electrotechnics n6

higher education & training Department; Higher Education and Training REPUBLIC OF SOUTH AFRICA NOVEMBER a NATIONAL CERTIFIGATES:» ~ v2 ELECTROTEGHNICS Né i > ra < P»Thisigquestion Paper consists of 6 pages and 5 formula sheets. 4 By. iy Copyright reserved Please turn over Downloaded fr (8080096) 2 TS40(E)(N26)T DEPARTMENT OF HIGHER EDUCATION AND TRAINING REPUBLIC OF SOUTH AFRICA NATIONAL CERTIFICATE ELECTROTECHNICS NG TIME: 3 HOURS MARKS: 100 ( ~ Yy INSTRUCTIONS AND INFORMATION 4. Answer ALL the questions. 2. Read ALL the questions carefully. 3. Number the answers according to the numbei 1g system u: + Yf this question paper. 4. 5. Round ALL calculations off to THREE de 6. Use the correct symbols wag on g Start each question on, ANEW pa Lagege. 8. Write neatly and sed cs Please turn over Copyright reserved e o.com

{ (8080098) 3 TS4O(E)N26)T QUESTION 1 4.4 Explain, with the aid of a neat diagram, the operation of a Ward-Leonard control system, controlling the speed and direction of a large DC shunt motor. (7) " 1.2 The following figures were obtained during a Hopkinson test an two similar 4 230 V DC shunt machines, °"““"Motor armature current § = Generator armature current ~Motor field current Generator field current Armature circuit resistance of each machine =0,33Q 3 aut (11) 118] QUESTION 2 An unbalanced three-phase, four-wire, sta rey I is connected to terminals R, Y and B of a 440 V three-phase, fourdyire > Supply. “pre load consists of the following impedances: zi aA Between terminals R and N: 20° jOe« Between terminals Y and N: 75+ jonni Between terminals Band Nv 12 os bs / 2.4 TRE total power ot ee \ Copyright reserved Please turn over '____ Downloaded fronm-hli (2080096) 4 T540(E)(N26)T ¥ QUESTION 3 3.4 Define voltage regulation of a transformer. (3) 3.2 What are the effects of harmonic voltages in transformers? (2) 3.3 The impedance refers to the primary of a 150 kVA, 2 000/500 V single-phase, 50 Hz transformer is (0,5 + j4) ohms. The power factor is 0,8 lagging. Calculate the following: 3.3.1 ‘The turns ratio (2) 2 The percentage resistance (3) 3.3.3: The percentage reactance (2) 3.3.4 The fullload copper loss 2) 8:35: The power factor at which maxi (2) 3.3.6 The voltage to be sine (2) current in the ee rd [18] 4 QUESTION 4 A 4.1 Define regulation of ou (3) 4.2 A12 KV, three- spdbe, 2M 1h, sar-connected altemator operates at a power factor of 0,8 lagging aad has aéynchronous impedance of (1,2 + j8) ohms. Calculate fhe folleying 421% % Yee open-circuit fine voltage to which the machine must be excited “to dag fulload O) The regulation of the alternator (2) The load angle at which’ thal faichine operates “ é @ Copyright reserved Please turn over e ayiso.com

(e080096) 5 T540(E)(N26)T ssh QUESTION 5 aw 5.4 State TWO advantages of a synchronous motor. (2) 5.2 A 11 ky, three-phase, star-connected synchronous motor has an armature . resistance of 1 ohm per phase and a synchronous reactance of 30 ohms per phase. It is loaded to have an angular retardation of 12 electrical degrees and is excited to an induced EMF of 13 kv. Calculate the following: 5.2.4 The input current to the motor (7) 6.2.2 The power factor of the motor (2) 623 The input power of the motor (2) 113] QUESTION 6 “eee 6.1 What is the function of an external woes nce Starter, coupled fo the slip- tings of a wound induction motor? ted induction has a rotor at standstill. The standstill EMF impedance of 0,16 + j1,1 ol 62 A 480 V, sixpole, three+ phése, _Bar-gor aS between slip-rings is ae ve AQ as at a fulload slip of 5% (7) Calculate the “a, Pe 6.2.1 The torqui ite, 6.2.2 athe fulload 11 Weer output if the friction and windage losses are A 880 (2) [14] Copyright reserved Please turn over Downloaded fro (080096) 6 QUESTION7 TA What are the effects of a low power factor? 12 A 190 KW, three-phase induction motor having an efficiency of 85% and a power factor of 0,8 lagging, is connected in parallel with a three-phase synchronous motor taking 250 kVA at 0,9 power factor leading. The supply voltage is 3 500 V. Calculate the following: 7.24 The line current of the total load P22! ‘The power factor of the total load Copyright reserved e TS40(E)(N26)T Q)

(080096) “1 TS40(E)(NZ6)T ELECTROTECHNICS N6 FORMULA SHEET GS-MASJIENE DC MACHINES SPOEDBEHEER TOETSING TESTING DIREKTE METODE DIRECT METHOD SWINBURNE- © | SWINBURNE METODE METHOD | [~ HOPKINSON- 2, HOPKINSON | RENDEMENTE EFFICIENCIES DIESELEDE » : THE SAME | YSTER= po, IRON LOSS | VERLIES i Le (P 1g)? Rat (+ fy LF? Rat (+h) ¥} | i ee Sr | generator yy +I) Rat dgV +S h | (hth) WG th~ hy Rath Ves motor “G+inv Copyright reserved Please turn over (8080096) Be WS-BELASTING AC LOADS STERSTELSELS STAR SYSTEMS ~ Ven = VERWYSING REFERENCE R-Y-B VOLGORDE SEQUENCE GEBALANSEERDE KRING In=0 sae nT eins DELTASTELSELS “DELTAS YSTEMS DRIEDRAAD- THREE-WIRE STELSELS SYSTEMS Copyright reserved Please turn over

wm / higher education Re training Department: Higher Education and Training REPUBLIC OF SQUTH AFRICA MARKING GUIDELINE Copyright reserved NATIONAL CERTIFICATE NOVEMBER EXAMINATION ELECTROTECHNICSN6 26 NOVEMBER 2013 This marking guideline consists of 13 pages. 5D ae turn “= = MARKING GUIDELINE T540(E)(N26)T “26 ELECTROTECHNICS N6 QUESTION 1 1 Lile Ls vt Gengratpr Fomor Motor M Ky (YIYVY + * The excitation of the motor is constant. « The control is obtained by adjusting the, field current of the genetator by means of a potentiometer.Y mei e One end is connected to the mid-point of the potentiometer and the other end to the adjustable point, which can be adjusted from zero to maximum The direction of rotation can be reversed by connecting K to the left M. qi ) Please turn over hlavi

MARKING GUIDELINE 3- TS40(E(N26)T MARKING GUIDELINE, ca ELECTROTECHNICS N6 ELECTROTECHNICS N6 se n= 29.5 A Generator fficioney = me x 100% q ean! See Pout + Generator Losses 08A _ $716 __ x 100% 6716 + 911615 a t 11615 at), q = 88,048 % ins} Generator QUESTION 2 24 Vp = WIN3 = 44013 = 254,034 VY | Zan =20 + j0 = 20,0°0 Ip = Van/ Zan Let C = iron, Friction and windage losses = 254,034L0° / 20 0° \ 2 2 = 12,702L0° A CHlk.vV- Ur +b lay .Ra + (ith Rat Ig. Vit b.V1 Zyy =5 458 = 7,071145°Q = (8,65 x 230) - [(87? x 0,33)+ (30x 0,33) * (0,86 x 230) + (0,8 x 230)] nn 2) ly = Ven /Z = 1980,5- (454,77 + 207 + 195,54 184) WW. 284,034l-"120° /7,071L45%Y = 35,9261-165° A = 861,23 Wi Zan = 2—)12 = 12,1661-80,538° C12 = 430,615 WY ~ aN 4 2468. Ip = Ven / Zan Sie ae = 954,034 120° / 12,1661-80,638°N = 29,2x 2 0 ~ ean 20,881L200,538° AY m, Generator Losses = (I) +s)" Ra + Is-V + C/2 22 IN=In* ly * Ie eg revi sees cls plgRe ss * G2,roa\o° + 36 926\-166° + 20,8811200,598" = 911,615 WY ee = -44,554- j16,624 Pd 6 = 44,7561-188,196°V (2) 2. @ @ Copyright reserved Please turn over Copyright reserved Please furn over eT = _

MARKING GUIDELINE 6- TB40(E)(N26)T MARKING GUIDELINE & TS40(E)(N26)T ELECTROTECHNICS N6 ELECTROTECHNICS NG 2.3 Method 1 3.3 3.3.4 NY Pr = Van -In. COS Ga N,V, = 254,034 x 12,702 x Cos 0° _ 2000 = 3,227 kW ~ 500 Py = Vyn. ly. Cos ey =4 = 264,034 x 36,926 x Cos 45° (2) = 6,453 kWy Pp = Von. |g. Cos 88 3.3.2 = S/V; = 150000/2000 = 75 AY = 254,034 x 20,881 x Cos 80,538° - = 0,872 KWY xR = LRe x 100% Method 2 4 5 x 100% Py = Tp. Re 3 = 12,7027 x 20 ‘ @) yo = 3,227 kW 3.33 a %X = LH x 100% Py = TR.Ry v, = 35928 x5 = BX4 x 100% Y= 6,453 kW 2000 5. =15% | 9) 2 (2) P, = The Ro = 2088F x2 3.3.4 Be= Ti Ra Vo = 0,872 kW (3) = 75? x 0,5 24. Pr= PrtPy+Py . = GES : ey = 3,227 + 6,453 + 0,872 BIO ,2527 NN aa 335 Ze: = 0,5 +)4 = 4,031 282,875°0N Method 1 QUESTION 3 Cos 62 = Rey /Ze1 i r F P f = 0,5/4,031 3.4 e It is the rise or fall in the secondary terminal voltage when the load is = 0124LaggingY removed and the primary voltage remaining constant. 5 'gaing OR Method 2 * Its the difference in the secondary voltage between no-load and full-load, Cos 02 = Cos 82,875 Q) expressed as a ratio of the no-load voltage with the primary voltage = 0,124 Leggingy(2) remaini aah 3 inin@onstant. WV a 8 Bale Veo # hy izes 3.2 * Increased dieleciric stresses = 75 x 4,031¥ * Electrostatic interference with communication circuits = 302,325 WY (2) « Resonance between the inductance of the transformer windings and the 118] capacitance of the feeder to which they are connected (Any 2 x 1) (2) satin Downloaded fronrhtayiso.com arte

