21 V,F1, xR, =0,025Ax 509 =1,25V ——» Answer” (1) 9.2.2 Current through shunt resistor: Fl ody = 1-0,025 = 0,975 A——» AnswerY (1) 9.2.3 Resistance of shunt: r= 4 = 1,257 v 0,975 A = 1,282.0 ——» Answer’ ; (2) Copyright reserved Please turn over Downloaded from hlayiso.com

MARKING GUIDELINE -8- T460(E)(M26)T ELECTRO-TECHNOLOGY N3 9.2.4 As a voltmeter: ‘ L _ 12 0,025 = 4800 ¥ —> Reis = Re - Ray = (480 — 50) OY =4300 —» Answer’ QUESTION 10 10.1 10.2 10.3 10.4 10.5 Copyright reserved e Common emitter e Common collector e Common base 3 valence electrons it has the ability to display types of wave form that must be studied Capacitor AN AND GATE WITH TWO INPUTS A_|B_ | Output Qo |0 |0 v Oo /1 {0 1 {1 /14 1 0 |[0 v TOTAL: Downloaded from hlayiso.com (3) [10] (3) (1) (2) (2) (2) [10] 100

higher education & training Department: Higher Education and Training REPUBLIC OF SOUTH AFRICA HOLL LLELOe T510(E)(N14)T . NOVEMBER EXAMINATION NATIONAL CERTIFICATE ELECTRO-TECHNOLOGY N3 (11040343) 14 November 2013 (X-Paper) 09:00-12:00 -osThis question paper consists of 6 pages and 3 formula sheets. Copyright reserved Please turn over Downloaded from hlayiso.com

(11040343) 2. TS10(E\(N14)T DEPARTMENT OF HIGHER EDUCATION AND TRAINING REPUBLIC OF SOUTH AFRICA NATIONAL CERTIFICATE ELECTRO-TECHNOLOGY N3 TIME: 3 HOURS MARKS: 100 INSTRUCTIONS AND INFORMATION 1. Answer ALL the questions. 2. Read ALL the questions carefully. 3. Number the answers according to the numbering system used in this question paper. ae 8 4. Keep subsections of questions together. 5. Sketches and diagrams must be done in pencil. The sketches/diagrams must be neat, reasonably large and fully labelled. 6. The answers must be rounded off to. THREE decimal places. 7. Use the correct units for answers. 8. Write neatly and legibly. Copyight resover Downloaded from hlayiso.com), ..4 sin ove:

(11040343) -3- TS510(E}(N14)T QUESTION 1 1.1 Define the term commutation as applicabie to a DC machine. (2) 4.2 Name ONE method that is used to wind an armature of a DC machine. (1) 1.3 Name TWO components in a DC machine, which can be damaged due to poor commutation. (2) 1.4 Make a neat labeled sketch showing the compensating and armature windings and also the relative directions of the current flowing through them. (8) : [10] QUESTION 2 21 Name TWO generators which are dependent on the excitation process in order to operate. : : (2) 2.2 Make TWO relevant sketches fo cistinguish between the two generators named in QUESTION 2.1. : . on (4) 2.3 List FOUR factors that determine the 5 magnitude of the induced EMF ina conductor. (4) [10] QUESTION 3 3.1 An eight-pole DC’ machine has 360 armature conductors and it rotates at 11t/s. The magnetic flux is 0,013 Wb per pole. Determine the following: 3.4 4 The speed i in r/min (2) 3.4.2 “EMF generated if it is a lap-wound machine (3) 3.14.3 EMF generated if it is a wave-wound machine (3) ‘ Name TWO variable factors that the torque of a DC motor is dependent upon. (2) oe [10] QUESTION 4 41 Briefly explain the operation of the bimetai-strip type overload as used in motor starters. (4) Copyright reserved Please turn over Downloaded from hlayiso.com

(11040343) -4- TS10(E)(N14)T 42 Label FIGURE 1 below. Write only the answer next to the letter (A-F) in the ANSWER BOOK. A he ro) —*——B ro) ro) ra) - Lo col +——_C E * D FIGURE 1 (6) [10] QUESTION 5 During a brake test performed on a DC motor, the effective load was 0,425 KN at a drum.speed of 420: rpm. The effective diameter of the drum is 0,6 m and the input to the motor is 33 000 mA at a supply voltage of 220 V. Determine the following: 5.4 Output power of the motor in kW (4) 6.2 Input power of the motor in kW (3) 5.3 The motor's efficiency (3) [10] Copyight resevea Downloaded from hlayiso.cony,,,. nove:

