gate – 2015 - career launchermedia.careerlauncher.com/gate-exams/gate actual paper...2 solved...

(Q. 1– 5) : Car r y One M ar k Each1. Which of the fol lowing combinat ions is incorrect?

(a) Acquiescence – Submission

(b) Wheedle – Roundabout

(c) Flippancy – L ightness

(d) Profligate – Extravagant2. Given set A = {2, 3, 4, 5} and Set B = {11, 12, 13,

14, 15}, two numbers are randomly selected, onefrom each set . What is probabil i ty that the sumof the two numbers equals 16 ?(a) 0.20 (b) 0.25(c) 0.30 (d) 0.33

3. Which of the fol lowing opt ions is the closest inmeaning to the sentence below?She enjoyed herself immensely at the par ty.(a) She had a ter r ible t ime at the par ty.

(b) She had a hor r ible t ime at the par ty.(c) She had a ter r i fic t ime at the par ty(d) She had a ter r i fying t ime at the par ty

4. Based on the given statements, select the mostappropr iate opt ion to solve the given quest ion.I f two floors in a cer tain building are 9 feet apart ,how many steps are there in a set of stairs thatextends from the first floor to the second floor ofthe building?Statements:I . Each step is ¾ foot high.I I . Each step is 1 foot wide.(a) St at emen t I al one i s su f f i ci en t , bu t

statement I I alone is not sufficient .(b) Statement I I alone is sufficient, but statementI

alone is not sufficient .(c) Both statements together are sufficient , but

neither statement alone is sufficient .(d) Statement I and I I together are not sufficient .

5. Didn’t you buy _______ when you went shopping?(a) any paper(b) much paper(c) no paper(d) a few paper

(Q. 6– 10) : Car r y Tw o M ar k s Each6. The given statement is folslowed by some courses

of act ion. Assuming the statement to be t rue,decide the cor rect opt ion.

Statement:

There has been a significant drop in the waterlevel in the lakes supplying water to the city.

Course of act ion:

I . The water supply author ity should impose apar t ial cut in supply to tackle the situat ion.

I I . The government should appeal to al l t heresidents through mass media for minimal useof water.

I II . The government should ban the water supplyin lower areas.

(a) Statements I and I I fol low.

(b) Statements I and I I I fol low.

(c) Statements I I and I I I fol low.

(d) Al l statements fol low.7. The number of students in a class who have

answered cor rect ly, wrongly, or not at temptedeach quest ion in an exam, are l isted in the tablebelow. The marks for each question are also listed.There is no negat ive or par t ial marking.

Answered Answered NotQ. NO. Marks

Correct ly Wrongly At tempted

1 2 21 17 62 3 15 27 23 1 11 29 4

4 2 23 18 35 5 31 12 1

What is the average of the marks obtained by theclass in the examinat ion?(a) 2.290 (b) 2.970(c) 6.795 (d) 8.795

8. The pie char t below has the br eakup of thenumber of students from different depar tmentsin an engineer ing col lege for the year 2012. Thepropor t ion of male to female students in eachdepar t ment is 5 : 4. Ther e ar e 40 males inElectr ical Engineer ing. What is the differencebetween numbers of female students in the Civi l

GATE – 2015EE : EL ECTRI CAL ENGI NEERI NG

Set - 1N o. of Quest i ons : 65 Maxi mum Mar k s : 100

GENERAL APTI TUDE (GA)

2 SOLVED PAPER – 2015 (SET - 1)

depar tment and the female students in t heMechanical department?

Electr ical20%

Mechanical10%

ComputerScience

20%

Civi le30%

9. Select the alternative meaning of the under l inedpar t of the sentence.

The chain snatchers took to their heels when thepolice par ty ar r ived.(a) took shelter in a thick jungle(b) open indiscr iminate fire(c) took to fl ight(d) uncondit ionally sur rendered

10. The pr obabi l i t i es t hat a st udent passes i nMathemat ics, Physics and Chemistry are m, pand c respectively. Of these subjects, the studenthas 75% chance of passing in at least one, a 50%chance of passing in at least two and a 40% chanceof passing in exact ly two. Following relations aredrawn in m, p.c:

I . p + m + c = 2720

I I . p + m + c = 1320

I II . (p) (m) (c) = 1

10(a) Only relat ion I is t rue.(b) Only relat ion I I is t rue.(c) Relat ions I I and I I I are t rue.(d) Relat ions I and I I I are t rue.

TECHNI CAL SECTI ON

(Q. 1– 5) : Car r y One M ar k Each1. A mov i ng aver age funct i on i s gi ven by

y(t) =T

1u( )d .

T

t

t I f input u is sinusoidal signal of

frequency1

H2T

z then in steady state the output

y wi l l lag u (in degrees) by .2. Consider a one-turn rectangular loop of wire place

in a uniform magnetic field as shown in the figure.The plane of the loop is perpendicular to the fieldl ines. The resistance of the loop is 0.4, and itsinductance is negligible. The magnetic flux density(in Tesla) is a funct ion of t ime, and is given byB(t) = 0.25 sint, where = 2 50 radian/second.

The power absorbed (in Watt) by the loop fromthe magnet ic field is __________.

10 cm

5 cm

3. I f the sum of diagonal elements of a 2 2 matr ixis – 6, then the maximum possible value ofdeterminant of matr ix is .

4. When the Wheatstone br idge shown is used to findvalue of r esistance Rx , t he Galvanometer Gindicates zero current when R1 = 50, R2 = 65 & R3

= 100. I f R3 is known with 5% tolerance on itsnominal value of 100, what is range of Rx in ohms?