MARKING GUIDELINE TS40(E)(N26)T Fe ELECTROTECHNICS N6 QUESTION 4 44 The rise or fall in terminal voltage when the full-load current is switched off, ware the speed and field excitation remains constant, is called regulation. FW OF Regulation is defined as the rise in voltage when the full-load is removed, while the speed and field excitation remains constant, divided by the rated voltage. VW¥ (3) 42 4.24 $= v1, foe 8 VB.V, = 2x 10° V3 x 12.000 = 96,225 4 ¥,, _ 12000 Vp = Eh = == = 6928,203V BOB LRa = 96,225x1,2 = 115,47 Vv IXs= 96,225x8 = 7698 ¥ Method 1 Ep =VpLe + I.Ral0° + Xs 190° = 6 928,203 |36,87° + 115,47LO° + 769,8190° = § 658,025 + j4 926,732 = 7 502,395141,048° VW E, = V3.Ep = v3 x 7502395 = 12 994,529 V Methed 2 E, = Y(V.Cos0 +LR,) + (V. Sind +L.X& = W([6 928,203 x 0,8]+ 115,47) + ([6 928,203 x 0,6 ]+ 7698)" . ¥ = 7502,4V ¢@ BE, = V3.E, = 73x 75024 = 12.994,538V L Pe Vv Copyright reserved Please turn over MARKING GUIDELINE 8. ELECTROTECHNICS NG Method 3 Ep =Vp + (I.Ra. Cos 9) + (Xs. Sin 6) 6 928,203 + (115,47 x 0,8) + (769,8x0,6) ¥ nou 7482459 V0 B, = V3.E, = V3 x7 482,459 = 12960 y 1) 422 For method 1 % Regulation = x 100% Pr y _ 7 502,395 6 928,203 6 928,203 Vv = 8,288 % For Method 3 % Regulation anal 100% af __ 7.482,459 — 6 928,203 7 6 928,203 =| =8% 4.2.3 a = 41,048°-36,87° = 4,178 Copyright reserved e x 100% x 100% (2) (1) [13] Please turn over

MARKING GUIDELINE a TS40(E)N26)T ELECTROTECHNICS NS QUESTION 5 5.4 e The power factor can be varied by changing the excitation current. « Itcan run at a constant speed. « While carrying a load, it can be used for power factor correction. * Hunting can be easily prevented. (Any 2 * 1) 2) 5.2 5.2.1 Vp = VLIV3 = 11.000/73 = 6 350,853L0°V Ep = E_/v3 = 13000/\3 = 7 505,554 Ep = 7 505,554|180°- 12° = 7505,554L168° VY Ep = Vp + Ep = 6 350,85310° + 7 505,554 L168" 990,687 + j1560,492V = 1848404112241 VV Zs = Ra + (Xs = 1+j30 = 30,017( 88,0970 y E, = 12, _ 1848,404.123,40° 30,017 288,097 61579234319 A @ a h 5.2.2 Power factor = Cos @ 5.2.3 Copyright reserved Cos 34,319 ¥ 0,826 Leadingv _ (2) ait Pp 213. V, IL. Cos 0 V3 x11 000x 61,579 0,826 V 969,094 kW (2) [13] ‘Downloaded fronrhtayiso.com MARKING GUIDELINE TS40(E)N26)T - =10- ELECTROTECHNICS NS QUESTION 6 61 « To reduce the starting current. « To increase the starting torque.’ (2) =N-0, £=™Y, _ 60.F, me OE _ 60x50. aA = 1000 r/min 6.2 6.2.1 Eo = E./¥3 = 290/13 = 167,432 V Ep = S, Eo = 0,05 x 167,432 = 8,372 W Zy = JRE +(S.Xo¥ = f0167 + (0,05 x 1,1) 169 V = 49,538 A Rotorcopperloss = 3.12. R, = 3 x 49,538 x 0,16 v =1177,926 W Rotor input = Rotor copper loss / Slip = 1 177,926 / 0,05 = 2355852 Vv "2x nx 1000 224,967 Nm q) a Please turn over

MARKING GUIDELINE 11- ELECTROTECHNICS N6- 6.2.2 No = Ny-S.Ny = 1000 - (0,05 x 1 000) 950 rmin.v Guwrerir = 22 Nee 60 _ 2x mx 950 x 224967 7 60 q = 22380,565 W Power output = Gross rotor output - Friction and windage losses 22 380,565 - 880 21 500,565 WY 6.2.3 S = Ro/Xo = 0,16/4,1 = 0145 No = Ni - S.Ni = 1000 - (0,145 x 4000) = 855 rmin.¥ QUESTION 7 7.4 Increases the total current for a given power and therefore decreases the efficiency of both apparatus and supply system. Limits the output for both generators and transformers. Causes a greater fall in terminal voltage Larger cableg-and switch gear needed for the load. Larger currerits cause greater copper losses in cables and transformers. Larger currents cause greater volt drop in cables. Larger cables needed on consumer's side to carry large currents. Cost is more for consumers. Consumers pay more for electricity. (Any 2 x1) eee oeeee Copyright reserved TS40(E)(N26)T (3) (2) 114] 2) Please turn over \________Downleaded fren:hlayise.cent MARKING GUIDELINE TS4O(L -12- ELECTROTECHNICS NG 2 7.2.4 Se 82 Py Pa 8 Ss Induction motor n= Pour / Pin = 223,529 KWY Cos 6; = 0,8 Lagging 1.04 = Cos’ 0,8 = -36,87° Gos 0; = P/S; Si = P1/Cos Ov = 223,529 / 0,8 = 279,411|-36,87° kVA Synchronous motor Cos 82 = 0,9 Leading..02= Cos" 0,9 = 25,842° S2 =250126,842° kVA Ss = So+ St = 250,25,842°+279,444|-36,87° 448,529 — j58,674 =452,361-7.453° kVA ¥ Method 1 8, =3.V1, = @ YBN, & _ 452,35%19" V3 x 3.500 y= T4619 A Copyright reserved Please turn over

MARKING GUIDELINE -13- T540(E)(N28)T ELECTROTECHNICS N6 Method 2 P, = V3. V,.1,. Cos® = PB BM, 4,.Cosd 448,529 10° ~ 3 x 350 x Cos7,453 = 74,619 A ¢ (8) 7:22 Power factor = Cos 6 = Cos 7,453 = 0,992 Lagging V (1) [11] TOTAL: 100 ’ ¢ Copyright reserved Feet wut Uwe Downloaded from hlayiso.com \ . . in 7 - ¥

higher ¢ education a aaog teercae Education and Training REPUBLIC OF SOUTH AFRIGA TSGO(E)(M23)T APRIL 2014 NATIONAL CERTIFICATE ELECTROTECHNICS N6 (8080098) 23 March (X-Paper) 08:00 —42:00 REQUIREMENTS: Graph paper Calculators may be used This question paper consists of § pages anda 6-page formula sheet, 2 TSSOLEKM DEPARTMENT OF HIGHER EDUCATION AND TRAINING REPUBLIC OF SOUTH AFRICA NATIONAL CERTIFICATE ELECTROTECHNIGS N6 TIME: 8 HOURS MARKS: 100 INSTRUCTIONS AND INFORMATION 1. 2. o aS Sf oR Answer ALL the questions, Read ALL the questions carefully. Number the answers carrectly according to the numbering system used in this question paper. Start each question on a NEW page. Keep subsections of questions together. Round ALL calculations off to THREE decimal places, Use the correct symbols and units. Write neatly and legibly. PTO

Ae TSBO(E)(M23)T fl 3 TS60(E)\(M23)T - QUESTION 1 E 3.2.4 The fulHoad copper loss 8) 44 Name THREE types of constant losses in a DC machine (3) 3.2.5 The power factor at which maximum regulation occurs @) 1.2 ASO KW, 400 V, DG ghunt:rioior has an armature resistance of 0,35 ohms. 3.28 The voltage to be applied to the primary to circulate ful-load When operating at full-load the armature takes 120 A. current on short-circuit na What resistance should be Inserted in the armature circuit to obtain a 20% speed reduction, when the motor is developing 75% of fulltoad torque? (8) QUESTION nq] 44 Explain, with the ald of a neat diagram, how an open-circult test is carried out QUESTION 2 w on-a three-phase, star-connected altemater. 6) 42 A three-phase, 11 KV, star-connected alternator has an armature resistance A thi Wil it “in fee phase, four whe, slat-conhecied eleenator, (wilt a/line Voltade of7600 My of 5 ohms per phase and a synchronous reactance of 25 ohms per phase. supplies a star-connected unbalanced load consisting of Calculate the percentage regulation for a load of 1 500 kW at a power factor = 45+]10 a = 20 +)o x of 0,8 lagging. a) Zen = 25-/45 [12] Take Van @s reference phasor and assume a phase rotation of R-Y-B. QUESTIONS Calauiats the following: 5.4 State TWO advantages that a power factor improvement has for the 24 The THREE line currents fo) songuiiers @ ; Th 5.2 A three-phase, 50 Hz, 2 200 V, star-connected induction motor has an output ae Scour Ii reset copier @) of 475 KW. ‘The efficiency of the motor is 87% and it Is running at a power 23 Thi isst it factor of 0,8 lagging. A bank of capacitors, connected in star, are connected e:power.dlesipatad inisanh Joad &) in parallel with the supply to improve the power factor to 0,9 lagging. Each 24 The total power X (1) bank of the capacitors consists of FOUR identical capacitors. 3} Determine the capacitance of each capacitor. (40) ¥i2] QUESTION 3 ‘ : 7 =e) sr 3.1 Name TWO types of variable losses in a transformer and state exactly where x l (e s each one occurs. Copper testa Handings, 4) y Ready Current Waser o pinch og aed coe . . 32 The percentage impedence of a 250 kVA, single-phase, 3 000/400 V trans- “ former is (1,2 + j5) per cent, at a power factor of 0,8 lagging. Calculates the following: 3.2.1 The tums ratio @) @ 3.2.2 ‘The percentage regulation a (2) @ a 3.2.3, ‘The secondary terminal volfage on full-load (2) PTO as) PIG “te te