(11040343) -5- T510(E)(N14)T QUESTION 6 The following ordinate points were read from the full cycle of an alternating quantity: (i) = 22 A; (i,) = 44. A; (1,) = 85 A; i) = 122 Al (i,) = 97 Al i.) = 37 AS ,) = 20A Determine the following from the data given above: 6.1 The type of alternating quantity stated above 6.2 Actual value (4) 63 Average value (3) 6.4 Form factor (1) 6.5 The type of wave form deduced from the value of the'crest factor, if the crest factor is 1,414 ee a (1) VEE [10] QUESTION 7 7.4 An 80 kW, 400 V, three-phase alternating-current motor has a power factor of 0,86 and an efficiency of 87%. : Determine the line current of the motor in kA. (4) 7.2 A 200 kW, three-phase, AC’ motor is connected in delta and the supply voltage is 440 V.: Determine the following if the power factor is 0,9: 7.2.1¢ ° Phase voltage in kV (2) 2:«.... Phase current of the motor in mA (4) ES [10] QUESTION 8 8.4 “Give FOUR differences between a single-phase transformer and an auto- transformer. Tabulate your answer. (8) 8.2 A single-phase transformer has 49 turns on the secondary winding and is connected to a 240 V supply. The output voltage is 80 V and the primary current is 0,218 A. Determine the primary number of turns. (2) [10] Copyright reserved Please turn over Downloaded from hlayiso.com

(11040343) -6- T510(E)(N14)T QUESTION 9 9.1 A galvanometer has a resistance of 60 0 and gives a full-scale deflection when 3 mA passes through. Calculate: 9.1.4 _ The series resistance required to use it as a voltmeter up to 3 V. (3) 9.1.2 The resistance required to use the instrument as an ammeter for a (3) 2A current flow. 9.2 List FOUR components which are found in a moving-iron dynamo- =meter instrument. (4) [10] QUESTION 10 10.1 Explain with the aid of a neat sketch the term forward bias. (2) 10.2 Change the following decimal numbers to binary and show ALL the necessary steps: 10.2.1 8,510 10.2.2 "4,251 10.2.3. Add the answers of QUESTION 10.2.1 and QUESTION 10.2.2 and leave the answer in binary number. (3 x 2) (6) 10.3 Draw the IEC symbols for the following logic gates: 10.3.1 NAND gate (1) 10.3.2... OR gate (i) [10] TOTAL: 100 - Copyight eevee DOwnloaded from hlayiso.com

(11040343) 4 ELECTRO-TECHNOLOGY N3 FORMULA SHEETS Any applicable formula may also be used. 1. E=V-IgRy 2, E=V+IgRq 3. E= 2p ZN. 60C A No K® 5. 7p 0:318 1a Zp Cc . VI 6. Efficiency/Rendement = z VI + da’ Ra + Is¥.+C . _ Ap? 7. Efficiency/Rendement = Vi = (at Ra - £ Ox 100 8. Efficiency/Rendement = 2a (W = St" x 100 9. Efficiency/Rendement = 10. E= Biv ll e= B,ySin2aft 12. a= 1, Sindaft 13, i - Cnveigein POV i specgem = 9,637 Em orfof im 14. Crnereak OFF inne wen =O 707 Em orfof Im $e, FO, te tly avelgem ~ 15. E n iti, tis tet, avesgem or/of 1 n Copyright reserved T510(E)(N14)T Please turn over Downloaded from hlayiso.com