R1

G

R2

R3 Rx

+ –

V

(a) [123.5, 136.5] (b) [125.898, 134.12]

(c) [117, 143] (d) [120.25, 139.75]

5. For a given circuit the thevenin equivalent is tobe determined. The thevenin voltage VTh(in volts),seen from terminal AB is .

A

B

1

20i

2V

– +

2

1

i

6. The impulse response g(t) of a system, G, is asshown in Figure (a). What is the maximum valueattained by the impulse response of two cascadedblocks of G as shown in Figure (b)?

g t( )

1

0 1( )a

tG G

( )b

(a)23

(b)34

(c)45

(d) 1

SOLVED PAPER – 2015 (SET - 1) 3

7. Base load power plants are

P wind farms

Q run-of-r iver plants

R nuclear power plants

S diesel power plants

(a) P, Q & S only (b) P, R & S only

(c) P, Q & R only (d) Q and R only

8. Of the four character ist ic given below, what aremajor requirements for an instrument amplifier?P: high common mode reject ion rat io

Q: high input impedance

R: high l inear ity

S: high output impedance

(a) P,Q & R only (b) P & R only

(c) P, Q & S only (d) Q, R & S only

9. A r andom var iable X has pr obabil i t y densit yfunct ion f(x) as given below.

f(x) =0 1

0 otherwise

a bx x

I f the expected value E[X] =23

, then Pr [X < 0.5]

is .

10. Consider a funct ion f = 2

1 ˆ,rr

where r is distance

from or igin and r̂ is unit vector. The divergenceof this funct ion over a sphere of radius R, whichincludes or igin is.(a) 0 (b) 2

(c) 4 (d) R.

11. A separ at el y exci t ed DC gener at or has anar matur e r esist ance of 0.1 and negl i gi bl earmature inductance. At rated field cur rent andrated rotor speed its open-circuit voltage is 200V.When this generator is operated at half the ratedspeed, wi t h hal f t he r ated f ield cur r ent , anun -char ged 1000µF capaci t or i s suddenl yconnected across the armature terminals. Assumethat the speed remains unchanged dur ing thetransient . At what t ime (in microsecond) afterthe capacitor is connected wil l the voltage acrossit reach 25V ?

(a) 62.25 (b) 69.3

(c) 73.25 (d) 77.3

12. I n the fol lowing chopper duty rat io of switch S is0.4. I f the inductor and capacitor are sufficient lylarge to ensure cont inuous inductor cur rent andripple free capacitor voltage, the charging current(in Ampere) of the 5V battery, under steady-state,is .

3

5V20V +

–

S L

+–

13. I f a cont inuous funct ion f(x) does not have a rootin the interval [a, b]. Then which one of thefol lowing statement is TRUE ?(a) f(a).f(b) = 0 (b) f(a).f(b) < 0(c) f(a).f(b) > 0 (d) f(a)/f(b) 0

14. The pr i mar y mmf i s l east af fect ed by t hesecondary terminal condit ions in a(a) power t ransformer(b) potent ial t ransformer(c) cur rent transformer(d) dist r ibut ion transformer.

15. Consider a HVDC link whcih uses thyr istor basedline-cooutated conver ters as shown in the figure.For a power flow of 750 MW from System 1 toSystem2, the voltages at the two ends, and thecur rent , are given by : V1 = 500kV, V2 = 485kVand I = 1.5kA. I f the direct ion of power flow is tobe reversed (that is, from System 2 to System1)without changing the electr ical connections, thenwhich one of the fol lowing combinat ions idfeasible ?

V = 560kV1V = 485kV = 15kA2

IV1 V2

+

–

+

–

I

If power is to be reversed(a) V1 = – 500kV, V2 = – 485kV and I = 1.5kA(b) V1 = – 485kV, V2 = – 500kV and I = 1.5kA(c) V1 = 500kV, V2 = 485kV and O = – 1.5kA(d) V1 = – 500kV, V2 = – 485kV and I = – .5kA

16. An i nductor is connect ed in par al l el wi t h acapacitor as shown in the figure.

i L

C Z

As the frequency of cur rent i is increased, theimpedance (Z) of the network var ies as

(a)

Inductive

fZ

Capaci t ive

(b)

Capaci t ive

Inductive

fZ


(c) Zf

InductiveCapaci t ive

(d) Zf

Inductive

Capaci t ive

17. For the signal– flow graph shown in the figure,which one of the fol lowing expressions is equal

to the t ransfer funct ion 1

2 X ( ) 0

Y( )X ( )

s

ss

.

Y(s)G21

–1

G1

X1(s) X2(s)

–1

(a)1

2 1

G1 G (1 G ) (b)

2

1 2

G1 G (1 G )

(c)1

1 2

G1 G G (d)

2

1 2

G1 G G

18. The vol tage developed across the 3 and 2r esistor s shown in the figur e are 6V and 2Vrespect ively, with the polar ity as marked. Whatis the power (in Watt) del ivered by the 5V voltagesource ?

3

Network 1 Network 25

2v

5v

6v +–

+

+

–

–

(a) 5 (b) 7(c) 10 (d) 14

19. The self inductance of the pr imary winding of asingle phase, 50 Hz, t ransformer is 800 mH, andthat of the secondary winding is 600 mH. Themutual inductance between these two windingsi s 480 mH. The secondar y winding of th i st ransformer is shor t circuited and the pr imarywinding is connected to a 50 Hz, single phase,sinusoidal voltage source. The current flowing inboth the winding is less than their respective ratedcur rents. The resistance of both windings can beneglected. In this connect ion, what is the effectiveinductance (in mH) seen by the source?(a) 416 (b) 440(c) 200 (d) 920

20. A Bode magnitude plot for the transfer functionG(s) of a plant is shown in the figure. Which oneof the fol lowing transfer functions best descr ibesthe plant?

