Page 2 GATE EE

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2 Page GATE EE SOL. Overall gain of the system is written as G = GG G 3 We know that for a quantity that is product of two or more quantities total percentage error is some of the percentage error in each quantity. so error in overall gain G is 3 G = ε + ε+ ε3 Hence (D) is correct option. MCQ.3 The following circuit has a source voltage VS through the circuit is also shown. as shown in the graph. The current The element connected between a and b could be SOL.3 MCQ.4 Figure shows current characteristic of diode during switching. Hence (A) is correct option. The two inputs of a CRO are fed with two stationary periodic signals. In the X-Y mode, the screen shows a figure which changes from ellipse to circle and back to ellipse with its major axis changing orientation slowly and repeatedly. The following inference can be made from this. (A) The signals are not sinusoidal (B) The amplitudes of the signals are very close but not equal

3 Page 3 GATE EE (C) The signals are sinusoidal with their frequencies very close but not equal (D) There is a constant but small phase difference between the signals SOL.4 MCQ.5 SOL.5 MCQ.6 SOL.6 MCQ.7 A circle is produced when there is a 90c phase difference between vertical and horizontal inputs. Hence (D) is correct option. The increasing order of speed of data access for the following device is (I) Cache Memory (II) CD-ROM (III) Dynamic RAM (IV) Processor Registers (V) Magnetic Tape (A) (V), (II), (III), (IV), (I) (B) (V), (II), (III), (I), (IV) (C) (II), (I), (III), (IV), (V) (D) (V), (II), (I), (III), (IV) The increasing order of speed is as following Magnetic tape> CD-ROM> Dynamic RAM>Cache Memory>Processor register Hence (B) is correct option. A field excitation of 0 A in a certain alternator results in an armature current of 400 A in short circuit and a terminal voltage of 000 V on open circuit. The magnitude of the internal voltage drop within the machine at a load current of 00 A is (A) V (B) 0 V (C) 00 V (D) 000 V Hence (D) is correct option. Given field excitation of = 0 A Armature current = 400 A Short circuit and terminal voltage = 00 V On open circuit, load current = 00 A So, Internal resistance = 000 = 5 Ω 400 Internal vol. drop = 5 # 00 = 000 V The current through the kω resistance in the circuit shown is

4 Page 4 GATE EE (A) 0 ma (C) ma (B) ma (D) 6 ma SOL.7 Hence (A) is correct option MCQ.8 In the bridge RR = RR 3 = So it is a balanced bridge I = 0 ma 4 Out of the following plant categories (i) Nuclear (ii) Run-of-river (iii) Pump Storage (iv) Diesel The base load power plant are (A) (i) and (ii) (C) (i), (ii) and (iv) (B) (ii) and (iii) (D) (i), (iii) and (iv) SOL.8 MCQ.9 Hence ( ) is correct Option For a fixed value of complex power flow in a transmission line having a sending end voltage V, the real loss will be proportional to (A) V (B) V (C) (D) V V

5 Page 5 GATE EE SOL.9 MCQ.0 SOL.0 Hence (C) is correct option. We know complex power S = P+ jq = VI( cos φ+ jsin φ) = VIe jφ I = S jφ Ve a Real Power loss = IR P S L = c jφ R = Ve m SR jφ # e V a SR jφ = Constant e So P L \ V How many 00 W/0 V incandescent lamps connected in series would consume the same total power as a single 00 W/0 V incandescent lamp? (A) not possible (B) 4 (C) 3 (D) Resistance of the bulb rated 00 W/0 V is R V = P ( 0 ) = = 4 00 Ω Resistance of 00 W/0 V lamp is R V ( T = 0 ) = = 484 P 00 Ω To connect in series R T = n# R 484 = n # 4 n = Hence (D) is correct option MCQ. A Linear Time Invariant system with an impulse response ht () produces output yt () when input xt () is applied. When the input xt ( τ) is applied to a system with impulse response ht ( τ), the output will be (A) y() τ (B) y( ( t τ)) (C) yt ( τ) (D) yt ( τ) SOL. Hence (D) is correct option. Let xt () L Xs () yt () L Ys () ht () L Hs () So output of the system is given as

6 Page 6 GATE EE MCQ. Ys () = XsHs () () Now for input xt ( τ) L -sτ e X() s (shifting property) ht ( τ) L sτ e H() s s s So now output is Y'( s ) = e X() s $ e H() s -sτ Y'( s ) = e X() s H() s -sτ Y'( s ) = e Y() s Or y'( t ) = yt ( τ) The nature of feedback in the op-amp circuit shown is SOL. (A) Current-Current feedback (B) Voltage-Voltage feedback (C) Current-Voltage feedback (D) Voltage-Current feedback Equivalent circuit of given amplifier Feedback samples output voltage and adds a negative feedback voltage ( v fb ) to input. So, it is a voltage-voltage feedback. Hence (B) is correct option. MCQ.3 The complete set of only those Logic Gates designated as Universal Gates is (A) NOT, OR and AND Gates (B) XNOR, NOR and NAND Gates (C) NOR and NAND Gates (D) XOR, NOR and NAND Gates

