ESE (Prelims) - Offline Test Series ELECTRICAL ENGINEERING. SUBJECT: ELECTRICAL & ELECTRONIC MEASUREMENTS AND POWER SYSTEMS Solutions

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TEST ID: 34 ESE- 9 (Pelims) - Offline Test Seies ELECTRICAL ENGINEERING Test-7 SUBJECT: ELECTRICAL & ELECTRONIC MEASUREMENTS AND POWER SYSTEMS Solutions.

1 TEST ID: 34 ESE- 9 (Pelims) - Offline Test Seies ELECTRICAL ENGINEERING Test-7 SUBJECT: ELECTRICAL & ELECTRONIC MEASUREMENTS AND POWER SYSTEMS Solutions. Ans: (c) Sol: In both given connection methods, eo caused by unknown esistance side mete i.e., in connection (a) ammete and in connection (b) voltmete. V in R a =. A R V =5k V R= If unknown esistance, R = R R a V, the eo caused in both methods ae equal Given: R a =. R V = 5 k R =. 5k = R = (given) Theefoe, any of the above two connection, as both of them give equal accuacy.. Ans: (b) Sol: ev kwh 6 ev min 3. Ans: (a) ev pm 6 Sol: In eed type fequency mete, the fequency at which the eeds esonates is twice the input fequency, i.e. if a eed is esonating at 5Hz, then the natual fequency of esonance fo the eed is Hz. The above explanation holds good only fo pue a.c cuents. But if the input cuent has a d.c offset along with a.c component then the eeds esonates at fequency equal to input fequency.

2 : : Electical Engineeing So in the given question it is said that the eed type fequency mete ange is 47 53Hz. So the eeds ae designed to esonate fo fequency ange of (47 53 ) = 94 6Hz. In the second case when dc bias is povided then the eed type fequency mete esponds to input fequency ange of 94 6Hz. 4. Ans: (c) Sol: Dynamomete instument is used as tansfe instument between ac and dc Themocouple based instument is a tue ms mete Ramp geneato (o) saw tooth geneato) is the time base geneato in CRO. Weston standad cell is standad of emf. 5. Ans: (a) Sol: In - electodynamomete type pf mete, the values of R and L ae so adjusted that the two moving coils cay same magnitude of cuent R = L. Hence - electodynamomete type mete indications ae dependent on both fequency and shape of wave fom (since any wavefom othe than sinusoidal contains hamonics) and will give ise to seious eos. But in 3- electodynamomete type pf mete, both the moving coils ae connected though esistos. Hence in 3- electodynamomete type pf metes, the indications ae independent of wavefom and fequency of supply. 6. Ans: (b) Sol: Dead zone is defined as the change in the physical vaiable to which the instument doesn t espond. Lag (minimum) is the time equied by the instument to begin to espond to a lage in the measuement. 7. Ans: (c) Sol: In moving Ion type instuments, flux poduced due to the cuent flowing though the instument don t follow a linea elationship in pactical case because of hysteesis. max c a Desendng path I b I max Ascendng path I

3 : 3 : ESE - 9 (Pelims) Offline Test Seies In ideal case we assume that I I = k But when hysteesis is taken in account then I = k does not hold tue. In the above figue, when cuent is inceased fom to I max, inceases fom to max. But when cuent is deceasing fom I max, follows a diffeent path along bc. Hence if we look closely, fo the same cuent I, flux poduced is geate in descending path [ > ] and hence it will indicate highe eading. 8. Ans: (c) Sol: Themocouple that can measue tempeatue in the ange 3 C to 4 C is R type of themocouple which has ange of 5 C to 768 C. Its composition is 87% platinum, 3% (platinum + Rhodium). 9. Ans: (a) Sol: (i) Given T d = I T C = fo sping contol T C = Sin fo gavity contol At balance position Fo sping: T d = T C = I I linea scale Fo gavity: T d = T C I s sin = I sin Sin I non linea (ii) Gavity contol can be used only in. Ans: (b) vetically mounted systems. Sol: Damping toque T D B R dbt B = flux density of damping magnets R = Radius of disc = speed of otation of the disc = esistivity of the disc t = thickness of the disc. Ans: (c) Sol: sh =. (fixed) Fo R =, full scale deflection (fsd) is obtained fo a cuent of 5 A Voltage acoss the shunt is V s = 5. =.5 V Cuent though the mete fo full scale deflection (i) = i =.5 ma Vs.5 =.5 ma R Now it is given that fo I = A deflection is 4%

