Electrical instrumentsQ2. An energy meter revolves 10 revolutions of disc for unit of energy. Find the number of revolutions made by it during an hour when connected across when connected 20A at 210V and 0.8 power factor leading. If energy meter revolves 350 revolutions, find the % error.

Answer.

Energy consumed in one hour = VI cos φ / 1000

       = 210 x 20 x 0.8 / 1000

       = 3.360 kwh.

The number of revolution the meter should make it is correct = 3.360 x registration const in revolution per kwh

    = 3.360 x 100

    = 336

Number of revolution actually made = 350

    % error = (350-336) x 100 / 350

    % error = 0.1466 %

Q4. Explain in brief why energy meter reads energy while wattmeter does not. An energy meter has a registration constant of 100rev/kwh if the meter is connected to a load drawing 20A at 230V and 0.8 power factor for 5 hours. Find the number of revolution should be made by it of it is actually made 1800 revolutions. find the age error and explain it from consumer point of view?Answer. An energy meter is fitted with some type of registration mechanism whereby all the instantaneous reading of power are summed over a definite period of time.

ENERGY = POWER X TIME

So, an energy meter can read energy.Whereas, a wattmeter indicates the value of power at a particular instant when it is read and hence it can not read energy.

Registration constant= 100 rev/Kwh I = 20 A,     V = 230 V,     cos φ = 0.8

Energy in 5 hours = VI Cos φ x 5 / 1000    = 20 x 230 x 0.8 x 5 / 1000     = 18.4 Kwh

The number of revolution of the meter should make it is correct = 18.4 x 100 = 1840 revolution. Number of revolution actually made = 1800 revolution % error = 1840 - 1800 x 100                     1840             = 0.0217% It would be better from the customer point of view % age error is very less.

Q7. A 50A, 230V meter on full load list makes 61 revolution in 37 seconds if the normal disc speed is 500 revolution per kwh find the age error? Answer.

Energy consumed per hour = VI cos φ / 1000 Kwh

Energy consumed in 37 seconds    = VI cos φ X     37     Kwh        1000       60 x 60

   = 230 x 215 x 1 X     37     Kwh (assuming PE to be unity)            1000           60 x 60

       = 0.11819 kwh.

Energy consumption registered by the parameter = Number of revolution made / meter constant        = 61 / 1000 = 0.122 kwh.

% error = (Actual registration - True energy consumption) / True energy consumption    = (0.122 - 0.11819) x 100 / 0.11819 % error = 3.22 %

Q12. Why shunt is usually used voltmeter and ammeter? A moving coil instrument has a resistance of 5 Ω and gives full deflection of 100mv. Show how the instrument may be used to measure:-

1. voltage upto 50V 2. current upto 10A

Answer.

Shunt is usually used in voltmeter and ammeter to extend the range of voltmeter and ammeters.

    Rm = 5Ω    Vm = 100mv    Im = Vm/Rm = 100mv/5Ω = 20mA

1. For measuring voltage upto 50V.

Series resistance is used with the instrument whose resistance is

    R = V/Im - Rm = 50/(20 x 10-3) - 5

    R = 2.5 x 10-3 - 5 = R = 2495 Ω

2. Such resistance of resistance Rf is used to be connected

    Rf = Rm/[I/Im - 1]

    = 5/[10/20 x 10-3 -1] = 5 x 2/998

    Rf = 0.01002004 Ω

Q13. A moving coil instrument gives full scale deflection with 15mA. The resistance of coil is 5Ω. It is desired to convert this instrument into an ammeter to read upto 2A. How to acheive it further how to convert this instrument to read upto 30V

    Im = 15mA , Rm = 5Ω

Answer. 1. Shunt of resistance R Ω is required to be connected with the instrument where

    R = Rm/[I/Im-1]

    = 5/[2/5 x 10-3-1]

    = 5 x 15/[2 x 10-3-15]

    = 5 x 15/ 1985

    R = 0.03778 Ω

2. Series resistance Rs Ω is required to be connected with the instrument.

    Rs = V/Im - Rm       = 30/15 x 10-3 - 5       = 2000 - 5     Rs = 1995 Ω

Q15. A moving coil instrument gives a full scale detection of 20mA. When a potential difference of 50mV is applied. Calculate the series resistance to measure 500V on scale?

Answer.