7 Page 7 GATE EE SOL.3 MCQ.4 NOR and NAND gates considered as universal gates. Hence (C) is correct Option The single phase, 50 Hz iron core transformer in the circuit has both the vertical arms of cross sectional area 0 cm and both the horizontal arms of cross sectional area 0 cm. If the two windings shown were wound instead on opposite horizontal arms, the mutual inductance will (A) double (C) be halved (B) remain same (D) become one quarter SOL.4 MCQ.5 Given single-phase iron core transformer has both the vertical arms of cross section area 0 cm, and both the horizontal arms of cross section are 0 cm So, Inductance = NBA (proportional to cross section area) When cross section became half, inductance became half. Hence (C) is correct option. A 3-phase squirrel cage induction motor supplied from a balanced 3-phase source drives a mechanical load. The torque-speed characteristics of the motor(solid curve) and of the load(dotted curve) are shown. Of the two equilibrium points A and B, which of the following options correctly describes the stability of A and B? (A) A is stable, B is unstable (B) A is unstable, B is stable (C) Both are stable (D) Both are unstable SOL.5 Given 3-phase squirrel cage induction motor.

8 Page 8 GATE EE At point A if speed -, Torque - speed., Torque. So A is stable. At point B if speed - Load torque. So B is un-stable. Hence (A) is correct option. MCQ.6 SOL.6 MCQ.7 An SCR is considered to be a semi-controlled device because (A) It can be turned OFF but not ON with a gate pulse. (B) It conducts only during one half-cycle of an alternating current wave. (C) It can be turned ON but not OFF with a gate pulse. (D) It can be turned ON only during one half-cycle of an alternating voltage wave. SCR has the property that it can be turned ON but not OFF with a gate pulse, So SCR is being considered to be a semi-controlled device. Hence (C) is correct option. The polar plot of an open loop stable system is shown below. The closed loop system is (A) always stable (B) marginally stable (C) un-stable with one pole on the RH s-plane (D) un-stable with two poles on the RH s-plane SOL.7 From Nyquist stability criteria, no. of closed loop poles in right half of s-plane is given as Z = P N P " No. of open loop poles in right half s-plane N " No. of encirclement of (, j0)

9 Page 9 GATE EE Here N = (` encirclement is in clockwise direction) P = 0 (` system is stable) So, Z = 0 ( ) Z =, System is unstable with -poles on RH of s-plane. Hence (D) is correct option. MCQ.8 SOL.8 MCQ.9 The first two rows of Routh s tabulation of a third order equation are as follows. 3 s s 4 4 This means there are (A) Two roots at s =! j and one root in right half s-plane (B) Two roots at s =! j and one root in left half s-plane (C) Two roots at s =! j and one root in right half s-plane (D) Two roots at s =! j and one root in left half s-plane Given Routh s tabulation. s 3 s 4 4 s 0 0 So the auxiliary equation is given by, 4s + 4 = 0 s = s =! j From table we have characteristic equation as s 3 + s+ 4s + 4 = 0 3 s + s+ s + = 0 ss ( + ) + ( s + ) = 0 ( s+ )( s + ) = 0 s =, s =! j Hence (D) is correct option. The asymptotic approximation of the log-magnitude v/s frequency plot of a system containing only real poles and zeros is shown. Its transfer function is

10 Page 0 GATE EE ( s + 5) (A) ss ( + )( s+ 5 ) 00( s + 5) (C) ss ( + )( s+ 5 ) (B) (D) 000( s + 5) s ( s+ )( s+ 5) 80( s + 5) s ( s+ )( s+ 5) SOL.9 Since initial slope of the bode plot is 40 db/decade, so no. of poles at origin is. Transfer function can be written in following steps: Slope changes from 40 db/dec. to 60 db/dec. at ω= rad/sec., so at ω there is a pole in the transfer function. Slope changes from 60 db/dec to 40 db/dec at ω = 5 rad/sec., so at this frequency there is a zero lying in the system function. The slope changes from 40 db/dec to 60 db/dec at ω 3 = 5 rad/sec, so there is a pole in the system at this frequency. Transfer function Ts () Ks ( + 5) = s ( s+ )( s+ 5) Constant term can be obtained as. Tj ( ω) = 80 at ω= 0. K() 5 So, 80 = 0 log (.) 0 50 K = 000 therefore, the transfer function is Ts () Hence (B) is correct option. # 000( s + 5) = s ( s+ )( s+ 5) MCQ.0 The trace and determinant of a # matrix are known to be and 35 respectively. Its eigenvalues are (A) 30 and 5 (B) 37 and (C) 7 and 5 (D) 7.5 and SOL.0 Hence (C) is correct option. a Let the matrix is A = > b c d H