(11040343) 2. T510(E)(N14)T 2 2 2 2 tes +e; +... 42% od 16. ‘ymas gk pe 32 2 32 ; P+i bi bth or/of Lnssvge = ne 17. Form factor/Vormfaktor = Ge ovlof aimee ‘ave gent ave gem 18. Crest factor/Kruinfaktor = En or/of Joy Ennsingh r pais raising 19, l= ¥ R Vv 20. X= 2nfk ; i= 1 = auf x, 2i. Xe = ; pa 2c Xc 22, Z=JR' 4x) ; Z=fR4X2 ; i=5 23. Tan@ = Xb : Tan 22S. R R 0 VR=IXR; Vi =IxX, 5 WHI KX, 2%. Za JRA, AXP 2a LR RY 26. Tan 0= Aya Xe ; Tan@ = Ae =X, R R : y3 2% P=¥KI; P=PR; Pa R 28. P=VICos @ 29. cose=* ; sos 9 = Hort KW zZ VA orlof KVA 30. Laaweatny = 1 00S 8 I =Isin 8 reactivereaktio€ Downloaded from hlayiso.cory,,.,,, .,.. Copyright reserved

(11040343) 3. 31. P=VICos @ Q=VISin 8 32. foe , aavLC L 33, Taj+l, s Tan 9 => Rx 34, feJfhen ; Tang =“ H 35. P=Ji+d,-Ie) ; Tand= <1 36. I= fh+de-L/ : Tan = "8 t 37. Cos 0 = Ja I 38. ¥o=¥, ; 1, -V3ip 39. ¥,=V3¥p ; 1, =1, 40. W=V3V1, CosOxa gy “eMeok mH oN, v4 3M 1000 43. Fan sina = Vncrer 7 ER = lB a 1, +1, 4. pat R, Copyright reserved Downloaded from hlayiso.com TS10(E}N14)T

higher education & training Department: Higher Education and Training REPUBLIC OF SOUTH AFRICA MARKING GUIDELINE NATIONAL CERTIFICATE NOVEMBER EXAMINATION ELECTRO-TECHNOLOGY N3 14 NOVEMBER 2013 This marking guideline consists of 7 pages. Copyright reserved Please turn over Downloaded from hlayiso.com

MARKING GUIDELINE -2- TSIO(E}(IN14)T QUESTION 1 ELECTRO-TECHNOLOGY N3 1.4 COMMUTATION is the reversal of the EMF and current, ¥ in a short-circuited coil, during its transfer from one commutator segment to the next. ¥ (2) 4.2 Lap winding or wave winding ¥ (1) 1.3 e Brushes “ e The commutator surface ¥ (2) 14 Compensating windings ¥ Armature windings ¥ Correct direction “ | Sketch ¥¥ (5) [10] QUESTION 2 21 * Separately excited generator ° Self-excited generator” (2) 2.2 SEPARATELY EXCITED GENERATOR - J Armature Pleid i ° . T sketch vv) Copyright reserved SELF-EXCITED GENERATOR { A é theostat 2) (om) field windings ; : Sketch VY (2) Downloaded from hlayiso.com ss. iis over

MARKING GUIDELINE 3. ELECTRO-TECHNOLOGY N3 2.3 The strength of the magnetic field ¥ The number of pairs of poles used ¥ The number of active conductors connected in series¥” T510(E)(N14)T The rate at which the magnetic flux is cut by the moving conductor” (4) [10] QUESTION 3 3.4 Given: p=4 Z = 360 conductors N = 11 rev/sec ® = 0,013 Wb 3.1.1 The speed in r/min (N] = speed x 60 =411x60 ¥ = 660 r/min ¥ (2) 3.1.2 EMF generated = 2ZpNb _2x360x4x 660 0,013 |, C60 2x4x 60 =5148V Vv (3) 3.1.3 EMF generated = 2ZpN¢ _2% 360.4 x 660 0,013), C60 2x60 = 205,92 VY (3) 3.2 e Armature current [I,,] © Flux per pole [>] an QUESTION 4 44 When the current reaches overload magnitude the heating coil will heat the bimetal strip” to the extent that it will bend far enough ¥to operate the tripping contacts ” (4) 4.2 A Trip-out mechanism” B Overload coil” Cc Oil dashpot” D Piston” E Oil” F Plunger” (6 * 1) (6) [10] Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE A. TS1O(E}(N14)T ELECTRO-TECHNOLOGY N3 QUESTION 5 Given: Effective load = 0,425 diameter = 0,6 m 1= 33000 =33A KN = 425N radius =0,3m V=220V N = 420 r/min 51 Output power of the motor = a wy _ 2x3142 x ~ A425 x03 yy = 5608.47 W ¥ = 5,609 kW (4) 5.2 input power of the motor = IV = 33 Ax 220 V ¥ = 7260 W ¥ = 7,26 kW Y (3) 5.3 Efficiency of the motor {n] = Output x 100% = 3609 x 100% ¥ input 7260 = 0°77259 x 100% 8) = 1% v 77,259% (10) QUESTION 6 6.1 Alternating current “ (1) 27 2 2 +2 i 4d +h tut i, 6.2 Actual Value = ,|J———--—_--—. v Hn . jaa +44? +85? +122? +97? +37? +207 | 7 =, 2 =V5101 % =71,421AV (4) 6.3 Average value [I,,] = 22 = 22A+44A + 85A 41224 497A 4+37A+20A v 7 221, 7 261A” (3) 6.4 Form factor = Fron _ Thd2dA Lave G1A =1471 ¥ (1) Downloaded from hlayiso.com esse ius over Copyright reserved

