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GENERAL APTI TUDE (GA)(Q.1 – 5) : Car r y One M ar k Each

1. The man who is now Municipal Commissionerworked as ____________________.(a) the secur i ty guard at a university(b) a secur i ty guard at the university(c) a secur i ty guard at university(d) the secur i ty guard at the university

2. Nobody knows how the I ndian cr icket team isgoing to cope wi th t he di ff i cul t and seamer -fr iendly wickets in Austral ia.Choose the opt ion which is closest in meaning tothe under l ined phrase in the above sentence.(a) put up with (b) put in with(c) put down to (d) put up against

3. Find the odd one in the fol lowing group of words.mock, der ide, praise, jeer

(a) mock (b) der ide(c) praise (d) jeer

4. Pick the odd one from the fol lowing opt ions.(a) CADBE (b) JHKIL(c) XVYWZ (d) ONPMQ

5. I n a quadrat ic funct ion, the value of the productof the roots (, ) is 4. Find the value of

n n

n n

(a) n4 (b) 4n

(c) 22n– 1 (d) 4n– 1

(Q.6 – 10) : Car r y Tw o M ar k s Each6. Among 150 faculty members in an inst i tute, 55

are connected with each other through Facebook®

and 85 ar e connect ed t hr ough What sApp®.30 faculty members do not have Facebook® orWhatsApp® accounts. The number of facul t ymember s connected only t hr ough Facebook ®

accounts is ______________.(a) 35 (b) 45(c) 65 (d) 90

7. Computers were invented for per forming onlyhigh-end useful computat ions. However, i t is nounderstatement that they have taken over ourwor l d t oday. The i n t er net , for exampl e, i subiquitous. Many bel ieve that the internet i tselfis an unintended consequence of the or iginalinvent ion. With the advent of mobile comput ingon our phones, a whole new dimension is nowenabl ed. One i s l ef t wonder i ng i f al l t hesedevelopments are good or, more impor tant ly,required.Which of the statement(s) below is/are logical lyval i d and can be i n fer r ed f r om t he aboveparagraph?(i ) The author bel ieves that computers are not

good for us.(i i ) Mobile computers and the internet are both

intended invent ions(a) (i ) only(b) (i i ) only(c) both (i ) and (i i )(d) neither (i ) nor (i i )

I NSTRUCTI ONS1. Total of 65 quest ions car rying 100 marks, out of which 10 quest ions car rying a total of 15 marks are in

General Apt i tude (GA)2. The Engineer ing Mathemat ics wi l l car ry around 15% of t he t ot al mar k s, the General Apt i tude sect ion

wil l car ry 15% of t he t ot al mar k s and the r emai n i ng 70% of t he t ot al mar k s.3. Types of Quest i ons

(a ) M u l t i p le Choi ce Quest i ons (M CQ) car rying 1 or 2 marks each in al l papers and sect ions. Thesequest ions are object ive in nature, and each will have a choice of four opt ions, out of which the candidatehas to mark the cor rect answer(s).

(b) Numer i cal Answ er Quest i ons of 1 or 2 marks each in al l papers and sect ions. For these quest ionsthe answer is a real number, to be entered by the candidate using the vir tual keypad. No choices wil lbe shown for these type of quest ions.

4. For 1-mar k mult iple-choice quest ions, 1/3 mar k s wil l be deducted for a wrong answer. L ikewise, for 2-mar k s mult iple-choice quest ions, 2/3 marks wil l be deducted for a wrong answer. There is no negat ivemarking for numer ical answer type quest ions.

GATE – 2016EE : EL ECTRI CAL ENGI NEERI NG

Set - 1No. of Quest i ons : 65 Maxi mum Mar k s : 100

2 SOLVED PAPER – 2016 (SET - 1)

8. All hi l l -stat ions have a lake. Ooty has two lakes.Which of the statement(s) below is/are logical lyval i d and can be i n fer r ed f r om t he abovesentences?(i ) Ooty is not a hi l l -stat ion.(i i ) No hi l l -stat ion can have more than one lake.(a) (i ) only(b) (i i ) only(c) both (i ) and (i i )(d) neither (i ) nor (i i )

9. I n a 2 4 rectangle gr id shown below, each cel l isa r ect angl e. H ow many r ect angl es can beobserved in the gr id?

(a) 21 (b) 27(c) 30 (d) 36

10.

Choose the correct expression for f(x) given in thegraph.(a) f(x) = 1 – | x – 1| (b) f(x) = 1 + | x – 1|(c) f(x) = 2 – | x – 1| (d) f(x) = 2 + | x – 1|

TECH NI CAL SECTI ON(Q.1 – 25) : Car r y One M ar k Each

1. The maximum value at tained by the funct ionf(x) = x(x – 1)(x – 2) in the interval [1, 2] is _____.

2. Consider 3 3 matr ix with every element beingequal to 1. I ts only non-zero eigenvalue is ____.

3. The Laplace Transform of f(t) = e2t sin(5t) u(t) is

(a) 2

54 29s s

(b) 2

55s

(c) 2

24 29

ss s

(d)5

5s

4. A funct ion y(t), such that y(0) = 1 and y(1) = 3e– 1,i s a sol u t i on of t he di f f er en t i al equat i on

2

22 0.

d y dyy

dtdt Then y(2) is

(a) 5e– 1 (b) 5e– 2

(c) 7e– 1 (d) 7e– 2

5. The value of the integral

2C

2 51

( 4 5)2

zdz

z z z

over t he cont ou r | z| = 1, t ak en i n t heant i-clockwise direct ion, would be

(a)2413

i(b)

4813

i

(c)2413

(d)1213

6. The t ransfer funct ion of a system is Y( )

.R( ) 2

s ss s

The steady state output y(t) is A cos(2t + ) fort he i npu t cos(2t ). The val ues of A and ,respect ively are

