General Aptitude. Q. No. 1 5 Carry One Mark Each

Transcription

1 EE GATE-6-PAPER- General Aptitude Q. No. 5 Carry One Mark Each. The man who is now Municipal Commissioner worked as. (A) the security guard at a university (B) a security guard at the university (C) a security guard at university (D) the security guard at the university (B). Nobody knows how the Indian cricket team is going to cope with the difficult and seamer-friendly wickets in Australia. Choose the option which is closest in meaning to the underlined phase in the above sentence. (A) put up with (B) put in with (C) put down to (D) put up against (D). Find the odd one in the following group of words. Mock, deride, praise, jeer (A) mock (B) deride (C) praise (D) jeer (C) 4. Pick the odd one from the following options. (A) CADBE (B) JHKIL (C) XVYWZ (D) ONPMQ (D) 5. In a quadratic function, the value of the product of the roots, is 4. Find the value of (A) (B) n n 4 n n n Given 4 n n n n n n n n n n n n n n n n ( ) 4 (B) n 4 (C) n (D) n 4 ICP Intensive Classroom Program egate-live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ Centers across India

2 EE GATE-6-PAPER- Q. No. 6 Carry Two Mark Each 6. Among 5 faculty members in an institute, 55 are connected with each other through Facebook and 85 are connected through WhatsApp. faculty members do not have Facebook or WhatsApp accounts. The number of faculty members connected only through Facebook accounts is. (A) 5 (B) 45 (C) 65 (D) 9 (A) F Facebook, W WhatsApp, E Total faculties given n(e) 5, n F W n F W n(e) N F W 5 n F W n f w n f n(w) n(f w n(f) 85 n(f) n(f) n F W n F w 55 n(f) 55 5 n(w) F n(f) 5 E, n(e) F w n F w W n(w) 6 F W 7. Computers were invented for performing only high-end useful computations. However, it is no understatement that they have taken over our world today. The internet, for example, is ubiquitous. Many believe that the internet itself is an unintended consequence of the original invention with the advent of mobile computing on our phones, a whole new dimension is now enabled. One is left wondering if all these developments are good or more importantly, required. Which of the statement(s) below is/are logically valid and can be inferred from the above paragraph? (i) The author believes that computers are not good for us (ii) Mobile computers and the internet are both intended inventions (A) (i) (B) (ii) only (C) both (i) and (ii) (D) neither (i) nor (ii) (D) 8. All hill-stations have a lake. Ooty has two lakes. Which of the statement(s) below is/are logically valid and can be inferred from the above sentences? (i) Ooty is not a hill-station (ii) No hill-station can have more than one lake. (A) (i) Only (B) (ii) Only (C) both (i) and (ii) (D) neither (i) nor (ii) (D) ICP Intensive Classroom Program egate-live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ Centers across India

3 EE GATE-6-PAPER In a 4 rectangle grid shown below, each cell is a rectangle. How many rectangles can be observed in the grid? (A) (B) 7 (C) (D) 6 (C) : (AEOK) : (AEJF), (FJOK) 4: (ABLK), (BCML), (CDNM), (DEON) : ACMK, ADNK :ECMD,EBLO :ACHF,ADIF : ECHJ, EBGJ :FHMK,FINK :JHMD,JGLO A B C D E F G H I J : BDNL : BDIG,GINL 8: ABGF, BCHJ, CDIH, EDI, FGLK, GHML, HINM Total = = K L M N O fx 4.5 X4.5 5 Chose the correct expression for f(x) given in the graph. (A) (C) f x x (B) f x x f x x (D) f x x (C) Substituting the coordinates of the straight lines and checking all the four options given, we get the correct option as C which is f(x)= x ICP Intensive Classroom Program egate-live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ Centers across India

4 EE GATE-6-PAPER- Electrical Engineering Q. No. 5 Carry One Mark Each. The maximum value attained by the function f x x x x in the interval f ( x) x( x )( x ) x But x only lies on the interval [,] f ( x) x x x f ( x) x 6x At x ;f ( x) 6x x is a point of minimum f(x) = x(x-)(x-) = at either ends x = & x = Max value =, is.. Consider a matrix with every element being equal to. Its only non-zero eigenvalue is. Consider A It s only non-zero Eigen value λ = order of the matrix = t. The Laplace Transform of f t e sin 5tu t 5 (A) s 4s 9 is 5 (B) s 5 (A) Consider 5 x(t) = sin5t u(t) s +5 X(S) X(S-S ) x By frequency shifting property, thus at S t f(t) = e sin5tu(t) 5 F(s) s 4s 9 5 (s-) +5 F(s) s (C) s 4s 9 ICP Intensive Classroom Program egate-live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ Centers across India (t)e St (D) 5 s 5 4

