Q. No. 1 5 Carry One Mark Each

Transcription

1 EE-GATE-4 PAPE- Q. No. 5 Carry One Mark Each. Choose the most appropriate phrase from the options given below to complete the following sentence. India is a post-colonial country because (A) it was a former British colony (B) Indian Information Technology professionals have colonized the world (C) India does not follow any colonial practices (D) India has helped other countries gain freedom Answer: (A). Who was coming to see us this evening? (A) you said (B) did you say (C) did you say that (D) had you said Answer: (B) 3. Match the columns. Column Column () eradicate (P) misrepresent () distort (Q) soak completely (3) saturate () use (4) utilize (S) destroy utterly (A) :S, :P, 3:Q, 4: (B) :P, :Q, 3:, 4:S (C) :Q, :, 3:S, 4:P (D) :S, :P, 3:, 4:Q Answer: (A) 4. What is the average of all multiples of from to 98? (A) 9 (B) (C) (D) Answer: (B) : 9 [() 9 + ] 9 : The value of is (A) (B) 3.93 (C) 4. (D) India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India

2 EE-GATE-4 PAPE- Answer: (C) let y + y y + y y (y 4)(y+ 3) y 4, y 3 Q.No. 6 Carry Two Marks Each 6. The old city of Koenigsberg, which had a German majority population before World War, is now called Kaliningrad. After the events of the war, Kaliningrad is now a ussian territory and has a predominantly ussian population. It is bordered by the Baltic Sea on the north and the countries of Poland to the south and west and Lithuania to the east respectively. Which of the statements below can be inferred from this passage? Answer: (B) (A) Kaliningrad was historically ussian in its ethnic make up (B) Kaliningrad is a part of ussia despite it not being contiguous with the rest of ussia (C) Koenigsberg was renamed Kaliningrad, as that was its original ussian name (D) Poland and Lithuania are on the route from Kaliningrad to the rest of ussia 7. The number of people diagnosed with dengue fever (contracted from the bite of a mosquito) in north India is twice the number diagnosed last year. Municipal authorities have concluded that measures to control the mosquito population have failed in this region. Answer: (D) Which one of the following statements, if true, does not contradict this conclusion? (A) A high proportion of the affected population has returned from neighbouring countries where dengue is prevalent (B) More cases of dengue are now reported because of an increase in the Municipal Office s administrative efficiency (C) Many more cases of dengue are being diagnosed this year since the introduction of a new and effective diagnostic test (D) The number of people with malarial fever (also contracted from mosquito bites) has increased this year 8. If x is real and x x + 3, then possible values of 3 x + x x include (A), 4 (B), 4 (C) 4, 5 (D) 4, 5 India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India

3 Answer: (D) x x + 3 (x 4)(x+ ) x 4, x 3 Values of x + x x For x 4 Value 5 for x Value 4 Option D 4,5 EE-GATE-4 PAPE The ratio of male to female students in a college for five years is plotted in the following line graph. If the number of female students doubled in 9, by what percent did the number of male students increase in 9? a tio of m al e to fe m al Answer: 4% m 3 f m.5 f m.5f m' 3 f m' 6f m' m m 3.5f %.5f % 4% India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India 3

4 EE-GATE-4 PAPE- At what time between 6 a.m. and 7 a.m will the minute hand and hour hand of a clock make an angle closest to 6? (A) 6: a. m. (B) 6:7 a.m. (C) 6: 38 a.m. (D) 6:45 a.m. Answer: (A) Angle by minute s hand 6 min 36 ο 36 min 6 6 o 8min 48 o Angle 48 with number 6 Angle by hours hand 6 min 3 3 min 6 o Total Angle48+59 o. 6.am ο India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India 4

5 EE-GATE-4 PAPE- Q.No. 5 Carry One Mark Each. Which one of the following statements is true for all real symmetric matrices? (A) All the eigenvalues are real. (B) All the eigenvalues are positive (C) All the eigenvalues are distinct (D) Sum of all the eigenvalues is zero. Answer: (A) Eigen values of a real symmetric matrix are all real. Consider a dice with the property that the probability of a face with n dots showing up is proportional to n. The probability of the face with three dots showing up is. Answer: /7 P( n) k.n where n to 6 x [ ] we know P x K K required probability is P( 3) 3K 7 3. Maximum of the real valued function 3 f x x occurs at x equal to Answer: ( A) ( B) ( C) ( D) (C) f ( x) ( ) is negative, x < or x in h, 3 x 3 is positive, x > or x in, + h h is positive & small f has local minima at x and the minimum value is '' 4. All the values of the multi-valued complex function i, where i, are Answer: (B) (A) purely imaginary (C) on the unit circle. cos( kπ ) + i sin ( kπ ) where k is int eger e i( kπ) i ( kπ) e All values are real and non negative (B) real and non-negative (D) equal in real and imaginary parts India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India 5