Im = 20mA

V = 250V

Vm = 50mV

R= V/Im - Rm

  = (500/20) x 103 - 2.5

  = 25000 - 2.5

R = 24997.5 Ω

3phase induction motorQues1: A 3 φ 4 pole 50 hz induction motor runs at 1460 r.p.m. find its %age slip. Solution N s = 120f/p = 120*50/4 = 1500r.p.m.Running speed of motor = n= 1460r.p.m.Slip S=( N s–N)/ N s*100 =(1500-1460) x 100 / 1500 = 2.667%

Ques2: A 12 pole 3 φ alternator driver at speed of 500 r.p.m. supplies power to an 8 pole 3 φ induction motor. If the slip of motor is 0.03p.u, calculate the speed. Solution Frequency of supply from alternator, f=PN/120 =12*500/120 = 50hz

where P= no of poles on alternatev N=alternator speed is r.p.m. Synchronous speed of 3 φ induction motor N=120f/Pm =120*50/8 = 750 r.p.m.Speed of 3 φ induction motor N=Ns (1-s)=750(1-0.03) = 727.5 r.p.m.

Ques3: A motor generates set used for providing variable frequency ac supply consists of a 3-φ synchronous and 24 pole 3 φ synchronous generator. The motor generate set is fed from 25hz, 3 φ ac supply. A 6 pole 3 φ induction motor is electrically connected to the terminals of the synchronous generator and runs at a slip of 5%. Find

  i) the frequency of generated voltage of synchronous generator  ii) the speed at which induction motor is runningSolution Speed of motor generator setNs=(120*f1(supply freq))/(no of pole on syn motor)=120*25/10 = 300 r.p.m.

(1) frequency of generated voltage fz=speed of motor gen set voltage *no of poles on syn gen/120= 300*24/120 = 60hz (2) Speed of induction motor , Nm=Ns(1-s)=120 fz /Pm(1-s) = 120*60/6(1-0.05) = 1140r.p.m.

Ques4: A 3-φ 4 pole induction motor is supplied from 3φ 50Hz ac supply. Find (1) synchronous speed (2) rotor speed when slip is 4%(3) the rotor frequency when runs at 600r.p.m. Sulution 1) Ns =120f/p=120*50/4 = 1500 r.p.m.

2) speed when slip is 4% or .04N=Ns (1-s)=1500(1-0.04) = 1440 r.p.m. 3) slip when motor runs at 600 r.p.m. S’=(Ns –N)/Ns

=(1500-600)/1500 = 0.6

Rotor frequency f’ = S’f = 0.6*50 = 30Hz.

Ques5: A 12 pole 3-φ alternator is coupled to an engine running at 500r.p.m. If supplied a 3φ induction motor having full speed of 1440r.p.m. Find the %age slip, frequency of rotor current and no of poles of rotor. Ans Frequency of supply from alternator f=Pa*Na/120=12*500/120 = 50Hz

Full load speed Nf =1440 r.p.m.The no of poles (nearest to and higher than full load speed of motor =1440) should be in even nos.P=120f/n = 120*50/1440 = 4Ns = 120f/Pm = 120*50/4 = 1500 r.p.m.% Slip s = (Ns-N)/Ns x 100 =(1500-1440) x 100 / 1500 = 4%

Rotor frequency f’ = sf = 0.04*50 = 2Hz

No A poles of the motor = 4

Ques6: The rotor of 3φ induction motor rotates at 900r.p.m. when states is connected to 3φ supply .find the rotor frequency. Solution Nr =980 r.p.m., f=50Hz, Ns=120f/pWhen P=2, Ns=3000r.p.m.,P=4, Ns=1500

P=6, Ns=1000, P=8, Ns=750r.p.m.

As we know that synchronous speed is slightly greater than rotor speed.

Ns=1000 r.p.m. P=6

Fr=Sf=(Ns-N)/Ns*f=Sf = (1000-980) x 50 / 1000

Ques7: A 3 φ 50Hz induction motor has a full load speed of 960 r.p.m(a) find slip(b) No of poles(c) Frequency of rotor induced e.m.f(d) Speed of rotor field w.r.t. rotor structure(e) Speed of rotor field w.r.t. Stator structure(f) Speed of rotor field w.r.t. stator fieldSolution: Given f = 50 Hz(supply frequency) N = 960r.p.mThe no. of pole will be 6 only(because at P=6, Ns = 1000 which is nearer nad greater then 960 r.p.m.)