QUESTION 6 connected induction moter: ‘The stator resistance per phase is 0,5 ohms. 61 Draw the Circle diagram using a scale of 1 om=4A 6.2 Datermine the following from the circle diagram at fulHoad: 6.2.1 The efficiency 6.2.2 The percentage slip 6.2.3 The power factor 6.2.4 The line current 6.2.6 The torque in synchronous watts ond T560(E)(M23)T The following are results of a test ona 15 KW, 500 V, 50 Hz, three-phase, 6 pole, star- No-load test 500 V 12A 2598 W Locked rotor test 125V 224 1667W PTO (10) (1) (1) a) 6 . TS60(E)(M23)1 QUESTION 7 aA Define the regulation of a transmission line, 7.2 A long three-phase transmission [ine delivers 60 MW at 0,8 power factor lagging and a line voltage at 110 kV, 50-Hz. Each conductor has a resistance of 20 ohms, an inductance of 0,18 henry and a capacitance fo neutral of 1,5 micro-farad. Calculate the following by using the 7 method: 7.2.1” ‘The sending end voltage 7.22 The sending end current 7.2.3 The sending end power factor 7.24 The efficiency of the line NOTE: Draw the x method circuit diagram. TOTAL: (8) (11) (2) (1) (2) [3]

education Department: Education REPUBLIC OF SOUTH AFRICA QUESTION 1 it © Friction losses v «Shunt field coppers ¢ Iron or core loss y~ 12 B= V-l Ba = 400 - (120% 0,35) = 358 VU No/Nr=0,8 and April 2011 ELECTROTECHNICS.N6 808006 April 2011 ELEKTROTEGNIEK NS: Te lad But $1: = ¢2 ‘Method EaN.¢ But hi = 23% a, 8) Ta/T1=0,75 aTaeh Method 2 hin h te Ie = xl = 0,75x 120 = 904 .EBaN @) (1 2@

QUESTIONS Method 2 Ve = Mele Pe TR = 500/ = 16,013? x15 = gers VV“. Zyu = 154]10 = 18,028133,69° Q Py =1R.Ry Ty = Vel Zen = 14434 x 20 = 288,675 Loe /18,028 L33,69° va = 4,167 EW = 16,0131-33,6° A =? Za = 20+ 50 = 20L0° 2 P= 15-Rp 5,608? x 25 Ty = Vie! Zn = 0,786 KW = 288,675 1-120° /20L0° ~~ cae ) = 144341207 A a 24 Zon = 25 -j45 = 51,478 1-60,945° Q 7 . Pr = Pat Py+ Pa Ja © Van/ Zoe - m= 3,846-+4,167 + 0,786 = 288,675 [120° /51,478 L-60.945" ~~ = 8,709 KW” Oy = 5,608 [180,945° A | : ) (13) ai : QUESTIONS k= htt 34 = 16,013 L-33,69° + 14,434 L-120° + 5,608 180,945" v or = 0,5 -j21,474 7 © Copper losses or I R losses windings = 21481 -98,666" A . @) © Biddy ourent losses or tray losses V~ veindings and ebre g ® 23 32 ‘Method 1 . 324 Pp = Vyur-Ja. Cos On N/R = Vi/'V2 288,675 x 16,013 x Cos (-33,69°) = 3000/4900 3,846 KW =7,5 Py = Vey. Ty. Cos Or pumsratio = 7,521 ~~ @ = 288,675 x 14,434 x Cos 0° = 467 WW 3.22 Pp = Van. In. Cos Op Cos 08 md Sind = 0,6 = 288,675 1,60! Cos (60,945! none We 0,945") 9% Regilation = %R.Cos0 + %X. Sine = 0,220,8)+6%0,9) \~ = 396% 2)

323 Ve = 400 -(0,0396% 400) = 384,16 V.~ @) 324 Methad 1 pce %RxS ~~ = 0,012 x250 000 |“ SEW Method 2 Th = S/Vi = 250 0007/3 000 = 93,333 A we %R = hRe/V Re = %RxVi/ Te = 0,012 x3 000/ 83,333 = 04922 Pao ~ Re 83,333" x 0,432 = 3kW a Method 3 I = 8/Va = 250 000/400 = 625 A %R=bRe/V2 | : . Re = %RxV2/b = 0,012 x 400 / 625 = 0,00768 2 |~ Poo = L?, Rez = 625" x 0,00768 =3 kw V7 8 3.25 Method 1 % WK = b. Kel V2 Ke = %XxV2/E = 0,05 x 400 / 625 = 0,822 | Ze ~ VRoP+ Kor = ¥0,00768" + 0,032" soos % Atmaximum regulation: Cos @2 = Cos Ge Cos Oe = Ree / Zee = 0,00768 /0,033 = 0,233 Lagging Metho’2 %X = hh. Xe/Vi Ker = %XxVi/h = 0,05 x3 000/ 83,333 Higa Zo, = YReP+ Ker = V048e + 1? = issia % Atmaximum regulation: Cos 61 = Cos Oe Cos Oo = Rei /Zen = 0,432 /1,852 = 0,233 Lagging Method 3 60 = Ta (%X/%R) = Tan’ (5/1,2)} = 16,804 )/ Cos Bo ~ Cos 76,504° = 0,233 Lagsing V~ 326 Method 1 ‘Vici = Psc/ Cos 80. = 3000/0233 x 83,333 ~~ = 154507 V7 8)

Method 2 Vee = Zex k= 0,033 x625 = 20,625 V Va ‘Veer / Vsca = Ni/ Neo e ‘Vsc. = Vsorx Nu/ Nz : {18} ‘QUESTION 4 41 ¥- (Ay or a ‘Tho alternator is driven at normal speed on open circuit A secies of readings are taken of corresponding values of field cureut and phase emi. Paes emia plied vray agnint eld caren y/~~ © 42 P= V3V,,1,.Cosd | eae eae ~V3.V,.Cos8 = 1500 x10" ¥3x11000 x 0,8 SOBAIIA YK Str: I, = Ip ¥,, _ 1000 g V, sk = SS = 6350883 0 et ner 8 Ca UR, = 98,412%5 = 492,06 ¥ LX; = 98412x25 = 24603 V Method 1 Ep = vple+LRalo*+Lxs190° = 6 350,853 |36,87° +492,0610° + 2 460,3 L90* = $572,736 + j6270,821 = 8 989,194 |48,373° V 77 Bp — Vy % Regulation = x 100% h, = £389,194 6350853 “ 6350853 3 100% = 32,096% ~ Method 2 Bee Aw Cos6 +I.R,)? + (V. Sin6 + IX)? = ((6350,853 x 0,8 ]+ 492,06)" + ([6350,853 x 0,6]+ 24603) ~~ = 8389,192'V ¥~ % Regulation =e x 100% - 8389,192— 6 350,853 6 350,853 = 32,096% ~~ x 100% ~~ Method 3 Ep = Vp-t (LRa. Gos 8)+ (Xs. Sin 0) = 6 350,853 + (492,06. 0,8)+ (2 460,3 0,6) ~~ = 820,681 Vy~ % Regulation = a x 100% 8 220,681 6 350853 6350853 a BAD x 100% 1 ia) [2]

QUESTION 5 Si 8 The consumer pays a lower tariff v @ The efficiency of their installation is higher V~ a2 ‘Cone, __ $45,977 , = eer . = 6820712 ~3687T EVA 9, = Cos" 0,9 lagging = 25,842" mee Ss ‘Cos8, _ 343,977 0,9 = 606,6412-25,867 kVA Q) ” Xen 5, =8,~8, = 606,6412~ 25,842 — 682,4712-36,87° = 0+ j145,055 = 145058790" kVA \~ 8 = Viv, 8 Lens BN, Vy 2200 = = SS = 1270171 V Were oanY y, Xo =a 1270,171 = "38,067 = 33367 1 Ba B.C, ‘ai i 2.1. Xeq : i © Fenn 50x 33,367 = 95396 uF | cut c, L ire: oe op Be C, G 1 CG 14 G « C =4xC, =4x 95,396 =38158¢uF |