11 Page GATE EE Trace of a square matrix is sum of its diagonal entries Trace A = a+ d = Determinent ad bc = 35 Eigenvalue A λi = 0 λ a λ c b d λ = 0 ( a λ)( d λ) bc = 0 ( a+ d) λ+ ( ad bc) = 0 λ ( ) λ+ ( 35) = 0 λ + λ 35 = 0 ( λ 5)( λ+ 7) = 0 λ λ = 5, 7, Q. to Q.60 carry two marks each. MCQ. The following circuit has R = 0 kω, C = 0 μf. The input voltage is a sinusoidal at 50 Hz with an rms value of 0 V. Under ideal conditions, the current Is from the source is (A) 0π ma leading by 90 % (B) 0π ma leading by 90 % (C) 0π ma leading by 90 % (D) 0π ma lagging by 90 % SOL. Let voltages at positive and negative terminals of op-amp are V + and V - respectively, then V+ = V- = V s (ideal op-amp) In the circuit we have, V- 0 V V () s = 0 R `Cs j ( RCs) V-+ V- V0 ( s) = 0 ( + RCs) V s = V0 () s

12 Page GATE EE Similarly current I s is, I s = V V R s 0 I s = RCs Vs R I s = jωcv s I s = ωcvs % -6 I s = πf # 0# 0 # 0-6 I s = # π # 50# 0# 0 # 0 I s = 0π ma, leading by 90 % Hence (A) is correct option. MCQ. In the figure shown, all elements used are ideal. For time t < 0, S remained closed and S open. At t = 0, S is opened and S is closed. If the voltage V c across the capacitor C at t = 0 is zero, the voltage across the capacitor combination at t = 0 + will be (A) V (C).5 V (B) V (D) 3 V SOL. For t < 0, S is closed and S is opened so the capacitor C will charged upto 3 volt. V C (0) = 3 Volt Now when switch positions are changed, by applying charge conservation C eq V C (0 ) = CV C (0 ) + CV C (0 ) + ( + ) # 3 = # 3+ #V C (0 ) + 9 = 3+ V C (0 ) + V C (0 ) = 3 Volt Hence (D) is correct option. MCQ.3 SOL.3 Transformer and emitter follower can both be used for impedance matching at the output of an audio amplifier. The basic relationship between the input power P in and output power Pout in both the cases is (A) Pin = Pout for both transformer and emitter follower (B) Pin > Pout for both transformer and emitter follower (C) Pin < Pout for transformer and Pin = Pout for emitter follower (D) Pin = Pout for transformer and Pin < Pout for emitter follower Input and output power of a transformer is same

13 Page 3 GATE EE Pin = Pout for emitter follower, voltage gain(a v ) = current gain( Ai ) > Power( Pout ) = AAP v i in Since emitter follower has a high current gain so P Hence (D) is correct option. out > P in MCQ.4 The equivalent capacitance of the input loop of the circuit shown is (A) μ F (C) 00 μ F (B) 00 μf (D) 4 μf SOL.4 Hence (A) is correct option. Applying KVL in the input loop 3 v i ( ) 0 + # ( 49 ) j C i + i = 0 ω Input impedance So, equivalent capacitance v = 0 i # + 50 jωc i Z C eq = v i 3 3 = 0 # + jω( C/ 50) C 00 nf = 50 = = 50 nf MCQ.5 In an 8085 microprocessor, the contents of the Accumulator, after the following instructions are executed will become XRA A

14 Page 4 GATE EE (A) 0 H (C) F0 H MVI B, F0 H SUB B (B) 0F H (D) 0 H SOL.5 MCQ.6 SOL.6 For the given instruction set, XRA A& XOR A with A & A = 0 MVI B, F0 H&B = F0 H SUB B &A = A B A = B = 0000 s complement of ( B) = A+ ( B) = A B = = 0 H Hence (D) is correct option. For the Y-bus matrix of a 4-bus system given in per unit, the buses having shunt elements are R V S W S W YBUS = j S W S W T X (A) 3 and 4 (B) and 3 (C) and (D), and 4 Y Bus matrix of Y-Bus system are given as R V S W S W Y Bus = js W S W S W T X We have to find out the buses having shunt element R V Sy y y3 y4w Sy y y3 y4w We know Y Bus = S y3 y3 y33 y W S 34W Sy 4 y4 y43 y44w T X Here y = y0 + y + y3 + y4 = 5j y = y + y + y + y = 0j 0 3 4