(a) Slip, S = (Ns-N)/Ns * 100 = (1000 – 960) / 1000 * 100 = 4%(b) No of poles = 6(c) Frequency of rotor induced emf = fr = SF = .04 * 50 = 2Hz(d) Speed of rotor field w.r.t rotor structure = 120fr/p = 120*2/6 = 40 r.p.m. (e) Speed of rotor field w.r.t. stator structure os actually the speed of stator filed w.r.t stator structure, Ns = 1000r.p.m(f) Speed of rotor field w.r.t stator field is zero

Ques8: A 3 φ, 400V wound rotor has delta connected stator winding and star connected rotor winding. The stator has 48 turns/phase while rotor has 24 turns per phase. Find the stand still or open circuited voltage across the slip ringsSolution Stator e.m.f/phase E1 = 400VStatur turns/phase N1 = 48Rotor turns/phase N2 = 24K= N2/N1 = 24/48 = 1/2

Rotor e.m.f/phase = KE1 = 1/2 * 400 = 200V

Voltage between slip rings = Rotor line voltage = √ 3 x 200 = 346 volt

Ques9: A 6 pole 3 φ 50Hz induction motor is running at full load with a slip of 4%. The rotor is star connected and its resistance and stand still reactance are 0.25 ohm and 1.5 ohm per phase. The e.m.f between slip ring is 100V. Find the rotor current per phase and p.f, assuming the slip rings are short circuited.Solution Rotor e.m.f./phase at stand still E2 = 100√3 = 57.7V Rotor e.m.f./phase at full load = sE2 = 0.04 * 57.7 = 2.31 VRotor reactance/phase at full Load = SX2 = .04 * 1.5 = .06 ohmRotor impedance/phase at full load = √ ((0.25)2 + (0.06)2) = .257 ohmFull load Rotor current/phase = 2.31/0.257 = 9ARotor P.f = 0.25/0.257 = 0.97 lag

Quest10: A 50 Hz, 8 pole induction motor has full load slip of 4%. The rotor resistance and stand still reactance are 0.01 ohm and 0.1 ohm per phase respectively. Find:i) The speed at which maximum torque occursii) The ratio of maximum torque to full load torqueSolution:Synchronous speed Ns = 120f/P = 120*50/8 = 750r.p.m. Slip at which maximum torque occurs = R2/X2 = 0.01/0.1 = 0.1

Rotor speed at maximum torque = (1-0.1) Ns = (1- 0.1) 750 = 675 r.p.m. Tm/Tf = (a2 + s2)/2asWhere s = Full load slip = 0.04

a = R2/X2 = 0.01/0.1 = 0.1Tm/Tf = ((0.1)2 + (0.04)2)/(2*0.1*0.04) = 1.45

Ques 11: An 8 pole 3 φ, 50 Hz induction motor has rotor resistance of 0.025 ohm/phase and rotor standstill reactance of 0.1ohm/phase. At what speed is the torque maximum? What proportion of maximum torque is the starting torque?Solution Ns = 120f/P = 120*50/8 = 750 r.p.m.R2 = SX2 ------------ for maximum torqueS = R2/X2 = 0.025/0.1 = 0.25Corresponding speed N = (1-s)Ns = (1 – 0.25)750 = 562.5 r.p.m.

ii) Ts/Tm = 2a/(a2+1) = 0.47 where a = R2/X2 = 0.025/0.1 = 0.25

Ques12: A 500 V, 3 φ, 50 Hz induction motor develops an output of 15 KW at 950 r.p.m. If the input p.f. is 0.86 lagging, Mechanical losses are 7.30 W and stator losses 1500W, Findi) the slip

ii) the rotor Cu lossiii) the motor inputiv) the line currentSolution:VL = 500V, motor output Pr = 15KWN = 950 r.p.m. P.f. = cos Ø = 0.86lagsMech. Loss = 730 WStator loss = 1500 WNs = 120f/P = 120 * 50/6 = 1000r.p.m. i) S = (Ns-N)/Ns * 100 = (1000 – 960)/1000 *100 = 0.05*100 = 5%ii) Rotor output = Motor output + Mechanical output = 15 + .730 watt = 15.73 KWattThere fore (Rotor Cu loss)/(Rotor output) = s/(s-1) Or Rotor Cu loss = 15.73 * (0.05)/(1-0.05) = 827.89 watt

