(B) 1 A 2 2 (C) 1 A (D) 2 A

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1 GATE SOLVED PAPER - EE 0 Q. No. 5 Carry One Mark Each Q. Roots of the algebraic equation x + x + x+ 0 are (A) ( +, + j, - j) (B) ( +, -, + ) (C) ( 000,, ) (D) (-, + j, -j) Q. With K as a constant, the possible solution for the first order differential equation dy e dx -x is x - (A) - e + K x - (C) - e + K x (B) - e + K - (D) - e x + K Q. The r.m.s value of the current it () in the circuit shown below is (A) A (B) A (C) A (D) A Q. 4 The fourier series expansion ft () a0 + / an cosnt w + bnsinnt w of n the periodic signal shown below will contain the following nonzero terms (A) a 0 and b, n 5,,,... (B) a 0 and a, n,,,... n n (C) a 0 a n and bn, n,,,... (D) a 0 and a n 5,,,... Q. 5 A 4 point starter is used to start and control the speed of a (A) dc shunt motor with armature resistance control (B) dc shunt motor with field weakening control (C) dc series motor (D) dc compound motor n

2 GATE SOLVED PAPER - EE 0 Q. 6 A three phase, salient pole synchronous motor is connected to an infinite bus. It is operated at no load a normal excitation. The field excitation of the motor is first reduced to zero and then increased in reverse direction gradually. Then the armature current. (A) Increases continuously (B) First increases and then decreases steeply (C) First decreases and then increases steeply (D) Remains constant Q. 7 A nuclear power station of 500 MW capacity is located at 00 km away from a load center. Select the most suitable power evacuation transmission configuration among the following options Q. 8 The frequency response of a linear system Gjw ( ) is provided in the tubular form below Gjw ( ) Gj ( w) -0c -40c -50c -60c -80c -00c Gain Margin and phase margin are (A) 6 db and 0c (B) 6 db and -0c (C) - 6dB and 0c (D) - 6dB and -0c Q. 9 The steady state error of a unity feedback linear system for a unit step input is 0.. The steady state error of the same system, for a pulse input rt () having a magnitude of 0 and a duration of one second, as shown in the figure is

3 GATE SOLVED PAPER - EE 0 (A) 0 (B) 0. (C) (D) 0 Q. 0 Consider the following statement () The compensating coil of a low power factor wattmeter compensates the effect of the impedance of the current coil. () The compensating coil of a low power factor wattmeter compensates the effect of the impedance of the voltage coil circuit. (A) () is true but () is false (C) both () and () are true (B) () is false but () is true (D) both () and () are false Q. A low-pass filter with a cut-off frequency of 0 Hz is cascaded with a high pass filter with a cut-off frequency of 0 Hz. The resultant system of filters will function as (A) an all pass filter (B) an all stop filter (C) an band stop (band-reject) filter (D) a band pass filter Q. The CORRECT transfer characteristic is

4 GATE SOLVED PAPER - EE 0 Q. A three phase current source inverter used for the speed control of an induction motor is to be realized using MOSFET switches as shown below. Switches S to S 6 are identical switches. The proper configuration for realizing switches S to S 6 is Q. 4 A point Z has been plotted in the complex plane, as shown in figure below. The plot of the complex number Y Z is

5 GATE SOLVED PAPER - EE 0 Q. 5 The voltage applied to a circuit is 00 cos( 00p t) volts and the circuit draws a current of 0 sin( 00pt + p/ 4) amperes. Taking the voltage as the reference phasor, the phasor representation of the current in amperes is (A) 0 - p/ 4 (B) 0 -p/ 4 (C) 0 + p/ 4 (D) 0 +p/ 4 Q. 6 In the circuit given below, the value of R required for the transfer of maximum power to the load having a resistance of W is (A) zero (C) 6 W (B) W (D) infinity Q. 7 Given two continuous time signals xt () e and yt () e which exist for t > 0 *, the convolution zt () xt () yt () is -t -t (A) e - e (B) e (C) e t (D) e -t e -t + Q. 8 A single phase air core transformer, fed from a rated sinusoidal supply, is operating at no load. The steady state magnetizing current drawn by the transformer from the supply will have the waveform