5 EE GATE-6-PAPER A function yt, such that y and y e, is a solution of the differential equation d y dt dy y dt. Then y is (A) 5e (B) 5e (C) 7e (D) 7e (B) The operator function of given D.E is A.E is D +D+ = D = -,- -x y = e C + C () - Given y()= & y() =e From ; y() = i.e, y = at x = = C (D +D+)y = - y() = e i.e, y = /e at x= e - C () e C C -x - - y = e [+x] y() = e [+4] = 5e 5. The value of the integral z 5 dz over the contour z. taken in the anti- C z z 4z 5 clockwise direction, would be (A) 4 i (B) Singular points are obtained by z z -4z+5 Out of these (B) 48 i z = & z = i z = only lies inside C. z = By Cauchy s integrated formula, z+5 z+5 (C) 4 dz = dz C = πi z -4z+5 z - C z -4z+5 z πi 4 5 (D) ICP Intensive Classroom Program egate-live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ Centers across India 5

6 EE GATE-6-PAPER The transfer function of a system is s, The steady state output x(t) is A cos Ys R s s the input cos t. The value of A and, respectively are t for (A), 45 O (B), 45 O (C) O, 45 (D) (B) S H(s) = S+ O, 45 H( ) = 9 tan +4 If input is COS t i.e., = H( ) = 45 y(t) = cos(t 45 ) by comparision A= & The phase cross-over frequency of the transfer function G(s) = s (A) (A) (B) Phase Crossover frequency : GH 8 PC = PC in rad/s is (C) (D) GH tan 8 tan w W tan PC PC 8. Consider a continuous-time system with input x(t) and output y(t) given by y t x t cos t. This system is (A) linear and time-invariant (C) linear and time-varying (C) Linearity y (t)= x (t)cost; y (t)= x (t)cost y (t) + y (t) = x (t)cost + x (t)cost y (t) + y (t) = [x (t) + x (t)]cost If x (t) + x (t) = x(t) then y (t) + y (t) = y(t) (B) Non-linear and time-invariant (D) Non-linear and time-varying ICP Intensive Classroom Program egate-live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ Centers across India 6

7 EE GATE-6-PAPER- Thus the system is linear Time-invariance consider y (t) x (t) cost ; If x (t) x(t - τ) y (t) x(t - τ)cost Define y(t - τ) x(t - τ)cos(t - τ) y (t) y(t - τ) system is time-varient t 9. The value of e t dt. where t is the Dirac delta function, is A B C e e e D (A). A temperature in the range of -4 O C to 55 O C is to be measured with a resolution of. O C. The minimum number of ADC bits required to get a matching dynamic range of the temperature sensor is (A) 8 (B) (C) (D) 4 (B) Vmax -Vmin Usually when voltage information is given we use the formula R = n Here based on the given information we can think the systems as following block diagram e Temp Temperature to voltage converter analog voltage nbit ADC.. Here the Vmax and V min So we can still use the above relation Tmax -Tmin n 5-(-4) n Resolution = = = 95 n. n log (95) 9.89 So minimum requirement is bits will be the equivalent of Tmax and T min respectively. Consider the following circuit which uses a -to- multiplexer as shown in the figure below. The Boolean expression for output F in terms of A and B is (A) A (C) A B (B) A B B (D) A B A Y S B F ICP Intensive Classroom Program egate-live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ Centers across India 7

8 EE GATE-6-PAPER- (D) We can redraw the max circuit as follows A A F B So the Boolean expression of F(A, B)=BA BA=A B=A B. A transistor circuit is given below. The Zener diode breakdown voltage is 5 V as shown The base to emitter voltage droop to be.6v. The value of the current gain is. V 4.7k 9 5. = I E IE.A - 5. I X = ma 4.7 I =.5.5mA I B E ( β)i B.. ( β).5 ( β) β 9.5. In cylindrical coordinate system, the potential produced by a uniform ring charge is given.5ma 5.V 47 f r,z, where f is a continuous function of r and z. Let E be the resulting electric field. Then the magnitude of E (A) increases with r (B) is (C) is (D) decreases with z (B) Since the charge is not varying with time the fielde is static So E ICP Intensive Classroom Program egate-live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ Centers across India 8