MARKING GUIDELINE 5- T510(E)(N14)T ELECTRO-TECHNOLOGY N3 65 Sine wave or peak wave or sinusoidal wave “ (1) [10] QUESTION 7 71 Line current (1, }= P = 80000 vv ¥3V,Cos0n V3 x 400V x 0,86 x 0,87 = 80000 518,368 = 154,331A Vv ; = 0,154 kA v (4) 7.2 Given: P = 200 kW; Delta connection; V ,= 440 / ; power factor = 0,9 7.2.1 Phase voltage = line voltage (Delta connection) = 440 Vv =O4dkV ¥ (2) 7.2.2 l= P _ 200000 v BY, xCos@ 1,732 4407 x09 = 291,599 Av But 1,=V3/,» 291,599A |, 224 PH V3 = 168,359 A (4) = 168 359 v : 6 mA [16] QUESTION 8 8.4 SINGLE-PHASE TRANSORMER AUTO-TRANSFORMER e Mutual induction’ e Self induction’ e No electrical connection” ° Are connected electrically” ° Step up or down” « Step down only” e Primary and secondary windings ° Primary and secondary have are apartY the same winding” (8) 8.2 Primary number of turns: i aM 2 2 = ae 49 v 2 = 147 turns ¥ 40) Copyright reserved Please turn over Downloaded from hlayiso.com

MARKING GUIDELINE -6- ELECTRO-TECHNOLOGY N3 QUESTION 9 9.4 Given: R ,,= 60 0251,, = 31 A = 0,003 A 9.1.4 As a voltmeter: V, 3¥V “7, 0,003 4 =1000Q v Roos Ry ~ Ry = 1000 O-60 OV =9400¥ 9.1.2 As an ammeter: Voltage across ammeter V = 1,R, = 0,003 A x 60.2 mn =O18V v Current through shunt: |,=1,-J,, =2 A-0,003 A T510(E)(N14)T =1,997A ¥ _h, _ 018V “7, 1,997A =0,090Q v (3) 9.2 MOVING-IRON DYNAMO-METER INSTRUMENT INSTRUMENT Uneven Scale Uneven scalew Pointer/Needle Pointer/NeedleY Balance weight Balance weight” Spring SpringY (4) [10] QUESTION 10 10.1 Movement of charge carriers (Current flows) Diagram” (2) FORWARS BIAS CONCE Copyright reserved own oaded from hlayiso.com,,......, over

MARKING GUIDELINE -T ~ TS10(E)(N14)T ELECTRO-TECHNOLOGY N3 10.2 10.2.1 > 8 remainder O 0,5x2=1 | 2 | 4 remainder 0 0,0x2=0 2 | 2 remainder 0 2 | 1 remainder 1 Showing steps” 8,5,,=1000,1, 10.2.2 0,25x2=0 O5x2 =14 | 2 | 4 remainder 0 2 | 2 remainder 0 2 | 1 remainder 1 Showing steps” 4,25,,= 100,01, ¥ 10.2.3 1000,1, +100,01, 1100,11, ¥ (3 2) 6) 10.3 10.3.1 v A c B ° a NAND gate (IEC symbol) (1) 10.3.2 A A\p- eo B nn v OR gate (IEC Symbol) (1) [10] -TOTAL; 100 Copyright reserved Downloaded from hlayiso.com

higher education & training Department: . Higher Education and Training REPUBLIC OF SOUTH AFRICA T540(E)(J27)T AUGUST 2012 NATIONAL CERTIFICATE ELECTRO-TECHNOLOGY N3 (11040343) 27 July (X-Paper) 09:00 — 12:00 Calculators may be used. This question paper consists of 7 pages and a 3 page-formula sheet. DHET Copyright reserved Please turn over Downloaded from hlayiso.com