15 Page 5 GATE EE y 33 = y30 + y3 + y3 + y34 = 9j y 44 = y40 + y4 + y4 + y43 = 8j y = y = y = j y 3 = y3 = y3 = 5. j y 4 = y4 = y4 = 0j y 3 = y3 = y3 = 5. j y 4 = y4 = y4 = 4j So y 0 = y y y3 y4 = 5j+ j+.5j+ 0j = 0.5j y 0 = y y y3 y4 = 0j+ j+. 5j+ 4j =. 5j y 30 = y33 y3 y3 y34 = 9j+. 5j+. 5j+ 4j = 0 y 40 = y44 y4 y4 y43 = 8j 0+ 4j+ 4j = 0 Admittance diagram is being made by as From figure. it is cleared that branch () & () behaves like shunt element. Hence (C) is correct option. MCQ.7 The unit-step response of a unity feed back system with open loop transfer function Gs ( ) = K/(( s+ )( s+ )) is shown in the figure. The value of K is (A) 0.5 (B) (C) 4 (D) 6 SOL.7 From the figure we can see that steady state error for given system is e ss = = 0. 5

16 Page 6 GATE EE Steady state error for unity feed back system is given by sr() s e ss = lim = s " 0 + Gs ( ) G s^s h = lim s " 0 + K ( s + )( s + ) = + > H ; () K = + K So, e ss = 0.5 = + K = K K = = 6 Hence (D) is correct option Rs = s (unit step input). s = -0 MCQ.8 The open loop transfer function of a unity feed back system is given by Gs () ( e )/ s. The gain margin of the is system is (A).95 db (B) 7.67 db (C).33 db (D) 3.9 db SOL.8 Open loop transfer function of the figure is given by, Gs () = e 0. s s Gjω ( ). = e jω j0 ω Phase cross over frequency can be calculated as, + Gj ( ω p ) = 80c b ω p # 90c π l = 80c So the gain margin (db) 0. ω 80 p # π c = 90c 0. ω p = 90 c # π 80 c ω p = 5. 7 rad/sec = 0 log e Gj ( ω p ) o = 0 log > b5. 7 l H = 0 log 5. 7 = 3. 9 db

17 Page 7 GATE EE Hence (D) is correct option. MCQ.9 Match the items in List-I (To) with the items in the List-II (Use) and select the correct answer using the codes given below the lists. List-I a. improve power factor b. reduce the current ripples c. increase the power flow in line d. reduce the Ferranti effect List-II. shunt reactor. shunt capacitor 3. series capacitor 4. series reactor SOL.9 MCQ.30 (A) a ", b " 3, c " 4, d " (B) a ", b " 4, c " 3, d " (C) a " 4, b " 3, c ", d " (D) a " 4, b ", c " 3, d " We know that Shunt Capacitors are used for improving power factor. Series Reactors are used to reduce the current ripples. For increasing the power flow in line we use series capacitor. Shunt reactors are used to reduce the Ferranti effect. Hence (B) is correct option Match the items in List-I (Type of transmission line) with the items in List-II (Type of distance relay preferred) and select the correct answer using the codes given below the lists. List-I a. Short Line b. Medium Line c. Long Line List-II. Ohm Relay. Reactance Relay 3. Mho Relay (A) a ",b ",c " 3 (B) a " 3,b ",c " (C) a ",b ",c " 3 (D) a ",b " 3,c " SOL.30 MCQ.3 We know that for different type of transmission line different type of distance relays are used which are as follows. Short Transmission line -Ohm reactance used Medium Transmission Line -Reactance relay is used Long Transmission line -Mho relay is used Hence (C) is correct option. Three generators are feeding a load of 00 MW. The details of the generators are

18 Page 8 GATE EE SOL.3 MCQ.3 Rating (MW) Efficiency (%) Regulation (Pu.) ( on 00 MVA base) Generator Generator Generator In the event of increased load power demand, which of the following will happen? (A) All the generator will share equal power (B) Generator-3 will share more power compared to Generator- (C) Generator- will share more power compared to Generator- (D) Generator- will share more power compared to Generator-3 Given that three generators are feeding a load of 00 MW. For increased load power demand, Generator having better regulation share More power, so Generator - will share More power than Generator -. Hence (C) is correct option. A 500 MW, kv, 50 Hz, 3-phase, -pole synchronous generator having a rated p.f = 0.9, has a moment of inertia of 7.5 # 0 3 kg-m.the inertia constant (H ) will be (A).44 s (B).7 s (C) 4.88 s (D) 5.4 s SOL.3 Given Synchronous generator of 500 MW, kv, 50 Hz, 3-φ, -pole 3 P.F = 0.9, Moment of inertia M = 7.5 # 0 kg-m Inertia constant H =? Generator rating in MVA G = P = cos 09. MW = MVA φ Stored K.E N = 0 # f = 0 # 50 = 3000 rpm pole M M 60 N = ω = π b l 3 = π 3000 MJ # # # b # 60 l = MJ Inertia constant ( H ) = Stored K.E Rating of Generator (MVA) H = =.44 sec Hence (A) is correct option. MCQ.33 fxy (,) is a continuous function defined over ( xy,)! [, 0] # [, 0]. Given the two constraints, x > y and y > x, the volume under fxy (,) is y = x= y y = x = (A) # # fxydxdy (,) (B) # # fxydxdy (,) y = 0 x= y y= x x= y