9 EE GATE-6-PAPER A soft-iron toroid is concentric with a long straight conductor carrying a direct current I. If the relative permeability r of soft-iron is, the ratio of the magnetic flux densities at two adjacent points located just inside and just outside the toroid, is. The field inside and outside the toroid is due to long straight conductor only Let the two points almost at same distance μ μ I The flux density inside toroid = r πr μi The flux density outside toroid = T πr μμri Ratio = μi μ r πr 5. RA and R B are the input resistances of circuits as shown below. The circuits extend infinitely in the direction shown. Which one of the following statements is TRUE? R A R B (A) RA RB (B) RA RB (C) RA RB (D) By comparing network on the input side A we can say that R =//R R B A R = B R A (D) R R R B A A 6. In a constant V/f induction motor drive, the slip at the maximum torque (A) is directly proportional to the synchronous speed (B) remains constant with respect to the synchronous speed (C) has an inverse relation with the synchronous speed (D) has no relation with the synchronous speed ICP Intensive Classroom Program egate-live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ Centers across India 9

10 (C) r r r SmT x jl jl.f SmT f N S f P SMT NS EE GATE-6-PAPER In the portion of a circuit shown, if the heat generated in 5 resistance is calories per second then heat generated by the 4 resistance, the calories per second, is. 4 6 Here the power information regarding the resistor is given because E Calorie P = =watt t sec P = 5 V 5Ω = 5 V 5Ω = 5 P is asked 4Ω V 4 4Ω P 4Ω = calorie 5 4 sec 5 8. In the given circuit, the current supplied by the battery, in ampere, is. ` l l V l.5 If we write KCL at node then ICP Intensive Classroom Program egate-live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ Centers across India

11 EE GATE-6-PAPER- I I = I I Write KVL in the outer boundary of network ( I ) ( ) I I I I =I J I.5A 9. In a bus power system, there are generators. In a particular iteration of Newton Raphson load flow technique (in polar coordinates). Two of the PV buses are converted to PQ type. In this iteration. (A) The number of unknown voltage angles increases by two and the number of unknown voltage magnitudes increases by two. (B) The number of unknown voltage angles remain unchanged and the number of unknown voltage magnitudes increases by two (C) The number of unknown voltage angles increases by two and the number of unknown voltage magnitudes decreases by two (D) The number of unknown voltage angles remains unchanged and the number of unknown voltage magnitudes decreases by two (B) Total no of Buses = Generator Buses = Load Buses = -=9 Slack Bus = If of the PV buses are converted to PQ type the no of on voltage magnitudes increases by with constant unknown voltage angles. The magnitude of three-phase fault current at buses A and B of a power system are pu and 8 pu, respectively. Neglect all resistance in the system and consider the pre-fault system to be unloaded. The pre-fault voltage at all buses in the system is. pu. The voltage magnitude at bus B during a three-phase fault as but. A is.8pu. The voltage magnitude at bus A during a threephase fault at bus B, in pu, is..84 Voltage at i th bus when fault is at k th bus is z V Vi E I z z X ik product f kk f p.u X.P.U XB xn 8 B.5P.U Z AB VB E Z AA A ICP Intensive Classroom Program egate-live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ Centers across India V I X I I

12 EE GATE-6-PAPER- Z. AB.8 ZAB. AB. VA.84 P.U BB.5. Consider a system consisting of a synchronous generator working at a lagging power factor, a synchronous motor working at an overexcited condition and a directly grid-connected induction generator. Consider capacitive VAr to be a source and inductive VAr to be a sink of reactive power. Which one of the following statements is TRUE? (A) Synchronous motor and synchronous generator are sources and induction generator is a sink of reactive power. (B) Synchronous motor and induction generator are sources and synchronous generator is a sink of reactive power. (C) Synchronous motor is a source and induction generator and synchronous generator are sinks of reactive power (D) All are sources of reactive power (A). A buck converter, as shown in figure (a) below, is working in steady state. The output voltage and the inductor current can be assumed to be ripple free. Figure (b) shows the inductor voltage V L during a complete switching interval. Assuming all devices are ideal, the duty cycle of the buck converter is V L V g M D VL C V R V TON TOFF t V.4 When M is ON, VS VL V VL VS V When M is OFF, VL V V V V V 5V S V V 5 D.4 S s a T a b ICP Intensive Classroom Program egate-live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ Centers across India