Power flow diagram for finding the motor input

Motor input = 15kw + 730 + 1500 + 827.89 = 18.058KWLine Current = √3V2I2CosφI2 = 24.25A

Ques13: A 6 pole 3φ induction motor develops 30hp including 2 hp mechanical losses at a speed of 950 r.p.m. on 550V, 50Hz Mains. The P.F. is 0.88 lagging.Find: 1) Slip2) Rotor Cu loss3) Total input if stator losses are 2kw4) η5) Line current

SolutionNs = 120f/P = 120 * 50/6 = 1000 r.p.m. 1) S = (Ns – N)/Ns = (1000 – 950)/1000 = 0.05Rotor output Pmech = 30hp = 30 * 735.5 = 22065 wattPower input to rotor = Pmech/(1-S) = 22065/(1-0.05) = 23,2262) Rotor Cu loss = s * rotor input = 0.05 * 23226 = 1161 Watt3) Total input = Power input to rotor + stator losses = 23226 + 2000 = 25226 Watt

Motor output = Rotor output – Mech loss = 30 – 2 = 28 HP = 28 * 735.5 = 20594 Watt4) η = (Motor output)/(Motor input) * 100 = 81.64%5) IL = (Motor Input)/( √3 * 550 * 0.88) = 30A

Ques14: A 4 pole 50 Hz 3 φ induction motor running at full load, develops a torque of 160N-m, when rotor makes 120 complete cycles per minute, find what power output Solution Supply frequency f = 50HzRotor e.m.f. frequency = f = 120/60 = 2Hz

Slip S = f’/f = 2/50 = 0.04Ns = 120f/p = 120 *50/4 = 1500 r.p.m. Shaft power output = Tsh * 2πN/160 = 160 * 2 π * 1440/60 = 24127W

Ques15: The power input to a 500V 50Hz, 6 pole, 3 φ squirrel case inductor motor running at 975 r.p.m. is 40kw. The stator losses are 1 kw and friction and windage losses are 2kw. Find:1) Slip2) Rotor Cu loss3) Brake hpSolution:i) Ns = 120f/P = 120*50/6 = 1000 r.p.m.S= ( Ns – N)/Ns = (1000 – 975)/1000 = 0.025Power input to station P1 = 40KwStator output power = P1 – stator losses = 40 -1 = 39kwPower input to rotor P2 = Stator output power = 39 KWii) Rotor Cu loss = sP2 = 0.025 * 39 = 0.975KWPmech = P2 – Pcu = 39 – 0.975 = 38.025iii) Motor output = Pmech – friction and windage loss = 38.025 – 2 = 36.025KW

1.If the applied voltage of a certain transformer is increased hr 50% and the frequency is reduced to 50% (assuming that the magnetic circuit remains unsaturated), the maximum core flax density will(a) change to three times the original value(b) change to 1.5 times the original value(c) change to 0.5 times as the original value(d) remain the same as the original value

2. The low—voltage winding of a core type transformer is subdivided into two equal halves, each of half the original width of the single winding will the high voltage winding in between (instead of having the usual construction of low-voltage winding adjacent to the core and surrounded by the high-voltage winding). Such an interlacing of coils would make the combined primary and secondary leakage reactance (in terms of primary) nearly.(a)twice(b) equal(c)half(d) one -fourth

3. Two 3-limb, 3-phase delta-star connected transformers are supplied from the same source. One of the transformers is of. Dy I and the other is of Dy II / connection. The phase difference between the corresponding phase voltages of the secondaries would he(a) 0°(b) 30°

(c) 60°(d) 120°

4. In a transformer fed from a fundamental frequency voltage source, the source of harmonics k the(a) overload(b) poor insulation(c) iron loss(d) saturation of core

5. A 40 kVA transformer/ a core loss of400 W and a full load copper of of 800 W. The proportion of full-load at maximum efficiency is(a)50%(b) 62.3%(c) 70.7%(d)100 %

6.A single phase transformer has rating of 15 KVA 600 / 120 V. It is recommended as an auto transformer to supply at 720 V from a 600 V primary source. The maximum load it can supply is ?(a)90 kVA(b)18 kVA(c)15 kVA(d) 12 kVA