4 : 4 : Electical Engineeing Fo fsd I = V sh = I R sh R = 4 / = 5 A = 5. = 5 V i =.5 ma Vsh 5 i.5 R =,. Ans: (b) 3 = 4 Sol: A ideal PMMC voltmete will ead aveage voltage, fo -, half wave voltage contolle, output voltage, V cos m V = V m = 3. Ans: (b) Vm = Sol: Uncetainty distibution is used fo analysis of single sample data. 4. Ans: (b) Sol: The disc should be conducting such that it povided path fo eddy cuents and at the same time it should be non-magnetic as it should not poduce any magnetic field which can affect the woking field. 5. Ans: (c) Sol: tan = W W 3 W W Reactive powe consumed pe phase, W W Q = 3 Q = tan = 6. Ans: (c) W W 3 59 = VAR 3 3 Sol: Rayleigh s cuent balance: This instument woks on the pinciple that if a cuent caying coil placed with its paallel to that of anothe cuent caying coil with thei axes coincident, thee will be a foce exeted between the coils. This foce is popotional to the poduct of cuents in the two coils and if the coils cay same cuent, the foce is popotional to the squae of the cuent. This foce can be measued if one of the coils is movable and is suspended. 7. Ans: (c) Sol: A swamping esistance of manganin having a esistance to 3 time the coil esistance is connected in seies with the coil and a shunt of manganin is connected acoss this

5 : 5 : ESE - 9 (Pelims) Offline Test Seies combination. Since coppe foms a small faction of the seies combination, the popotion in which the cuents would divide between the mete and the shunt would not change appeciably with change in tempeatue. 8. Ans: (c) Sol: Clamp on metes ae used to measue cuent flowing in a line with out baking the cicuit. 9. Ans: (c) Sol: Phase angle eo in cuent tansfome is given by = ImCos IeSin ad. ni s Fo inductive budens is positive. is positive fo small values of and is negative fo lage values of Fo capacitive budens is negative. is positive iespective of magnitude of. Wattmete eading, W = V ms I ms cos = = 5 W. Ans: (d) Sol: Linea Vaiable Diffeential Tansfome has one pimay and two seconday coils connected in phase opposition i.e., 8 out of phase.. Ans: (c) Sol: At bidge is balanced R eq = R// R R / /R I S = E R eq = R E.. () R At balanced detecto is shot cicuited I S R R/3 R/ R/3 R/3. Ans: (c) E Sol: 3 3 i L ef. V i & i L = 6 R eq = R 3 S S ; 3E S.. () R I I E I 3E V i = 3

6 : 6 : Electical Engineeing

7 : 7 : ESE - 9 (Pelims) Offline Test Seies 3. Ans: (d) Sol: The least suitable tansduce fo static pessue measuement is piezoelectic tansduce 4. Ans: (b) Sol: V Y =cos34t V y > V x f y = f x = 9 V X =sin34t 8. Ans: (c) Sol: We know T = nts... fo noise ejection f f f lck f clk n 5Hz n clkmax clk max 5kHz n min 5kHz 5Hz = 5kHz Vetical Ellipse 5. Ans:(c) Sol: R eff = NR i = M = M 6. Ans: (b) Sol: A dual tace oscilloscope usually offes two modes-chop and altenate. Altenate mode is usually used fo displaying high fequency wave foms while chop mode is used fo displaying low fequency wave foms. 7. Ans: (d) Sol: V steps 4 9. Ans:(a) Sol: Electonic DC voltmete (like FET input voltmete) offes vey high esistance and high sensitivity than electonic AC voltmete (like ectifie AC voltmete). 3. Ans: (d) Sol: In full wave ectifie type voltmete, deflection () Aveage voltage V If an ac voltage of 5 V ms voltage is applied, then aveage voltage is V avg = V m sin d V = m sin d m = cos V.V = mv 5 = = 3