20 log| G(j2 )f

200

– 20

0.1 1 10 100 1k 10k 100k f(Hz)

(a)1000( 10)

1000s

s

(b)

10( 10)( 1000)

ss s

(c)1000

10 ( 10)ss s

(d)1000

10( 10)s

s

21. I n the 4 1 mult iplexer, the output F is given byF = A B. Find required inputs I 3I 2I 1I 0.

A B

F4×1MUX

I 0

I 1

I 2

I 3S1 S0

(a) 1010 (b) 0110

(c) 1000 (d) 1110

22. I n the given circuit the si l icon t r ansistor has = 75 col lector voltage VC = 9V. The rat io ofRB and RC is .

RCRB

15V

VC

23. A(0– 50A) mov i ng coi l ammet er has avol tage dr op of 0.1V acr oss i t s ter minals atfu l l scal e def l ect i on . The ex t er nal shun tresistance (in mil l iohms) needed to extend itsrange to (0 – 500A) is .


24. Consider the circuit shown in the figure. In thiscircuit R = 1k, and C = 1F. The input voltage issinusoidal with a frequency of 50 Hz, representedas phasor with magnitude V i and phase angle0 r adian as shown in the figure. The outputvol t age i s r epr esen t ed as a phasor w i t hmagnitude V0 and phase angle radian. What isthe value of output phase angle (in radian)relat ive to the phase angle of the input voltage?

R

C

C

R

+

–

vi i = V 0 vo o = V

(a) 0 (b)

(c)2

(d) –2

25. A steady cur rent I is flowing in the – x direct ionthr ough each of two infinitely long wi r es at

L2

y as shown in the figure.

The per meabi l i t y of the medium i s 0. The

B

– field at (0, L , 0) is

(a) 04 Iˆ

3 Lz

(b) 04 Iˆ

3 Lz

(c) 0

y = – L /2 y = L /2

0

x

Cur rent=I Cur rent=I

z

(d) 03 Iˆ

4 Lz

26. Consider a discrete t ime signal given byx[n] = (– 0.25)n u[n] + (0.5)n u[ – n – 1]

The region of convergence of its Z-transform wouldbe(a) the region inside the circle of radius 0.5 and

centered at or igin.(b) the region outside the circle of radius 0.25 and

centered at or igin.(c) the annular region between the two circles,

both centered at or igin and having radii 0.25and 0.5.

(d) the ent ire Z plane.

27. Two players A and B, alternately keep rol l ing afair dice. The person to get a ‘six’ fir st wins thegame. Given that player A star ts the game, theprobability that A wins the game is .

(a) 511

(b) 12

(c) 713

(d) 611

28. The circuit shown in meant to supply a resist iveload RL from two separate DC voltage sources.The switches S1 and S2 are controlled so that onlyone of them is ON at any instant . S1 is turned onfor 0.2 ms and S2 is turned on for 0.3 ms in a0.5 ms switching cycle t ime per iod. Assumingcont inuous conduction of the inductor current andnegl igible r ipple on the capacitor voltage, theoutput voltage Vo (inVolt ) across RL is ________.

+–

+–

S1

10V 5V

S2

L

CRL Vo

+

–

29. Determine the cor rectness or otherwise of thefol lowing Asser t ion [a] and the Reason [r ].Assert ion: Fast decoupled load flow method givesapproximate load flow solut ion because it usesseveral assumpt ions.Reason : Accu r acy depends on t he powermismatch vector tolerance.(a) Both [a] and [r ] are t rue and [r ] is the cor rect

reason for [a].(b) Both [a] and [r ] are t rue and [r ] is not the

cor rect reason for [a].

(c) Both [a] and [r ] are false.(d) [a] is false and [r ] is t rue.

30. I n the given circuit parameter k is posit ive andpower dissipated in 2 resistor is 12.5W. Thevalue of k is .

5A4V

2 5

10

kV0

V0


31. I n the signal flow diagram given in the figure, u1

and u2 are possible inputs whereas y1 and y2 arepossible outputs.When would the SISO systemder ived fr om this diagram be contr olable andobservable ?

5

u1

1

1

1/sx1 y1

u2

y2

x21/s

1

2

– 1

1

(a) When u1 is the only input and y1 is the onlyoutput

(b) When u2 is the only input and y1 is the onlyoutput

(c) When u1 is the only input and y2 is the onlyoutput

(d) When u2 is the only input and y2 is the onlyoutput

32. In a l inear two-port network, when 10 V is appliedto Por t 1, a cur rent of 4 A flows through Por t2 when it is shor t -circuited. When 5V is appliedto Por t1, a cur rent of 1.25 A flows through a 1resistance connected across Por t 2. When 3V isappl ied to Por t 1, then cur r ent (in Amper e)t hr ough a 2 r esi st ance connect ed acr ossPor t 2 is _________.

33. A self commutat ing switch SW, operated at dutycycle is used to control the load voltage as shownin the figure.

DVL

Vdc

L

SW C VCRL

Under steady state operat ing condi t ions, theaver age vol tage acr oss the inductor and thecapacitor respect ively, are

(a) VL = and VC = 1

V1 dc

(b) VL = V2 dc

and VC =

1V

1 dc

(c) VL = 0 and VC = 1

V1 dc

(d) VL = V2 dc

and VC = V

1 dc

34. The figure shown a digital circuit constructedusing negat ive edge t r igger ed J-K fl ip flops.Assume a star t ing state of Q2Q1Q0 = 000. Thisstate Q2Q1Q0=000 wi l l r epeat aft er number of cycles of the clock CLK

1

1 1 1Clk

J0 Q0

K0

J1

K1

J2 Q2

K2 Q2

Q1

Q0Q0

Clock

35. The signum funct ion is given by

sgn(x) = ; 0

0 ; 0

xx

x

x

The four ier ser ies expansion of sgn(cos(t)) has(a) only sine terms with al l harmonics