7. Equalizing pulses in TV are placed during the(a) vertical blanking period(b) horizontal blanking period(c)serration(d) horizontal retrace

8. The most useful approach to radar system for monitoring the speed of moving vehicles is:(a) Pulsed radar(b) Monopulse(c)Doppler radar(d) Auto tracking radar

9. A. dc shunt generator, when driven at its rated speed, is found to he not generating any voltage. Which of the following would account for this?I. There is no residual magnetism2. The connection of the field winding is not proper with respect to the armature terminals.3. The resistance of the field circuit is greater than the critical field resistance.4. The load resistance is less than the critical armature resistance.Select the correct answer using the codes given below:

Codes:(a)3 and 4(b)l,2 and 4(c)1,2 and 3(d)l.2,3 and 4

10. To have spark/as commutation, the armature reaction effect in a dc machine is neutralised by(a) using compensating winding and commutating poles(b) Shifting the brush axis from the geometrical neutral axis to the magnetic neutral axis(c) fixing the brush axis in line with the pole axis(d) increasing the field excitation

11.In a dc shunt general or working on load, the brushes are moved forward in the direction of rotation, as a result of this, commutation will(a)improve but terminal voltage will fall -(b)worsen and terminal voltage will fall(c) improve and terminal voltage will rise(d) worsen and terminal voltage will rise

12. Consider the following statements:The maximum range of radar can be increased by1. increasing the peak transmitted power2. increasing the gain of the receiver3. increasing the diameter of the antenna4. reducing the wavelength usedOf these statements(a) 1, 3 and 4 are correct(b) 1, 2, 3 and 4 are correct(c) 2 and 4 are correct(d)1 and 3 are correct

13. Consider the following statements about broadband communication using submarine cables:1. A submarine cable repeater contains filters for the two directions of transmits2. Armored submarine cable is used for the shallow-shore ends of the cable.3. Fibre optic submarine cable is used to prevent inadvertent ploughing in of the cableOf these statements(a) 1 and 2 are correct(b) 2 and 3 are correct(c) 1 and 3 are correct(d) 1, 2 and 3 are correct

14. A dc overcompounded generator was operating satisfactorily and supplying power to an infinite bus when the prime mover failed to supply any mechanical power. The machine would then run as a ?

(a) cumulatively compounded motor with speed reversed(b) cumulatively compounded motor with direction of rotation as before(c) differentially compounded motor with speed reversed(d) differentially compounded motor with direction of speed as before.

15 To couple a coaxial line to a parallel wire line, It is best to use a(a) slotted line(b) balun(c)directional coupler(a) quarter-wave transformer

16. When a synchronous motor is running at synchronous speed, the damper winding produces(a) damping toruqe(b) eddy current torque(c)torques aiding the developing torque(d) no torque

17. For a given developed power, a synchronous motor operating from a constant voltage and constant frequency supply, will draw the minimum and maximum armature currents, Imin and Imax respectively, corresponding to?(a)Imin at unity pf, but Imax at zero pf(b) Imax at unity p, but Imin at zero p1(c) both Imin and Imax at unity pf(d )both Imin and Imax at zero pf

18. Consider the folio wing statements regarding an RC phase-shift oscillator:1. The amplifier gain is positive2. The amplifier gain is negative.3. The phase shift introduced by the feedback network is 1804. The phase shift introduced by the feedback network is 360Of these statements(a) 1 and 3 are correct(b) 2 and 3 are correct(c) 2 and 4 are correct(d) 1 and 4 are correct

19 While conducting a “slip “ test for determination of direct and quadrature axis synchronous reactance ‘Xd’ and ‘Xq’ of salient pole synchronous machine, the rotor of the machine is run with a slip ‘s’ and stator supply frequency ‘f’. The frequency of1. voltage induced across open field terminals,2. envelope of the armature terminal voltage.3. envelope of the armature current, and4. armature currentwill be respectively(a) sf, sf,sf and f

(b) sf, f,sf and f(c) f, sf,f and af(d) f, (1-s)f,(2-s)f and sf

20. If two induction motors A and B are identical except that the air-gap of motor ‘A ‘ is 59% greater than that of motor ‘B’, then(a) the no-load power factor of A will be better than that of B.(b) the no-load power factor of A will be poorer than that of B.(c) the core losses of A will be more than those of B(d) the operating flux of A will be smaller than that of B.