8 : 8 : Electical Engineeing Usually = 3.5 =.9 5 V dc =.9 V ac (fo a full wave ectifie) V dc =.9 5 V dc = 3.5 V = = 9 3. Ans: (b) 3.58 Sol: Fo successive appoximation ADC convesion time is same Convesion time fo 5.V input = 4s L mutual = L mutual due to ext = Ans: kh/m ln d ( st tem is independent of diamete) 34. Ans: (c) Sol: D 4 (d)(d D s ) ( d).. () 9 s ( dd)( dd)( dd) (d ) () 3 d d d 3. Ans: (c) d Sol: V s = AV R + B I R I s = CV + D I R But at no load, I R = V R Vs A 44.4 kv.9 D 6 4 s ( dd d).9 4 d 3 (3) 33. Ans: (b) d Sol: L a 8 ln ln d d d L self = L self due to int + L self dut to ext = ln 8 d

9 : 9 : ESE - 9 (Pelims) Offline Test Seies 35. Ans: (d) Sol: Fo DC, insulato capacitances ae ineffective and voltage acoss each unit is same. 36. Ans: (b) Sol: Suge impedance fo a given tansmission line is constant and is independent of length of the tansmission line and fequency of suge. It depends only on magnitude of inductance/km and capacitance/km. Z 37. Ans: (a) Inductan ce/ km capacitan ce/ km Sol: Coona losses ae significantly lowe in case of HVDC tansmission. 38. Ans: (a) Sol: j.5 A 3 V By taking C as efeence and neglecting voltage dop in BC I BC = 4[ + j] A I BC = 4 A I AB = I BC + 3[.8.6] A = j8 = (64 j8) A B 3 A.8 lag j.35 C 4 A upf Voltage AC is V AC = I Z I AB AB BC = (64 j8)j.5 + (j.35)(4) = 6j j V AC = j V 39. Ans: (a) Sol: By poviding seies compensation to the line stability gets impoved. By poviding shunt compensation to the line Feanti effect can be educed. By poviding shunt compensation line cuent can be contolled, powe loss can be educed and efficiency can be impoved. 4. Ans: (d) Sol: Capacitance pe phase = 3C C + C S Whee C C = capacitance between any two conductos C S = capacitance between any conducto and sheath (neglected) C ph = 3C. 4. Ans: (b) Sol: Voltage of the bus is contolled by eactive powe of the bus. If at a bus Reactive powe supply > Reactive powe demand voltage Reactive powe supply < eactive powe demand voltage Z BC

10 : : Electical Engineeing Moeove fequency of a bus is a function of the active powe at the bus. I f Z Z 3Z Z (Z Z ) 4. Ans: (c) Sol: In LLG fault, Ia Ea Z Z Z Since Z, Z and Z ae in pu, E a = I I a a ZZ Z Z Z (Z Z).. () Z Z Z (Z Z ) Connection of sequence netwoks in LLG fault is given by: Fom the cicuit, I I a a I a ZZ Z Z Z Z Z (Z [fom()] Z ) Fault cuent in LLG fault is given by I f = I b + I c I f = 3I a + I a I a I a Z Z Z E a 43. Ans: (b) Sol: In impedance elay, the elay opeates when impedance seen by it is less than a set value Z < Z C. By univesal toque equation,.. () I V I V V I Z Z < Z C Fom () opeating toque is poduced by cuent; estaining toque is poduced by voltages. 44. Ans: (d) Sol: Since I a + I b + I c = 3Ia, a single eath elay is enough to detect eath faults[since all eath faults contain zeo sequence cuents] but to potect the entie 3-phase line against phase faults at least two phase elays ae equied. 45. Ans: (a) Sol:. Insulation esistance of a cable is given by R = R ln. So insulation l