(b) only cosine terms with al l harmonics

(c) only sine terms with even numbered harmonics

(d) only cosine terms with odd numbered harmonics

36. A DC motor has the following specifications: 10hp,37.5 A, 230V; flud/pole = 0.01Wb, number ofpoles = 4, number of conductors = 666, numberof par al l el paths = 2. Ar mat ur e r esi st ance= 0.267. The armature reaction is negligible androtat ionallosses are 600W. The motor operatesfrom a 230V DC supply. I f the motor runs at1000 rpm, the output torque produced in (in Nm) is

37. Find the t ransfer funct ion Y( )X( )

ss of the system

given below.+

G1

G2

H

–+

+

+–

X(s)Y(s)

Y(s)

(a)1 2

1 2

G G1 HG 1 HG

(b)

1 2

1 2

G G1 HG 1 HG

(c)1 2

1 2

G G1 H(G G )

(d)

1 2

1 2

G G1 H(G G )

38. The t ransfer funct ion of a second or der r ealsystem with a per fect ly flat magnitude responseof unity has a pole at (2 – j3). L ist al l the polesand zeroes.(a) Poles at (2 ± j3), no zeroes(b) Poles at (±2 – j3), one zero at or igin(c) Poles at (2 – j3), (– 2 + j3), zeroes at (– 2 – j3),

(2 + j3)(d) Poles at (2 ± j3), zeroes at (– 2 ± j3)


39. Two single-phase t ransformers T1 and T2 eachr ated at 500 k VA ar e oper ated in par al lel .Percentage impedances of T1 & T2 are (1 + j6) and(0.8 + j4.8) r espect ively. To shar e a l oad of1000 k VA at 0.8 l agging power fact or, t hecontr ibut ion of T2 (in kVA) is .

40. A paral lel plate capacitor is par t ial ly fi l led withglass of dielectr ic constant 4.0 as shown below.The dielect r ic st r engths of ai r and glass ar e30 kV/cm and 300 kV/cm, r espect ivel y. Themaximum voltage (in ki lovolts), which can beappl i ed acr oss t he capaci t or w i t hout anybreakdown, is _______.

10mm

Air, r = 1.0 5mm

Glass, r = 4.0

41. A sustained three-phase fault occurs in the powersystem shown in the figure. The cur r ent andvoltage phasors dur ing the fault (on a commonreference), after the natural transients have dieddown, are also shown. Where is the fault located?

V1

I 1 I 3

I 4I 2

V2

V1

V2

I 3

I 1

I 2

Transmission l ine

R

Transmission l ine

I 4

Q S

P

(a) Locat ion P

(b) Locat ion Q

(c) Locat ion R

(d) Locat ion S

42. The maximum value of "a" such that the matr ix

3 0 2

1 – 1 0

0 2a

has three linearly independent

real eigenvectors is

(a)2

3 3(b)

1

3 3

(c)1 2 3

3 3

(d)

1 3

3 3

43. The open loop poles of a third order unity feedbacksyst em ar e at 0, – 1, – 2. Let t he f r equencycor responding to the point where the root locus ofthe system transits to unstable region be K. Nowsuppose we introduce a zero in the open looptransfer function at – 3, while keeping all the earlieropen loop poles intact . Which one of the followingis TRUE about the point where the root locus ofthe modified system transits to unstable region?(a) I t cor responds to a frequency greater than K(b) I t cor responds to a frequency less than K(c) I t cor responds to a frequency K(d) Root locus of modified system never t ransits

to unstable region44. A 200/400V, 50 Hz, two-winding t ransformer is

rated at 20 kVA. I ts windings are connected as anauto-t ransformer of rat ing 200/600V. A resist iveload of 12 is connected to the high voltage (600V)side of t he aut o-t r ansfor mer. The value ofequivalent load resistance (in Ohm) as seen fromlow voltage side is _________.

45. Consider the economic dispatch problem for apower plant having two generating units. The fuelcosts in Rs/MWh along with the generation l imitsfor the two units are given below:

C1 (P1) = 21 10.01P 30P 10 ; 100MW

P1 150MW

C2 (P2) = 22 20.05P 10P 10 ; 100MW P2 180MW

The incremental cost (in Rs/MWh) of the powerplant when it supplies 200 MW is _____.

46. An unbalaced DC Wheatstone br idge is shown inthe figure. At what value of p wil l the magnitudeof V0 be maximum?

(a) 1 x

(b) (1 + x)

(c)1

(1 )x

(d) 1 – x

47. A separately excited DC motor runs at 1000 rpmon no load when it s ar matur e terminals ar econnected to a 200V DC source and the ratedvol t age is appl ied to the field winding. Thear mature resistance of this motor is 1. Theno-load armature cur rent is negligible. With themot or devel oping i t s fu l l l oad t or que, t hearmature voltage is set so that the rotor speed is500 rpm. When the load torque is reduced to 50%of the ful l load value under the same armaturevoltage condit ions, the speed r ises to 520 rpm.Neglect ing the rotat ional losses, the ful l loadarmature cur rent (in Ampere) is _______.


48. A solut ion of or dinar y di ffer ent ial equat i on2

2

56 0

d y dyy

dt dt i s such that y(0) = 2 and

y(1) = 3

1 3ee

. The value of (0)dydt

is .