6 GATE SOLVED PAPER - EE 0 Q. 9 A negative sequence relay is commonly used to protect (A) an alternator (B) an transformer (C) a transmission line (D) a bus bar Q. 0 For enhancing the power transmission in along EHV transmission line, the most preferred method is to connect a (A) Series inductive compensator in the line (B) Shunt inductive compensator at the receiving end (C) Series capacitive compensator in the line (D) Shunt capacitive compensator at the sending end Q. An open loop system represented by the transfer function ( s - ) Gs () is ( s + )( s + ) (A) Stable and of the minimum phase type (B) Stable and of the non minimum phase type (C) Unstable and of the minimum phase type (D) Unstable and of non minimum phase type Q. The bridge circuit shown in the figure below is used for the measurement of an unknown element Z X. The bridge circuit is best suited when Z X is a (A) low resistance (C) low Q inductor (B) high resistance (D) lossy capacitor Q. A dual trace oscilloscope is set to operate in the ALTernate mode. The control input of the multiplexer used in the y-circuit is fed with a signal having a frequency equal to (A) the highest frequency that the multiplexer can operate properly (B) twice the frequency of the time base (sweep) oscillator (C) the frequency of the time base (sweep) oscillator (D) haif the frequency of the time base (sweep) oscillator Q. 4 The output Y of the logic circuit given below is (A) (B) 0 (C) X (D) X

7 GATE SOLVED PAPER - EE 0 Q. No. 6 5 Carry Two Mark Each Q. 5 Circuit turn-off time of an SCR is defined as the time (A) taken by the SCR turn to be off (B) required for the SCR current to become zero (C) for which the SCR is reverse biased by the commutation circuit (D) for which the SCR is reverse biased to reduce its current below the holding current Q. 6 Solution of the variables x and x for the following equations is to be obtained by employing the Newton-Raphson iterative method. equation () 0xsin x equation () 0x -0xcos x Assuming the initial values are x 00. and x 0., the jacobian matrix is (A) > H (B) 0 0 > 0 0 H (C) > H (D) 0 0 > 0-0 H Q. 7 The function fx () x-x- x+ has (A) a maxima at x and minimum at x 5 (B) a maxima at x and minimum at x -5 (C) only maxima at x and (D) only a minimum at x 5 Q. 8 A lossy capacitor C x, rated for operation at 5 kv, 50 Hz is represented by an equivalent circuit with an ideal capacitor C p in parallel with a resistor R p. The value C p is found to be 0.0 μf and value of R p.5 MW. Then the power loss and tan d of the lossy capacitor operating at the rated voltage, respectively, are (A) 0 W and (B) 0 W and (C) 0 W and 0.05 (D) 0 W and 0.04 Q. 9 Let the Laplace transform of a function ft () which exists for t > 0 be F () s and the Laplace transform of its delayed version ft ( - t) be F () s. Let F *( s) be the complex conjugate of F () s with the Laplace variable set s s+ jw. If F() s F*() s Gs (), then the inverse Laplace transform of Gs () is F () s (A) An ideal impulse d() t (B) An ideal delayed impulse d( t - t) (C) An ideal step function ut () (D) An ideal delayed step function ut ( - t) Q. 0 A zero mean random signal is uniformly distributed between limits - a and + a and its mean square value is equal to its variance. Then the r.m.s value of the signal is (A) a (B) a (C) a (D) a

8 GATE SOLVED PAPER - EE 0 Q. A 0 V, DC shunt motor is operating at a speed of 440 rpm. The armature resistance is 0W. and armature current is 0 A. of the excitation of the machine is reduced by 0%, the extra resistance to be put in the armature circuit to maintain the same speed and torque will be (A) 79W. (B) W. (C) 8. 9 W (D) W. Q. A load center of 0 MW derives power from two power stations connected by 0 kv transmission lines of 5 km and 75 km as shown in the figure below. The three generators G, G and G are of 00 MW capacity each and have identical fuel cost characteristics. The minimum loss generation schedule for supplying the 0 MW load is P (A) P P P (C) P P 80 MW + losses 0 MW 0 MW 40 MW 40 MW 40 MW + losses P (B) P P P (D) P P 60 MW 0 MW + losses 0 MW 0 MW + losses 45 MW 45 MW Q. The open loop transfer function Gs () of a unity feedback control system is given as Ks b + l Gs () s ( s+ ) From the root locus, at can be inferred that when K tends to positive infinity, (A) Three roots with nearly equal real parts exist on the left half of the s-plane (B) One real root is found on the right half of the s-plane (C) The root loci cross the jw axis for a finite value of KK! ; 0 (D) Three real roots are found on the right half of the s-plane Q. 4 A portion of the main program to call a subroutine SUB in an 8085 environment is given below. h LXI D, DISP LP : CALL SUB LP+ h It is desired that control be returned to LP+DISP+ when the RET instruction is executed in the subroutine. The set of instructions that precede the RET instruction in the subroutine are

9 GATE SOLVED PAPER - EE 0 (A) (C) POP DAD PUSH POP DAD PUSH D H D H D H (B) (D) POP DAD INX INX INX PUSH XTHL INX INX INX XTHL Q. 5 The transistor used in the circuit shown below has a b of 0 and I CBO is negligible If the forward voltage drop of diode is 0.7 V, then the current through collector will be (A) 68 ma (B) 08 ma (C) 0.54 ma (D) 5.6 ma Q. 6 A voltage commutated chopper circuit, operated at 500 Hz, is shown below. H D H H H H D D D If the maximum value of load current is 0 A, then the maximum current through the main ( M ) and auxiliary ( A ) thyristors will be (A) i M max A and i A max 0 A (B) (C) (D) M A and i A max A i max M 0 A and i max A i max A M 0 A and i A max 8A i max