Method 2 Qu = KVAr before power factor comection Qu = PinxTan 0) = 545,977 x Tan 36,87° = d0984 LAr V7 Qa = EVA after power factor correction -Q = PinxTan 82 = 545977 x Tan 25,8422 ~~ = 26443 KAR 7 «Qe = EVAr of the capacitor bank Gee or Qc = Pin (Tan «Tm 04) = 409,484 — 264,43 7 = 145954 kVAr J Qe oa 1 ONE Eon . . “ 145,055x 10° é Wx2200x sone. ~~ =95397 pF Cx 4xCr i = 4295,397 pF = 381,588 pF QUESTION 6 64 Cos Oe, = P/V3.Vi.Te = 2598 /¥3x500x12 = 025 On = Cost 0,25 = 75,522° a Cos Osc = P/V3. Vi. = 1667 (V3 x125 x22 = 035 Qsc = Cos 0,35 = 69,513" a = 545,977 (Tan 36,87° - Tan 25,842") |“ = 145055 VAR (qo) [2] Isc ™ Test current x Rated Volinge / Test Voltage = 22 x 500/125 =A FD onthe circle diagram = Stator copper losses at standstill FD = 3. ise. Rs/V3, Vb. Soule - : = 3x B87x 05/43 x500x4 = 3353 em |~ (10) Seale: Lom = 4A Im = 12/4 = 3 om Isc = 88/4 = 22am 6.2 6.2.1 7, = Pout/Pin x 100% = GH/GK x 100% = 433/555 x 100% = 7ap1s% 7 @ 622 % Sip = HI/ GI x 100% = 025/46 x 100% "= EAI % a 623 Power factor = Cos @ = (0s36,5° = 0,804 Lagging = a) 624 _ 1 = Distance x Seale = 0Gx4 a . = 695 %4 aMBA @ Pin = V3. Vi. Distance (GI). Scale V3 x 500 x46 x4 = 15934867 Syn.Watts | a é

on QUESTION 7 TL ‘The regulation of a transmission lins is the rise pr fall in voltage at the receiving end when the full-load is thrown offf while the sending end Voltage remains constant. .// @ 1 Td X= 2meL = 2xnx50x0,18 = 5655 2 Zap = RF Ky = 2) + 556,55 = spel a ~~ Se C2 = 15/2 = 0,75 pe i * Omt.C2 - 1 Dunx50x 0,15 x10" = 424433 2° vm Oh = 0-}4 244,13 = 4mas[90° 2 ~~ oo Vy _ 110x10° By a = 63 508,53.20° V ~ ys Downloaded from hlayiso.com WO:

ace 2g = 835085320" | 4:244,132-90" = 1496290 A Tag = P = V3.¥,.1,.Cose : > Yeh = ?R™ © By, Cosd 60108 Wx it0 x10 x08 = 393,65 £—-36,87° A va daa = Int lbp = 393,65 |-36,87° + 14,96 L90' = 314,92 -j221,231 = 384,8611-35,088° A ‘Vas = Tan. Zan = 384,861 [..35,088° x 59,98170,52° = 23:083,963 [35,432°-V > Vs = Vat Van = 63 508,53 L0° +23 083,963 135,432" =°82 317,439 +15 382,612 = 8339817219234 ¥ Vi, = V3 x 83 398172 = 146,45 KV | 722 Vy ae Lp Za. =. 33398,172.29,234" 4244,132— = 19,65.299,234* A va ty = wth = 384,861 |-35,088° + 19,651.99,234° = 31,767 —j201,836 = 371,398 |.a2.919° A“ @ > RRL ay 723 Power factor = Cos 6 = Cos (32,919° + 9,234°) = 0,741 Lagging |// T24 Py = V3-V,-1,.Cosé = V3 x 144,45 x 10° x 371,398 x 0,741 = 68,855 MW 7 n= 22 x 100% _ 60 68,855 = and x 100% to) @ [19] TOTAL: 100

=» higher education y Straining Na (a —) Department: 4 Higher Education and Training WA } h i id Trai Cana” REPUBLIC OF SOUTH AFRICA T570(E)(M23)T APRIL 2042. NATIONAL CERTIFICATE ELECTROTECHNICS N6 (g080096) 23 March (X-Paper) 09:00 — 12:00 Calculators may be used. ‘This question paper consists of 6 pages and a. 5-page formula sheet, 2 TSTOE(M23)T DEPARTMENT OF HIGHER EDUCATION AND TRAINING REPUBLIC OF SOUTH AFRICA NATIONAL CERTIFICATE ELECTROTECHNICS NG TIME: 3 HOURS MARKS: 100 INSTRUCTIONS AND INFORMATION 1. Answer ALL the questions, 2. Read ALL the questions carefully. 3. Number the answers correctly according to the numbering system used In this question paper. 4. Start each question on a NEW page, ai Keep subsections of questions together, Round ALL calculations off to THREE decimal places. Use the correct symbols and units, P&S & Write neatly and legibly. PTO Downloaded from hlayiso.com

3 TS7O(E}(M23)T QUESTION 4 44 Name THREE types of variable losses In a DC machine, Br A 100 KW, 480 V, DG shunt motor takes a current of 10 A at no-load, The resistance of the armature circuit is 0,2 ohm and that of the shunt field Is 240 ohms, . Calculate the following: 1.2.4 ‘The input current at fulkload 1.22 The efficiency at fulHoad Aes ‘The armature current at maximum efficiency . ~b+ yb" —4a0 HINT: I, — QUESTION 2 An unbalanced, three-phase, delta-connected load consists of the following impedances: Zpy = 160 - j120 ohms Zys = 200 + J250 ohms Zer = 150 + JO ohms ae fine vortege is 380 V. Take Ver as reference phasor and assume a phass rotation of R-Y-B. Calculate the following: 24 ‘The current In each phase of the load 22 The current in each line 8) (9) @ 2) [16] @) f2] Ae TS7O(EKM23)T QUESTION 3 3.4 Name FOUR of the most common three-phase transformer connections, 32 A 500 KVA single-phase transformer working at a power factor of 0,8 lagging has a maximum efficiency of 95 per cent If the secondary current at maximurn efficiency Is 85 per cant that of the full-lead, calcutate the following: 3.2.1 The iron loss at maximum efficiency 3.22 ‘The copper loss at maximum efficiency 3.2.3 The copper loss at fullload 3.2.4 ‘The full-toad efficiency QUESTION 4 44 Explain, with the ald of a neat diagram, how a short-circuit test is carted out ona three-phase, star-connected altamator. 42 The armature of a 12 pole alternator with a flux per pole of 0,05 Wb has 72 slots. The form factor ts 1,11. Calculate the following: 4.2.4. The alternator speed If the frequency is 60 Hz, 422 The terminal EMF if the altemator is wound with a three-phase, star-connected, full-pitched winding having 10 conductors per slot. 4.2.3 The terminal EMF if the coll pitch is % of the pole pitch. “6 QUESTION 5 5.4 Explain, with the aid of phasor dlagrams, what happens when the load torque of a synchronous motor Is increased. 5.2 The phase impedance of a 2,5 kV, three-phase, star-connected synchronous motor is 1,5 + j10 ohms. Calculate the Induced EMF and the load angle In e electrical degreas when the machine Is loaded to take 75 amps Be power factor of 0,8 leading. is PTO

aa TS70(E\(M23)T QUESTION 6 64 Name TWO methods to siart an induction motor. 82 ‘The input of a sk-pole, three-phase, 50 Hz induction motor is 50 KW at a speed cf 970 vmin. The stator losses are 1 KW and the frlction and windage losses are 1,6 KW. Calculate the following: 624 Thestip 62.2 The rotor copper losses 6.2.3 The output 6.2.4 ‘The ful-load efficiency QUESTION 7 7A Define the regulation of a transmission line. T2 Apply the z-method to calculate the sending end voltage, current and power factor of a 150 km transmisston line, ‘The line delivers a three-phase load of 15 MW at a power factor of 0,8 lagging and a lina Voltage of 90 KV, 50 Hz. Gach conductor has a resistance of 0,285 ohm/km, an inductance of 41,845 milllhenny/km and a capacitance of 0,00853 micro-farad/km to neutral, NOTE; Draw the n-method circuit diagram. TOTAL: Q a) 112] @) (13) [16] 100 ELECTROTECBNICS N6 GS-MASJIENE E=¥-laRa Au No. E, Ny %, Hoh Se hh ht SPOEDBEHEER povenl 22 oe R+ Ree EV ~ Ia Ra~ ke Rse TORTSING _2nNr ( ~S) DIREKTE METODE 1 OW STINBURNE- 9 _W-U@Ra+ JV +1s¥) METODE sto ‘7 1 ra ” generator” IF + 1a? Rat Ia, V+ 1sP [~ HOPKINSON- DIESELFDE YSTER- VERLIES fost +h = hv {0 45) Rat G+ h- ly)? Ret tl} =o " AV generator — AF eh thy Rat 7 +S ‘ Ger -fth +h,~ Ly)? Roth, al motor Gti? s Downloaded _ from hlayiso.com ‘TS70(E}(M23)T DC MACHINES SPEED CONTROL TESTING DIRECT METHOD METHOD HOPEINSON EFFICIENCIES THE SAME IRON Loss PTO

MARKING GUIDELINE 2 ‘ EBLECTROTECHNICS NG i ’ Faves " Question4 higher education | . TRAE . * Armature copper loss training | ® intepole copper lose i * Series winding copper loss Department: : i © Compensating winding copper loss Higher Education and Training i * Brush resistance AnyTHREE (3) REPUBLIC OF SOUTH AFRICA - E 4.2 1.24 Ien= WiRoy= 480/240 = 2A Noto: lov'= lp and Row = Re * Way = ln toy = 10-2=8 A No-load Losses ” Pag. Ra = 8 x 0,2 = 12,80 Isu V = 2X 480 = 960 Wor ’eu. Row = 2? x 240 = 960 W C= lag. V~Fagay. Ra= (8x 480)+ 12,8 = 3 827,2W = 8 Full-oad Losses Pa. Ra 20,2 By , lou. VE 2 X480 = 960 W C=3827,2W NATIONAL CERTIF ICATE J oy Total full-load losses = 0,2 I2, +4 787,2 Piy=V Ua + ley) = 480 (I, + 2) = 4801 +960 * Pw = Pour + Total fulload losses } 480 Ia + 960 = 100 x 10° + 0,2 1, +.4:787,2 ‘ fa ELECTROTECHNICS N6 rs 0,2 I?,- 480 Ia + 103 788,2 = 0 an i’ . a I _ bt yb? = 4ac 5 . a . 2a ea i _ - dt ay ° (A480) of(-480)? - (4 x 0,2 x 103788,2) 23 APRIL 2012 ~ o, ia 2x02 s _ 480 4: 383,887 wo ary ce a = 240,283 A * The lasger value (2 159,717 A) Is rejected because of unstable operation of the motor, This marking guideline consists of 11 pages. , - ‘Hore? t Sint nen anata BA ‘@ | 4.2.2 Pra= Vx I= 480 x 242,283 = 116 295,84 W= 116,296 kW n= Pour/Py 100% . = 100/116,296 x 100% = 85,987 % (o) 123 At maximum efficiency PaRa= Igy. V+C Pq X0,2= 960 +3 827,2 1, = 154,713 A mC) & Hele, Copyright reserved Ploase tum over const record : Please tum over’