13 EE GATE-6-PAPER- A steady dc current of A is flowing through a power module (S.D) as shown in Figure (a). The V-I characteristics of the IGBT (S) and the diode (D) are shown in Figures (b) and (c), respectively. The conduction power loss in the power module (S, D), in watts, is Is A I A S D V V dv dl. V.7V dv dl. 69 to 7 A a VS Volt Vl characteristic of GBT b VS Volt Vl characteristic of diode c 4. A 4pole, lap-connected, separately excited dc motor is drawing a steady current of 4 A while running at 6 rpm. A good approximation for the waveshape of the current in armature conductor of the motor is given by 4A (A) (B) I I A t t (C) I A T 5ms (D) I A T 5ms T 5ms A T 5ms A (C) no of parallel paths = 4 Armature current/conductor = 4 A 4 For linear commutation, the change from A to A is straight line N = 6rpm Time for revolution =-. sec. For pole-pitch,. t 5m sec 4 poles ICP Intensive Classroom Program egate-live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ Centers across India

14 EE GATE-6-PAPER If an ideal transformer has an inductive load element at port as shown in the figure below, the equivalent inductance at port is ` n : L (A) nl (B) n L (C) (B) The property of an ideal transformer is of port is terminated by an impedance ZL then the ZL impedance seen from port is N / N In the given problem Port Port Z in Z L /n L n L n L (D) n L Q. No. 6 5 Carry Two Mark Each 6. Candidates were asked to come to an interview with pens each. Black, blue, green and red were the permitted pen colours that the candidate could bring. The probability that a candidate comes with all pens having the same colour is..6 Total possible options = 4 Favorable choices {BBB, BlBlBl, GGG, RRR} = 4 4 Probability = n 7. Let S n where. The value of in the range. Such that S is..9 S = n = n n α n α α α + α ]\ α α[ α α ] α α[ α] if α ( α) α α.9 ICP Intensive Classroom Program egate-live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ Centers across India 4

15 EE GATE-6-PAPER Let the eigenvalues of a x matrix A be, - with eigenvectors x and x respectively. Then the eigenvalues and eigenvectors of the matrix A A 4l would, respectively, be (A),4; x,x (B),4; x x,x x (C),; x,x (D),; x x, x x (A) Matrix A Eigen values, - Matrix A -A+4I () 4,( ) ( ) 4 Eigen values, 4 respectively A & P(A) = ai + aa + aa have same eigen vectors 9. Let A be a 4 real matrix with rank. Which one of the following statement is TRUE? (A) Rank of A T A is less than (B) Rank of A T A is equal (C) Rank of A T A is greater than (D) Rank of A T A can be any number between and (B) ρ A Given that 4 T From the properties of Rank; we have ρaa = ρ(a) T T Rank of AA = Rank of A A = Rank of A. Consider the following asymptotic Bode magnitude plot is in rad s. magnitude db db dec db.db Which one of the following transfer function is best represented by the above Bode magnitude plot? s (A).5s.5s s (C) s 4s.5 8 (B) 4.5s s.5s 4s (D) s 4s ICP Intensive Classroom Program egate-live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ Centers across India 5

16 EE GATE-6-PAPER- (A) By looking to the plot we can say that since the initial slope is + there must be a zero on the origin If we find we can get the answer by eliminating options magnitude db M- M Slope = log - log db db dec 4 log 8 - log.db log 8- log log = log 8- =4 4 So one of the corner frequency is = 4s at this frequency poles should exist because the change in slope is -4db From this we can say option A satisfies the condition (i) A zero at origin s (ii) one of corner frequency 4H term will be + having poles 4. Consider the following state-space representation of a linear time-invariant system 6 T x t xt, yt c xt, c and x The value of y(t) for t log is.. Loop transfer function of a feedback system is GsHs e s s s the clockwise direction. Then the Nyquist plot of G(s) H(s) encircles + j (A) Once in clockwise direction (C) Once in anticlockwise direction (A) S+ GH = S(S-).Take the Nyquist contour in (B) Twice in clockwise direction (D) Twice in anticlockwise direction / ( +a) GH / ( +a) GH tan 8 8 tan tan ICP Intensive Classroom Program egate-live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ Centers across India 6