11 : : ESE - 9 (Pelims) Offline Test Seies esistance is invesely popotional to length of cable.. Gadient of cable is g = V. R x ln So, dielectic stess is maximum at the conducto suface and minimum at the inne adius of sheath. 3. Gading of cable is meant to distibute the dielectic mateial such that the diffeence between the maximum gadient and the minimum gadient is educed. 4. Capacitance pe phase is given by 46. Ans: (c) Sol: Y C ph = 3C C + C S. j.5 Y = j Y 3 j. Y 3 = 5j Y 3 j.5 Y 3 = 4j Y = Y +Y 3 = 7j Y = Y 3 +Y = 6j Y 33 = Y 3 +Y 3 = 9j 47. Ans: (c) Sol: Bus type Paametes specified Load Geneato Slack 48. Ans: (d) P, Q P, V V, Sol: When an unloaded tansfome is switched on, it daws a lage initial magnetizing cuent which may be seveal times the ated cuent of the tansfome. This initial magnetizing cuent is called the magnetizing inush cuent in which second hamonic is moe pedominant. To avoid unnecessay tipping of the tansfome due to magnetizing inush cuent second hamonic estaining coil should be used in conjunction with diffeential potection to distinguish between fault cuent and magnetic inush cuent. 49. Ans: (c) Sol: An inteconnected powe system:. Fequency will be same at all buses in the system.. Voltages can be diffeent at diffeent buses. 5. Ans: (d) Sol: I f = A 5 I seconday of CT = 4

12 : : Electical Engineeing Relay setting (I s ) = PSM = 5. Ans: (b) I I f s If 5 5 A 5 fault cuent cuent setting 5.5.5A Sol: Fo a definite time static ove cuent elay to wok as an invese time ove cuent elay an integato cicuit is added such that the chaging RC cicuit takes place though a souce of vaiable voltage magnitude depending upon the seveity of fault and capacito gets chaged in diffeence times. 5. Ans: (c) Sol: SF 6 gas has excellent insulating stength because of its affinity fo electons and has excellent heat tansfe popeties because of its high molecula weight which makes it suitable fo use in cicuit beakes. 53. Ans: (d) Sol: Reactance of each altenato (X S ) =. pu Shot cicuit fault MVA flowing in each altenato is S A. 5 MVA = 5 MVA Shot cicuit fault bus ba Numbe of altenatos Fault MVA of each altenato 54. Ans: (a) = 4 5 = MVA Sol: Total load, P load = 5 MW PG G P 5.. () Cost cuves C 5P C P G G.P G G.P Most economical load scheduling dc dp dc dp G G +. PG 5.4P G..4P 4 PG G PG 4P G PG G Equation () and () P P G G PG P P 3P P G G G G 4 P.. () 5 3 4MW

13 : 3 : ESE - 9 (Pelims) Offline Test Seies

14 : 4 : Electical Engineeing 55. Ans: (c) Sol: = df dp n n P. L P n Fo economic load dispatch the geneation with highest positive incemental tansmission loss will opeate at the lowest incemental cost of poduction. 56. Ans: (b) EV Sol: Steady state stability P max = X By using Bundle conductos o double cicuit lines, the eactance is educed than single cicuit line. Hence steady state stability limit inceases 57. Ans: (b) df Sol: Cost of eceived powe L dp P MW Whee, L is the penalty facto. Also, L P P L 8 8 C (. () 3) 8 4 5Rs 8 / MWh 58. Ans: (a) Sol: Flat fequency egulation: Let S and S (stations) connected though a tie line. If the change in load is eithe at S o at S, but only the geneation of S is egulated to take cae the load change, then it is called flat fequency egulation. Eg: S opeating at base load, i.e., S would maintain a constant output. Advantage: S can be opeated with maximum output with moe efficiency. Disadvantage: Tie-line is subjected to absob the load changes at S. Paallel fequency egulation: If the change in the load is shaed by both the stations to maintain a constant fequency. Flat Tie-line contol: Load change in a paticula aea is taken cae by the geneatos of that aea only. In this case Tie-line loading emains constant. 59. Ans: (a) Sol: Given data, Rated nomal cuent = 5 A S = MVA, V = 33kV Then, Beaking cuent = 3 33 = ka