6 EE-GATE-4 PAPE- d y dy 5. Consider the differential equation x + x y. Which of the following is a solution dx dx to this differential equation for x >? (A) Answer: (C) x e (B) d y dy x + x y is cauchy Euler equation dx dx d ( θ ).y where θ and z log x, x e dz x (C) /x (D) ln x z A.E : m m, Z Z C Solution is y Ce + Ce + Cx x is a solution x 6. Two identical coupled inductors are connected in series. The measured inductances for the two possible series connections are 38µ H and 4µ H. Their mutual inductance in µ H is Answer: 35µ H Two possible series connections are. Aiding then L equation L + L + M.. Opposing then L equation L + L M L + L + M 38 H L + L M 4 H From &, M 35µ H () () 7. The switch SW shown in the circuit is kept at position for a long duration. At t +, the switch is moved to position Assuming V > V, the voltage VC ( t ) across capacitor is '' SW V V C VC t/c t /C ( A) vc ( t) V ( e ) V B v t V e + V c t/c t/c ( C) vc ( t) ( V + V )( e ) V D v t V + V e + V c India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India 6

7 EE-GATE-4 PAPE- Answer: (D) When switching is in position ϑ C t z V t Initial final + final value VC ( t) V e t C When switch is in position Initial value is VC ( t) V e Final value is V t C e V C ϑ c VC ( t) V [ V V ] e t C C ϑ c 8. A parallel plate capacitor consisting two dielectric materials is shown in the figure. The middle dielectric slab is place symmetrically with respect to the plates. Volt ε ε ε Answer: If the potential difference between one of the plates and the nearest surface of dielectric interface is Volts, then the ratio ε : ε is (A) : 4 (B) : 3 (C) 3: (D) 4 : Q (C) CV C V cons tan t C V Aε C d ε ε V V ε V ε ε V V 6 ε ε + ε ε + ε V ε + ε ε ε ε : ε 3: ε 3 d / d V 8V V V ε ε ε India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India 7

8 9. Consider an LTI system with transfer function EE-GATE-4 PAPE- H`( s) s s 4 ( + ) If the input to the system is cos(3t) and the steady state output is Asin( 3t + α ), then the value of A is (A) /3 (B) /5 (C) 3/4 (D) 4/3 Answer: (B) Given H( s) s s 4 ( + ) H( jω ) ω ω +6 cos ( ω t) ( ω) A H j ω 3 H( jω) ωω A ( ω) ( ω + θ) y t H j cos t ωω ( ω) where θ H j ω 5t. Consider an LTI system with impulse response h(t) e u( t) t 5t y( t) e u( t) e u( t) then the input, x(t), is given by 3t ( A) e u( t) 3t B e u ( t ) 5t C e u ( t ) 5t D e u ( t ) Answer: (B) x ( t) 5t h ( t) e u( t) H( s) s + 5 3t 5t y( t) e e u( t) Y( s) s + 3 s + 5 Y s H( s) X s 3t h ( t ) y ( t) ( + ) ( + ) ( 5 3)( s 5) Y s 5 5 s 3 X( s) H s + + s + 3 s + 5 x t e u t. If the output of the system is India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India 8

9 EE-GATE-4 PAPE- Assuming an ideal transformer,. The Thevenin s equivalent voltage and impedance as seen from the terminals x and y for the circuit in figure are sin ( ωt) Ω x ( A) sin ( ωt ), 4Ω ( ω ) ( C) sin ( ωt ), Ω ( ω ) Answer: A ϑ xy Voc ϑ ϑ in xy ϑ ϑ ω xy 4 ϑ th sin ω t th 4Ω xy oc sin t : B sin t, Ω D sin t,.5ω. A single phase, 5kVA, V/V two winding transformer is connected as an autotransformer as shown in the figure. y V V V The kva rating of the autotransformer is. Answer: 55kVA Given, V 5 kva, V 3 5 I 5 kva 5 55 kva A.TFr V I 5A V India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India 9