10 GATE SOLVED PAPER - EE 0 Q. 7 The matrix [ A] > 4 - H is decomposed into a product of a lower triangular matrix [ L ] and an upper triangular matrix [ U ]. The properly decomposed [ L ] and [ U ] matrices respectively are (A) 0 > 4 - H and > 0 - H (B) 0 > 4 - H and > 0 H (C) 0 > 4 H and > 0 - H (D) 0 5. > 4 - H and > 0 H Q. 8 The two vectors [,,] and [, aa, ] where a c- + j m, are (A) Orthonormal (B) Orthogonal (C) Parallel (D) Collinear Q. 9 A three-phase 440 V, 6 pole, 50 Hz, squirrel cage induction motor is running at a slip of 5%. The speed of stator magnetic field to rotor magnetic field and speed of rotor with respect of stator magnetic field are (A) zero, - 5rpm (B) zero, 955 rpm (C) 000 rpm, - 5rpm (D) 000 rpm, 955 rpm Q. 40 A capacitor is made with a polymeric dielectric having an e r of.6 and a dielectric breakdown strength of 50 kv/cm. The permittivity of free space is 8.85 pf/m. If the rectangular plates of the capacitor have a width of 0 cm and a length of 40 cm, then the maximum electric charge in the capacitor is (A) μc (B) 4 μc (C) 8 μc (D) 0 μc Q. 4 The response ht () of a linear time invariant system to an impulse d () t, under -t t initially relaxed condition is ht () e + e -. The response of this system for a unit step input ut () is -t -t (A) ut () + e + e - (B) ( e t - t + e ) u( t) -t - t (C) ( 5. -e -05. e ) u() t - (D) e t - t d () t + e u() t Q. 4 The direct axis and quadrature axis reactances of a salient pole alternator are. p.u and.0 p.u respectively. The armature resistance is negligible. If this alternator is delivering rated kva at upf and at rated voltage then its power angle is (A) 0c (B) 45c (C) 60c (D) 90c Q. 4 A 4 digit DMM has the error specification as: 0.% of reading + 0 counts. If a dc voltage of 00 V is read on its 00 V full scale, the maximum error that can be expected in the reading is (A)!0. % (B)!0. % (C)!0. % (D)!0. 4%

11 GATE SOLVED PAPER - EE 0 Q. 44 A three bus network is shown in the figure below indicating the p.u. impedance of each element. The bus admittance matrix, Y -bus, of the network is R V R S W V S W (A) js W (B) js W S W S W RT X T X V R S W V S W (C) js W (D) js W S W S W T X T X Q. 45 A two loop position control system is shown below The gain K of the Tacho-generator influences mainly the (A) Peak overshoot (B) Natural frequency of oscillation (C) Phase shift of the closed loop transfer function at very low frequencies ( w " 0) (D) Phase shift of the closed loop transfer function at very high frequencies ( w " ) Q. 46 A two bit counter circuit is shown below It the state Q A Q B of the counter at the clock time t n is 0 then the state Q A Q of the counter at tn + (after three clock cycles) will be (A) 00 (B) 0 (C) 0 (D) B

12 GATE SOLVED PAPER - EE 0 Q. 47 A clipper circuit is shown below. Assuming forward voltage drops of the diodes to be 0.7 V, the input-output transfer characteristics of the circuit is Common Data questions: 48 & 49 The input voltage given to a converter is v i 00 sin(00 pt) V The current drawn by the converter is ii 0 sin(00 pt- p/) + 5 sin(00 pt+ p/4) + sin(500 pt -p/6) A Q. 48 The input power factor of the converter is (A) 0. (B) 0.44 (C) 0.5 (D) 0.7 Q. 49 The active power drawn by the converter is (A) 8 W (B) 500 W (C) 707 W (D) 887 W

13 GATE SOLVED PAPER - EE 0 Common Data questions: 50 & 5 An RLC circuit with relevant data is given below. Q. 50 The power dissipated in the resistor R is (A) 0.5 W (B) W (C) W (D) W Q. 5 The current I C in the figure above is (A) - ja (B) -j A (C) + j A (D) +ja Linked Answer questions: Q.5 to Q.55 Carry Two Marks Each Statement for Linked Answer Questions: 5 & 5 Q. 5 Two generator units G and G are connected by 5 kv line with a bus at the mid-point as shown below G 50 MVA, 5 kv, positive sequence reactance X G 5% on its own base G 00 MVA, 5 kv, positive sequence reactance X G 0% on its own base L and L 0 km, positive sequence reactance 0.5 W/km X L