17 EE GATE-6-PAPER- GH = tan at, GH = at, GH = 8 at, GH = 9 9 So the plot start at and goes to 8 through 9 Since there are poles on origin we will get radius semicircle those will start where the mirror image ends and will terminate where the actual plot started in clockwise direction. So the plot will be -+io w = So the Nyquist plot of G(s) H(s) encircles + j once in clockwise direction M= - w =. Given the following polynomial equation s 5.5s 8.5s, the number of roots of the polynomial, which have real parts strictly less than, is The polynomial is S +5.5S +8.5S+=, since we are interested to see the roots wrt S = - so in the above equation replace S by z- then the equation is (Z-) +5.5(Z-) +8.5(Z-)+= Z -Z +Z-+5.5(Z +-Z)+8.5Z-8.5+= Z +Z ( Using RH table Z +.5Z.5 Z.5 Z.5 Z.9 Z 5.5)+Z(+8.5-)+( )= The single sign change in st column indicate that out of roots root lie on the right half of S = - plane if memory remaining lies on left half of S = - plane. ICP Intensive Classroom Program egate-live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ Centers across India 7

18 EE GATE-6-PAPER Suppose xt and xt have the Fourier transforms as shown below. X X Which one of the following statements is TRUE? (A) x t and x t are complex and (B) x t and x t are real and x tx t is also real (C) x t and x t are complex but x tx t is real (D) x t and x t are imaginary but x tx t is real (C) x (t) & x (t) are complex functions x ( ) = x (- ), x (t) = x (-t) x (t) x (t)will be real x t x t is also complex with nonzero imaginary part 5. The output of a continuous-time, linear time-invariant system is denoted by T{x(t)} where xt is the input signal. A signal z(t) is called eigen-signal of the system T, when Tz t z t,, where is a complex number, in general, and is called an eigen value of T. suppose the impulse response of the system T is real and even. Which of the following statements is TRUE? (A) cos(t) is and eigen-signal but sin(t) is not (B) cos(t) is and sin(t) are both eigen-signal 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-signal with identical eigenvalues (D) Consider the Eigen signal Z(t) = Cos t For Cost, y(t) cos(t-τ)h(τ)dτ (cos t cos τ +sin t sin τ)h(τ)dτ = cos t cos τh(τ)dτ sin t sin τh(τ)dτ h(τ)is an even signal sinτh(τ)dτ = ICP Intensive Classroom Program egate-live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ Centers across India 8

19 EE GATE-6-PAPER- y(t) =cos t cos τh(τ)dτ Thus the integration value will be an Eigen value γ. Similarly consider the Eigen signal Z(t) = sin t For sint, y(t) sin(t-τ)h(τ)dτ =sin t cos τh(τ)dτ cos t sin τh(τ)dτ Thus y(t)= sin t cos τh(τ)dτ Eigen value γ is same for both the Eigen functions sint & cost 6. The current state Q A Q B of a two JK flip-flop system is. Assume that the clock rise-time is much smaller than the delay of the JK flip-flop. The next state of the system is 5V J Q A J Q O K Q V K CLK (A) (B) (C) (D) (C) Logic J A QA J B QB KA QA KB QB J = K = A A J = K =Q B B A It is given initially Q A Q B = Since it is a synchronous counter, when clock is applied both flipflop will change there state simultaneously based on JK FF state table J =, K =, Q = Q = A A A A J =, K =, Q = Q = B B B B So next state QA QB is ICP Intensive Classroom Program egate-live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ Centers across India 9

20 EE GATE-6-PAPER A -bit flash Analog to Digital Converter (ADC) is given below. The input is VIN Volts. The expression for the LSB of the output B as a Boolean function of X,X, and X is V X X Digital circuit B B X (A) X X X (B) X X X (C) X X X (D) X X X (A) 8. Two electric charges q and -q are placed at (,) and (6,) on the x-y plane. The equation of the zero equipotential curve in the x-y plane is (A) x (B) y (C) x y (D) q q The potential due to Q 4-Q at (x, y) is k x + y k (x-6) + y If potential at (x, y) = where q k x + y q k (x-6) + y (x-6) + y 4(x + y ) x y - 4x- = Option:D (x+) + y 6 k= 4πε V IN x y 6 (D) (x-6) + y x + y x +6 -x + y = 4x 4y x +4x +4+ y = 6 x y + 4x- = ICP Intensive Classroom Program egate-live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ Centers across India