10 EE-GATE-4 PAPE A three-phase, 4pole, self excited induction generator is feeding power to a load at a frequency f. If the load is partially removed, the frequency becomes f. If the speed of the generator is maintained at 5 rpm in both the cases, then Answer: (C) (A) f f > 5Hz and f > f (B) f < 5Hz and f > 5Hz (C) f f < 5Hz and f > f (D) f > 5Hz and f < 5Hz Initially self excited generator supply power to a load at f. If load is partially removed then slightly speed increase, also frequency f f > f But both cases f f < 5Hz 4. A single phase induction motor draws MW power at.6 lagging power. A capacitor is connected in parallel to the motor to improve the power factor of the combination of motor and capacitor to.8 lagging. Assuming that the real and reactive power drawn by the motor remains same as before, the reactive power delivered by the capacitor in MVA is. Answer: 7MVA Given, φ Induction motor draws mw at.6pf, lag Let P mw cosφ.6 pf To improve pf, cosφ.8 Q? c del bycapacitor P cosφ S MVA 6 S S.6 4 eactive power, Q S P 6 MVA When capacitor is connected then cosφ P S.8 S S 6 5MVA ( eal power drawn issame) e active power, Q S P 9MVA But motor should draw the same reactive power. Q 6 9 7MVA c delby capacitor India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India

11 EE-GATE-4 PAPE A three phase star-connected load is drawing power at a voltage of.9 pu and.8 power factor lagging. The three phase base power and base current are MVA and A respectively. The line-to line load voltage in kv is. Answer: 7- Given, mva, A V L L (kv)? We know that, S 3 V L.IL 6 3.V L.IL V V L L kV But it is drawing power at a voltage of.9 pu V Vpu V actual Base V V V V actual L L pu B kV 6. Shunt reactors are sometimes used in high voltage transmission system to (A) limit the short circuit current through the line. (B) compensate for the series reactance of the line under heavily loaded condition. (C) limit over-voltages at the load side under lightly loaded condition. (D) compensate for the voltage drop in the line under heavily loaded condition. Answer: (C) 7. The closed-loop transfer function of a system is due to unit step input is. Answer: Steady state error for Type- for unit step input is. 4 T s. ( s +.4s + 4) The steady state error 8. The state transition matrix for the system x x u x + x is t e A e t e t ( B) t e e t e t t ( C) te t e t t e ( D) t t e te t e Answer: (C) India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India

12 EE-GATE-4 PAPE- Given A B s s [ SI A] ( s ) ( s ) ( s ) [ SI A] The state transition matrix At e L SI A e At t e t t te e 9. The saw-tooth voltage wave form shown in the figure is fed to a moving iron voltmeter. Its reading would be close to V ms 4ms Answer: V m sec 4 m sec t Moving iron meter reads MS value only MS value of saw-tooth waveform is Meter reads volts ϑmax 3 India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India

13 EE-GATE-4 PAPE- While measuring power of a three-phase balanced load by the two-wattmeter method, the readings are W and 5 W. The power factor of the load is. Answer:.8 In two-wattmeter method, The readings are W & 5 W Power factor cos φ 3 ω ω cos tan ω + ω 3 5 cos tan Which of the following is an invalid state in an Binary Coded Decimal counter Answer: (D) (A) (B) (C) (D) In binary coded decimal (BCD) counter the valid states are from to 9 only in binary system to only. So, in decimal it is which is invalid state in BCD counter.. The transistor in the given circuit should always be in active region. Take VCE ( sat).v. VEE.7 V. The maximum value of C in Ω which can be used is. s kω 5V + C β + 5V Answer:.3Ω I B C 5.7.5mA k I.5A 5. C.3Ω.5 India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India 3