14 GATE SOLVED PAPER - EE 0 Q. 5 In the above system, the three-phase fault MVA at the bus is (A) 8.55 MVA (B) 85. MVA (C) 70.9 MVA Statement for Linked Answer Questions: 54 & 55 (D) 8.8 MVA A solar energy installation utilize a three phase bridge converter to feed energy into power system through a transformer of 400 V/400 V, as shown below. The energy is collected in a bank of 400 V battery and is connected to converter through a large filter choke of resistance 0 W. Q. 54 The maximum current through the battery will be (A) 4 A (B) 40 A (C) 80 A (D) 94 A Q. 55 The kva rating of the input transformer is (A) 5. kva (B) 46.0 kva (C).6 kva (D) 7.5 kva Q. No Carry One Mark Each Q. 56 There are two candidates P and Q in an election. During the campaign, 40% of the voters promised to vote for P, and rest for Q. However, on the day of election 5% of the voters went back on their promise to vote for P and instead voted for Q.5% of the voters went back on their promise to vote for Q and instead voted for P. Suppose, P lost by votes, then what was the total number of voters? (A) 00 (B) 0 (C) 90 (D) 95

15 GATE SOLVED PAPER - EE 0 Q. 57 Choose the most appropriate word from the options given below to complete the following sentence: It was her view that the country s problems had been by foreign technocrats, so that to invite them to come back would be counter-productive. (A) Identified (B) ascertained (C) Texacerbated (D) Analysed Q. 58 Choose the work from the options given below that is most nearly opposite in meaning to the given word: Frequency (A) periodicity (B) rarity (C) gradualness (D) persistency Q. 59 Choose the most appropriate word from the options given below to complete the following sentence: Under ethical guidelines recently adopted by the Indian Medical Association, human genes are to be manipulated only to correct diseases for which treatments are unsatisfactory. (A) Similar (B) Most (C) Uncommon (D) Available Q. 60 The question below consists of a pair of related words followed by four paris of words. Select the pair that best expresses the relation in the original pair: Gladiator : Arena (A) dancer : state (B) commuter : train (C) teacher : classroom (D) lawyer : courtroom Q. No Carry Two Mark Each Q. 6 The fuel consumed by a motorcycle during a journey while traveling at various speeds is indicated in the graph below.

16 GATE SOLVED PAPER - EE 0 The distances covered during four laps of the journey are listed in the table below Lap Distance (kilometers) Average speed (kilometers per hour) P 5 5 Q R S 0 0 From the given data, we can conclude that the fuel consumed per kilometer was least during the lap (A) P (B) Q (C) R (D) S Q. 6 Three friends, R, S and T shared toffee from a bowl. R took / rd of the toffees, but returned four to the bowl. S took 4 / th of what was left but returned three toffees to the bowl. T took half of the remainder but returned two back into the bowl. If the bowl had 7 toffees left, how many toffees were originally there in the bowl? (A) 8 (B) (C) 48 (D) 4 Q. 6 Given that fy () 6@ y/ y and q is any non-zero real number, the value of fq ()-f(- q) is (A) 0 (B) - (C) (D) Q. 64 The sum of n terms of the series is (A) (4/8) [ 0 n + - 9n - ] (B) (4/8) [ 0 n - - 9n - ] (C) (4/8) [ 0 n + - 9n - 0] (D) (4/8) [ 0 n - 9n - 0] Q. 65 The horse has played a little known but very important role in the filed of medicine. Horses were injected with toxins of diseases until their blood build up immunities. Then a serum was made from their blood. Serums to flight with diphtheria and tetanus were developed this way. It can be inferred from the passage that horses were (A) given immunity to diseases (B) generally quite immune to diseases (C) given medicines to flight (D) given diphtheria and tetanus serums

17 GATE SOLVED PAPER - EE 0 SOLUTION Sol. x + x + x+ 0 Sol. x ( x+ ) + ( x- ) 0 ( x+ )( x + ) 0 or x + 0 & x - and x + 0 & x -j, j x -, -j, j Hence (D) is correct option. dy dx -x e -x dy e dx by integrating, we get x - y - e + K, where K is constant. Hence (A) is correct option. Sol. The frequency domains equivalent circuit at w rad/sec. Since the capacitor and inductive reactances are equal in magnitude, the net impedance of that branch will become zero. Equivalent circuit Current it () sin t ( sin t) A W rms value of current