2. TS40(E)(V27)T DEPARTMENT OF HIGHER EDUCATION AND TRAINING REPUBLIC OF SOUTH AFRICA NATIONAL CERTIFICATE ELECTRO-TECHNOLOGY N3 TIME: 3 HOURS MARKS: 100 INSTRUCTIONS AND INFORMATION 10. 11. Answer ALL the questions. Read ALL the questions carefully. Number the answers correctly according to the numbering system used in this question paper. Keep subsections of questions together. The correct information must be copied from the question paper and substituted with the correct symbol. Rule off after the completion of each question. Sketches and diagrams must be done in pencil. The sketches/diagrams must be neat, reasonably large and fully labelled. The answers must be rounded off to THREE decimal places after a comma, for example 3,142. Use the correct units for answers. Write neatly and legibly. DHET Copyright ese OWNIoaded from hlayiso.cons wn ove:

3. 7540(E)(J27)T QUESTION 1 1.4 Choose the correct word(s) from those given in brackets. Write ‘only the word(s) next to the question number (1.1.1 — 1.1.3) in the ANSWER BOOK. 4.1.4 The (field winding/yoke/pole shoes) is/are that part of the DC machine which protects the inner parts. 1.1.2 (Field poles/Pole shoes/Carbon brushes) increase the efficiency of the magnetic path. 1.1.3 (Laminated plates/Thin micas/Wrought irons) are used to reduce eddy currents. 1.2 List TWO methods which are used to wind the armature of a DC machine and give ONE function of each. QUESTION 2 2.1 Name ONE disadvantage of the brush-shifting method in order to achieve optimal commutation in a DC machine. 2.2 Indicate THREE methods which can be employed to counter the effects of armature reaction, (Except the one mentioned in QUESTION 2.1.) 2.3 Which TWO main properties of carbon brushes make them suitable for use in a DC machine? : 24 List THREE classifications of generators according to the type of field-winding used. QUESTION 3 3.1 An 8-pole DC motor is connected to a 480 V supply and takes an armature current of 64 A. The resistance of the armature circuit is 0,6 0. The armature is wave wound with 520 conductors and the useful flux per pole is 44 mWb. Determine the following: 3.1.4 The torque developed by the armature in KN.m. 3.1.2 The back EMF of the motor in kV. 3.2 Name THREE components of a DC motor starter that can be affected by a high starting current if it is not limited. DHET Copyright reserved Please turn over Downloaded from hlayiso.com (2) (2) (4) [10] (2) (2) (3) [10] (3) (3) [10]

-4- T540(E)(27)T QUESTION 4 41 Tests were performed on two identical 380 V, 190 kW generators. The current between the machines is equal to the full-load current and 65 Ais taken from the supply. Determine the approximate efficiency of each machine. 4.2 Briefly explain the term embedded brushes as applicable to direct-current machines. QUESTION 5 5.1 Define the term period as applicable to alternating current. 5.2 The following ordinate points were read from the full cycle of an alternating quantity: e , = 20V,e, = 42 V;e3=83V;e,=120V,e5= 95 Vi e6536V; e,=18V. Determine the following from the above data: 5.2.1 _ What kind of alternating quantity is mentioned in the given points? 5.2.2 Average value 5.2.3 Actual value 5.2.4 Form factor 5.2.5 What type of wave form is deduced from the value of the crest factor, if the crest factor is 1,414? Also refer to the result of QUESTION 5.2.4. Downloaded from hlayiso.cgnj.,,, ..«, DHET Copyright reserved (7) (3) [10] (1) (1) (3) (3) (1) (1) [10]