* MARKING GUIDELINE 3. ELECTROTECHNICS Ne QUESTION 2 2.8 Zny = 160 - j120 = 2001-36,87° 2 Zyp = 200 + j250 = 320,156 L51,34°2. Zax ~ 150+}0 = 150L0° 0 = 1,9.236,87' A Xe 2yp — 3802-120" "32055625136 = 11872-17134" A “ lp = 3802120" 150.20" = 2,$33.2120° A * 220 Iy= day —Ine = 19.236,87° - 2,533.20" = 2,787 —j1,054 = 2,98 2~20,716" A ly = Typ — Ty = 11872-17134" - 19.236,87" 2,693 ~ j4319 : = 2,999.2-153,905" A” Ip = Tye Ty = 2,533 2120" - 1,187 2-171,34" /378.292,268" A Copyright reserved “TS1O(E\(M23}T (6) Copyright reserved MARKING GUIDELINE + ELECTROTECHNICS NE QUESTION 3 31 Delta - Delta Delta - Star Star - Star Star - Delta sees 3.2 3.24 Method 1 _ __ k.8.C0s0 "ae = 5S.Cos6 + 2.PO 0.95 = 085% 500 «9.8 = ©B5 x 500x 0,8) + 2P, 340 095 = sy 2Pp Py = 8947 kW - x 100% x 100% Method 2 ke [Pe a Py =k? xP Po = 0,85? xPy. Py = 0,723P,. > At maximuni efficiency k.8.Cos® k.8.Cos@ + Po + k?.Pse 095 = 0,85-x 500-x 0,8 (0,85 x 500 x 0,8) + 0,723P,. + 0,857 Py. 340 095 a2 0s LMP e Pyc = 12375 kW Taw = x 100% . Py = 0,723P.e = 0,723 x 12,375 = 8947 kW 3.2.2 + At meximum efficiency y= 12,383 kW Downloaded from hlayiso.com TS10(&\(M23)T oo) Substitute in the formula below ~ @ (2) Please tum over

MARKING GUIDELINE ca TorUTe sees) ELECTRCTECHNICS NS 32.3 Peg = 12,375/0,85 = 14,568 KW () 3.24 $.Cos8 . = x 100% "= SGrsarPosbes “1% 500 x 0,8 = x 100% Goo x 08) +8,947 412,383 “1” = 94,937 % (2) , [13] QUESTION 4 4d (nm) ar] ‘The short-circuit curve fs obtained by short-circuiting the alternator terminals through ammeters and taking readings of short-circuit current and field current, The short-circuit current is plotted against the fie and itis a straight line through the origin. a 42 4.24 p= Ne 60 wa tx6o P = 50x 60 6 N= 500 r/min Copyright reserved . : Please turn over | MARKING GUIDELINE 6. | ELECTROTECHNIC NO Number of slots per pole = 72/12 =6 n = Number of slots per pole per phase = 6/3 = 2 = 180°/Number of slots per pole = 180°/6 =30° axsin( kd = = 0,966 Z-= Total conductors in serles per phase = (72x 103 = 240 Ep= 2kf. kd kp. f@.Z * =2x 1,11 x 0,966 x 1x50 xX 0,05 x 240 = 1 286,712 V Ey V3 x1 286,712 = 228,651 V @) 423 w= MY x 180° 60° kp = Cos wi2= Cos 60°72 = 0, Ep = 2.kf.kd.kp.f.0.2 = 2x 1,11 x 0,966 x 0,866 x 50 x 0,05 x 240 = 1114,293 V Ey = ¥3 x 1114293 = 1930,012 V (4) (20) @, ‘Copyright reserved Piease tum over Downloaded from h layiso.com

MARKING GUIDELINE ba i ELECTROTECHNICS N6 QUESTION 5 5.4 E Ex DLO. I T810(B)(M23)T If the load torque is increased, angle a also increases. but E remains the same If the excitation is constant. If E retards further it forces ER to retard. Angle @, the impedance angle remains constant therefore |_becomes more lagging, 52 vy = xv, 7 vp -# = 2500 3 . = 1443376 V Copyright reserved 4) a Please tum over MARKING GUIDELINE TS10(E)(M23)T 8. ELECTROTECHNICS NS Method LRg = 75x1,5=112,5V LXs = 75x10=750V Ep = Vpl0+ IRal 180° +1XsL-90° = 1 443,376 |-36,87° + 112,51.180° + 750L-90° = 1 042,199 -j1 616,028 = 1922,9471-57,181° V = V3Ep * = V3 x 1922,947 = 3.330,642 V 6, = 57,181" - 36,87" = 20,311" Method 2 7 Zs =R+jXs ng =1,5 +j10 = 10,112|81,469° 2 Cos @= 0,8 leading Er=1Zs = 75136,87° x 10,1121 81,469° = 758,4[118,339° V Ep =Er-V, - 758,41 118,330° = 1443,376 Lo” = - 1 803,379 + j667,509 = 1 922,952 |159,688° V EL=13. Ep = 3.x 1 922,952 = 3330,651 V G, = 180° ~ 159,688" = 20,312" 5 8 = 3687" 1 =75136,87° A 7 @) {1}. QUESTION 6 64 * — Direct switching » Stardelta starting «© Auto-transformer starting *_ Stip-ring motor starting 62 621 woe _ 60x50 3 = 1000 r/min 6 Copyright reserved Plonse tum over Downloaded from hlayiso.com

MARKING GUIDELINE ES ELECTROTECHNICS NO ga Nira x 100% es = 1000-970 1000 =3% x 100% 6.2.2 — Py= Stator output + Stator losses Stator output = Py, - Stator losses = 50-1 = 49 kw Stator output = Rotor input = 49 kW ‘ 2.2N,.T oR eoehul cotor input 7 p ~ 60x Rotor input 2X, 60 x 49 000 2x xx! 000 = 467,916 Nm Rotor copper loss = 220, =N)T ‘Ts10(e)(M23yT 1) (4) = 2%%x (1000 - 970) x 467,916 = ZE%% (1000 - 970) x 467,916 60 =1470W = 147 kw 4 6.2.3 Pe = Pour + Stator losses + Rotor copper losses + Friction and windage losses Pour = Pay = Total losses =50~(1+1,6+ 1,47) =48,93 kW — 1 = BE x 100% 45,93 A * 100% = 91,86 % Copyright reserved a @) ay [2] Please turn over MARKING GUIDELINE a1 ELECTROTECHNICS NE QUESTION 7 bs) The regulation of a transmission line is the rise or fall in voltage at the receiving end when the full-load fs thrown off, while the sending end voltage remains constant. 7.2 Rr= 0,285 x.150= 42,750 Lr= 1845x109 x 150= 0277 Xi=2.nfLy = 2xex50x 0,277 ” = 87,0222 Zap Rr +jX “= 42,75 + 87,022 = 96,956 163,837° 2 Cy = 0,00863 x 10x 150 = 1,295 pF C2 = 1,295/2 = 0,648 pR ! Xe = S407 . 1 » 2x wx 50 x 0,648 x10 491219 2 25e = 0-jXc + = 0-34912,19 . = 4912,19|.590°O \Le Va = Vi E = 90x 10°83 = 51961,524 10° V Becomign reserved, e, 3) Please tum over Downloaded from hiayiso.com r

MAKING GUIDELINE i ELECTROTECHNICS NB A Ind = Ve/Zpe 6 sNt = 51 961,524 0°/4 912,191-90° e > = 10,5781 90° A ~ Ne NE PNBVi.Cos 8 ‘ “15x 10°N3 x90 x 10x 0,8 = 120,281 1-36,87° A ah ) ). ( Tan=In+Ipe ‘= 120,281 L-36,87°+ 10,578 L90° = 96,225 - 61,591 = 114,248 [32,6227 A *Vaa = Tap.Zap = 114,248 |-32,622° x 96,956 L63,837° = 11. 077,029 |31,215° V Vs = Vat Van = 51 961,524 0° + 11 077,029 [31,2159 = 61 434,916 +35 740,681 = 6f 702,54715,338° V Vi = ¥3x 61 702,547 = 106,872 KV NP Ino = Ve/Zun & = G1 702,547 L5,338°/ 4 912,19 -o0° “A = 12,561 95,338" 4. sw fo b= latho 5 NS = 114,248 |-32,622° + 12,561 95,338" 3 = 95,056 - 549,083 = 106,98 L-27,31° A Power factor = Cos@ = Cos (5,338 + 27,31°) = 0,842 Lagging Copyright rests Downloaded from hlayiso.com Answer Force jes) TOTAL: (13) ne] 100

higher education actraining Hichar Eduoation and Training REPUBLIC OF SOUTH AFRICA TS50(E)(J21)T AUGUST 2011 NATIONAL CERTIFICATE ELECTROTECHNICS NG (8080096) 24 July (X-Paper) - 09:00 ~ 12:00 REQUIREMENTS: Graph paper Calculators may be used. This question paper consists of 6 pages and a 5-page formula sheet, HOU ae DEPARTMENT OF HIGHER EDUCATION AND TRAINING REPUBLIC OF SOUTH AFRICA NATIONAL CERTIFICATE ELECTROTECHNICS N6 TIME: 3 HOURS MARKS: 100 INSTRUCTIONS AND INFORMATION 1. Answer ALL tha questions. Read ALL tho ueelors carefully. Number the. answers correctly according to the numbering system used In this question paper, ’ Start each question on a NEW page. Keep subsections of questions together. Round ALL calculations off to THREE decimal places. Use the correct symbols and units. SX On sp Write neatly and legibly, PTO