18 GATE SOLVED PAPER - EE 0 Sol. 4 i rms Hence (B) is correct option. A ft () a + ( a cos wt+ b sin nwt) 0 / n The given function ft () is an even function, therefore b 0 n ft () is a non zero average value function, so it will have a non-zero value of a 0 a 0 T/ n n # ftdt () (average value of ft) () ^ T/ h 0 a n is zero for all even values of n and non zero for odd n T a n ft () cos( ntd w ) ( wt) T # 0 So, fourier expansion of ft () will have a 0 and a n, n 5f,, Sol. 5 Sol. 6 Sol. 7 Hence (D) is correct option. The armature current of DC shunt motor I V Eb a - Ra at the time of starting, Eb 0. If the full supply voltage is applied to the motor, it will draw a large current due to low armature resistance. A variable resistance should be connected in series with the armature resistance to limit the starting current. A 4-point starter is used to start and control speed of a dc shut motor. Hence (A) is correct option The Back emf will go to zero when field is reduced, so Current input will be increased. But when Field increases (though in reverse direction) the back emf will cause the current to reduce. Hence (B) is correct option. Sol. 8 Gain margin is simply equal to the gain at phase cross over frequency ( w p ). Phase cross over frequency is the frequency at which phase angle is equal to - 80c. From the table we can see that + Gj ( w p) - 80c, at which gain is 0.5. GM 0 log 0e Gj ( w ) o 0 logb0.5 l 6 db p

19 GATE SOLVED PAPER - EE 0 Phase Margin is equal to 80c plus the phase angle f g at the gain cross over frequency ( w g ). Gain cross over frequency is the frequency at which gain is unity. From the table it is clear that Gj ( w ), at which phase angle is -50c f PM 80 c + + Gj ( w g ) c g Sol. 9 Sol. 0 Hence (A) is correct option. We know that steady state error is given by sr() s e ss lim s " 0 + Gs () where Rs () " input Gs ()" open loop transfer function For unit step input Rs () s s b s l So e ss lim 0. s " 0 + Gs () + G( 0) 0 G() 0 9 Given input rt () 0[ m( t) -m( t-)] or Rs () 0 : s - s e -s D 0 s e s - - : D So steady state error ( e ) s 0 -s - # el ss lim s s " 0 + Gs () 0 0( - e ) Hence (A) is correct option. The compensating coil compensation the effect of impedance of current coil. Hence (A) is correct option. Sol.

20 GATE SOLVED PAPER - EE 0 Sol. So, it will act as a Band pass filter. Hence (D) is correct option. The first half of the circuit is a differential amplifier (negative feedback) V a -( V i ) Second op-amp has a positive feedback, so it acts as an schmitt trigger. Since V a - V i this is a non-inverting schmitt trigger. Threshold value V TH 6V -6V V TL

21 GATE SOLVED PAPER - EE 0 Sol. Hence (D) is correct option. Only option C allow bi direction power flow from source to the drive Hence (C) is correct option. Sol. 4 Z is Z 0 where q is around 45c or so. ` Z Z 45c where Z < Y -45c Z Z 45c Z Y > [ a Z < ] Sol. 5 So Y will be out of unity circle. Hence D is correct option. Voltage in time domain vt () 00 cos( 00pt) Current in time domain it () 0 sin( 00pt + p/ 4) by applying the following trigonometric identity sin( f ) cos( f -90c) So, it () 0 cos(00 pt + p/4 -p/)

22 GATE SOLVED PAPER - EE 0 Sol. 6 0 cos(00 pt -p/4) In phasor form ` I 0 -p/ 4 Hence (D) is correct option. Sol. 7 Power transferred to the load P I R L 0 br R R L th + l L For maximum power transfer R th, should be minimum. R th R R R 0 Note: Since load resistance is constant so we choose a minimum value of R th Hence (A) is correct option. xt () e -t Laplace transformation Xs () s + yt () e -t Ys () s + Convolution in time domain is equivalent to multiplication in frequency domain. zt () xt ()) yt () Zs () XsYs () () bs + lbs + l by partial fraction and taking inverse laplace transformation, we get Zs () - s + s + t ` zt () e -e - -t Sol. 8 Hence (A) is correct option. An air-core transformer has linear B -H characteristics, which implies that magnetizing current characteristic will be perfectly sinusoidal. Hence (C) is correct option.

23 GATE SOLVED PAPER - EE 0 Sol. 9 Sol. 0 Sol. Negative phase sequence relay is used for the protection of alternators against unbalanced loading that may arise due to phase-to-phase faults. Hence (A) is correct option. Steady state stability or power transfer capability E V P max X To improve steady state limit, reactance X should be reduced. The stability may be increased by using two parallel lines. Series capacitor can also be used to get a better regulation and to increase the stability limit by decreasing reactance. Hence (C) is correct option. Transfer function having at least one zero or pole in RHS of s-plane is called nonminimum phase transfer function. Gs () s - ( s + )( s + ) In the given transfer function one zero is located at s (RHS), so this is a nonminimum phase system. Poles -, -, are in left side of the complex plane, So the system is stable Hence (B) is correct option. Sol. Let Z R jwc so admittance Y + jwc Z R Z R Z 4 R 4 Let Z X R + jwl (Unknown impedance) X For current balance condition of the Bridge ZZ 4 Z Z ZX X Y Let Z X ZZY R X + jwl X 4 X RR 4b + jwc R l by equating imaginary and real parts R X RR 4 R and L X RRC 4 Quality factor of inductance which is being measured Q R L X w wrc X From above equation we can see that for measuring high values of Q we need a large value of resistance R 4 which is not suitable. This bridge is used for measuring low Q coils.r Note: We can observe directly that this is a maxwell s bridge which is suitable for low values of Q (i.e. Q < 0) Hence (C) is correct option.