21 EE GATE-6-PAPER In the circuit shown, switch S has been closed for a long time. A time t = switch S is closed At t, the rate of change of current through the inductor, in amperes per second, is. S S V H V At - t = Network is in steady state with S opens S (closed) So we can say - i L( )=.5A + At t = indicator behaves as ideal current source of.5a if we draw the network at both switch closed + V L ( ) Writing Nodes equation at V ( ) node V L( )=.5 V L ( ) = + di( ) L dt + di L( ) A/sec dt L + t =, 4. A three-phase cable is supplying 8kW and 6kVAr to an inductive load, It is intended to supply an additional resistive load of kw through the same cable without increasing the heat dissipation in the cable, by providing a three-phase bank of capacitors connected in star across the load. Given the line voltage is.kv, 5Hz, the capacitance per phase of the bank, expressed in microfarads, is. 47 to A MVA, -phase, 5Hz.8kV, star-connected synchronous generator has positive, negative and zero sequence reactance, 5%,5% and 5% respectively. A reactance X is connected between the neutral of the generator and ground. A double line to ground fault takes place involving phases b and c, with a fault impedance of j. p.u. The value of X V limit the positive sequence generator current to 47 A is..8 X.5 P.U.5 + at t = n n V in p.u. that will ICP Intensive Classroom Program egate-live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ Centers across India

22 EE GATE-6-PAPER- 6 Ibase X X Xf I a.5p.u.5p.u.p.u I 47 I 55.9 actual.4p.u I P.U base Ea Z Z Z z z O f n.4.5 Z Z z z f n f n Z z z.44 Z z zln.44 f z z.675 f n zn Zn.8P.U 4. If the star side of the star-delta transformer shown in the figure is excited by a negative sequence voltage, then (A) V AB leads V ab by 6 O A a (B) V AB lags V ab by 6 O (C) V AB leads V ab by O B N c (D) V AB lags V ab by O C b (D) 4. A single-phase thyristor-bridge rectifier is fed from a V, 5Hz, single-phase, AC mains. If it is delivering a constant DC current A, at firing angle of O, then value of the power factor at AC mains is (A).87 (B).9 (C).78 (D).45 (C) IPF =.9cos.9 cos.78 ICP Intensive Classroom Program egate-live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ Centers across India

23 EE GATE-6-PAPER The switches T and T are shown in Figure (a). Vdc 5V T C i L.8 XL 6 at R Hz Vdc 5V T t V m They are switched in a complementary fashion with sinusoidal pulse width modulation technique. The modulating voltage t.8sin t V and the triangular carrier voltage m are as shown in Figure (b). The carrier frequency is 5kHz. The peak value of the Hz component of the load current i, in ampere, is. V Vcarrier m ma.8 L Vds Vl max ma..8 5 V L 6 IL max A 45. The voltage s a across and the current g l through a semiconductor switch during a turn-on transition are shown in figure. The energy dissipated during the turn ON transition, in mj, is. b c v s 6V t i s 5A A T s T s t ICP Intensive Classroom Program egate-live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ Centers across India

24 75 EE GATE-6-PAPER- 6 E Pdt Vt tdt mJ 46. A single-phase 4V, 5Hz transformer has an iron loss of 5W at the condition. When operated at V, 5Hz, the iron loss is W. When operated at 46V, 5Hz, the value of the hysteresis loss divided by the eddy current loss is.44 Since V 4 8 is constant, Bm is constant f 5.5 Pcore P P k f k f P h e h kf 5 5k 5. k 5k 5k P e Kf 5k 5k 5 k 5k 65k k 5k k.8 k 5k 8 k 6 Pcore = 6f.8f Ph e 6f P.8f When f 5Hz, Ph 65 Pe Ph.44 P 6. e 47. ADC shunt generator delivers 45 A at a terminal voltage of V. The armature and the shunt field resistance are. and 44 respectively. The stray losses are 75W. The percentage efficiency of the DC generator is Pout 45 99W I f 5A 44 Ia IL If A Armature losses = I r 5. 5W Field losses = a a I R 5 44 W f f Stray losses = 75W Total losses = 5W losses P out I f 44 IL 45A ra. V ICP Intensive Classroom Program egate-live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ Centers across India 4