19 Page 9 GATE EE y = x = y= x x= y (C) # # fxydxdy (,) (D) fxydxdy (,) y = 0 x = 0 SOL.33 Given constraints x > y and y > x # # y = 0 x = 0 h Limit of y : y = 0 to y = Limit of x : x = y to x = y & x = y So volume under fxy (,) # y = x= V = fxydxdy (,) y = 0 Hence (A) is correct option # x= y y MCQ.34 Assume for simplicity that N people, all born in April (a month of 30 days), are collected in a room. Consider the event of at least two people in the room being born on the same date of the month, even if in different years, e.g. 980 and 985. What is the smallest N so that the probability of this event exceeds 0.5? (A) 0 (B) 7 (C) 5 (D) 6 = n SOL.34 No of events of at least two people in the room being born on same date C three people in the room being born on same date = n C3 similarly four for people = n C4 n n n n probabilty of the event, 0.5 C C C Cn $ $ 3 $ 4 g N 7 N = Hence(B) is correct option. MCQ.35 A cascade of three Linear Time Invariant systems is causal and unstable. From this, we conclude that (A) each system in the cascade is individually causal and unstable (B) at least on system is unstable and at least one system is causal (C) at least one system is causal and all systems are unstable (D) the majority are unstable and the majority are causal

20 Page 0 GATE EE SOL.35 Let three LTI systems having response H(), z H () z and H3 () z are Cascaded as showing below Assume H () z = z + z + (non-causal) 3 H () z = z + z + (non-causal) Overall response of the system Hz () = H() z H() z H3() z 3 Hz () = ( z + z + )( z + z + ) H3( z) To make Hz () causal we have to take H3 () z also causal Let H3 () z = z + z Hz () = ( z + z + )( z + z + )( z + z + ) Hz ()" causal Similarly to make Hz () unstable atleast one of the system should be unstable. Hence (B) is correct option. MCQ.36 The Fourier Series coefficients of a periodic signal xt () expressed as 3 jπkt/ T xt () = / ae k are given by a- = j, a = j0., a0 = j, k =-3 a = 05. j0., a = + j and ak = 0 for k > Which of the following is true? (A) xt () has finite energy because only finitely many coefficients are non-zero (B) xt () has zero average value because it is periodic (C) The imaginary part of xt () is constant (D) The real part of xt () is even SOL.36 Given signal xt () = 3 / k k =-3 ae jπkt/ T Let ω 0 is the fundamental frequency of signal xt () xt () a T π = ω 0 = / 3 jkω0t ae k k =-3 -jω0t -jω0t jω0t jω0t - 0 xt () = a e + a e + a + a e + a e - - jω t -jω t 0 0 = ( je ) + ( je ) + j+ ω + (0.5 0.) e + ( + j) e -jω t jω t jω t -jω t j t jω t = 6e + + j6e e j ω t e j t 0.j e j t e j t 0 - ω0 + ω0 - ω j

21 Page GATE EE = (cos ω0t) + j(jsin ω0t) + 0.5( cos ω0t) 0. j( jsin ω 0t) + j = [4 cos ω t sin ω t+ cos ω t+0.4 sin ω t] + j Im[ xt ( )] = (constant) Hence (C) is correct option MCQ.37 The z-transform of a signal xn [ ] is given by 4z + 3z + 6z + z It is applied to a system, with a transfer function Hz () = 3z - Let the output be yn. [ ] Which of the following is true? (A) yn [ ] is non causal with finite support (B) yn [ ] is causal with infinite support (C) yn [ ] = 0; n > 3 Re[ Yz ( )] ji z e = Re[ Yz ( )] -ji = z= e (D) Im[ Yz ( )] ji = Im[ Yz ( )] -j i; π # θ < π z= e z= e SOL.37 Z-transform of xn [ ] is -3-3 Xz () = 4z + 3z + 6z + z Transfer function of the system Hz () = 3z - Output Yz () = HzXz () () Yz () = (3z )(4z + 3z + 6z + z ) - z 4-9z - 6z 8z 6z - = z 3-6z 3 4+ z 4z - z 4-8z 3 - = + 9z 4 8z+ 8z 4z 3 Or sequence yn [ ] is yn [ ] = δ[ n 4] 8δ[ n 3] + 9δ[ n ] 4δ[ n] 8 δ[ n+ ] + 8 δ[ n+ ] 4 δ[ n+ 3] yn [ ] =Y 0, n < 0 So yn [ ] is non-causal with finite support. Hence (A) is correct option. MCQ.38 A cubic polynomial with real coefficients (A) Can possibly have no extrema and no zero crossings (B) May have up to three extrema and upto zero crossings (C) Cannot have more than two extrema and more than three zero crossings (D) Will always have an equal number of extrema and zero crossings SOL.38 Assume a Cubic polynomial with real Coefficients Px () = ax ax 3 + ax + a3 a0, a, a, a3 are real P'( x ) = 3ax+ ax+ a 0