24 GATE SOLVED PAPER - EE 0 Sol. In the alternate mode it switches between channel A and channel B, letting each through for one cycle of horizontal sweep as shown in the figure. Sol. 4 Hence (C) is correct option. Y X5 X XX+ XX XX + X X X+ X Sol. 5 Sol. 6 Hence (A) is correct option. Once the SCR start conducting by an forward current, the gate has no control on it and the device can be brought back to the blocking state only by reducing the forward current to a level below that of holding current. This process of turn-off is called commutation. This time is known as the circuit turn-off time of an SCR. Hence (C) is correct option. f 0x sin x -0. 8

25 GATE SOLVED PAPER - EE 0 f 0x -0x cos x -0.6 Jacobian matrix is given by R V S f f x x 0x cos x J f W S W S f W > 0x sin x S x xw T X for x 0, x 0 sin x 0x - 0 cos x H J > H Hence (B) is correct option. Sol. 7 Sol. 8 fx () x- x + fl () x - x 0 x fm () x - fm () x is negative for x, so the function has a maxima at x. Hence (C) is correct option. ( ) Power loss Vrated 5 0 # 0W R 6 p 5. # 0 For an parallel combination of resistance and capacitor tan d wcr p p p # 50 #. 5 # Hence (C) is correct option. Sol. 9 L ft () F () s L - ft ( - t) e s t F () s F () s Gs () ) F() s F () s F () s -st ) e F() s F () s F () s

26 GATE SOLVED PAPER - EE 0 e -se e -st F () s F () s Taking inverse laplace transform - -st gt () L [ e ] d( t-t) ) " a F() s F () s F() s Hence (D) is correct option. Sol. 0 Let a signal px () is uniformly distributed between limits - a to + a. Sol. Variance s x p() x dx p -# -# a a a a x : a dx a x : D - a a a 6 It means square value is equal to its variance a p a rms sp p a rms Hence (A) is correct option. Initially E b V-I a R a 0 - # 0 0 V Now the flux is reduced by 0% keeping the torque to be constant, so the current will be f I a f I a I a E b f I f ` f 0.9f a 0 #. 09. A \ Nf & Eb N f E N f 0.9 N N b E b 0.9E V b #

27 GATE SOLVED PAPER - EE 0 Now adding a series resistor R in the armature resistor, we have E b V- I ( R + R) a ( + R) R.79 W a Sol. Sol. Hence (A) is correct option. We know that loss \ P G loss \ length Distance of load from G is 5 km Distance of load from G& G is 75 km generally we supply load from nearest generator. So maximum of load should be supplied from G. But G& G should be operated at same minimum generation. Hence (A) is correct option. Ks b + l Gs () s ( s+ ) Steps for plotting the root-locus () Root loci starts at s 0, s 0 and s - () n > m, therefore, number of branches of root locus b () Angle of asymptotes is given by ( q + ) 80c, q 0, n- m (I) (II) ( # 0 + ) 80c 90c ( - ) ( # + ) 80c 70c ( - ) (4) The two asymptotes intersect on real axis at centroid / Poles - / Zeroes x n- m - - b- l - - (5) Between two open-loop poles s 0 and s - there exist a break away point. s ( s+ ) K - s + b l dk 0 ds s 0 Root locus is shown in the figure

28 GATE SOLVED PAPER - EE 0 Sol. 4 Three roots with nearly equal parts exist on the left half of s-plane. Hence (A) is correct option. LXI D, DISP LP : CALL SUB LP + When CALL SUB is executed LP+ value is pushed(inserted) in the stack. POP H & HL LP + DAD D & HL HL + DE LP + + DE PUSH H & The last two value of the stack will be HL value i.e, LP + DISP + Sol. 5 Hence (C) is correct option. Zener Diode is used as stabilizer. The circuit is assumed to be as Sol. 6 We can see that both BE and BC Junction are forwarded biased. So the BJT is operating in saturation. Collector current I C ma k. Note:- In saturation mode IC -Y bib Hence (D) is correct option. Maximum current through main thyristor I M ( max) I + V s C L 0. # 0 # 0-6