(a)1

, 452

(b)1

, 452

(c) 2, 45 (d) 2, 45 7. The phase cross-over frequency of the t ransfer

funct ion G(s) = 3

100( 1)s

in rad/s is

(a) 3 (b)1

3

(c) 3 (d) 3 38. Consider a cont inuous-time system with input x(t)

and output y(t) given byy(t) = x(t) cos(t)

This system is(a) l inear and t ime-invar iant(b) non-l inear and t ime-invar iant(c) l inear and t ime-varying(d) non-l inear and t ime-varying

9. The value of (2 2) ,te t dt

where (t) is the

Dirac delta funct ion, is

(a)12e

(b)2e

(c) 2

1e

(d) 2

12e

SOLVED PAPER – 2016 (SET - 1) 3

10. A temperature in the range of – 40 C to 55 C isto be measured with a resolut ion of 0.1 C. Theminimum number of ADC bits required to get amatching dynamic r ange of t he temperaturesensor is(a) 8 (b) 10(c) 12 (d) 14

11. Consider t he fol l owing ci r cui t which uses a2-to-1 mult iplexer as shown in the figure below.The Boolean expression for output F in terms ofA and B is

(a) A B (b) A B(c) A + B (d) A B

12. A t ransistor cir cui t is given below. The Zenerdiode breakdown voltage is 5.3 V as shown. Takebase to emit ter voltage drop to be 0.6 V. The valueof the cur rent gain is _________.

13. I n cyl indr ical coordinate system, the potent ialproduced by a uniform r ing charge is given by = f(r , z), where f is a cont inuous funct ion of rand z. Let E

be the result ing elect r ic field. Then

the magnitude of E

.

(a) increases with r .(b) is 0.(c) is 3.(d) decreases with z.

14. A soft-iron toroid is concentr ic with a long straightconductor car r ying a di r ect cur rent I . I f t herelat ive permeabil i ty r of soft -iron is 100, ther at i o of t he magnet i c f l ux densi t i es at t woadjacent point s locat ed just i nside and justoutside the toroid, is _______.

15. RA and RB are the input resistances of circuits asshown below. The circuits extend infinitely in thedi r ect ion shown. Which one of t he fol lowingstatements is TRUE?

(a) RA = RB (b) RA = RB = 0(c) RA < RB (d) RB = RA /(1 + RA)

16. I n a constant V/f induct ion motor dr ive, the sl ipat the maximum torque(a) is direct ly propor t ional to the synchronous

speed.(b) r emai ns const an t wi t h r espect t o t he

synchronous speed.(c) has an inverse relat ion with the synchronous

speed.(d) has no relat ion with the synchronous speed.

17. I n the por t ion of a ci r cui t shown, i f the heatgenerated in 5 resistance is 10 calor ies persecond, then heat generated by the 4 resistance,in calor ies per second, is _______.

18. I n the given circuit , the cur rent suppl ied by thebat tery, in ampere, is _______.

19. I n a 100 bus power syst em, t her e ar e10 generators. In a part icular i terat ion of NewtonRaphson l oad f l ow t echn i que (i n pol arcoordinates), two of the PV buses are conver tedto PQ type. I n this i terat ion.(a) t he number of unk nown vol t age angl es

increases by two and the number of unknownvoltage magnitudes increases by two.

(b) t he number of unk nown vol t age angl esr emai ns unchanged and t he number ofunknown voltage magnitudes increases by two.

(c) t he number of unk nown vol t age angl esincreases by two and the number of unknownvoltage magnitudes decreases by two.

(d) t he number of unk nown vol t age angl esr emai ns unchanged and t he number ofunknown vol tage magnitudes decreases bytwo.


20. The magnitude of three-phase fault cur rents atbuses A and B of a power system are 10 pu and8 pu, respect ively. Neglect al l resistances in thesystem and consider the pre-fault system to beunloaded. The pre-fault voltage at al l buses inthe system is 1.0 pu. The voltage magnitude atbus B dur ing a three-phase faul t at bus A is0.8 pu. The voltage magnitude at bus A dur ing athree-phase fault at bus B, in pu, is ________.

21. Consider a system consist ing of a synchronousgenerator working at a lagging power factor, asynchronous motor working at an overexci tedcondit ion and a direct ly gr id-connected induct iongenerator. Consider capacit ive VAr to be a sourceand induct ive VAr to be a sink of react ive power.Which one of the fol lowing statements is TRUE?(a) Synch r onous mot or and synch r onous

generator are sources and induct ion generatoris a sink of react ive power.

(b) Synchronous motor and induct ion generatorare sources and synchronous generator is asink of react ive power.

(c) Synchronous motor is a source and induct iongenerator and synchronous generator aresinks of react ive power.

(d) Al l are sources of react ive power.22. A buck conver ter, as shown in Figure (a) below,

is working in steady state. The output voltageand the inductor cur rent can be assumed to ber ipple free. Figure (b) shows the inductor voltageV L dur i ng a compl et e sw i t ch i ng i n t er val .Assuming al l devices are ideal, the duty cycle ofthe buck conver ter is ________.

23. A steady dc cur rent of 100 A is flowing through apower module (S, D) as shown in Figure (a). TheV-I character ist ics of the IGBT (S) and the diode(d) are shown in Figures (b) and (c), respect ively.The conduct ion power loss in the power module(S, D) in wat ts, is ________.