49. The op-amp shown in the figure has a finite gainA = 1000 and an infinite input resistance. A step-voltage V i = 1 mV is applied at the input at t imet = 0 as shown. Assuming that the operat ionalampli fier is not satur ated, the t ime constant(in mil l isecond) of the output voltage Vo is

+–

–

+V i

R

1k A=1000+

–Vo

1mV

t=0s

1 F

C

(a) 1001 (b) 101

(c) 11 (d) 150. A 3-phase 50 Hz square wave (6-step) VSI feeds a

3-phase, 4 pole induct ion motor. The VSI l inevoltage has a dominant 5th harmonic component.I f the operat ing sl ip of the motor with respect tofundamental component voltage is 0.04, the sl ipof t he mot or wi t h r espect t o 5 t h har mon i ccomponent of voltage is ________.

51. An 8 bi t un ipolar successive appr oximat ionregister type ADC is used to convert 3.5V to digitalequal output . The reference voltage is +5V. Theoutput of ADC at end of 3rd clock pulse after thestar t of conversion is .

(a) 1010 0000 (b) 1000 0000

(c) 0000 0001 (d) 0000 0011

52. The si ngle-phase ful l -br i dge vol t age sour ceinver ter (VSI ), shown in figure, has an outputfrequency of 50Hz. I t uses unipolar pulse widthmodulat ion with switching frequency of 50kHzand modulat ion index of 0.7. For Vm =100V DC,L = 9.55mH, C = 63.66 mF, and R = 5W, theamplitude of the fundamental component in theoutput voltage V0 (in volt ) under steady-state is

Vo

+

–

+

–

C

L

Ful l Br idge

VSI+–

Vn

V in

R

For the given 1-phase, ful l br idge VSI , switchingfrequency is 50 kHz and single pulse modulatedwith modulat ion index m = 0.7. The fundamentaloutput voltage is .

53. f(A,B,C,D) = m (0,1,3,4,5,7,9,11,12,13,14,15) is amaxterm representat ion of a Boolean funct ionf(A,B,C,D) where A is the MSB and D is the LSB.The equivalent minimized representat ion of thisfunct ion is

(a) (A C D)(A B D)

(b) ACD ABD

(c) ACD ABCD ABCD

(d) (B C D)(A B C D)(A B C D) 54. A 50H z gener at i ng un i t has H -const ant of

2 MJ/MVA. The machine is init ial ly operat ing insteady state at synchronous speed, and producing1 pu of real power. The init ial value of the rotorangle is 5, when a bolted three phase to groundshor t circuit fault occurs at the terminal of thegenerator. Assuming the input mechanical powerto remain at 1 pu, the value of in degrees, 0.02second after the fault is ________.

55. The cir cuit shown in figur e has two sour cesconnect ed i n ser i es. The i nst an t aneousvol t age of AC sour ce (i n vol t ) i s gi ven byv(t) = 12sin t. I f the circuit is in steady-state,then the rms value of cur rent (in Ampere) flowingin circuit is .

1

1H8v

v(t)

+

–


GENERAL APTI TUDE

2. Here the given sets areA = {2, 3, 4, 5}

and B = {11, 12, 13, 14, 15}desired outcome = {(2, 14), (3, 13), (4, 12), (5, 11)} n(E) = 4and total outcome = 4 5 = 20 n(s) = 20 Probabil i ty of an event

=(E)( )

nn s

= 4

20 = 1

5 = 0.2

7. Total number of students in class = 44

Average of marks,

= (2 21) (3 15) (1 11) (2 23) (5 31)

42

= 42 45 11 46 155

44

= 299

6.79544

Opt ion (c) is cor rect .8. According to quest ion,

There are 40 males in Electr ical Engg. Total number of females in Electr ical Engg.

= 72 – 40 = 32

EXPL ANATI ONS

Total number of students in Electrical Engg. = 7220% = 72 30% = Total number of students in Civi l Engg. Total number of students in Civi l Engg.

= 30 72

20

= 36 × 3 = 108

Number of female students in Civi l Engg.

= 4 108

9

= 4 × 12 = 48Now, total number of students in Mechanical Engg.

= 10 72

20

= 36

Number of female students in Mechanical Engg.

= 4 36

9

= 16

Difference between female students in Civi land Mechanical depar tments

= 48 – 16 = 32

ANSWERS

GENERAL APTI TUDE

1. (b) 2. (a) 3. (c) 4. (a) 5. (a) 6. (a)

7. (c) 8. (32) 9. (c) 10. (d)

TECHNI CAL SECTI ON

1. (90) 2. (0.192) 3. (9) 4. (a) 5. (3.36) 6. (d)

7. (c) 8. (a) 9. (0.25) 10. (c) 11. (b) 12. (1)

13. (c) 14. (c) 15. (b) 16. (b) 17. (b) 18. (a)

19. (a) 20. (d) 21. (b) 22. (105.1) 23. (0.22 to 0.23) 24. (d)

25. (a) 26. (c) 27. (d) 28. (7) 29. (d) 30. (0.5)

31. (b) 32. (0.545) 33. (a) 34. (6) 35. (d) 36. (57 to 58)

37. (c) 38. (d) 39. (555) 40. (18.75) 41. (b) 42. (b)

43. (d) 44. (1.3 to 1.4) 45. (20) 46. (a) 47. (8) 48. (– 3)

49. (a) 50. (1.16 to 1.22) 51. (a) 52. (60 to 64) 53. (a) 54. (5.7 to 6.1)

55. (10)


TECHNI CAL SECTI ON

1. u() = sin ()

= 2f = 1

2 .2T

T =

y(t) =T

T

1 cos( )sin( )

T T

tt

tt

d

= 1[cos ( T) cos ]t t

= 1[cos cos T sin sin T cos ]t t t

y(t) =2 2

cos sin(90 )t t

x(t) = sin t = 90

2. P = 2emfVR

Vemf = ddt

= 1

B. S B.S. sin800S

d t

Vemf = 1

cos8

dt

dt

p =2

2 1cos

64 Rt

p = 2 1 cos2

0.4 64 2t

pavg = 2 2

cos220 0.4 64 0.4 64 2

t

pavg = 2

20 0.4 64

= 0.192 W

3. Sum of the diagonals elements is – 6 for 2 2matr ix The possible eigen value are

1, 5 5, 1 8,22, 3 4, 2 9,3

3, 1 3, 3 10,4

M aximum possibl e value of deter minant i s– 3 – 3 = 9.