29 GATE SOLVED PAPER - EE 0 Sol. 7 Sol. 8 Sol. 9 A Maximum current through auxiliary thyristor I A ( max) I0 0 A Hence (A) is correct option. We know that matrix A is equal to product of lower triangular matrix L and upper triangular matrix U. A 6L@ 6U@ only option (D) satisfies the above relation. Hence (D) is correct option. Let the given two vectors are X [,, ] X [, aa, ] Dot product of the vectors R V S W T X$ X XX 8BSa W + a+ a Sa W T X Where a - + j / -p so, + a+ a 0 X, X are orthogonal Note: We can see that X, X are not orthonormal as their magnitude is! Hence (B) is correct option. The steady state speed of magnetic field n s 0 # rpm 6 Speed of rotor n r ( - Sn ) s S 005. n r 0.95 # rpm In the steady state both the rotor and stator magnetic fields rotate in synchronism, so the speed of rotor field with respect to stator field would be zero. Speed of rotor which respect to stator field nr-ns rpm None of the option matches the correct answer.

30 GATE SOLVED PAPER - EE 0 Sol. 40 Charge Q 0 r CV ` C eea d 0 r eea V d ( eea) V d 0 r Q Q max Sol. 4 Sol. 4 Maximum electrical charge on the capacitor when V V 50 kv/ cm d b d l Thus, max - Q 8.85 # 0 #.6 # 0 # 40 # 50 # 0 8mC Hence (C) is correct option. t ht () e + e - -t 4 Laplace transform of ht () i.e. the transfer function Hs () + s + s + For unit step inputinput rt () m() t or Rs () s Output Ys () RsHs () () s: s+ + s+ D By partial fraction Ys () - - s s + bs + l Taking inverse laplace e u() t yt () ut () e ut () -t -t ut () 5. e t -t Hence (C) is correct option. Power angle for salient pole alternator is given by Vtsin f + IaXq tan d Vtcos f + IaRa Since the alternator is delivering at rated kva rated voltage pu I a

31 GATE SOLVED PAPER - EE 0 V t pu f 0c sin f 0, cosf X q pu,. pu X d () tan d # d 45c Hence B is correct option. Sol. 4 Sol digit display will read from to So error of 0 counts is equal to! 0.0 V For 00 V, the maximum error is e! ( 00 # )! 0. V Percentage error! % 00 #! 0. % of reading Hence (C) is correct option. The admittance diagram is shown below here y 0-0j, y - 5j, y. 5j, y -0j 0 Note: y is taken positive because it is capacitive. Y y0 + y -0j- 5j -5j Y Y - y 5j Y Y - y 0 Y y0 + y + y 0 + (- 5j) + (. 5j) 75. j Y Y - y -. 5j Y y0 + y + y

32 GATE SOLVED PAPER - EE 0-0j j 5. j Sol. 45 Sol. 46 So the admittance matrix is R Y Y Y V S W Y SY Y YW SY Y YW R T X S -5j 5j S 5j 75. j S j Hence (B) is correct T option. 0 V W -. 5jW 5. j W X The system may be reduced as shown below Ys () Rs () ss ( + + K) + s + s( + K) + ss ( + + K) This is a second order system transfer function, characteristic equation is s + s( + K) + 0 Comparing with standard form We get s x Peak overshoot M p + xwns+ wn 0 + K e - px/ - x So the Peak overshoot is effected by K. Hence (A) is correct option. The characteristics equation of the JK flip-flop is Q n + JQ + KQ Q n+ is the next state n n From figure it is clear that J Q ; K Q The output of JK flip flop Q An ( + ) Q Q + Q Q Q ( Q + Q ) B Q B Output of T flip-flop B B A B A B A A

33 GATE SOLVED PAPER - EE 0 Q B( n+ ) Q A Clock pulse Q A Q B Q An ( ) + Q Bn ( + ) Initially(t n ) 0 0 tn tn tn Sol. 47 Hence (C) is correct option. We can obtain three operating regions depending on whether the Zener and PN diodes are forward biased or reversed biased.. vi #- 0.7 V, zener diode becomes forward biased and diode D will be off so the equivalent circuit looks like The output v o -0.7 V. When v # 57., both zener and diode D will be off. The circuit is i Output follows input i.e vo vi Note that zener goes in reverse breakdown(i.e acts as a constant battery) only when difference between its p-n junction voltages exceeds 0 V.. When vi > 5.7 V, the diode D will be forward biased and zener remains off, the equivalent circuit is v o V Hence (C) is correct option.