24. A 4-pole, lap-connected, separately exci ted dcmotor is drawing a steady cur rent of 40 A whilerunning at 600 rpm. A good approximat ion forthe waveshape of the cur rent in an armatureconductor of the motor is given by

(a) (b)

(c) (d)

25. I f an ideal t ransformer has an induct ive loadelement at por t 2 as shown in the figure below,the equivalent inductance at por t 1 is

(a) nL (b) n2L

(c)Ln

(d)2

Ln


(Q.26 – 55) : Car r y Tw o M ar k s Each26. Candidates were asked to come to an interview

with 3 pens each. Black, blue, green and red werethe permit ted pen colours that the candidatecould br ing. The probabi l i t y that a candidatecomes with al l 3 pens having the same colouris________.

27. Let 0

S where 1.n

n

n

The value of in

the range 0 < < 1, such that S = 2 is _______.28. Let the eigenvalues of a 2 2 matr ix A be 1, – 2

with eigenvectors x1 and x2 respect ively. Then theeigenval ues and ei genvect or s of t he mat r i xA2 – 3A + 4I would, respect ively, be(a) 2, 14; x1, x2

(b) 2, 14; x1 + x2, x1 – x2

(c) 2, 0; x1, x2

(d) 2, 0; x1 + x2, x1 – x2

29. Let A be a 4 3 real matr ix with rank 2. Whichone of the fol lowing statement is TRUE?(a) Rank of AT A is less than 2.(b) Rank of AT A is equal to 2.(c) Rank of AT A is greater than 2.(d) Rank of AT A can be any number between 1

and 3.30. Consi der t he fol l owi ng asympt ot i c Bode

magnitude plot ( is in rad/s).

Which one of the fol lowing t ransfer funct ions isbest represented by the above Bode magnitudeplot?

(a) 2

2(1 0.5 )(1 0.25 )

ss s

(b)4(1 0.5 )(1 0.25 )

ss s

(c)2

(1 2 )(1 4 )s

s s

(d) 2

4(1 2 )(1 4 )

ss s

31. Consider the following state-space representat ionof a l inear t ime-invar iant system.

1 0( ) ( ),

0 2x t x t

y(t ) = CT x(t ), C = 1

1

and

x(0) = 1

1

The value of y(t) for t = loge 2 is __________.32. Loop t ransfer funct ion of a feedback system is

G(s)H (s) = 2

3( 3)s

s s

. Take the Nyquist contour

in the clockwise direct ion. Then the Nyquist plotof G(s) encircles – 1 + j0(a) once in clockwise direct ion(b) twice in clockwise direct ion(c) once in ant iclockwise direct ion(d) twice in ant iclockwise direct ion

33. Given the fol lowing polynomial equat ions3 + 5.5 s2 + 8.5 s + 3 = 0,

the number of roots of the polynomial, which havereal par ts st r ict ly less than – 1, is ________ .

34. Suppose x1(t ) and x2(t ) have t he Fou r i ert ransforms as shown below.

Which one of the fol lowing statements is TRUE?(a) x1(t) and x2(t) are complex and x1(t)x2(t) is also

complex with nonzero imaginary par t(b) x1(t) and x2(t) are real and x1(t)x2(t) is also real(c) x1(t) and x2(t) are complex but x1(t)x2(t) is real(d) x1(t) and x2(t) are imaginary but x1(t)x2(t) is

real35. The output of a cont inuous-t ime, l inear t ime-

invar iant system is denoted by T{x(t)} where x(t)is the input signal. A signal z(t) is cal led eigen-signal of the system T , when T{z(t)} = z(t), where is a complex number, in general, and is cal ledan eigenvalue of T. Suppose the impulse responseof the system T is real and even. Which of thefol lowing statements is TRUE?(a) cos(t) is an eigen-signal but sin(t) is not(b) cos(t) and sin(t) are both eigen-signals but

with different eigenvalues(c) sin(t) is an eigen-signal but cos(t) is not(d) cos(t) and sin(t) are both eigen-signals with

ident ical eigenvalues


36. The cur rent state QA QB of a two JK fl ip-flopsystem is 00. Assume that the clock r ise-t ime ismuch smaller than the delay of the JK fl ip-flop.The next state of the system is

(a) 00 (b) 01(c) 11 (d) 10

37. A 2-bit flash Analog to Digital Conver ter (ADC)is given below. The input is 0 VN 3 Volts. Theexpression for t he LSB of the output B0 as aBoolean funct ion of X2, X1, and X0 is

(a) 0 2 1X [X X ] (b) 0 2 1X [X X ]

(c) 0 2 1X [X X ] (d) 0 2 1X [X X ]

38. Two elect r ic charges q and – 2q are placed at(0, 0) and (6, 0) on the x-y plane. The equat ion ofthe zero equipotent ial curve in the x-y plane is(a) x = – 2 (b) y = 2(c) x2 + y2 = 2 (d) (x + 2)2 + y2 = 16

39. I n the circuit shown, switch S2 has been closedfor a long t ime. At t ime t = 0 switch S1 is closed.At t = 0+, the rate of change of cur rent throughthe inductor, in amperes per second, is _____.

40. A three-phase cable is supplying 800 kW and600 kVAr to an induct ive load. I t is intended tosupply an addi t ional resist ive load of 100 kWthrough the same cable without increasing theheat dissipat ion in the cable, by providing a three-phase bank of capacitors connected in star acrossthe load. Given the l ine voltage is 3.3 kV, 50 Hzthe capacitance per phase of the bank, expressedin microfarads, is ________.

41. A 30 M VA, 3-phase, 50 H z, 13.8 k V, st ar -connected synchronous generator has posi t ive,negat ive and zero sequence reactances, 15%,15% and 5% respect ively. A reactance (Xn) i sconnected between the neutral of the generatorand ground. A double l ine to ground faul t takesplace involving phases ‘b’ and ‘c’, wi th a faul timpedance of j0.1 p.u. The value of Xn (in p.u.)that wi l l l imit the posit ive sequence cur rent to4270 A is _________.