22 Page GATE EE P''( x ) = 6ax+ a P'''( x ) = 6a 0 iv P () x = 0 Hence (C) is correct Option 0 MCQ.39 Let x 7 = 0. The iterative steps for the solution using Newton-Raphon s method is given by (A) x x 7 k+ = k+ b x l (B) x x 7 k+ = k x k (C) x x xk k+ = k (D) x x x 7 k+ = k k+ 7 b x l k k SOL.39 An iterative sequence in Newton-Raphson s method is obtained by following expression fx ( k) x k+ = xk f'( x ) k fx () = x 7 f'( x ) = x So fx ( k ) = xk 7 f'( x k ) = xk = # 7 So x 7 k+ x xk = k x Hence (D) is correct option. = x x 7 k k : + x D k MCQ.40 F(,) xy = ( x+ xy) atx+ ( y+ xy) at y. It s line integral over the straight line from (,) xy = (, 0) to ( xy,) = (, 0) evaluates to (A) 8 (B) 4 (C) 8 (D) 0 k SOL.40 Equation of straight line y = 0 ( x 0) 0 y = x F$ dl = [( x + xy) atx+ ( y + xy) aty][ dxatx+ dyaty+ dzatz] = ( x + xy) dx+ ( y + xy) dy Limit of x : 0 to Limit of y : to 0 0 # F$ dl = ( x + xy) dx+ ( y + xy) dy # 0 #

23 Page 3 GATE EE Line y dy = x = dx MCQ.4 0 So # F$ dl = [ x + x( x)] dx+ y + ( y) y dy = 4 4 = 0 Hence (D) is correct option. # # 0 0 = # xdx + # y dy 0 x y = : + D ; E 0 0 An ideal op-amp circuit and its input wave form as shown in the figures. The output waveform of this circuit will be

24 Page 4 GATE EE SOL.4 This is a schmitt trigger circuit, output can takes two states only. VOH =+ 6 volt V OL = 3 volt Threshold voltages at non-inverting terminals of op-amp is given as VTH 6 + VTH 0 = 0 3V TH 6 = 0 V TH = V (Upper threshold) Similarly VTL ( 3) + VTL = 0 3V TL + 3 = 0 V TL = V (Lower threshold) For Vin < Volt, V0 =+ 6 Volt Vin > Volt, V0 = 3 Volt Vin < Volt V0 =+ 6 Volt Vin > Volt V0 = 3 Volt Output waveform Hence(D) is correct option. MCQ.4 A 00 V, 50 Hz, single-phase induction motor has the following connection diagram and winding orientations as shown. MM is the axis of the main stator winding (MM ) and AA is that of the auxiliary winding(aa ). Directions of the winding axis indicate direction of flux when currents in the windings are in the directions shown. Parameters of each winding are indicated. When switch S is closed the motor

25 Page 5 GATE EE (A) rotates clockwise (B) rotates anti-clockwise (C) does not rotate (D) rotates momentarily and comes to a halt SOL.4 MCQ.43 Hence ( ) is correct Option The circuit shows an ideal diode connected to a pure inductor and is connected to a purely sinusoidal 50 Hz voltage source. Under ideal conditions the current waveform through the inductor will look like.

26 Page 6 GATE EE SOL.43 Hence (D) is correct option. Current wave form for i L v L i L = Ldi L dt = vdt L # for 0 < ω t + π, v L = v 0 sin t di L in = ω = dt i L = vdt L = cos 00πt+ C # at 00πt = π/, i L = 0, C = 0 i L = 00 cos πt = Amp for π < ω t v = v = 0 i L( peak) L in

27 Page 7 GATE EE MCQ.44 The Current Source Inverter shown in figure is operated by alternately turning on thyristor pairs (T,T ) and (T 3,T 4 ). If the load is purely resistive, the theoretical maximum output frequency obtainable will be (A) 5 khz (C) 500 khz (B) 50 khz (D) 50 khz SOL.44 In CSI let T 3 and T 4 already conducting at t = 0 At triggering T and T, T 3 and T 4 are force commulated. Again at t = T, T and T are force commutated. This completes a cycle. Time constant τ = RC = 4# 0. 5 = μ sec frequency f = = τ = khz # 0 Hence (C) is correct option.

28 Page 8 GATE EE MCQ.45 In the chopper circuit shown, the main thyristor (T M ) is operated at a duty ratio of 0.8 which is much larger the commutation interval. If the maximum allowable reapplied dv/ dt on T M is 50 V/ μ s, what should be the theoretical minimum value of C? Assume current ripple through L 0 to be negligible. (A) 0. μ F (C) μ F (B) 0.0 μf (D) 0 μf SOL.45 Hence (A) is correct option. duty ratio T M = 08. maximum dv on TM dt = 50 V/ μsec minimum value of C =? Given that current ripple through L 0 is negligible. current through T M = Im = duty ratio# current = 0.8 #.5 = 0 A a I m = C dv dt 0 = C 50 # 6 0 C = # = μ F MCQ.46 Match the switch arrangements on the top row to the steady-state V -I characteristics on the lower row. The steady state operating points are shown by large black dots.