21 A 6-pole, 50 Hz, 3-phase synchronous motor and an 8-pole, 50 Hz. 3—phase slipring induction motor are mechanically coupled and operate on the some 3-phase, 50 Hz supply system If they are left open-circuited, then the frequency of the voltage produced across any two slip rings would be?(a) 12.5 Hz(b) 25.0 Hz(c) 37.5 Hz(d) 50.0 Hz

22. Which of the following statements regarding skewing of motor bars in a squirrel-cage induction motor are correct?I. It prevents cogging;2. It produces more uniform torque.3. it increases starting torque.4. It reduces motor ‘hum’ during its operation.Select the correct answer using the codes given below.Codes:(a)2,3 and 4(b) l,2 and 3(c)1,3 and 4(d) l,2 and 4

23. The rotor power output of a 3-phase induction motor is 15kW and the corresponding slip is 4%. The rotor copper loss will be(a) 600 W(b) 625 W(c) 650 W(d) 700 W

24. If an input signal with non-zero dc component is applied to a low pass RC network, the dc component in the output will be ?(a) the same as that in the input(b) less than that in the input(c) more than that in the input(d) zero

25. A 3 phase wound rotor induction motor, when started with load connected to its shaft, was found to start but settle down at about half synchronous speed. If the rotor winding as well as the stator winding were star connected, then the cause of the malfunction could be attributed to(a) one of the stator phase windings being short–circuited(b) one of the supply fuses being blown(c) one of the rotor phases being open-circuited(d) two of the rotor phases being open circuited

26. Consider the following statements regarding fractional horse power shaded-pole motor:1. Its direction of rotation is from unshaded to shaded portion of the poles.2.. Its direction of rotation is from shaded to unshaded portion of the poles.3. It can remain stalled for short periods without any harm.4. It has a very poor power factor.Of these statements(a) 1, 3 and 4 are correct(b) 2, 3 and 4 are correct(c) 2 and 4 are correct(d) 1 and 3 are correct

27. In the case of a converter-inverter speed control arrangement f an induction motor operating with v/f constant and with negligible stator impedance.(a) the maximum torque is independent of frequency(b) the maximum torque is proportional to frequency(c) the slip at maximum torque is proportional to frequency(d) the starting torque is proportional to frequency

28 An amplifier with mid-band gain A = 500 has negative feedback beta = 1/100 the upper cut-off without feedback were at 60 kHz, then with feedback it would become(a) 10kHz(b) 12kHz(c) 300 kHz(d) 360 kHz

29. If the discharge is 1 m cube and the head of water is 1 m then the power generated by the alternator in one hour (assume 100% efficiency of generator and turbine) will be(a) 10kW(b) 73/75kW(c) 746/75kW(d) 100kw

30. Control rods used In unclear reactors are made of(a) zirconium(b) boron

(c) beryllium(d) lead

31. if alpha = 0.98, Ico=6 microA I beta =100 microA for a transistor, Then the value of Ic will he(a) 2.3mA(b) 3.lmA(c) 4.6mA(d) 5.2mA

32. In a 3-core extra-high voltage cable, a metallic screen around each core-insulation is provided to(a) facilitate heat dissipation(h) give mechanic strength(c) obtain radial electric stress(d) obtain longitudinal electric stress

33. Galloping In transmission line conductors arises generally due to(a) asymmetrical layers of ice formation(b) vortex phenomenon in light winds(c) heavy weight of the line conductors(d) adoption of horizontal conductor configurations

34.. In a 3- phase rectifier circuit, thyristor number 1, 2 and 3 are connected respectively to R, Y and B phases of the star-connected transformer secondary. When the current is being commutated front thyristor No. 1 to No. 2, the effect of the transformer leakage and the ac system Inductance will be such that it will?