42. I f the star side of the star -del t a t r ansformershown in t he f igure is exci t ed by a negat ivesequence voltage, then

(a) VAB leads Vab by 60 (b) VAB lags Vab by 60(c) VAB leads Vab by 30 (d) VAB lags Vab by 30

43. A single-phase thyr istor -br idge rect i fier is fedfrom a 230 V, 50 Hz, single-phase AC mains. I f i tis del iver ing a constant DC current of 10 A, atfi r ing angle of 30, then value of the power factorat AC mains is(a) 0.87 (b) 0.9(c) 0.78 (d) 0.45

44. The switches T1 and T2 in Figure are switchedin a complementary fashion with sinusoidal pulsewidth modulat ion technique. The modulat ingvoltage vm(t) = 0.8 sin (200t) V and the tr iangularcar r ier voltage (vc) are as shown in Figure (b).The car r ier frequency is 5 kHz. The peak valueof the 100 Hz component of the load cur rent (i L),in ampere, is ________ .


45. The voltage (vs) across and the current (i s) througha semi conduct or swi t ch dur i ng a t ur n-ONt r ansi t i on ar e shown i n f i gur e. The ener gydissipated dur ing the turn-ON t ransit ion, in mJ,is _______.

46. A single-phase 400 V, 50 Hz t ransformer has aniron loss of 5000 W at the rated condit ion. Whenoperated at 200 V, 25 Hz, the iron loss is 2000 W.When operated at 416 V, 52 Hz, the value of thehysteresis loss divided by the eddy cur rent lossis ______.

47. A DC shunt generator del ivers 45 A at a terminalvoltage of 220 V. The armature and the shuntfield resistances are 0.01 and 44 respect ively.The st r ay losses are 375 W. The per centageefficiency of the DC generator is ____________.

48. A three-phase, 50 Hz sal ient -pole synchronousmotor has a per-phase direct -axis reactance (Xd)of 0.8 pu and a per -phase quadr at ur e-axi sreactance (Xq) of 0.6 pu. Resistance of the machineis negl igible. I t is drawing ful l-load cur rent at0.8 pf (leading). When the terminal voltage is1 pu, per-phase induced voltage, in pu, is _________.

49. A single-phase, 22 kVA, 2200 V/ 220 V, 50 Hz,dist r ibut ion t ransformer is to be connected as anauto-transformer to get an output voltage of 2420 V.I ts maximum kVA rat ing as an autot ransformeris(a) 22 (b) 24.2(c) 242 (d) 2420

50. A single-phase full-br idge voltage source inverter(VSI ) is fed from a 300 V bat tery. A pulse of 120durat ion is used to tr igger the appropr iate devicesi n each hal f -cycl e. The r ms val ue of t hefundamental component of the output voltage,in volts, is(a) 234(b) 245(c) 300(d) 331

51. A si ngl e-phase t r ansmi ssi on l i ne has t woconductors each of 10 mm radius. These are fixedat a cent er -t o-cent er di st ance of 1 m i n ahor izontal plane. This is now converted to a three-phase t ransmission l ine by int roducing a thirdconductor of the same radius. This conductor isfixed at an equal distance D from the two single-phase conductors. The three-phase l ine is ful lytransposed. The posit ive sequence inductance perphase of the three-phase system is to be 5% morethan that of the inductance per conductor of thesingle-phase system. The distance D, in meters,is _______.

52. I n the circuit shown below, the supply voltage is10 sin(1000t) volts. The peak value of the steadystate current through the 1 resistor, in amperes,is ______.

53. A dc voltage with r ipple is given by v(t) = [100 +10 sin(t) – 5 sin (3t)] volts.M easurement s of t hi s vol t age v(t ), made bymoving-coi l and moving-iron voltmeters, showreadings of V1 and V2 respect ively. The value ofV2 – V1, in volts, is _________.

54. The circuit below is excited by a sinusoidal source.The value of R, in , for which the admit tance ofthe ci r cui t becomes a pure conductance at al lfrequencies is _____________.

55. I n the circuit shown below, the node voltage VA

is ___________ V.


ANSWERS

GENERAL APTI TUDE

1. (b) 2. (a) 3. (c) 4. (d) 5. (b) 6. (a) 7. (d) 8. (d)

9. (c) 10. (c)

TECH NI CAL SECTI ON

1. (0) 2. (3) 3. (a) 4. (b) 5. (b) 6. (b) 7. (a)

8. (c) 9. (a) 10. (b) 11. (d) 12. (19) 13. (b) 14. (100)

15. (d) 16. (c) 17. (2) 18. (0.5) 19. (b) 20. (0.84) 21. (a)

22. (0.4) 23. (170) 24. (c) 25. (b) 26. (0.01818) 27. (0.2928) 28. (a)

29. (b) 30. (a) 31. (6) 32. (a) 33. (2) 34. (c) 35. (d)

36. (c) 37. (a) 38. (d) 39. (2) 40. (48) 41. (1.1) 42. (d)

43. (c) 44. (10) 45. (75) 46. (1.44) 47. (86.84) 48. (1.61) 49. (c)

50. (a) 51. (1.44) 52. (1) 53. (0.31) 54. (14.14) 55. (11.42)

EXPL ANATI ONSGENERAL APTI TUDE

2. From the given under l ined phrase copewi thmeans put up with.

3. Gi ven wor ds ‘mock , der i de and j eer ’ ar esynonyms which means mockery. So, the givenword praise is odd one.

4. C A D B E ,

+1 +1

+1

J H K I L

+1 +1

+1

X V Y W Z,

+1 +1

+1

O N P M Q

+1 +1

– 1

So, opt ion (d) is odd one from the given group.