14 EE-GATE-4 PAPE A sinusoidal ac source in the figure has an rms value of V. Considering all possible values of L, the minimum value of S in Ω to avoid burnout of the Zener diode is. S V ~ 5V L / 4W Answer: 3Ω Vm V Pz Pz Vz Iz Iz 5mA V 5 S ( min) 3Ω 5mA z 4. A step-up chopper is used to feed a load at 4 V dc from a 5 V dc source. The inductor current is continuous. If the off time of the switch is µ s, the switching frequency of the chopper is khz is. Answer: 3.5 khz V 4v, Vs 5 v, Toff µ sec, F? Given chopper in step up chopper Vs Vo D D D 4 D but T D T off ( 3 ) 6 T T 3µ sec 8 Then f 3.5 Hz 6 T 3 f 3.5kHz India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India 4

15 EE-GATE-4 PAPE In a constant V/f control of induction motor, the ratio V/f is maintained constant from to base frequency, where V is the voltage applied to the motor at fundamental frequency f. Which of the following statements relating to low frequency operation of the motor is TUE? Answer: (B) (A) At low frequency, the stator flux increases from its rated value. (B) At low frequency, the stator flux decreases from its rated value. (C) At low frequency, the motor saturates. (D) At low frequency, the stator flux remains unchanged at its rated value. During constant V control, at low frequency, the voltage also applied to the induction motor f is low. Hence the stator flux also decreases from its rated value. Q.No Carry Two Marks Each 8 ( y/) + x y 6. To evaluate the double integral dx dy y/, we make the substitution x y u and y v. The integral will reduce to Answer: (B) 4 4 ( A) ( u du ) dv ( B) ( ) 4 4 ( C) ( u du ) dv ( D) ( ) x y y u... and V... y y x u ; x + u y v ; y 8 v 4 + from and,x u v... 3 and y v... 4 x x u v Jacobian transformation; J y y u v y x y dx dy ( u ) J du dv y v u 4 u du dv u du dv u du dv India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India 5

16 EE-GATE-4 PAPE Let X be a random variable with probability density function Answer:.4., for x f ( x)., for < x 4, otherwise The probability p(.5 < x < 5) is 5 P.5 < x < 5 f x dx Opposite f ( x) dx + f ( x) dx f ( x) dx Hypotenuse 4. x +. x The minimum value of the function f ( x) x 3x 4x + in the interval [-3. 3] is Answer: (B) (A) (B) 8 (C) 6 (D) 3 f x x x 8 x, 4 [ 3,3] ( ) ( ) Now f 3 8 ; f 3 8 and f 8 ; f 4 44 f x is min imum at x 3 and the min imum value is f Assuming the diodes to be ideal in the figure, for the output to be clipped, the input voltage v i must be outside the range k V i ~ k V V V o Answer: (B) (A) V to V (B) V to 4V (C) + V to V (D) + V to 4V vi When both diodes are FF, vo (Not clipped). For the clipped, vi must bt ouside the range V to 4V India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India 6

17 EE-GATE-4 PAPE The voltage across the capacitor, as sown in the figure, is expressed as v t A sin ω t θ + A sin ω t θ t Ω H sint ~ VC ( t) F sin 5t The value of A and A respectively, are (A). and.98 (B). and 4. (C).5 and 3.5 (D) 5. and 6.4 Answer: (A) By using super position theorem,. ϑ ( t) C When sin t voltage source is acting, Network function H( j ) ϑ c. ϑ ( t) c jωc ω + j jωc ( t) sin t tan ( + ) When sin 5t current source is acting ϑ c.j.j ϑ c j.j ϑ ( t ).sin 5t θ +. c ϑ c ( t).98 sin ( 5t θ ) ( θ ) + ( θ ) V t sin t.98 5t C A. By comparing with given expression, A The total power dissipated in the circuit, show in the figure, is kw. A A Ω Xc XL ac source ~ X c V V Load The voltmeter, across the load, reads V. The value of X L is. India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India 7

18 Answer: 7.34 Ω Total power dissipated in the circuit is kw. P kw I.+ I Ω V Z I Z + X L X Z L L X 9.96 EE-GATE-4 PAPE- XL 7.34 Ω 3. The magnitude of magnetic flux density ( B) at a point having normal distance d meters from µ an infinitely extended wire carrying current of l A is I (in SI units). An infinitely nd extended wire is laid along the x-axis and is carrying current of 4 A in the +ve x direction. Another infinitely extended wire is laid along the y-axis and is carrying A current in the +ve y direction µ is permeability of free space Assume ˆ I, J, ˆ K ˆ to be unit vectors along x, y and z axes respectively. y I amps d µ I B πd A 4A (,, ) z Assuming right handed coordinate system, magnetic field intensity, H at coordinate (,,) will be 3 ˆ π ( A) k weber / m ( C) 3 k ˆ A / m π ( B) ( D) 4 ˆ i A / m 3π A / m India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India 8