29 Page 9 GATE EE (A) P-I, Q-II, R-III, S-IV (C) P-IV, Q-III, R-I, S-II (B) P-II, Q-IV, R-I, S-III (D) P-IV, Q-III, R-II, S-I SOL.46 Characteristics are as Hence (C) is correct option MCQ.47 For the circuit shown, find out the current flowing through the Ω resistance. Also identify the changes to be made to double the current through the Ω resistance.

30 Page 30 GATE EE (A) (5 A; PutVS = 30 V) (B) ( A; PutVS = 8 V) (C) (5 A; Put IS = 0 A) (D) (7 A; Put IS = A) SOL.47 Voltage across X resistor, V S = V Current, I VS Ω = = 4 = A To make the current double we have to take V S = 8 V Hence (B) is correct option. MCQ.48 The figure shows a three-phase delta connected load supplied from a 400V, 50 Hz, 3-phase balanced source. The pressure coil (PC) and current coil (CC) of a wattmeter are connected to the load as shown, with the coil polarities suitably selected to ensure a positive deflection. The wattmeter reading will be (A) 0 (C) 800 Watt (B) 600 Watt (D) 400 Watt SOL.48 MCQ.49 Hence (C) is correct option. Wattmeter reading P = VPC ICC V PC " Voltage across potential coil. I CC " Current in current coil. V PC = V bc = c I CC = I ac = + c = 4+ 0c 00 Power P = c# 4+ 0c = c = 600 # = 800 Watt An average-reading digital multi-meter reads 0 V when fed with a triangular wave,

31 Page 3 GATE EE symmetric about the time-axis. For the same input an rms-reading meter will read (A) 0 (B) (C) 0 3 (D) 0 3 SOL.49 Hence (D) is correct option. Average value of a triangular wave V av = V 3 m rms value V ms Given that V V m av = = 0 V 3 = Vm 3 So V rms Vm = = 3Vav = 0 3 V 3 MCQ.50 Figure shows the extended view of a -pole dc machine with 0 armature conductors. Normal brush positions are shown by A and B, placed at the interpolar axis. If the brushes are now shifted, in the direction of rotation, to A and B as shown, the voltage waveform V AB ' ' will resemble

32 Page 3 GATE EE SOL.50 Wave form V AB l l Hence (A) is correct option. Common Data Questions Common Data for Questions 5 and 5: The star-delta transformer shown above is excited on the star side with balanced, 4-wire, 3-phase, sinusoidal voltage supply of rated magnitude. The transformer is under no load condition MCQ.5 With both S and S open, the core flux waveform will be (A) a sinusoid at fundamental frequency (B) flat-topped with third harmonic (C) peaky with third-harmonic (D) none of these

33 Page 33 GATE EE SOL.5 MCQ.5 SOL.5 When both S and S open, star connection consists 3 rd harmonics in line current due to hysteresis A saturation. Hence (B) is correct option. With S closed and S open, the current waveform in the delta winding will be (A) a sinusoid at fundamental frequency (B) flat-topped with third harmonic (C) only third-harmonic (D) none of these Since S closed and S open, so it will be open delta connection and output will be sinusoidal at fundamental frequency. Hence (A) is correct option. Common Data for Questions 53 and 54 : The circuit diagram shows a two-winding, lossless transformer with no leakage flux, excited from a current source, it, () whose waveform is also shown. The transformer has a magnetizing inductance of 400/ π mh. MCQ.53 SOL.53 The peak voltage across A and B, with S open is (A) 400 V (B) 800 V π (C) 4000 V (D) 800 π π V Peak voltage across A and B with S open is V = m di = m ( slope of I t) dt #

34 Page 34 GATE EE = π # #: 3 = 5 # 0 D V π Hence (D) is correct option. MCQ.54 If the wave form of it () is changed to it ( ) = 0 sin(00 πt) A, the peak voltage across A and B with S closed is (A) 400 V (B) 40 V (C) 30 V (D) 60 V SOL.54 Hence ( ) is correct Option Common Data for Question 55 and 56 : A system is described by the following state and output equations dx() t = 3x() t + x() t + u() t dt dx() t = x( t) + u( t) dt yt () = x() t when ut () is the input and yt () is the output MCQ.55 SOL.55 The system transfer function is (A) s + (B) s + 3 s + 5 s 6 s + 5 s + 6 (C) s + 5 (D) s 5 s + 5 s + 6 s + 5 s 6 Given system equations dx() t = 3 x( t) + x( t) + u( t) dt dx() t dt = x () t + u() t yt () = x() t Taking Laplace transform on both sides of equations. sx () s = 3X() s + X() s + U() s ( s+ 3) X( s) = X() s + U() s...() Similarly sx () s = X () s + U() s ( s+ ) X( s) = Us ()...() From equation () & () Us () ( s+ 3) X( s) = + Us () s + X () s = Us () + ( s + ) s + 3; s + E