5.1 1

n n n n n nn n

n n n n

n n n n

n n = ( )n = (4)n

So, the value of n n

n n

is 4n.

6. From quest ionTotal number of faculty members = 150The number of faculty members having facebookaccount = (FB) = 55The number of f acu l t y member s hav i ngwhatsapp = (W) = 85The number of faculty members do not have facebook or Whats App accounts = 30The number of facul ty members having anyaccount = 150 – 30 = 120The number of faculty members having both theaccounts = (FB + W) – 120 = (55 + 85) – 120 = 20 The number of faculty members connectedonly through Facebook accounts = 55 – 20 = 35

So the number of faculty members connected onlythrough facebook accounts is 35.


9. I n the given 2 4 rectangle gr id, the fol lowingtype of rectangles are present .One figured rectangles = 8Two figured rectangles = 10Three figured rectangles = 4Four figured rectangles = 5Six figured rectangles = 2Eight figured rectangles = 1 Total number of rectangles

= 8 + 10 + 4 + 5 + 2 + 1 = 30So the number of rectangles observed in the givengr id is 30.

10. From the given graph funct ion f(x) must beequals to zero at x = 3.Fr om Opt i on A :

f(x) = 1 – | x – 1|at x = 3

f(x) = 1 – | 3 – 1| = 1 – 2 = – 1So, i t is falseFr om Opt i on B :

f(x) = 1 + | x – 1|at x = 3

f(x) = 1 + | 3 – 1| = 1 + 2 = 3So, i t is falseFr om Opt i on C :

f(x) = 2 – | x – 1|at x = 3

f(x) = 2 – | 3 – 1| = 2 – 2 = 0So, i t is t rue.Fr om Opt i on D :

f(x) = 2 + | x – 1|at x = 3

f(x) = 2 – | 3 – 1| = 2 + 2 = 4So, i t is false.Hence the cor rect expression for f(x) given in thegraph is 2 – | x – 1| .

TECH NI CAL SECTI ON

1. Given funct ion f(x) = x(x – 1)(x – 2) in the interval[1, 2]

f (x) = 3x2 – 6x + 2 = 0

x = 26 ( 6) 4 3 2

2 3

= 6 36 24

6

= 6 2 3

6

x = 3 3

3

f (x) = 6x – 6

f 3 3

3

= 3 3

6 63

= 6(3 3) 18

3

= 2(3 3) 6

= 2 3= 3.46 > 0 minimum

f 3 3

3

= – 3.46 < 0 maximum

f(1) = 0, f(2) = 0Max value = 0So, the maximum value at tained by the givenfunct ion is 0.

2. Given A =

1 1 1

1 1 1

1 1 1

Character ist ics equat ion is | A – I | = 0 – 3 + 32 = 0

2(3 – ) = 0 = 3, 0, 0So, i t has three eigen values.

3. Given laplace t ransform isf(t ) = e2t sin(5t)ut

sin 5t u(t) LT

2 2

55s

e2t sin 2t u(t) LT

2

5( 2) 25s

LT( ) ( )ate f t F s a

= 2

54 2 2 25s s

= 2

54 4 25s s

= 2

54 29s s


e2t sin 2t u(t) LT

2

54 29s s

So, the laplace tansform of f(t) is 2

54 29s s

4. The different ial equat ion is given by

2

22

d y dyy

dtdt = 0

m2 + 2m + 1 = 0 (m + 1)2 = 0 y(t) = (c1 + c2t)e

– t

Given y(0) = 1 c1 = 1Also given y(1) = 3e– 1

3e– 1 = (1 + c2)e– 1

3 = 1 + c2

c2 = 2 y(t) = (1 + 2t)e– t

y(2) = 5e– 2

So, the value of y(2) is 5e– 2.

5. Given integral is 2C

2 51

( 4 5)2

z

z z z

is

f(z) = 2

2(1/2) 5(1/2) 4(1/2) 5

=6 24

13/4 13

= 2i [sum of residues]

=24 48 i

2 i13 13

So, the value of integral is 4 i13

.

6. Here A =22

jj

= 2 2

2 2 1

2 2 22 2

=22

jj

= 90 – tan – 1 22

= 90 – tan – 1 (1)= (90 – 45)= 45

So, the value of A and are 1

2, +45 respect ively

7. Given t ransfer funct ion G(s) = 3

100( 1)s

3

100180

( 1)pc

j

– 3 tan – 1 pc = – 180°

pc = 3So, the phase cross-over frequency of the given

transfer funct ion is 3 .8. Given y(t) = x(t) cos(t)

I t sat isfies both addi t ivi t y and Homogent i t ypr inciples, so i t is l inearI f the input is delayed by t0, then

y1(t) = x(t – t0) cos(t)I f the output is delayed by t0, theny(t – t0) = x(t – t0) = x(t – t0) cos(t – t0)here, y1(t) y(t – t0) , So, i t is t ime varying.Hence the gi ven system is l inear and t ime-varying.

9. Let I = (2 2)te t dt

= (2( 1))te t dt

=1

( 1)2

te t dt

=1

( 1)2

te t dt

=12

te = 12e

10. From the given quest ion

T0.1

2n

Here T = Change in temperature T2 – T1

T1 = – 40 CT2 = – 55 C

55 ( 40)

2n

0.1


952n 0.1

952n

110

950 2n

The minimum ‘n’ value which can sat isfy theabove equat ion is 10. The minimum number of bits are 10.

11. From the given circuit boolean expression foroutput F is given by

F = 0 1SI SI

= BA BA= A B= A B

12. Zener diode is in breakdown, replace i t with avoltage source of value V2 = 5.3 V and VBE = 0. 6 V

From the above t ransistor circuitApplying KCL, we get

I B = 1 – 0.5 = 0.5 mA

I E =4.7470

= 10 mA

I E = ( + 1) I B

+ 1 =E

B

I 10I 0.5

= 20

= 19So, the value of the cur rent gain () is 19.