3 TSSO(EKU21)T QUESTION 4 14 Draw a neat diagram of a pair of identical shunt motors mechanically coupled to the same load and connected in parallel. 2 Draw the speed and torque characteristics with respect to the armature current of the above-mentioned shunt motors. 13 A 440 V, DC series motor with an armature circuit resistance of 0,2 ohm runs at 600 rfmin when taking a current of 50 A. Calculate the value of the resistor which, when inserted in series with the amature, will reduce the speed to 400 r/min. The gross torque is half its previous value. Assume the flux to be proportionat tothe field current. ~ QUESTION 2 7 A three-phase, star-connected altemator with a line voltage of 500 V, suppiies an unbalanced star-connected load with no neutral connestior. The load consists of the following impedances: Za = 5 +)0 Z = 10+)0 Zp = 15+jo Take Vrw @s reference between phasor and assume a phase rotation of R-Y-B. Calculate the following: 21 The potential difference betwesn the star point of the load and the neutral point of the aliemator 22 ‘The potential difference across each phase of the load 23 The current in each line a (a nz + TSSO(EXIZ1)T QUESTION 3 3.41 Explain, with the aid of a diagram, how a short-vircuit test Is carried out on a single-phase transformer 3.2 A 25 KVA, single-phase, 50 Hz, 2 200/380 V transformer gives the following fest results: 380 V at normal frequency epplied to the 380 V winding. The power Input Is 150 W, Open-cirouit test: Short-circuit test: 60 V at normal froquency applied to the 380 V winding and fuil-load current circulating in the 2200 V winding. The power input is 200 W. Calculate the following: . e 3.2.41 The resistance, reactance and impedance referred to the secondary side 3.22 “The percentage regulation at ful-load end 0,8 power factor lagging 3.2.3 / The efficiency at full-oad and 0,8 power factor lagging QUESTION 4 44 Show, with the aid of phasor diagrams, how the regulation of an altemator Is affected by the following power factors of the load at: 44.4 Unity power factor 41.2 Legging power factor i 4A3 Leading power factor (8 x 2) 42 A_ three-phase, eight-pole, star-connected altemator has a_single-layer winding with 16 conductors per slot. ‘the winding is full-pitched. The armature has a total of 48 slois. The altemator is Funning at 750 r/min with a flux of 0,04 Wh per pole, Calculate the following: 4.2.1 ‘The frequency 4.2.2 The distribution factor 4.23 The induced EMF per phase «4 PTO 6) 6) @ (2) H7] (6) @) (3) [16]

ca TSSO(E|I24)T ye QUESTION 5 54 52 Explain what is meant by Aunting in a synchronous motor and how tt Is minimised. The intemal resistive and reactive volt drops of a 700 KVA, 3 KV, three-phase synchronous motor that draws 550 kW at a leading power factor are 70 V and 720 V respectively, when fully-loaded, Calculate the following: 5.2.4 The power factor 5.22 The resistance, reactance and impedance 5.23 The line EMF , 5.2.4 The load angle’ QUESTION 6 61 6.2 Draw the circle diagram of a 15 KW, 440 V, 50 Hz, 4 pole, three-phase induction motor using the following data: No-load test: 440V 112A Cos = 0,138 Locked rotor test: 160V 40A Cos6= 0379 Assume the stator and rotor Capper losses are equal at standstill Use ascale 1 cm = 6A, Determine the following from the circle diagram: 6.2.4 The percentage slip at maximum torque 6.2.2 The efficiency at maximum torque 6.2.3 The power factor at maximum torque 8.2.4 The line current at maximum ie 62.5 The maximum torque in N.m 6.2.5 ‘The rotor copper losses at standstill TO a a (a 4 (9) (i) a) (4) (1) (2) (1) [16] HOOUU eH) QUESTION 7 Two 2200V, star-connected altemators operating in parallel, supply the following loads: = 50 kW atunity power factor - 400 KW at 0,9 power factor lagging - 360 kW at 0,8 power factor lagging - 100 KW at 0,8 power factor leading One machine supplies current of 190A at a power factor of 0,8 lagging. Calculate the following: 7.4 ‘The armature current of the second machine (8) Tz The power factor of the second machine (2) 7.3 The output of the second machine in kilowatts (2) q12] TOTAL: 100 6 a e

education Se QUESTION! Department: : P + a Education T : T REPUBLIC OF SOUTH AFRICA Cw OQ J 2 Ses ie ae ee 12 i : : "Torque Characteristics Speed Characteristics : 5 Torque Speed vw August 2011 | v ELECTROTECHNICS N6 i F * 8080095 . * e) -Augustus 2011 re &LEKTROTEGNIEK N6- ° Te = 05.01 Th/Th = 05 BG ss 30 wv = 43097 i Toho But @ Ih. “n.Te RB “De T ty . tee fan, vv : T = f0 5x50" = 353554 \~ Ba No Bat bec 2 BeN a, a ol ‘ ds Downloaded from hlayiso.com

Ey _ Nada BH Nida Nale-E g, = Mele 2 Na _ A00x 35,355 x 430 ~ 600 x 50 = 202,702 V a = V-I_,R,+R) -YoE Rn / Tg = 440 202,702 _ 35,355 = 6522 \/ z, R 0,2 @) 1) = 0,220" + O20" + 0,067.20" = 0,367 + jo = 036720" 1% New. 2886520" «59 73520" = 57,735 + j0- % seat ow Voy _, 288,6752-120" _ aCe -y . Fe gr 7 PRR6RL 10" = —14494— 25 ~ Ve _ 288,675.2120" = 7 Many B8B:SISL120" = 19.045 2120" = — 9,623 + §16,667 Z ao 9,245, 4+ 516,667 Mer... Vie. Max. 33,678 ~ $8,333 = 34,694 213,898 |” 6 Downloaded from h we SA6044 18,898" "0,367.20" = 94,8342-13,898" V 22 Vas = Vax Vex = 288,675.20" — 94,5342-~-13,898" = 196,908 + 22,707 = 198,213.26,578 V Ves = Vir — Vax 288,675 2-120" — 94,534.2—13,898" 236105 — 5227,293 = 327,7312~-136,089' V Vas = Vou — Var = 288,675.2120" — 94,5342 ~13,898" = 236,105 + 272,707 ° = 360,714.2130,885 V 23 a 4 =e 19821326578, $£L0° = 39,643.26,578" A we 6 -com

= Yn ed = 327,73.2-136,089"° 1020 = 32,773 £—136,089" A v Nos, Ez = 360,714.2130,885" 1520" = 240482130805 4 \/ Ip = QUESTION 3 L 8) 13} 32 ~ 324 S=Vi.b h=8/¥V. ~~ = 25 000/380 = 65,789 A Psc = 12, Re, Re, = Pso/Iz = 200 165,789" = 0,046 & Vso = I. Zea Zen = Veal VY = 60/ 65,789 ~o9n2 / Ze, = JRE +H X= (ZR Y . = 09127 — 0,046? =091 a \/ ® 3.2.2 Method 7, Cos 62= 0,8 «Sin =0,6 % Regulation = 226 Gos8, + Fe, S063) sop, 2 65,789 ([0,046 x 0, r me x08]+[0911%06]) . sogy, 380 = 161% wf ‘Method 2 Cos8, = —Pee_ T-Vin 200 "65,789 x 60 = 6,051 Og = Cos 0,051= 87,077" 8, = Cos 0,8.= 36,87" Regulation = Yie:Cos@a~ 82) «100% - 60 x Cos(87,077" - 36,87*) = SOX 605(87,077" — 36,87") ‘380 * 100% = 10,106% v oO) al

323 ne 2 Cm. 1008 $.Cos@ + Po +Psc 25000 x 08 100% +1504220 ~ v @ 17) QUESTION 4 4d Unity power factor Ep Is v v a Ve Rm r ‘Logging power factor © 4.2 42d *E Glow 50 = 50 Bz \/ 42.2 Number of slotsper pole = 488 = 6 n = Nomber of slots per pole per phase = 6/3=2 V~ o. = 180% Nuuober of stots per pole = 18076 = 30° | ~ = 0965 wieneae , Z = Total conduotors in series per phase = (48x 16)/3 = 256 |” we ead nossecl es0x40x10%x 286° W/ = 1097,994.7,/ Method 2 T= Tums /phase ~ 256/2 = 128 V~ Ep = 4.k£.4d.4p.f.0.7 = 4x 111 x0,966x 1x 50x40x 10" x 128 = 1097,994 V7 6 @ © @) [16]