15 : 5 : ESE - 9 (Pelims) Offline Test Seies Now, Making cuent = Ans: (b) Boile Atmosphee Flue gases Supe Heate Chimney = 89.kA Sol: aplan tubine is best suitable fo low headhigh flow conditions. 6. Ans: (b) Sol: Modeato in nuclea eacto is used to slow down the neutons, by absobing some of the kinetic enegy of the neutons by diect collision, theeby inceasing the chances of fission. 6. Ans: (b) Sol: Economize Induced Daught Fan Ai pe- Heate Electostatic Pecipitato 64. Ans: (a) Sol: Given data: f = 5Hz, f = 6Hz, =.9 P loss Then, P P loss loss P P 65. Ans: (c) loss loss f f 5 f Sol: Both equivalent T and cicuits given same esult but fom pactical point of view, equivalent- is pefeed because no need to ceate a new bus. The natue of eactive powe compensation changes as the load changes. Feanti effect means eceiving end voltage is geate than sending end voltage. This phenomenon occus in medium and long tansmission lines which ae at lightly loaded o no-load condition 63. Ans: (b) Sol: In multi machine inteconnected system, tansient stability can be studied by knowing the solution of swing equation whee as Equal aea citeia is used study the tansient stability of two machine inte connected system but not fo multi machine system. 66. Ans: (a) Sol: An oveexcited synchonous moto opeated unde no-load condition called Synchonous condense which is used to delive the eactive powe only such that powe facto can be coected, fo this pupose diamete of shaft need not be lage.

16 : 6 : Electical Engineeing 67. Ans: (c) Sol: Fo lage powe systems, NR method is found to be moe efficient and pactical fom the view point of computational time and convegence chaacteistics. The convegence chaacteistics of NR method ae not affected by the selection of a slack bus when compaed to GS method, and In case of NR method, the numbe of iteations is moe o less independent of the size of the system and vay between 3 to 5 iteations 68. Ans: (a) Sol: Lighting aestes ae connected between the line and gound at the substation to divet o dischage the suge to the gound and always act in shunt with the equipment to be potected. Shunt capacitos ae used in conjunction with lighting aestes fo the potection of equipment to be potected against suges. These shunt capacitos will absob some pat of the ove voltages. (i.e. educe the cest of the suges). 69. Ans: (d) Sol: Given maximum demand = 5 MW Load facto 5 Aveage demand Maximum demand Aveage demand 5 Aveage demand = 5 MW Actualenegy thatis geneated Capacityfacto Totalenegy thatcould' vebeen geneated 4 Aveage demand t Plant capacityt 5 Plant capacity 5 Plant capacity = 65 MW Reseve capacity = Plant capacity Maximum demand = 65 5 Reseve capacity = 5 MW 7. Ans: (a) Sol: Both statement (I) and statement (II) ae individually tue and statement (II) is the coect explanation of statement (I) 7. Ans: (d) Sol: Ai coed electodynamomete type instuments ae potected against extenal magnetic fields by enclosing them in a casing of high pemeability alloy. 7. Ans: (a) Sol: Shunts used with measuing instuments should have the following popeties: (i) The esistance tempeatue coefficient of shunt should be low and as nealy as possible to that of the instument. (ii) Resistance of shunts should be time invaiant

17 : 7 : ESE - 9 (Pelims) Offline Test Seies (iii)they should have low themal emfs with coppe. (iv)they should cay cuent without excessive tempeatue ise. Manganin has a low tempeatue coefficient of esistance which is 4-6 C. Hence it is a pefeed shunt mateial in DC instuments. Constantan s themal emf with coppe, even though high, is unidiectional and hence it is used as a shunt in AC instuments. 73. Ans: (a) Sol: Both statement (I) and statement (II) ae individually tue and statement (II) is the coect explanation of statement (I) 74. Ans: (d) Sol: The otating disc is aluminium disc and it is not a magnetic mateial. 75. Ans: (d) Sol: Statement (I) is false, but statement (II) is tue.

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Experiment I Voltage Variation and Control

Experiment I Voltage Variation and Control ELE303 Electicity Netwoks Expeiment I oltage aiation and ontol Objective To demonstate that the voltage diffeence between the sending end of a tansmission line and the load o eceiving end depends mainly