13. A uniformly charged r ing can be considered asstat ic. A stat ic electr ic charge produces an electr ic

field for which E 0.

14.

From the above figure

B (P– ) due to st raight conductor = 0IT

2 r

B (P+) due to st raight conductor = 0100 IT

2 r

B (P+) / B (P– ) = 100

Thus the rat io of the magnet ic flux densit ies attwo adjacent point located just inside and justoutside the toroid is 100.

15. Here the equivalent circuit of the given diagramis given by

From the above circuit diagram, we have

RB = A

A

1 R1 R

16. The sl ip at the maximum torque = 2

2

rx

x2 = 2L 2 f

Slip at the maximum torque = 2

22 Lr

f

Now; i f frequency is changed x2 also changes. Sosl i p for max i mum t or que i s i nver sel ypropor t ional to frequency. Synchronous speed isdirect ly propor t ional to frequency. Hence sl ip ofmaximum torque has an inverse relat ion withsynchronous speed.

17.

From the above circuit diagram.Heat generated by 5 = 10 cal/s1 cal/s = 4.184 W


So, power dissipated in5 = 4.184 10 = 41.84 W

So, 2V

5 = 41.84

V = 5(41.84) = 209.2 V = 14.464So, Voltage across 4 is

=4

14.463710

= 5.785 VHeat generated by the 4 resistance

(P4) = 2(5.785)

4 = 8.368 Watt

(Since 1 calor ie = 4.184 Joule) = 2 calor ies per second

18.

From the above circuit diagramApply KVL (Kirchnoff’s voltage Law); we get

– 1 + I 1 + 2I 2 = 0I 1 + 2I 2 = 1 ...(i )

also I 1 = 2I 2 ...(i i )So, cur rent through bat tery means

I 1 =1

A2

19. From given dataTotal number of buses = 100Generator bus = 10 – 1 = 9Load busses = 90Slack bus = 1From the given quest ion i f 2 buses are conver tedto PQ from PV it wi l l add 2 unknown voltagesfor i t er at i on but unknown angl es r emai nsconstant .

20. Post fault voltage at bus B for fault at bus A isgiven by

VBAF = VBBF – ZABIFA = 0.81 – ZAB. 10 = 0.8

ZAB = 0.02

Post fault voltage at bus A for fault at Bus ‘B’ isgiven by

VAAF = VABF – ZAB I FB

= 1 – 0.02 8= 1 – 0.16 = 0.84 pu

So, the vol tage magnitude at bus A dur ing athree-phase fault at bus B is 0.84 pu.

22. From the given figure :I n steady state, area of inductor voltage for oneswitching cycle is zero30 TON – 20 TOFF = 030 TON = 20 TOFF

ON

OFF

TT

= 2030

Duty cycle D =

ON ON

ON OFFON ON

T T3T T T T2

OFF ON

3since T T

2

= ON

ON

T 20.4

5T 52

23. From the given fig., no cur rent flows through theIGBT. So cur rent flows only in Diode Conduct ion loss = V tI av + I 2

rms Ron

= 0.7 100 + 1002 0.01= 70 + 100 = 170 W

Therefore, the conduct ion power loss in the powermodule is 170 W.

24. With lap winding, and 4 poles, number of paral lelpaths = 4, with a total armature cur rent of 40 A;cur rent in each path and hence cur rent in eacharmature conductor is 10 A. I t remains constantat 10 A as long as the conductor is in one path.When i t goes i n t o t he next pat h (due t ocommutator act ion) the current in it reverses andbecomes – 10 A . Assumi ng st r ai gh t l i necommut at i on, t he change f r om (+ 10 A) t o(– 10 A) is l inear.With 600 RPM; t ime for 1 revolut ion = 0.1 sec.Time of a conductor to cover 1 pole-pitch = 0.1/4= 25 ms. This is the width of one half cycle ofconductor cur rent .

25. From the given figure, the equivalent inductanceat por t 1 is n2L.

26. Probabil i ty (3 pens having the same colour)

= 3 3 3 3

3 3 3 312

3

C C C CC

=

33

123

4 C 4220C

= 0.01818


27. Given S = 0

n

n

n

S = 0 + 1 + 22 + 33 + ....... S = + 22 + 33 + .........

S = 2(1 ) ( 0 < < 1)

but given, S = 2

2(1 ) = 2

12

= (1 – )2

(1 – ) = 1

2

(1 – ) = 1

2 or (1 – ) = –

1

2

1 – = 1

2

= 1 – 1

2 = 1 – 0.707 = 0.2928

and (1 – ) = – 1

2

= 1 + 1

2 = 1 + 0.707 = 1.707

but given range is 0 < < 1, = 0.2928

28. Given matr ix isA2 – 3A + 4I

Now the eigen values of a 2 2 matr ix forA 1, – 2

A2 1, 4– 3A – 3, 6

4I 4, 4 (Since I = I dent i ty matr ix) A2 – 3A + 4I 2, 14So, the eigen value is 2, 14And eigen vectors do not change.