19 EE-GATE-4 PAPE- Answer: (C) H Hx + Hy I 4 H aφ a a a πρ π π x x y z I H aφ a a a πρ π π y y x z 3 H a π π z 33. A discrete system is represented by the difference equation It has initial condition X k + a a X k X k + a + a X k X ; X. The pole location of the system for a, are (A) ± j (B) ± j (C) ± + j (D) ± j Answer: (A) from the given difference equation, a a A a + a The pole locations of the system for a. Then A SI A. ( s ) S ± j 34. An input signal x ( t) + 5sin ( πt) is sampled with a sampling frequency of 4 Hz and applied to the system whose transfer function is represented by N Y z z X z N z where, N represents the number of samples per cycle. The output y(n) of the system under steady state is (A) (B) (C) (D) 5 India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India 9

20 Answer: (C) x( t) + 5sin( π t) x n Ts + 5sin π n. 4 π + 5sin n, N 8 4 Ω jω j Y e jωn e H e jω X e N e [ ] [ ] + [ ] [ n] x n x n x n due to x jω [ ] [ ] y n H e x n r [ ] y n EE-GATE-4 PAPE- jω π y n H e sin n H e 4 jω jω [ ] π + jω H e π Ω 4 [ ] [ ] + [ ] [ ] y n y n y n y n Ω π 4 Ω 4 Thus at steadystate y[n] 35. A khz even-symmetric square wave is passed through a bandpass filter with centre frequency at 3 khz and 3 db passband of 6 khz. The filter output is (A) a highly attenuated square wave at khz (B) nearly zero. (C) a nearly perfect cosine wave at 3kHz. (D) a nearly perfect sine wave at 3kHz. Answer: (C) KHz even symmetric square wave have frequency component present KHz, 3KHz, 5KHz, 7KHz [only odd harmonics due to half wave symmetry] Since bandpass filter is contered at 3KHz, 3KHz component will pass through filter output is nearly perfect cosine wave at KHz Cosine in due to reason that signal in even signal. 36. A 5 V dc shunt machine has armature circuit resistance of.6ω and field circuit resistance of5 Ω. The machine is connected to 5 V supply mains. The motor is operated as a generator and then as a motor separately. The line current of the machine in both the cases is 5 A. The ratio of the speed as a generator to the speed as a motor is. India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India

21 EE-GATE-4 PAPE- Answer:.7 Given: A 5A A 5A 48A 5A 5Ω 5V 5Ω 5V E V I Ng? N m b a a V N g g We know that ( flux is constant ) N Motor m E E b 8. Ng.7. N m Generator E V + I b a a V 37. A three-phase slip-ring induction motor, provided with a commutator winding, is shown in the figure. The motor rotates in clockwise direction when the rotor windings are closed. 3 phase ac, fhz Pr ime mover Slip ing Induction Motor f f r If the rotor winding is open circuited and the system is made to run at rotational speed f r with the help of prime-mover in anti-clockwise direction, then the frequency of voltage across slip rings is f and frequency of voltage across commutator brushes is f. The values of f and f respectively are (A) f + f r and f (B) f - f r and f (A) f - f r and f+ f r (D) f - f r and f Answer: (A) Whenever the otor winding is open circuited and rotating in anti-clockwise direction then the ( Ns + Nr ) P frequency of voltage across slip rings is f f f + f r At the same time frequency of voltage across commutator brushes if f N P s f f India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India