4. For balanced wheat stone br idgeR1RX = R2R3

RX = 2 3

1

R RR

...(i)

R3 has tolerance of 5%

Value of R3 can be = 100 5% of 100

= 100 5 = 95, 105

Value of RX i f R3 = 95

RX =65 95

50

= 123.5

Value of RX i f R3 = 105

RX =65 105

50

=136.5

Range of RX in ohms = 123.5, 136.5

5.

– +

i

2V 1 2

A

B

I 1

1( + I ) i 120i

Apply KVL

2 = 11 I 1i i

2 = 2i + I 1 ...(i)

2 = 1 11 I 20 2Ii i

2 = – 19i + 3I 1 ...(i i)Solving equat ions, (i ) & (i i )

I1 = 1.68 AmpThevenin’s voltage,

V th = 2I 1 volt = 2 1.68 V th = 3.36 volt

6.

0

1

1

g t( )

G H( )sGImpulse Response

Impulse Response,

H(s) = G(s) · G(s) ...(i)

Taking Inverse Laplace t ransform both sides,

h(t) = g(t) g(t) ...(i i)

where g(t) = u(t) – u(t – 1)

h(t) = ( ) ( 1) ( ) ( 1)u t u t u t u t

= u(t) u(t) – u(t) u(t – 1)

– u(t) u (t – 1) + u(t – 1) u(t – 1)

= r (t) – r (t – 1) – r (t – 1) + r (t – 2)

h(t) = r (t) – 2r (t – 1) + r (t – 2)


t210

1h t( )

Therefore, Maxm value of Impulse Response

max( ) 1h t

7. Base Load power plants are

(i) Wind farms

(ii) Run of r iver plants

(iii)Nuclear power plants8. M ajor r equ i r emen t for an I nst r umen t i on

Amplifier

(i) High Common Mode Reject ion Rat io

(ii) High Input Impedance

(iii) High L inear ity

9. ( )f x dx

= 1

So1

0( )a bx dx = 1

2b

a = 1

2a + b = 2 ... (1)given E[X] = 2/3

=1

0[ ]x a bx dx

23

=2 3a b

3a + 2b = 4 (2)from (1) and (2)

a = 0b = 2

pr[X < 0.5] = 0.5 0.5

0 0( ) 2 0.25f x dx x dx

11.2

1

EE

b

b =

2 2 1 1

1 1 1 51

N 0.5N 0.5N N

Eb2 = 0.25 Eb1 = 0.25 200 = 50 = R C = 0.1 1000

50 = 5000e– t/100 10– 6

= 69.3 sec

12. V0 = DVS = 0.4 20 = 8 V

I 0 = 0V – E 8 5 3R 3 3

= 1A

13. From intermediate value theroem,I f f(a) f(b) < 0 then f(x) wil l posses one root in (a, b)Thus for cont inuous funct ion f(x) which do nothave a root in (a, b), the condit ion should be

f(a) f(b) > 0

14. The pr imary mmf is least affected by the secondaryterminal condit ions in a cur rent t ransformer. I tis required for voltage stabil i ty.

15.

V2V1

+

–

+

–

I

I = 1 2V VR

For power to be reversed

I = 2 1V V( Ve)

R

V1 = – 485kVV2 = – 500kV

I = 1.5kA

16.i

C

L

Since both the impedances are in paral lel

Z = 1 2

1 2

Z ZZ Z

where Z1 = jXL = jL

Z2 = – jXc = jc

Z = L

L

jj

cj

jc

= 2

LCLC 1

jc

Z = 2

LLC 1

j

= 2

C1

LC

j

Resonant frequency of the circuit

r = 1

rad/secLC

Case (i ) : I f < 1

LC

Z = 21

LC

cj

I nduct ive


Case (i i ) : I f > 1

LC

Z = – j2 1

LC

c

Capacitive

Not e : I t is infinite at Resonant frequency

Z at r = 1

LCInductive

Capacitive

O f

z

w (fr)r

17.X (s)1

X (s)2

Y(s)

–1–1

1 G1 G2

when X 1(s) = 0

X (s)2

Y(s)

–1–1

1 G1 G2

For war d pat h gain ,

P1 = G2

Loop gain,

L 1 = – G1

L 2 = – G1 G2

= 1 – (L 1 + L 2) + L 1L 2

= 1 + G1+ G1G2 1 2L L 0 = 1 + G1(1 + G2)

Tr ansfer funct ion,

2

Y( )G ( )

ss = 1 1P

= 2

1 2

G .11 G (1 G ) 1 1

1

2

2 1 2G ( ) 0

GY( )G ( ) 1 G (1 G )

s

ss

18.I 1

N/w N/w1 2

5V– +

(Source)

6V– +

2V –+I 2

I 3

3

2

I 1 = VR

= 63

= 2 Amp

I 2 = 22

= 1 Amp

Apply KCL,

I 1 = I 2 + I 3

2 = 1 + I 3

I 3 = 1 Amp

Power delivered by 5V source,

P = V I 3 = 5 1P = 5 wat t s

19.