QUESTION 5 , 523 54 vy, = ey, ‘Whan a synchronous motor is uscd for driving a varying load, then a y, = condition kuowmn as hunting is produced. =e ‘To minimise the build up of these oscillations, dampers or damping . : oat grids ere used, |/ : aB = 1732,051 V va ‘These dampers consist of short circuited copper bars embedded in the faces ofthe field poles ofthe motor. |// ‘Method 1 ‘The oscillatory motion of the rotor sets up eddy currents in the dampers 8 = Cos? 0,786 = 38,187" ‘Which flow in euch a way as to suppress these oscillations. . @ . Sea ve Ep = Vpl-0+ ial 180° + LXs L-90° S21 = 1732,051 L-38,187° + 70L180° +720 L-90° Cos0 =F = 1293,387-j1790,806 = 550. = 2.207,865 L-s4,204° V // 700 = 0,786 Leading By = ¥3.Ep = ¥3x2 207,865 = 3 824,134 ¥ ; o Method 2 eee Br = [.2s , ’ s = 134,715138,187° x 5,37184,443° s= Vo = 72342 [122,63 V7 5 . , : Eee tO NB, Ep = Er- Vp = 700 x10? = 723,42 122,63 — 1 732,031 LO" 3 «3.000 : = 145A = —2122,128 +-j609,243 R=Vx/ I= 70/134;715 = 0,520 “ , = 2207,851163,982° VV = Vx/1=720/134,715 =5345 2 BL = 13.Ep 5 as . i eal = ¥3x2 207,851 ( Zs = ofR?4K? or Zs =R+jKy =384l1V | ® 7 = 0,527 45,3457 = 0,52 + 55,345 Ere yen TST 52a nw oa om =5372 = 537284443 2 |” v @ or ce = 180°. 163,982 = 16,018" | w@ G4] « . * « " Downloaded from hlayiso.com

QUESTION 6 6A Cos'Oe. = 0,138 Ou. = Cos 0,138 = g2° ~ Cos Gsc = 0,379 @se = Cos 0,379 = 68° Isc Test current x Rated Voltage / Test Voltage = 40 x 440/160 “110A / Scale: lem = 6A Tn = 12/6 = 2 cm Isg = 110/6 = 18,3 cm 62 624 % Slip-* BL/GI x 100% = 1,6/7,2 x 100% = 22,02 % \/ 622 11 = Pout/Pin x 100% = GH/GK x 100% = 5,65/9 x 100% = 19% 623 Power factor = Cos 8 = Cos 46° = 0,695 Lagging 62.4 I = Distance x Scale : = 0Gx6 = 128x6 = 6A 6.2.5 Pin = V3. VL. Distance (GI). Scalo =V3 x 440 x 72x 6 = 32,923 KW @) ® oO @ @) Ni = 60.£/p = 60x50/2 = £500 r/min Ps 2.2n.N,.T 60 __60.P ~ 20.Ny _ 60x 32.923 ~ 2xnx 1500 = 209,594 Nan v 6.2.6 Pin = 13. Vi. Distance (RF). Seale = 3 x 440 x33 x6 = 1509 KW 7 QUESTION TA Method 1 For the 50 KW load: 8, = P/Cos8 = 50/1 = 5020" LVA For the 400 XW load: 8, = P/Cos8 = 400/09 = AB AMELIA VA ‘For the 350 KW load: 8, = P/Cos0 = 350 /0,8 : = 43752-3687 kVA a * For the 100 kW load: : 8, = P/Cosd = 100/08 = 1823687 KVR | /K Q) {i6]

S, =8, +8, +8, +8 = 50Z0" 4 486,444 2~ 25,842" + 437,5.2—36,87" + 125236,87° = 899,999 j381,229 4 = 97TAIL- 22,957 VA 8, = B.VG 1 == 7 ABN, 977,812-« 10 OB x 2200 = 2565042-22,957 A Method 2 Bor the 50 kW load: ba Tu = Bry, Cost _ 50x10 “ Y5x2200%1 * = 13,1220" A For the 400 KW loads” e la = "y,ood 400°x 10° * ox 2 200% 09 = 1166352-25,842" AV” Yor the 350 KW load: flee ee ea. BV, Cos a 50x10" © Fax 2 200% 0,8 4 = 14842-3687 A or the 100 kW load: facet V3. V, .Cos8 10010? 3 *2200 x 08 = 32,86.23687 A el+h+h+h 13,122 20" + 116,636 Z—25,842° + 114,814 £~36,87" + 32,804 236,87" 236188 — j100,048 = 256,5042-22,987" KVAy” For thefirst machine: I,, = 1902-3687" A For the second machine: ly =Int+ly Tasty -Iy = 256,504.2-22,957" —190.2—36,87", = 84,18B + J13,953 »% = 8533629," A ® 12 f Power Factor = Cos 0 = Cos 941° = 0987 // @) 13 P, = 3. Vz. Tpq.Cos0, = 3 x 2.200 x 85,336 x 0,987 = 320,947 LW @) 2) TOTAL:100 . al

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higher education & training Depariment: Higher Education and Training REPUBLIC OF SOUTH AFRICA JELALLLIOZ Ce T660{E)(N10)T NOVEMBER 2011 NATIONAL CERTIFICATE @ ELECTROTECHNICS Né (8080096) 410 November (X-Paper} 09:00 -— 12:00 Calculators may be used. This question paper consists of 6 pages and a 5-page formula sheet. pes e 2 TSSO(ENN10)T DEPARTMENT OF HIGHER EDUCATION AND TRAINING REPUBLIC OF SOUTH AFRICA NATIONAL CERTIFICATE ELECTROTECHNICS N6 . TIME: 3 HOURS MARKS: 100 INSTRUCTIONS AND INFORMATION 1. Answer ALL the questions. 2s Read ALL the questions carefully. 3. Number the answers correctly according to the numbering system used in this question paper. 4, Start each question on a NEW page. 5. Keep subsections of questions together. 6. Round ALloaloulations, off ) THREE decimal places. Pa Use the os iteymbols al 8. Write neally and legibly; p Bag Downloaded from hlayiso.com

-3- TESO(E\(N1O)T QUESTION 1 141 A three-wire DC system has two operating voltages of 440 V and 220 V. Show by means of neat diagrams, how the armatures of two shunt motors are connected fo this three-wire system so that the speed of one motor is twice the speed of the other. (4) 1.2 Two similar DC shunt machines were tested by means of the Hopkinson method. The following readings were obtained: ‘Terminal voltage of each machine = 600V Armature circuit resistance of each machine 0,052 Generator output current 1200A Input current from the mains = 400A Motor field current = 20A Generator field current = 225A Calculate the efficiency of the machine operating as a motor, assuming: (10) Equal iron and friction losses [14] QUESTION 2 oo An EMF represented by e = 100 Sin (raise ) + 50 Sin (942t ) voitis applied toa series circuit having a resistance of 10 ohms’and a capacitance of 80 mierofarad., aid Calculate the following: Levinas 21 An expression for the instantaneous value of the current . 3 (8) 2.2 The RMS value of the current and voltage (4) 23 © The power absorbed by the circuit w @ 24 \cthe energy dissipated in the circuit during 5 milliseconds . i Aq + TSSO(E)(N10)T QUESTION 3 3.4 What are the effects of harmonic currents in transformers? 3.2 Define voltage regulation of a transformer. 3.3 A 200 kVA,, 2 000/250 V, single-phase transformer has a percentage impedance of (1,2 + j3,5)%. Calculate, the following at a power factor of 0,8 lagging: 3.3.1 The full-oad efficiency if the iron losses amount to 2 kW i 3.3.2 The maximum efficiency 3.3.3 The applied primary voltage to circulate fullload current on short- circuit @ 3.3.4 The percentage regulation ~ QUESTION 4 44 Explain what is meant by jnfinite bus-bars as applied to a number of altemators running in parallel. 4.2 A 11 kV, 50 Hz, three-phase alternator, working at a power factor of 0,9 lagging, has a percentage synchronous impedance of (5 + j30) %. Calculate the following: 4.24 The EMF of the alternator if the stator is star-connected 4.2.2 The percentage voltage regulation @ 423 The load angle at which the machine operates PTO 2) (3) 6) (3) @) (2) [18] 6) @) 2) a) 116]

T660(E)(N10)T «4 QUESTIONS 5.1 Explain, with the aid of phasor diagrams, that the power factor of a synchronous motor working on a constant mechanical load, depends on its excitation. 5.2 The impedance of a three-phase, 50 Hz, star-connected synchronous motor is (1 + j5) ohms per phase. The motor takes a current of 196,8 A from the supply. If the terminal EMF of the motor is 3.338 V at a power of 0,8 leading, calculate the following: 5.2.4 The terminal voltage when the load angle is 28 electrical degrees 5.2.2 The input to the motor in kilowatts 5.2.3 The mechanical output in kilowatts, if the efficiency is 88 % QUESTION 6 A four-pole, three-phase, 50 Hz induction motor is delta-connected on the stator with 520 conductors per phase. The rotor is star-connected with 106 conductors per phase. The standstill impedance is (0,2 + j1,8) ohms per phase. The supply voltage is 525 V. Calculate the following: 6.1 The rotor EMF and current at standstill. 6.2 The rotor EMF and current at a slip of 4%. Vv, _N, Hint: <=) E 0 2 QUESTION 7 : Each conductor of a 200 km transmission line has a resistance of 0,3 ohms per km, an inductance of 4,95 millihenry per km and a capacitance of 0,0093 microfarad per km to neutral. The line delivers a three-phase load of 20 MW at a power factor of 0,8 lagging and a line voltage of 132 kV, 50 Hz. ‘@ Calculate the following by using the T-method: a 7A The sending current (6) (6) (1) (1) [13] (11) 7.2 7.3 Downloaded froim hlayiso.com : -6- TS6O{E)(I The sending voltage The power factor at the sending end Important: Draw the T-method circuit diagram

education Bepartment: Education REPUBLIC OF SOUTH AFRICA November 2011 ELECTROTECHNICS Né 8080096 Movember 2011 ELEKTROTEGNIEK NG QUESTION 1 LA A sv . 220.V vw | 220V we ¥v ® 12 = 4004 : Y=1 200A it aii LtbhY 1600 - ‘a rd L=WAy L+h-hy 15804 Ltb{ 125A Lhasa sry Pstotor Generator |” Ra 0,052 aS 0,052" i Let © = Iron, Friction and Windage losses CHk.V-(Gith-L).Re+ (hth WRat LV + bY] = (400 x 600) - [(1 5807 0,05) + (1 2257x 0,05) + (20 x 600) + 25 x 600)} See an ad aah oa = rama w Ci2 = 6 574,375 W Va as | Downloaded from hlayiso.com