22 EE-GATE-4 PAPE A -pole alternator is having 8 identical stator slots with 6 conductors in each slot. All the coils of a phase are in series. If the coils are connected to realize single-phase winding, the generated voltage is V. If the coils are reconnected to realize three-phase star-connected winding, the generated phase voltage is V. Assuming full pitch, single-layer winding, the ratio V / V is ( A) Answer: (D) Given poles, P 3 Total slots 8 4 Total no. of conductor ( B) ( C) 3 D 4 the ratio of voltage generated when the coils are connected in φ to when the coils are connected in 3 φ, Y-connection. ( V ) i.e., ( V ) φ 3 φ 39. For a single phase, two winding transformer, the supply frequency and voltage are both increased by %. The percentage changes in the hysteresis loss and eddy current loss, respectively, are Answer: (A) (A) and (B) - and (C) and (D) - and Given φ Transformer V and f are increased by % % W? e n % W? Here V is constant f W n n f as 'f ' increased by % W also % n W W n e f f W e e f e W.f W by % 4. A synchronous generator is connected to an infinite bus with excitation voltage Ef.3 pu. The generator has a synchronous reactance of. pu and is delivering real power (P) of.6 pu to the bus. Assume the infinite bus voltage to be. pu. Neglect stator resistance. The reactive power (Q) in pu supplied by the generator to the bus under this condition is. India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India

23 EE-GATE-4 PAPE- Answer:.9 Given, E f.3 P.u X s.p.u P. 6pu V. pu Q? We know that,.3.6 sin δ. δ 3.5 s EV P sin δ X V Q [ Ecosδ V].9 X s 4. There are two generators in a power system. No-load frequencies of the generators are 5.5 Hz and 5Hz, respectively, and both are having droop constant of Hz/MW. Total load in the system is.5 MW. Assuming that the generators are operating under their respective droop characteristics, the frequency of the power system in Hz in the steady state is. Answer: 5 Given, two generators in a power system has no load frequency of 5.5 & 5 Hz. drop constanthz/mw Total load.5 mw for generator '', f x generator '' f x + 5 x x + 5 x x.5...() Total load x + x.5...() 3 By solving () & () x.5 f Hz 4. The horizontally placed conductors of a single phase line operating at 5 Hz are having outside diameter of.6 cm, and the spacing between centers of the conductors is 6 m. The permittivity of free space is The capacitance to ground per kilometer of each line is (A) F (B) F (C) 4. - F (D) F India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India 3

24 Answer: (B) Given, diameters of conductor.6m radius, r.8cm Spacing between conductors, d6m Permitivity capacitance to ground per km? π π 8.85 d 6 ln ln r.8 o C C / km 8.4 F EE-GATE-4 PAPE A three phase, MVA, 5 kv generator has solidly grounded neutral. The positive, negative, and the zero sequence reactances of the generator are. pu,. pu, and.5 pu, respectively, at the machine base quantities. If a bolted single phase to ground fault occurs at the terminal of the unloaded generator, the fault current in amperes immediately after the fault is. Answer: 55 Single line to ground fault, Fault current in If 3Ia positive sub transient circuit, Ea Ia z + z + z + jo j. + j. + j.5 j.3pu j.45 Fault current Base current I 3 I f a 3 j. j6.666 pu pu Fault circuit pu fault circuit in pu x Base circuit in Amp If 5396A 44. A system with the open loop transfer function: G s K s s s s ( + )( + + ) is connected in a negative feedback configuration with a feedback gain of unity. For the closed loop system to be marginally stable, the value of K is India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India 4

25 Answer: 5 The characteristic equation + G(s) k + s s s s ( + )( + + ) 4 3 s + 4s + 6s + 4s + k H Arry: S S 4 3 S 4k S S 6 k k 5 k For marginally stable, 4k 4 k k 5 EE-GATE-4 PAPE For the transfer function G s ( + ) 5 S 4 s s.5 s 4s 5 ( + )( + + ) The values of the constant gain term and the highest corner frequency of the Bode plot respectively are (A) 3., 5. (B) 6., 4. (C) 3., 4. (D) 6., 5. Answer: (A) G( s) ( + ) 5 s 4 s s.5 s 4s 5 ( + )( + + ) If we convert it into time constants, G s s s 4 s s[.5] s s s 4 s s +.s G s Constant gain term is 3. ω n 5 highest corner frequency India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India 5

26 EE-GATE-4 PAPE The second order dynamic system dx PX + Qu dt y X has the matrices P, Q and as follows: P Q 3 [ ] The system has the following controllability and observability properties: (A) Controllable and observable (B) Not controllable but observable (C) Controllable but not observable (D) Not controllable and not observable Answer: (C) Given P Q 3 For controllability: QC [ Q PQ] 3 QC controllable For observability: T T T Q P. 3 Q Not observable. 47. Suppose that resistors and are connected in parallel to give an equivalent resistor. If resistors and have tolerance of % each., the equivalent resistor for resistors 3Ω and Ω will have tolerance of (A).5% (B) % (C).% (D) % Answer: (B) 5 ± % T + E T % T T T T ± + 3 ± % T 36.36Ω.....5; ± ± % India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India 6