Zin = 1

1

VI

= (R1 + jX1)2 2

2 2 L

MR jX Z

Given, L 1 = 800 mH

L 2 = 600 mH

M = 480 mH

W = 314 rad/sec

ZL = 0

R1 R2 neglected

Zin =2 2

12

MjX

jX

=2 2

12

Mj X –

jX

=2 2314 0.48

j 314 0.80.6 314

= j[251.32 – 120.576]= j| 30.744 = jw L eff = j314.L eff

L eff = 0.416 = 416 mH

20. I nit ial slope is zeroI t is type ‘0’ system(No pole at or igin)

Star ting gain = 20 log K db20 log K = 20

K = 10


Given T/F

G(s) = 1000

10( 10)s+

s = 1000 1

1000

10 10 110

s

s

10 11000

G( )1

10

s

ss

21. I 0

I 1

I 2

I 3s1 s0

F

A B

4 × 1MUX

We know that ,

for 4 1 MUX

F = 0 1 2 3A.B.I + AB.I + A.B.I + A.B.I

But required output

F = A B = AB + ABTherefore

I 0 = I 3 = 0

I 1 = I 2 = 1

Requi r ed I /P

I 3I 2I 1I 0 = 0110

22. 15 volt

RC

= 75

IC

I + BC I

V = 9 voltC

V = 0.7 voltBE

RB IB

I C = BI B

I C = 75 I B

I C + I B = 75 I B + I B = 76I B

I C + I B = C

15 9R

76I B = C

6R

...(i )

I B = B

9 0.7R

= B

8.3R

...(i i )

Equat i on (i )/Equat ion (i i ),

B

B

76II

= C

B

6R

8.3R

= B

C

R6R 8.3

B

C

RR =

76 8.36

= 105.1

B

C

R105.13

R

23. 50 Amp

0.1 V(0–50)A

Voltage dr op across i ts terminals at ful l scaledeflect ion = 0.1 volt

0.1 VRsh

50 Amp

450 A 500 A

450 × Rsh = 0.1

Rsh =0.1450

= 0.22 m

Required shunt resistance,Rsh= 0.22m

25.

H = H 1 + H 2

= 1 1( ) ( )

3L2 (L /2) 22

z za a

=1 2 2

( )2 L 3Lza

=4I

( )3 L za


26. x[n] = (–0.25)n u[n] + (0.5)n u[– n – 1]= x1[n] + x2[n]

x1[n] = (– 0.25)n u[n] (+ve sided sequence)

x1[z] = – (– 0.25) 0.25z z

z z

I m

Rc

– 0.25

z-plane

ROC1 :| z| > | – 0.25|| z| > 0.25

x2[n] = (0.5)n u[– n – 1](– ve sided sequence)

x2(z) =– 0.5z

zIm

Rc

z-plane

0.5

ROC2 :| z| < | 0.5|| z| < 0.5

Therefore required ROC for given signal

0.5 | | 0.25z

27. The person who get ‘six’ fir st win the game Probabil i ty of get t ing ‘6’ as outcome is throw

of dice = 16

i .e. Probabil i ty of A wins the game P(A) = 16

Probabil i ty the A do not win the game

P(A) =1 5

1 –6 6

Similar ly

Probabil i ty of B win the game P(B) = 16

Probabil i ty that B do not win the game

P(B) =1 5

1 –6 6

Now according to quest ion, A star ts the game Probabil i ty that A wil l win the game

= P(A) P(A).P(B).P(A)

P(A)P(B)P(A)P(B)P(A) ...

= 1 5 5 1 5 5 5 5 1

...6 6 6 6 6 6 6 6 6

= 2

1 16 5

16

= 1 366 36 25

= 611

28.

5V

10V

0.2 0.2 t(msec)

V0 = 10 0.2 5 0.3

0.5

= 7V

30.

+–

2 5

10

KV0

4 volt 5 Amp

I

+ –V0

Power dissipated in 2P = I2R

12.5 = I 2 × 2I = 2.5 Amp

V0 = I × 2 = 2.5 × 2V0 = 5 volt

Apply KCL,I + KV0 = 5

2.5 + K × 5 = 5

K 0.5


32. Case (i ):

V1 = 10V V2 = 0

I 2 = 4 Amp

Two PortNetwork

(S/C)

We know that ,I2 = Y21V1 + Y22V2

4 = Y21 × 10 + Y22 × 0

Y21 =4

0.410

Case (i i ):

V1 = 5V 1

1.25 Amp

Two PortNetwork +

–

V = –1.25 volt2

I2 = Y21V1 + Y22V2

1.25 = 0.4 × 5 + Y22 (– 1.25)– 0.75 = – 1.25 Y22

22Y 0.6

Case (i i i ):

V1 = 3V 2Two PortNetwork +

–V = –22 2I

I 2

I2 = Y21V1 + Y22V2

I2 = 0.4(3) + 0.6 (– 2I 2)I2 = 1.2 – 1.2I 2

2.2I 2 = 1.2

I2 =1.22.2

2I 0.545 Amp

34. First flip flop acts as mod-2 counter Second 2 flipflops from mod (2n-1) Johnson counter = mod counter overal l modulus = mod – 6 counter

35. The signum function is given by,

sgn(x) =;

0; 0

xx 0

| x|x

Similar ly,

Sgn(cos t) =cos

; cos 0cos

0; cos 0

tt

| t |

t

cos| cos

tt |

=1, cos 0

– 1; cos 0

t

t

+ve

–ve

+ve

–ve

+ve

t

– 1

Cost

t

Sgn (cost)

1

I t i s even funct i on and show hal f wavesymmetr y. I t contains only cosine term withodd numbered harmonics.

0 0a

0nb

an =0; even

0; odd

n

n

Sgn(Cost) = a1cost + a3cos3t + a5cos5t + ...