25 EE GATE-6-PAPER % 48. A three-phase, 5Hz salient-pole synchronous motor has a per-phase direct-axis reactance X d of.8pu and a per-phase quadrature-axis reactance X q of.6pu. Resistance of the machine is negligible. It is drawing full-load current at.8pf (leading)j. When the terminal voltage is pu, per-phase induced voltage, in pu, is..68 Xd.8P.U P.f.8 leading Ra Xq.6PU Vt P.U cos Ia 6.86 Ia q cos Ia R a sin Ia q cos.6.8 tan.59 V I q sin I R cos V Ia q sin.6.6 t a a a O I I sin sin d q a I I cos a q a d t Ef Vcos Id d I R Vcos I d.cos pu EF.68PU 49. A single-phase, kva, V/, 5Hz, distribution transformer is to be connected as an autotransformer to get an output voltage of 4 V. Its maximum kva rating as an auto-transformer is (A) (B) 4. (C) 4 (D) 4 (C) A Rated LV current = kva A KVA Or max 4kVA 4 4kVA A VLOV V A V 4V 5. A single-phase full-bridge voltage source inverter (VSI) is fed from a V battery. A pulse of O duration is used to trigger the appropriate devices in each half-cycle. The rms value of the fundamental component of the output voltage, in volts, is (A) 4 (B) 45 (C) (D) (A) ICP Intensive Classroom Program egate-live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ Centers across India 5

26 EE GATE-6-PAPER- VO t cos sin n t n,,s 4V 4Vs n n 6 S V cos 6.9 4V 5. A single-phase transmission line has two conductors each of mm radius. These are fixed a center-to-center distance of m in a horizontal plane. This is now converted to a three-phase transmission line by introducing a third conductor of the same radius. This conductor is fixed at an equal distance D from the two single-phase conductors. The three-phase line is fully transposed. The positive sequence inductance per phase of the three-phase system is to be 5% more than that of the inductance per conductor of the single-phase system. The distance D, in meters, is..49 For single phase m Dm L.n Ds. n D s For three phase DM D L.n. n DS Ds L.5L / D..5. n DS DS D.49 mts D m D D D.D. D m 5. In the circuit shown below, the supply voltage is sin t volts. The peak value of the steady state current through the resistor, in amperes, is. 4 F 4 F 5mH 4mH W =, the various impedance at this frequency are ~ sin t 5 ICP Intensive Classroom Program egate-live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ Centers across India 6

27 EE GATE-6-PAPER- -j 5 Z5μf Z 4mH (j 4 ) 6 ( j4) (j4) open circuit -j 5 Z4f Z 5mH (j 5 ) 6 ( j5) (j5) open circuit Since both LC pair parallel combination becomes open then the circuit can be redrawn as sint IΩ sint 4++5 So peak value of IΩ A 4 V 5 sint 5. A dc voltage with ripple is given by t sin t 5sin t this voltage volts. Measurements of t, made by moving-coil and moving-iron voltmeters, show readings of V and V respectively. The value of V V, in volts, is.. V V 5 V.V V V.V 54. The circuit below is excited by a sinusoidal source. The value of R, in, for which the admittance of the circuit becomes a pure conductance at all frequencies is. F R.H R ~ 4.4 Admittance becomes pure conductance means the imaginary part of Y must be zero which imply resonance condition. Let first get Y expression interms of L,C then by equalising imaginary part we will ICP Intensive Classroom Program egate-live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ Centers across India C L R R 7

28 EE GATE-6-PAPER- get the answer. j R+ R-jL Y C m gy [ eq ] R+jL R- R +( L) C R+ C L C R +( L) R + C Cross multiplying ( L)R L R +( L) C C C R L- L- C C C L R L- C C Now by looking into above equation we can say that if R then it will have no depending on frequency C L for resonancer C L L. So R C 55. In the circuit shown below, the node voltage V A is V. I A 5 5 5A 5 5 I V.4 All the branch currents are expressed interval of VA now writing KCL at node A V V I V A A A ICP Intensive Classroom Program egate-live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ Centers across India 8

29 EE GATE-6-PAPER- VA I V 5 A VA 6 VA VA 8 VA.4V 7 V A 5 5 5A V I A 5 VA I 5 5 I 5 VA - ICP Intensive Classroom Program egate-live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ Centers across India 9