35 Page 35 GATE EE ( s + 5) X () s = Us () ( s+ )( s+ 3 ) From output equation, Ys () = X() s ( s + 5) So, Ys () = Us () ( s+ )( s+ 3 ) System transfer function Ys () ( s + 5) T.F = = Us () ( s + )( s + 3) ( s + 5) = s + 5s + 6 Hence (C) is correct option. MCQ.56 The state-transition matrix of the above system is -3t e 0 (A) = -t -3t -tg (B) e -3t e -t e -3t = -t G e + e e 0 e -3t -t -3t 3t -t -3t (C) e e + = e -t G (D) e e = e -t G 0 e 0 e SOL.56 Given state equations in matrix form can be written as, xo 3 x > xo H = > u() t 0 H> + x H > H dx() t dt = AX() t + Bu() t State transition matrix is given by φ () t = L 6 Φ() Φ () s = ( si A) s ( si A) = 0 3 > 0 s H > 0 H s ( si A) = 3 > + 0 s + H ( si A) s + = ( s 3 )( s ) > s + 3 H R V S W ( s 3) ( s 3)( s ) So Φ () s = ( si A) = S W S W S 0 ( s + ) W T 3t t 3t X e e e φ() t = L [ Φ()] s = > t H 0 e Hence (B) is correct option

36 Page 36 GATE EE Statement for Linked Answer Questions 57 and 58 : The figure above shows coils- and, with dot markings as shown, having 4000 and 6000 turns respectively. Both the coils have a rated current of 5 A. Coil- is excited with single phase, 400 V, 50 Hz supply. 400 MCQ.57 The coils are to be connected to obtain a single-phase, 000 V, auto-transformer to drive a load of 0 kva. Which of the options given should be exercised to realize the required auto-transformer? (A) Connect A and D; Common B (B) Connect B and D; Common C (C) Connect A and C; Common B (D) Connect A and C; Common D SOL.57 Hence (A) is correct option. N = 4000 N = 6000 I = 5 A V = 400 V, f = 50 Hz Coil are to be connected to obtain a single Phase, 400 V auto transfer to drive 000 Load 0 kva Connected A & D common B MCQ.58 In the autotransformer obtained in Question 6, the current in each coil is (A) Coil- is 5 A and Coil- is 0 A

37 Page 37 GATE EE (B) Coil- is 0 A and Coil- is 5 A (C) Coil- is 0 A and Coil- is 5 A (D) Coil- is 5 A and Coil- is 0 A SOL.58 Given 3-phase, 400 V, 5 kw, Star connected synchronous motor. Internal Resistance = 0 Ω Operating at 50% Load, unity p.f. So kva rating = 5 # 400 = 000 Internal Resistance = 0 Ω So kva rating = 000 # 0 = 0000 kva Hence (D) is correct option Statement for Linked Answer Question 59 and 60 : MCQ.59 SOL.59 For the circuit given above, the Thevenin s resistance across the terminals A and B is (A) 0.5 kω (B) 0. kω (C) kω (D) 0. kω To obtain equivalent thevenin circuit, put a test source between terminals AB By applying KCL at super node VP V V 5 + P + S = IS VP 5+ VP+ VS = I S

38 Page 38 GATE EE VP+ VS = I s + 5 VP+ VS = I S () VP VS = 3V S & V P = 4V S So, 4VS + VS = I S +.5 5V S = I S +.5 V S = 0.I S () For thevenin equivalent circuit V S = IR S th+ Vth...(3) By comparing () and (3), Thevenin resistance R th = 0. kω Hence (B) is correct option MCQ.60 SOL.60 For the circuit given above, the Thevenin s voltage across the terminals A and B is (A).5 V (B) 0.5 V (C) V (D) 0.5 V Hence (D) is correct option From above V th = 0.5 V. Answer Sheet. (A) 3. (C) 5. (D) 37. (A) 49. (D). (D) 4. (C) 6. (C) 38. (C) 50. (A) 3. (A) 5. (A) 7. (D) 39. (D) 5. (B) 4. (D) 6. (C) 8. (D) 40. (D) 5. (A) 5. (B) 7. (D) 9. (B) 4. (D) 53. (D) 6. (D) 8. (D) 30. (C) 4. (*) 54. (*) 7. (A) 9. (B) 3. (C) 43. (D) 55. (C) 8. (*) 0. (C) 3. (A) 44. (C) 56. (B) 9. (C). (A) 33. (A) 45. (A) 57. (A) 0. (D). (D) 34. (B) 46. (C) 58. (D). (D) 3. (D) 35. (B) 47. (B) 59. (B). (B) 4. (A) 36. (C) 48. (C) 60. (D)

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