29. (A4 3) = 2;

T3 4A = 2

(Where A be a 4 3 real matr ix whose rank 2)(A B) min[(A), (B)]AAT of order 4 4 whose rank 2 ATA is of order 3 3 whose rank 2

30. From the given Bode plot the corner frequenciesare 2 rad/sec and 4 rad/sec respect ively.

Transfer funct ion (TF) = 2

Ks

1 12 4s s

20logK + 20 log = 0 dBAt = 0.5, we get

20logK + 20 log0.5 = 0

20logK = 1

20 log2

20logK = 20log (2) K = 2

TF = 2

2(1 0.55)(1 0.25 )

ss

31. Given y(t) = cTx(t )x(t) = eAt x(0)eAt = L – 1[(sI – A)– 1]

Now sI – A =

1 0 1 0s

0 1 0 2

(Where I = I dent i ty matr ix)

=0 1 0

0 0 2

s

s

=1 0

0 2

s

s

eAt = 2

0

0

t

t

e

e

x(t) = 2

1010

t

t

e

e

= 2

t

t

e

e

y(t) = 2

1, 1t

t

e

e

= [et + e2t] t = 0.693

t = loge 2 = 0.693= [2 + 4] = 6

y(t) = 6

32. Given t ransfer funct ion G(s).H(s) = 2

3( 3)s

s s

CE = 1 + G(s).H(s) = 2 2

31 0

3s

s s

s3 – 3s2 + s + 3 = 03

2

1

0

S 1 1S 3 3

S 2

S 3


System is unstable with two r ight half of s-planepoles Z = 2, P = 1

N = P – ZN = 1 – 2 = – 1 (i t shows once in clockwise

direct ion.33. Given polynomial equat ions is

s3 + 5.55s2 + 8.5s + 3 = 0(Z – 1)3 + (5.5)(Z – 1)2 + 8.5(Z – 1) + 3 = 0

(Here s = Z – 1)Z3 – 3Z2 + 3Z – 1 + 5.5Z2 – 11Z + 5.5

+ 8.5Z – 8.5 + 3 = 0Z3 + 2.5Z2 + 0.5Z – 1 = 0

3

2

1

0

+Z 1 0.5+Z 2.5 1

2.25+Z

2.5+Z 1

So, the number of roots of the polynomial whichhave real par ts less than – 1 is 2.

34. From the given Fig.

Atr i(t /T) ATsa2 T2

From dual i ty proper ty

2 TATs 2 At r i

2 Ta

t

2 TTSa 2 tr i

2 Tt

T = 12

21 TSa 2 t r i 2

2 2t

21 Tsa tr i 2

4 2t

Assume, x(t) = 21 Tsa ,

4 2t

X() = t r i ()x(t) is real funct ion

X1() = 1 1 1 1

X X 0.3X2 2 3 2

x1(t) = 1 3 12 2 21

( ) ( ) 0.3 ( )2

j t j t j te x t e x t e x t

x1(t) is complex funct ionX2() = x1(– )x1(t) = x1(– t)

x2(t) = 1 3 12 2 21

( ) ( ) 0.3 ( ) ( )2

j t j t j te x t e x t e x t x t

= x(t)

x2(t) = 1 3 12 2 21

( ) ( ) 0.3 ( )2

j t j t j te x t e x t e x t

x2(t) is complex funct ion

x1(t)x2(t) = 2 2 21( ) ( ) 0.3 ( )

2j t j tx t e x t e x t

2 2 2 21 1( ) ( ) 0.15 ( )

2 4j t j te x t x t e x t

+ 0.3e– j t x2 (t) + 0.15e– j2t x2 (t) + 0.09 x2(t)x1(t)x2(t) = x2(t) [1.34 + cost + 0.6cost + 0.3cos2t ]x1(t)x2(t) is real funct ion

36. From the given fig :

KA KB

The next state of the system QA QB = 1 1

37. Truth Table Input Output

X2 X1 X0 B1 B0 0 0 0 0 0 0 0 1 0 1 0 1 1 1 0 1 1 1 1 1

B0 = 2 1 0 2 1 0X X X X X X

= 2 1 2 1 0[X X X X ] X

2 1 0 0 2 1(X X )X X X X

38.

Potent ial at P due to q (at or igin)

P1 = 2 204

q

x y V (ref : )

Potent ial at P due to – 2q at (6, 0)

P2 = 2 2

0

2V

4 ( 6)

q

x y


Net potent ialP = (P1 – P2)

= 2 2 2 20

1 24 ( 6)

q

x y x y

= 0

4(x2 + y2) = (x – 6)2 + y2 = x2 + 36 – 12x + y2

3x2 + 3y2 = 36 – 12xx2+ y2 = 12 – 4x(x + 2)2 + y2 = 16

(Equat ion of zero equipotent ial curve).So, the equat ion of the zero equipotent ial curvein the x-y plane is (x + 2)2 + y2 = 16

39. From the given circuit diagramat t = 0–

at t = 0+

Nodal Analysis

L L(V (0 ) 3) 3(V (0 ) 3)1 2

= 0

L L2V (0 ) 6 3 V (0 ) 32

= 0

3VL(0+) = 6

VL(0+) = 2

So, LV (0 )di (0 ) 2dt L 1

= 2 A/sec

Hence, the rate of change of cur rent through theinductor is 2 Amp.

40. From given quest ionInit ial load = (800 + j600)load after modificat ion = (900 + j600) to maintainsame heat dissipat ion magnitude of power shouldbe same

Load with compensat ion= (900 + j600 + Compensat ion)= 900 + jx)

Equat ing magnitude of power8002 + 6002 = 9002 + x2

82 + 62 = 92 + x2

100 = 81 + x2

x = 19 = 4.3588Require react ive power = 435.8 kVArAfter compensat ionReact ive power to be compensated by capacitorto achieve this is 164.11 kVAr

2phV

X /Phasec= QC / Phase

2ph(3.3 / 3) Ck =

164.113

k

Cph = 48 FSo, the capacitance per phase is 48 F.