QUESTION 6 Refer to FIGURE A below and determine the following: X, =62,830 Q A TS40(EKI27)T Vv X_ =31,8302 FIGURE A 6.1 Impedance of the circuit 6.2 Total current flow in the circuit 6.3 The phase angle 64 Voltage across the resistor DHET Copyright reserved > Vg =250V/50 Hz (3) (2) (3) [10] Pleas: Downloaded from hlayiso.com

6 T540(E)(J27)T QUESTION 7 A coil is connected in series with a 75 watts, 140 V incandescent lamp so that it can operate from a 230 V, 60 Hz alternating-current supply. (HINT: Resistance of the coil is negligible.) Determine the following: 7A Current flow in the circuit 7.2 Resistance of the lamp 7.3 The total impedance of the circuit 7.4 The inductance of the coil QUESTION 8 8.1 A 100 kW 440 V three-phase alternating-current motor has a power factor of 0,82 and an efficiency of 88%. Calculate the line current of the motor in kA. 8.2 A 180 KW three-phase alternating-current motor is connected in delta and the supply voltage is 400 V. If the power factor is 0,8 determine the following: 8.2.4 - Phase current of the motor in mA 8.2.2 Phase voltage in kV QUESTION 9 9.1 Name THREE mechanisms which are found in electrical measuring instruments. 9.2 A single-phase transformer has 300 windings on the primary coil which is connected to a 400 V supply. Determine the following: 9.2.1 The number of windings on the secondary side if the output voltage Lis 44V 9.2.2 The primary current in mA if the secondary current is 8A 9.2.3 Calculate the rating of the transformer in VA DHET Copyright reed OWNIOaded from hlayiso.ceERE tas ove: (2) (2) (2) (4) [10] (2) [10] (3) (2) (3) (2) [10]

7 T540(E)(U27)T QUESTION 10 10.1 With the aid of a neat labelled sketch, show the concept of reverse bias. Show ALL the components which affect this concept. 10.2 Change the following decimal numbers to binary and show ALL necessary steps: 10.2.4 14,510 10.2.2 15,2510 10.2.3 Subtract the answer of QUESTION 10.2.1 from QUESTION 10.2.2 and leave the answer in binary number TOTAL: DHET Copyright reserved Downloaded from hlayiso.com (4) (2) (2) [70] 100

FORMULA SHEET Any applicable formula may also be used. I. E=V- IgRq 2. E=V+tlQRq ZN 3. E =2p@ & 60C Vv N=—— 4, KB _ 0,318 Iq Zp ® 5. Tr Cc iency/1 I 00 6. Efficiency/Rendement = Viste Rak ibVre . 2 vec 7 Efficiency/Rendement = Vi = da’ Ra < +O) 100 2nN (WW - S)xr ene’ KID 8. Efficiency/Rendement GOTT - fi 9. Efficiency/Rendement = x100 . f, +1) 10. E=Blv It. e =&,,Sin2rft 12, f =1,Sin2zft 13. offvor i. , 0,637 Em or/of Im 14. Cras age OVE brmeuge =O,707 Em orsof ln 1s. E _& +e, +e, +... 4, ave/gem n al th th te he ave/ gem or/of | n TS40(E)(27)T OHET Copyright eseehoWnloaded from hlayiso.com

9 T540(E\(27)T 2 2 2 ? _ fe; te, +e; +... +2, 16, ee i n 2, 72g ad “2 _ ftp ti ti te. ta, or/of Fins web _ n Firms weh Lins wgh 17, Form factor/Vormfaktor = ="""*= or/of ——S— fave xem ave gem E, L, 18. Crest factor/Kruinfaktor = 7, "— orlof ——™ rms wek pas wek ; v . i= 19 R Vv X, =27 ,i=— 20, h a x, i v Xo see Ss 21. © 2fC Xe 22, ZafR 4X7; Z=fR 4X2 5 1 = x X, 23. Tan 6 = ; Tan 6 =2E R R 24, VR=IXR ; V,=IKNX, | Me HEX 25. Z=fR 4X, XeP oo ZR? +X XP X,-° X. X.- X 26. Tan @ == ©} Tan @ =k R R 2 ve 27. P=¥KI; P= PR; Ps 28. P=VICos @ R + VF or/of KW os : Cos @ =O A 29. Cos 0 yf C08 6 TT oriof KVA 30. Lactivsasney = 4 COS O Treactiverreattier 1 Sin © 31. P=¥ICos @ DHET Copyright reserved Please turn over Downloaded from hlayiso.com