+ MotorInpui (Pin) = y+ B).V = 1600 x 600 = 960 KW” Motor Losses = (+ h- 1). Rat. V +02 = 124 820+ 12.000 +6 574,375 = 143,394 LW vo MotorBticiecy = 22% x 100% = ieee = ; x 100% Pin _ 960-143,394 so 99% 960 = B5063% \/ ay [4] QUESTION 2 r R c JL 10a 1150 uF 21 * 34x50 x 10° . = 63,6942 Z, = R-jXq =10-j63,604 = 64,474.2-B1077 2 a _ 100260" * 64,474 2-81077° fa ent = 1,5512141,077" A 4 Third Harmonic Xo = Keil = 6369413 = 21,231 Q a . y = R-jXqq = 10~J21231 = 23,468.2-64,719"° 2 | So 502-30" ° , 23,4682-64,779" = 2,131.234,779" A i= 1,551 Sin(B14¢ + 141,077) + 2,131 Sin(942t +34,779%) A . 22 fag HE : : 2 _ [1007 +50" uv 2 = 79,05TV ~~ Eas ee Ta tNis poss 4 raat we 7 = 1,864 A ey ® _ 23 ‘Method 1 : Py = Tne RK = 1,864? x10 = 34,745 -W Downloaded from h ayiso.com

«Method 2 Tyg: Ena 0080), ,, Ing-Fro-Co8ts B= ee 2 2 _ 1,551. 100 x Cos81,077" , 2,131x50 x Cos 64,779" a 2 2 =34,73 W \/ 24 =Pyxt = 34,745 x 5x 10° = O174T =i4ml | QUESTION 3 3d Any two e Increase incoreloss * «Increase in YR loss due to circulating currents and protective devices’ : 32 seis tho ise o fl inthe secondary ena voll when fhe Toad ie remy andthe primary voliage remaining consent“ It is the difference in the scoondary Voltage between no-load and «/ full-load, expressed as a ratio ‘of the no-load voliage with the primary ‘voltage remaining constant. |“ 33 334 Method 2 ; s=V-h 8 hes Ly _ 200x10" 2 000 = 100 A q@ @ [4} > 6 we RM y _ 0012 x 2.000 Ra Be = GR = 800*x 3,725.0 107 = 24kW ____ 8.Cost = "§Cos0 + Po + Bsc _ 20x 08 200 x08) +2424 = 97324% via x 100% x 100% ~ Downloaded from hlayiso.com

Py = Bye = [Po k = ype . E . V24 = 0913 / a Method 1 : = K.S.Cos0 .S.Cos + Po + k*.Pso 0,913 x 200 x 0,8 x 100% a * Q915 x 200 x 0,8) +24 (0915 «2,4) = 97334% | Method 2 1 x 100% .__ E5.Cos0 &.S.Cos@ + 2.Po + 0,913 » 200 x 0,8 (G915 x 200 x 0,8) + @xz) * 1% 2 = 97334% V7 x 100% e) 333 L * = 0035 2 000 100 =0,7 2 ~ 5 Method 4 : Buy = Ry, FjXp = 0244507 = 0742-71015 A Method 2 Zp = fR3, 4X, = 024 + 0,7 ~0,749 a . =4V \/ : 334 Method 1 Voltage Reguation = 8m :0280+ Xn. SiN8) 4996 A = 20000,24 x 081 + [0,7 x06) patna x 100% = 306% V7 Method 2 Voltage Regulation = %R.CosO + %X. Sin8 = (1,2%x0,8)+ (5% x00) ~~ = 3,06% 7 s § ; : QUESTION 4 - ° 4. ‘The synchronous i of all eltemators feeding into the system in parallel is regarded.as zed. Tho bus-bem voltage remains consient under nil condiicas aloes Sat power fist. 42 a 424 Vp = Va/%8 = 1000/3 = 6350853 V V~ IRA = %RLV = 0,05 x6 350,853, = 317543 ¥ LXs= %X.V = 0,3 x6350,853 = 1905,256 V Va 8) @ [18] 6) Downloaded _from hlayiso.com —

Method 8 = Cost 0,9 = 25,842" Ep = vie + IRaLo? + LXsL90° = 6 350,853 125 842° + 317,54310° + 1905,256|90° = 6 033,307 + j4673,535 = 7631,692|37,762° V if By = ¥3x7 631,692 = 13218478 V y/ ‘Method 2 Cos 0 = 0,9 Sind = 0,436 Ep = f(V.Cos6+ER,) +(V.Sind0+LX,) = ¥{([6350,853 x 0,9] + 317,543) + ([6350,853 x 0,436] 1905256)" Va = 7632,119 vy HE, = ¥8x7 632,119 = 13 219,218 V\//” ‘Method 3 Ep = Vp-+ (LRa. Cos6)+ (LXs. Sind) __ = 6 350,853 + (317,543 x 0,9) + (1 905,256x 0,436)“ ” = 7467333 V\/ ‘ Ey, = ¥3x7 467,333 = 129338 Vy ® 422 Regulation = PP=VP_ x 100 9% Vp = 7631,692 — 6350853 . roqo, 6350853 : = 018% @ 423 ~ 2 9 = 37,762° ~ 25,842° =19r 7 @ {16} &, | | al s QUESTION 5 51 ‘Lagging power factor z Wa ak a a Ve . qT, v Increase in excitation increases Ep end causes the resultent current to lead. ‘Leading power factor 52 - 524 T= 1968A “ =» Ata power factor of 0,8 leading = 196,8136,87° A Ep = B,/V3 = 3338/13 ‘ = 1927195 V > —At28 electrical degrees = 1927,195 |180 -28° ‘ = to27aesltsz° VK Method = Zs = 1445 = 5,099178,69 2 Fa = 128 = 196,8136,87° x 5,099 178,69° = 1003483 L1i5,56° V & ®-

Ex = Vp + Ep Vp = Ex - Ep = 1003,483 115,56? - 1927,1951152° = 1268,653 + 40,513 / = 126a653lo" V7 Vy =03 «Vp = V3 x1 268,653 = 2397371 V // Method 2 Yn LRa = 196,8x1 = 1968 V LXs = 196,8x5 = 984.V Ep = VpL-$+LRy |180°+LXs L-90° Ve = Ep|-36,87° - 28°= LR, L180? ~ LXs |-90° “ = 1927,195 L-64,87° - 196,81180° - 9841-90" = 1015,229 —§760,779 ~ = 1268,651 |.36,847° VL“ V, =V3 x Vp = 3 x1 268,651 = 2197368V 1 Se. Pp = V3. Vz.3..Cos® = V3 x2 197,368 x 196.8 x 0,8 = 599,209 KW | ‘ oO @ @ © | 5.23 | ' n= Pour! Pa . Pour = 7) x Pry = 0,88 x 599,21 | = 527,305 kW / QUESTION 6 64 Sie NL E, N, aN, / Ba _ 525 x 106 520 =107,019 VV” Ey =V3xE, . = V3 x 107,019 = 185,362 V Method 1 Zo= Horan = 02+j1, = 1pitl36e°-n ~~ Method 2 Z, = fREVK? = 02 +18" =18110 Ey L-z _ 107,019.20" 18H 283,66" = 59,094.2-83,66" A 7 Downloaded fr wo [33] om hlayiso.com

ol % % fl 62 E, = 8. E> = 0,04 x 107,019 = 42a VO By = V3" B, = 3x 4281 = 7415 27 BT = 0,24 40,072 = oars, Method 2 Za = YRIFKE = fo? + 0,072? 4 =o20 \% En Z, __ 428120" 0.213.219,799" =20,099.2-19,799 A ~~ i= QUESTION 7 Dik R/2 X/2 °° ATR R/2 MI z ® {10} | [ fi fio] ts Load. Ya B & Wl Rr = 03200 = 690 wf Lg = 195x107 2200 = 0,391 X= 2.2.f.1e = 2x2x50x0,39 = 12522 \/ a Zag = By /24 5K IZ _ = 30+j 61,261 \ = 68,212 163,909" a eed Cr = 0,0093 x 10°x200 = 1,86 pF 1 Xo F5tG = 1 = = Jem x 50x 1,86 x 10* = 171388 - Va = Ves 8 = 132x108 = 76210236L0°V ~~ "pe JE; dy.Co38 I= —_P we ABN, Cos _ 20x 10° OB x 152 x10? «0,8 = 109,3472-3687 A wv > Vas = In-Zas = 109,347 |-36,87° x 68,212 163,909° = 7488,777L27,039° VV” = Van + Ve 1 458,177 (27,039° + 76 210,236 Lo® = 82.853,749 +} 3 390,737 = 92.923,10212,343° V NM = A Zap = 0-j Xo = 0-j1 711,343 21 7i13a3190° 2 Ae > Nt hk

Tan © Van Zan 7 f = 62,923,102 [2,343°/ 1 721,343 L-0° 48,455 L92,343° A Jy = Tan tp » 48,455 L92,343° + 109,347 -36,87° 85,496 —j17,194 = 87,208 L-11,371° A 7 12 ‘Vos = Is -Zpa = 87,208 L-11,371° x 68,212 163,909" = 5948,63252,538° V \/ Vs “Vos + Van = 5 948,632 |52,538° + 82 923,102 |2,343° = 86 471,945 +8 111,802 = 86 881,59 |5,359° v ‘Vi = 3x86 851,59 = 350431 KV | 3 Power factor = Cos = Cos (11,371° +5,359) . 0958 Lain 7 ay) @) TOTAL: 100 <ul Downloaded from hlayiso.com POAOKE NAO) T