34 GATE SOLVED PAPER - EE 0 Sol. 48 v i 00 sin(00 pt) V Fundamental component of current i i 0 sin(00 pt -p/) A Input power factor pf I ( rms ) I rms cos f I ( rms ) is rms values of fundamental component of current and I rms is the rms value of converter current. Sol. 49 Sol. 50 Sol. 5 pf 0 cos p/ Hence (C) is correct option. Only the fundamental component of current contributes to the mean ac input power. The power due to the harmonic components of current is zero. So, P in V I cos f rms rms 00 # 0 cos p/ 500 W Hence (B) is correct option. Power delivered by the source will be equal to power dissipated by the resistor. P VIcos / 4 s s p # cos p/ 4 W Hence (B) is correct option. I C Is - IRL p/ 4 - -p/ 4 I C $ ^cos p/ 4+ jsin p/ 4h-^cos p/ 4-jsin p/ 4h. j sin p/ 4 j Hence (D) is correct option. Sol. 5 For generator G Xm pu G # 50 For generator G Xm G # 0. pu 00 X L X W L # For transmission lines L and L Z kv base base 5 # 5.5 W MVAbase 00 ` Xm L ( pu) 5. pu 5.

35 GATE SOLVED PAPER - EE 0 Xm L ( pu) 5. pu 5. So the equivalent pu reactance diagram Sol. 5 Sol. 54 Hence (A) is correct option. We can see that at the bus, equivalent thevenin s impedance is given by X th ^j0. + j0. h ^j0. + j0. h j. j. j0.55 pu Fault MVA Base MVA Xth MVA 055. Hence (D) is correct option. Output voltage of -phase bridge converter V 0 V p Maximum output ( ) ph V 0 max Vph cos a cos a p 400 p # # V Resistance of filter choke is 0 W, So ( V 0 ) max E+ IR chock I( 0) I - 4 A Hence A is correct option. Sol. 55 Sol. 56 kva rating VI L L # # p #4 7.5 kva Hence (D) is correct option. Let us assume total voters are 00. Thus 40 voter (i.e. 40 %) promised to vote for P and 60 (rest 60 % ) promised to vote fore Q. Now, 5% changed from P to Q (5 % out of 40)

36 GATE SOLVED PAPER - EE 0 Sol. 57 Sol. 58 Sol. 59 Sol. 60 Sol. 6 Sol. 6 Changed voter from P to Q # Now Voter for P Also, 5% changed form Q to P (out of 60%) Changed voter from Q to P # Now Voter for P Thus P P got 49 votes and Q got 5 votes, and P lost by votes, which is given. Therefore 00 voter is true value. Hence (A) is correct option. The sentence implies that technocrats are counterproductive (negative). Only (C) can bring the same meaning. Hence (C) is correct option. Periodicity is almost similar to frequency. Gradualness means something happening with time. Persistency is endurance. Rarity is opposite to frequency. Hence (B) is correct option. Available is appropriate because manipulation of genes will be done when other treatments are not useful. Hence (D) is correct option. A gladiator performs in an arena. Commutators use trains. Lawyers performs, but do not entertain like a gladiator. Similarly, teachers educate. Only dancers performs on a stage. Hence (A) is correct option. Since fuel consumption/litre is asked and not total fuel consumed, only average speed is relevant. Maximum efficiency comes at 45 km/hr, So least fuel consumer per litre in lap Q Hence (B) is correct option. Let total no of toffees be x. The following table shows the all calculations. Friend Bowl Status R x - 4 x + 4 S x 4: + 4 D- x + 4- x + 6 x + - x x + 6 T Now, or x a + 6 k- x + 6- x - 4 x + 4 x x x x # 4 48 Hence (C) is correct option.

37 GATE SOLVED PAPER - EE 0 Sol. 6 Sol. 64 Sol. 65 fy () y y y Now f( - y) - -f() y y or fq ()-f(- q) fq () Hence (D) is correct option ( ) Hence (C) is correct option. (...) [( ) ( )...] [ ( ) n] n 9 4 n 0 : # D n 8 4 n Option B fits the sentence, as they built up immunities which helped humans create serums from their blood. Hence (B) is correct option.

38 GATE SOLVED PAPER - EE 0 Answer Sheet. (D). (C) 5. (C) 7. (D) 49. (B) 6. (A). (A) 4. (D) 6. (B) 8. (B) 50. (B) 6. (C). (b) 5. (D) 7. (C) 9. (*) 5. (D) 6. (D) 4. (d) 6. (A) 8. (C) 40. (C) 5. (A) 64. (C) 5. (A) 7. (A) 9. (D) 4. (C) 5. (D) 65. (B) 6. (B) 8. (C) 0. (A) 4. (B) 54. (A) 7. (A) 9. (B). (A) 4. (C) 55. (D) 8. (A) 0. (A). (A) 44. (B) 56. (A) 9. (A). (C). (A) 45. (A) 57. (C) 0. (A). (C) 4. (C) 46. (C) 58. (B). (D). (C) 5. (D) 47. (C) 59. (D). (D) 4. (A) 6. (A) 48. (C) 60. (D) *******