41. From the given quest ion al l quant i t ies are givenin ‘pu’

Posit ive sequence cur rent (in pu) = B

4270I

=4.27 3 13.8

30

= 3.4 puEquivalent circuit

Posit ive sequence cur rent = eq

13.4

X

Xeq =1

3.4 = 0.2941 pu

0.15(0.35 3X )0.15

3X 0.5n

n

= 0.2941

0.15(0.35 3X )3X 0.5

n

n

= 0.1441

0.0525 + 0.45Xn = 0.4323 Xn + 0.072050.0177 Xn = 0.01955

Xn = 1.104 puHence the value of Xn is 1.1 pu.


42.

With – ve sequence voltages,

ANV = V0,

BNV = V120

and CNV = V– 120

ABV = 3 V– 30from figure abV = E0

ABV lags abV by 30.43. Power factor at AC mains

= 2 2

cos = 0.9 cos 30

= 3 0.9 1.732

0.92 2

= 0.78

44. Modulat ion index, (ma) = ˆ 0.8ˆ 1

m

tr i

vv

= 0.8

Ampli tude of the fundamental output voltage,

AO 1

Vˆ

2dc

av m = 0.8 250 = 200 V

From the given modulat ing voltage equat ion,1 = 200

fundamental component frequency = 100 HzNow Load impedance at 100 Hz frequency,

(Z1) = 2 2 2 2R X 12 16 = 20

L1I = AO1

1

V 200Z 20

= 10 A

So, the load cur rent (i L) is 10 Amp.45. From given fig.

Energy loss dur ing

T1 =1 1T T

0 0

. 600v i dt i dt = 600 area under cur rent curve

= 61600 150 1 10

2

= 300 150 10– 6

= 45 103 10– 3

= 45 10– 3

= 45 mJ

Energy loss dur ing

T2 = 2 2T T

0 0

. 100v i dt vdt

= 100 area under voltage curve

= 61100 600 1 10

2

= 300 600 10– 6

= 30 103 10– 6

= 30 10– 3

= 30 kJTotal energy loss = (45 + 30) = 75 mJ

46. I f V/f is kept constant , then maximum core fluxdensity is constant .From the given quest ion, we have

(400/50) = (200/25) = 8 (416/52).I f V/f is kept constant . So the maximum core fluxdensity is constant .Let the magnet ic field be B.Hyster isis loss Wh = kh f B

n

Eddy cur rent loss Ww = ke f2 B2

Total core loss = Wn + We.At 400 V and 50 H z,

kh 50 Bn + ke 502 B2 = 5000 kh B

n + ke 50 B2 = 100 ...(i )At 200 V and 25 H z,

kh 25 Bn + ke 25 B2 = 2000kh B

n + ke 25 B2 = 80 ...(i i )Solving equat ion (i ) and (i i ), we get

ke 25 B2 = 20;kh B

n = 60At 416 V and 52 H z;

(kh Bn) 52 = 60 52

ke 522 B2 = 0.8 522

Hysteresis lossEddy current loss = 2

60 52 75520.8 52

= 1.44.

So, the rat io of hysteresis loss to the eddy currentloss is 1.44.

47.

From the given circuit diagramTotal copper losses = 52 44 + 502 0.01

= 1100 + 25= 1125 W


Other losses = 375 WTotal losses = Total copper losses + other losses

= (1125 + 375)W =1500 W Output = 220 45 = 9900 W

Efficiency = 9900

100 %11400

= 86.84%

So, the percentage efficiency of the DC generatoris 86.84%.

48. Phasor diagram for E is constructed in the figurewhich is shown below.

AC2= AB2 + BC2 – 2(AB) (BC) cos 126.87 = 1 + 0.36 + 1.2 (0.6) = 2.08

AC = 2.08 = 1.4422

1.44220.8

=1

sin

sin =0.8

1.4422 = 0.5547

= 33.7, cos = 0.832CF = 0.1664AF = E = 1.61 p.u.

Hence the value of per phase induced voltage is1.61 p.u.

49.

Rated cur rent =22000

100A220

Maximum kVA rat ing (auto-t ransformer)

=2200 110

1000

=2420 100

1000

= 242 kVA.So, t he maxi mum kVA r at i ng for an aut o-t ransformer is 242 kVA.

50. Pulse width (2d) = 120 d = 60

Vrms =2 2

V sindc d

= 0.9 300 sin 60

3

702

135 1.732

= 233.8 VH ence t he r ms val ue of t he fundament alcomponent of the output voltage is 234 (approx).

51. From the given quest ion the required fig. is givenby

1.05 L 1– = L 3

1.05 0.2 ln

12 31000 (1000D )

0.2 ln0.788 10 0.788 10

11.05 2 31000 (1000D )

7.88 7.88

D = 1,438 mm = 1.44 mSo, the distance (D) is 1.44

52.


From the above fig, we have

I 1 =V

4 1 5

=10sin(1000 )

10t

I 1 = 1 sin (1000t)AHence the peak value of the steady state cur rentthrough 1 resister is 1 Amp.

53. For PMMC; V1 = 100 VFor M.I ;

V2 =

2 22 10 5

(100)2 2

= 10000 50 12.5 = 100.31 V

V2 – V1 = (100.31 – 100)V= 0.31 V

So, the value of V2 – V1 is 0.31 V.54. From the given circuit diagram

R = L 0.02C 100

(Since L = 0.02H and 100)

=

6

40.02 100.02 10 200

100

= 14.14 .So, the value of R is 14.14 .

55. From the given circuit diagram

A A 1 AV (V 10I ) (V 10)5

5 5 10

= 10

2VA – 50 + 2VA + 20 I 1 + VA – 10 = 05VA + 20I 1 = 60 ...(1)

Also, I 1 =A(V 10)10

...(2)

So, A

A

(V 10)5V 20

10

= 60

7VA = 80

VA =807

VA = 11.42 VoltsSo, the node voltage is 11.42 Volts.

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