“10 T540(E)(J27)T 32. r= oeJEG 33. ati: Tan @ nit 34, =ylh > Tan @ -" I-Ie 35, raft d,- Ie? : Tan o “So R Inet 36, F=Nlnt de> L,)? ¢ Tan => R f 37. Cos 0 =" rf 38. i =V3 Ip 39. VW, =V3 Vp; I, =I, 40. W =J3 V1, Cos 0 xn V; N, i gy eo ee Wy Ne Ly 42 RV. _v3 VL, 1 000 43, Panunraun Ver 6 Ase Ln Rn 44, Fy at, + 4, V, qT, 2— 45. r=, Downloaded from hlayiso.com uno: DHET Copyright reset

higher education & training Department: Higher Education and Training REPUBLIC OF SOUTH AFRICA MARKING GUIDELINE NATIONAL CERTIFICATE AUGUST EXAMINATION ELECTRO-TECHNOLOGY N3 27 JULY 2012 This marking guideline consists of 7 pages. Please turn over Cooyiant reserved Downloaded from hlayiso.com

MARKING GUIDELINE -2- ELECTRO-TECHNOLOGY N3 TS40(E)(J27)T QUESTION 1 14 144 Yoke v v (2) 1.1.2 Poles shoes Vv ¥ (2) 1.1.3 Laminated plates Vv v (2) 1.2 - Lap winding v¥- is used for high current machine or low voltage machine. ~ - Wave winding “ — is used for high voltage machine or low current machine. ¥ (4) [10] QUESTION 2 2.1 Is to move brushes forward and backward in the case of generator and motor respectively when current changes. (2) 2.2 - Byusing interpoles ~ - By using compensating winding ¥ (3) - Increasing the fietd flux. v 2.3 - Carbon brushes are cheap. ¥ - Carbon brushes have a low current density. (2) 2.4 - Series — Parallel or compound generator. - Series generator ¥ - Shunt generator ¥ (3) [10] QUESTION 3 3.1 3.1.1 Torque developed [T] — 03187, ZpP. C = 0.318 x644 “520 x4 x0,044 = 931,308 Nm ¥ = 0,931 kNm ¥ ———-* Answer (4) 3.1.2 Motorback EMF =V-I.2, = 480 V— (64Ax 0,60) ¥ = 4416VV = 0,442 kV ¥ ———> Answer (3) 3.2 - Brushes or Brushgear ¥ - Armature winding ~ -Commutator ” (3) [10] Copyiahtssere’ DYgwnloaded from hlayiso.co . Please turn over

MARKING GUIDELINE -3- ELECTRO-TECHNOLOGY N3 TS40(E)(J27)T QUESTION 4 41 Power [P =Vi, , = 190 xlo? 380V =500A VY _, Therefore:- Efficiency = 500A x 100% Vv in] 5004 +654 i x100% itl, = (08849 x100% VW = 94,07 % v——_*_ Answer (7) 42 It is when brushes are ground with emery paper/sand paper ¥ so that brushes take the arc of the commutator “ to increase total contact area of brushes to ensure maximum current flow. rol QUESTION & 5.1 PERIOD — is the time taken to complete one full cycle (in seconds) ” (1) §.2 5.2.1 Alternating voltage. ¥ (1) 2.2 5.2 Average Value [E aru] = O72 4% FO: 48 48 taf ~ (224444 854122-+9737 42017 Vv = 427 , 0% =61V ¥ -—» Answer (3) §.2.3 2, 2 7,2 2 Actual [Exc] = yor te, te te, Fey Fe, n - = +4d? +85? 41222 4972 4377 +207) 7 = [35707 _ 7 = f5101 = 71,421V vv —-—» = Answer (3) Conytgnt reserved PyYgwnioaded from hiayiso.com. Please turn over