37. + G1

G2

H

–

++

+–

X(s)Y(s)

Y(s)

Y2(s)

Y1(s)

Y(s) = Y1(s) + Y2(s)

Y1(s) = (X – HY)G1 = XG1 – HG1Y

Y2(s) = (X – HY)G2 = XG2 – HG2Y

Y(s) = XG1 – HG1Y + XG2 – HG2Y

Y = X(G1 + G2) – H(G1 + G2)Y

Y + Y(G1 + G2)H = (G1 + G2)X

Y(1 + (G1 + G2)H) = (G1 + G2)X

Y(s)X(s) =

1 2

1 2

G G1 H(G G )

38. System is second order

I t means, number of poles = 2

System has per fect ly flat magnitude response.

I t means, I t is al l pass system

In al l pass system,

Poles and zeros are mirror image about Imaginaryaxis,


One pole = (2 – j3)

Other pole = (2 + j3)

(– 2+j3) (2+j3)splene

Re

(2– j3)(– 2– j3)

I m

Therefore

poles at (2 ± j3)

zeros at (– 2 ± j3)

39. ST2 = 1

1 2

Sz

z z

= 6.08 80.531000

10.94 80.53

= 555KVAA

40.

C1 =0A

d

C2 =04A

d

Ceq =01 2

1 2

4AC CC C 5d

Dn =eqvCQ CV

A A As

Dn =04A

V5 Ad

Dn =04

V5 Ad

E1 =0

D 4V

5n

d

30 105 =4V4d

V =3 530 5 5 10 10

4

V = 18.75 kV

42. The character ist ic equat ion of A is| A-XI | = 0

f(x) = x3 + 6x2 + 11x + 6 + 2a= (x + 1)(x + 2)(x + 3) + 2a = 0

f(x) cannot have al l 3 real roots (i f any) equalfor i f f(x) = (x – k)3, then compar ing coefficients,we get

6 = – 3k, 3k 2 = 11No such k exists(a) Thus f(x) = 0 has repeated (2) roots (say) ,,

or(b) f(x) = 0 has real roots (distance)(say) ,,Now f (x) = 0

x1 =6 3

2.577;3

x2 =6 3

1.4223

At x1, f(x) has relat ive max.At x2, f(x) has relat ive min.The graph of f(x) wil l be as below

x1

x2

xMax Min

(x1 is repeated root )

xx2 x1

Max

(x2 is repeated root )

Min

yy

case (a) r ep eat ed r oot s ( , , )

Note that the graph of f(x) cannot be l ike the onegiven below

Thus in al l possible cares we have

f(x2) 0 3

2 09

a

1

3 3a


43. Open loop t ransfer function of a third order unityfeedback system

G(s) =K

( 1)( 2)s s s

Now zero is int roduce at – 3.Modified t ransfer funct ion,

G(s) =K ( 3)

( 1)( 2)s

s s s

Closed loop character ist ic equat ion,1 + G(s) = 0

K ( 3)1

( 1)( 2)s

s s s

= 0

s(s + 1)(s + 2) + K(s + 3) = 0s3 + 3s2 + 2s + Ks + 3K = 0s3 + 3s2 + (K + 2)s + 3K = 0

Routh ar r ay cor r esponding to char acter ist icequat ion,

3

2

1

0

1 (K 2)

3 3K2 0

3K 0

s

sss

For every value of K > 0I st column of the ar ray are posit ive.Therefore, Root locus of modified system nevertransits to unstable region.

45. 1

1

CP

dd

= 2 0.01P1 + 30

= 0.02 P1 + 30

2

2

CP

dd

= 2 0.05P2 + 10

= 0.1P2 + 10

1

1

CP

dd

= 2

2

CP

dd

0.02P1 + 30 = 30 = 0.1P2 + 102P1 + 3000 = 10P2 + 10002P1 + 2000 = 10P2

P1 + P2 = 200P2 = 200P1 = 0

1

1

CP

dd

= 20 Rs/Mwh

46.

V0(– ) =RE E

R PR 1 P

V0(+)=R.

.E ER PR

y yy p y

V0 = V0(+) – V0(– ) = 1

E1

yp y p

0Vddp

= 2 2

1E

( ) (1 )y

p y p

= 0

2

1(1 )p = 2( )

yp y

11 p =

yp y

p + y = y p y

(1 )p y = y y = [1 ]y y

p = y = 1 x

48. Roots, – 3, – 2y(t) = C1e

– 3t + C2e– 2t

y(0) = C1 + C2 = 2

y(1) = 3

1 3ee

= – e– 3 + 3e– 2 = C1e– 3 + C2e

– 2

so, C1 = – 1, C2 = – 3so,

y(t) = – e– 3t + 3e– 2t

( )dy tdt

= 3e– 3t – 6e– 2t, (0)dydt

= 3 – 6 = – 3

51. 8 bit successive Approximation register type ADC

Reference voltage = +5V

If MSB = 1 2.5 volt

V i > 2.5 volt (yes) MSB = 1

If next bit = 1 2.5 + 1.25

= 3.75 volt

V i > 3.75 volt (no) Bit = 0

If next bit = 1 2.5 + 0 + 0.625

= 3.125 volt

V i > 3.125 volt (yes) Bit = 1

Output of ADC at the end of 3rd clock pulse

= 1010 0000


55.

1

1H8v

v t t( ) = 12 sin

+

–

Applying superposit ion theorem,

Consider D.C. source

1

X = L 0

8 vol t+

–

I

I = V 8R 1

= 8 Amp

Consider A.C. source

1

X = 1L

v t t( ) = 12 sin

i t( )

XL = L = 1 1 = 1

Vrms =V 12

6 22 2m volt

Z = 2 2LR X = 2 2(1) (1)

= 2

I rms =rmsV 6 2Z 2

= 6 Amp

We know that ,(I rms)

2 = (I dc)2 + (I ac)

2

= (8)2 + (6)2

= 64 + 36 = 100I rms = 10 Amp

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