(a) Prolong the conduction in No. 1 and delay the turn on of No. 2 correspondingly.(b) stop the conduction in No. 1 at the scheduled time, but delay the turn on of No. 2(c) produce conduction in both No. 1 and No. 2 in parallel for an overlapping period through a transient(d) double the voltage output through a commutation transient

35. The incremental generating costs of two generating units are given byIC1=0.10 X + 20 Rs /MWhrIC1=0.15 Y + 18 Rs /MWhrwhere X and V are power generated by the two units in MW.For a total demand of 300 MW, the value (in MW) of X and Y will be respectively(a) 172 and 123(b) 123 and 172(c) 175 and 125(d) 200 and 100

36. Consider the following statements:To provide reliable protection for a distribution transformer against over voltages using

lightning arrestors, it is essential that the1. lead resistance is high.2. distance between the transformer and the arrestor is small3. transformer and the arrestor have a common inter-connecting ground.4. Spark over voltage of the arrestor is greater than the residual voltage.Of these statements(a) 1, 3 and 4 are correct(b) 2 and 3 are correct(c)2, 3 and 4 are correct(d) 1 and 4 are correct

37. The reflection coefficient of a short-circuited line is(a) -1(b) 1(c) 0.5(d) zero

38.Iif an intrinsic semiconductors is doped with a very small amount of born then in the extrinsic semiconductor so farmed, the number of electrons and holes will(a)decrease(b) increase and decrease respectively(c)increase(d) decrease and increase respectively

39. Hollow conductors are used in transmission lines to(a) reduce weight of copper(b) improve stability(c) reduce corona(d) increase power transmission capacity

40. In the solution of load-flow equation, Newton-Raphson (NR) method is superior to the Gauss- Seidel (GS) method, because the(a) time taken to perform one iteration in the NR method is less When compared to the time taken in the OS method(b) number of iterations required in the NR method is more when compared to that in the GS method(c) number of iterations required is not independent of the size of the system in the NR method(d) convergence characteristics of the NR method are not affected by the selection of slack bus

41. In a synchronous generator, a divided winding rotor is preferable to a conventional winding rotor because of(a) higher efficiency(b) increased steady-state stability limit

(c) higher short-circuit ratio(d) better damping

42. Consider the following statements regarding speed control of induction motors by means of external rotor resistors:1. Reduction in speed is accompanied by reduced efficiency.2. With a large resistance in the rotor circuit, the speed would vary considerably with variation in torque3. The method is very complicatedThe Disadvantages of such a method of speed control would include(a) l and 2(b) 2 and 3(c) l and 3(d) 1,2 and 3

43. Zero sequence currents can flow from a line into a transformer bank if the windings are in ?(a) grounded star/delta(b) delta/star(c) star/grounded star(d) delta/delta

44. When a line to ground fault occurs, the current in a faulted phase is 100 A. The Zero Sequence current in this case will be(a) zero(b) 33.3 A(c) 66.6A(d) 100 A

45. Consider the following statements:Switched mode power supplies are preferred over the continuous types, because they are

1. suitable for use in both ac and dc.2; more efficient,3. suitable for low-power circuits.4. suitable for high-power circuits.Of these statements(a) 1 and 2 are correct(b) 1 and 3 are correct(c) 2 and 3 are correct(d) 2 and 4 are correct

46. The power generated by two plants are P1 = 50MW, P2= 40 MW.If the loss coefficients are B11 = 0.001, B22 = 0.0025 and B12 = -0.0005, the power loss will be(a)5.5 MW

(b) 6.5 MW(c) 4.5Mw(d) 8.5 MW

47. In dc choppers, per unit ripple is maximum when then duty cycle is?(a) 0.2(b) 0.5(c) 0.7(d) 0.9

48. The following data pertain to two alternators working in parallel and supplying a total load of 80 MW:Machine 1 : 40 MVA with 5% speed regulationMachine 2: 60 MVA with 5% speed regulationThe load sharing between machines 1 and 2 will be(a)P1/48MW, P2/32MW(b) 40MW. 40MW(c)30MW,50MW(d) 32 MW. 48MW

49. The per unit impedance of a synchronous machine is 0.242. If the base voltage is Increased by 1.1 times, the per unit value will be(a)0.266(b) 0.242(c)0.220(d) 0.200

50. A 3-pulse converter feeds a pure resistive load at a firing angle of alpha = 60°. The average value of current flowing in the load is 10 A, If a very large inductance is connected in the load circuit, then the(a) average value of current will remain as 10 A(b) average value of current wilt become greater than 10 A(c) average value of current will become less than 10 A(d) trend of variation of current cannot be predicted unless the exact value of the inductance connected is known

Answers

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