MARKING GUIDELINE -4- ELECTRO-TECHNOLOGY N3 9.2.4 = Eps - 71,421 Form factor = (a My =1,171 “ —-> Answer 5.2.5 Sinusoidal wave/Sine wave/ Peak wave QUESTION 6 6.1 Impedance of the circuit [2] = (2? +(%, - ¥.)’ /t5S? + (62.830- 31.830" Vv = 1186 = 34,438. Q ~—» Answer 6.2 . we _ 250V Total current flows in the circuit [IJ 34.4380 v =7,259A v~—»> Answer — 6.3 The phase angle [tan6] = Xy> Xe _ 62,830- 31,830 ~ 15 8 =Tan~'2,066 =64,18° v——» Answer vv OR 6.3 R Cos@ = Z = 15 34,438 vv 0,436 Cost 0,436 64,18° “¥ -—» Answer Cos Hou 6 9 6.4 Voltage across resistor [V2] =/, X® connected in series : = 7,259Ax 160 ¥ = 108,885 V ¥ —> Answer Copyiahtreeered Downloaded from hlayiso.com TS40(E)(J27)T (1) (1) [10] (3) (3) (2) [10] Please turn over

MARKING GUIDELINE 5- T540(E\J27)T ELECTRO-TECHNOLOGY N3 QUESTION 7 7.1 Current flows in the circuit: P = VI Is TSW ~ 140V 72 7.3 7.4 Copyrentresened Downloaded from hlayiso.com =0,536A v ——»* Answer Resistance of the lamp:- V = IR R= 140V ~ 0.5364 = 261,194 Q ¥ ——» Answer vy Z, _ 230% ”~ 0,536A4 = 429,105Q v —* Answer Total impedance of the circuit:- Z = The inductance of the coit- X,7=Z?- R? X, =¥ 429,105? - 261,194" = 3404540 v X= 2afl Le 340,4542 ~ 23,142 x60 =0,903H v —* Answer (2) (4) [10] Please turn over

MARKING GUIDELINE -6- TS40(E)(J27)T ELECTRO-TECHNOLOGY N3 QUESTION 8 8.1 Power = V3V,1,Cos@ x1) I= 100 000 FY “V3 X440 V X0,82 X0,88 = 181,840A v =0,182kA ¥Y — > Answer 8.2 8.2.1 Phase current of the motor in mA:- P= /37,7,Cos@ _ 180 000W 1.5 400 X08 = 324,759A v But], =v3/,,, delta connection —_ L, Ira V3 324,759 sO v3 = 187,499A =187499mA ¥ —» Answer 8.2.2 Phase voltage in kV:- IndeltaV, =Yrn Veg =400V¥ 0 ¥——> Answer QUESTION 9 9.1 - Deflecting mechanism ~ - Controlling mechanism “ -Damping mechanism “ 9.2.1 N, XV, Nj =o y = 300 x44 Y 400 =33 windings “ —» Answer 9.2.2 yo yh 2 i 4a XB 400 F =0,88A v =B80mMA ¥ -——> Answer ial (4) 2) (3) Conviahtresened DYQwnioaded from hlayiso.com °"""*"

9.2.2 9.2.3 QUESTION 10 10.1 vv vv 10.2 10.2.1 10.2.2 10.2.3 MARKING GUIDELINE -7- ELECTRO-TECHNOLOGY N3 OR Ni che N, 4, I= 33x8 | 300 =0,88A v =880 mA “ —> ~~ Answer | Rating =V,>Z, = 400 Vx 088A ~ =352VA v ——* Answer (Ho current flows) 7 P it ~19QO19OE!- To ————_ OC OG @@i-} —___o 109 010 09;- a \ Increased depletion bayer REVERSE BIAS Sketch = 2 Labels = 2 11,510. = 1011,10. ¥ remainder 1101 2411 remainder 1 2|5 remainder 1 v 2|2 remainder 0 2/1 remainder 1 15,2510. =1111,01. > Answer __ 2115 _ remainder 1 v 2 \7 remainder 1 213 remainder 1 2/1 remainder 1 1111,012 4011,102 W412 Vv — > Answer Copyright reserved TOTAL: Downloaded from hlayiso.com T54O(E)(J27)T (3) [10] (2) [10] 100