27 EE-GATE-4 PAPE Two ammeters X and Y have resistances of.ω and.5 Ω respectively and they give full scale deflection with 5 ma and 5 ma respectively. The ranges have been extended by connecting shunts so as to give full scale deflection with 5 A. The ammeters along with shunts are connected in parallel and then placed in a circuit in which the total current flowing is 5A. The current in amperes indicated in ammeter X is. Answer:.57 X and Y ammeters are connected in parallel Shunt egistration of X and Y meters:. 5 5 shx 3 shx. Ω shy 3 shy.54ω Current through X ammeter is.54 5 (. +.54).57 ampers 5A shx. Ω Imx 5mA.5 Ω Imy 5 ma shy 49. An oscillator circuit using ideal op-amp and diodes is shown in the figure C + 3kΩ 5V + 5V Vo kω kω The time duration for +ve part of the cycle is t and for-ve part is t e t t C will be..the value of India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India 7

28 Answer:.3 + ( ) V t V V V e C max initial max t EE-GATE-4 PAPE- t τ where 5 τ UTP + Vsat + LTP Vsat e UTP t τ e 5 4 LTP e e.5 e t τ t.69 τ sat (.69τ+.98τ) τ t τ t τ LTP V + LTP + V e sat 5 5 t τ e (.5) e t τ t τ 7.5.5e e 3 t.98τ e t τ t τ 5. The SOP (sum of products) form of a Boolean function is Σ(,,3,7,), where inputs are A,B,C,D (A is MSB, and D is LSB). The equivalent minimized expression of the function is Answer: (A) ( A) ( B + C)( A + C) ( A + B)( C + D) ( B) ( B + C)( A + C) ( A + C)( C + D) ( C) ( B + C)( A + C) ( A + C)( C + D) ( D) ( B + C)( A + B) ( A + B)( C + D) CD AB ( C + D) ( B + C) ( A + B) ( A + C) The equivalent minimized expression of this function is ( B + C)( A + C)( A + B)( C + D) India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India 8

29 EE-GATE-4 PAPE A JK flip flop can be implemented by T flip-flops. Identify the correct implementation. ( A) J K Clk T T flip flop Q n ( B) J K Clk T T flip flop Q n Q n Q n ( C) J K Clk T T flip flop Q n ( D) J K Clk T T flip flop Q n Answer: (B) Q n Q n Q J K n Q n + T Q n JK T Q J + Q k n n Analysis: If you will observe the combinational circuit output expression which is the input for T flip flop is not matching directly, so you should go through the option. If you will solve the combinational circuit of option (B) then ( n ) ( n ) T J + Q. K + Q ( ) J.K + JQ + K.Q + Q Q J.K + JQ + K.Q + Q.Q J.K + JQ + K.Q n n n n n n n n n n Now, according to consensus theorem J-K will become redundant term, so it should be eliminated. Hence, T JQn + K.Qn, which in matching with our desired result and option-(b) is correct answer. India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India 9

30 EE-GATE-4 PAPE In an 885 microprocessor, the following program is executed Address location Instruction H XA A H MVI B,4H 3H MVI A, 3H 5H A 6H DC B 7H JNZ 5 AH HLT At the end of program, register A contains (A) 6H (B) 3H (C) 6H (D) 3H Answer: (A) Address location Instruction Operation H XA A [ A] H, CY, Z H MVI B, 4H [ B] 3H MVI A, 3H [ A] 4H 3H 5H A otate accumulator right with carry 6H DC B Decrement content of B register by one 7H JNZ 5H Jump on no zero to location 5H AH HLT Accumulator CY Initial: value A [ B] 3H A [ B] H India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India 3

31 EE-GATE-4 PAPE- A B H A [ B] H Nowloopwillbeover andhltwillexecuteand program will be over 53. A fully controlled converter bridge feeds a highly inductive load with ripple free load current. The input supply ( v ) to the bridge is a sinusoidal source. Triggering angle of the bridge converter is O α 3 s. The input power factor of the bridge is. + V ~ S i s Load Answer:.78 For fully controlled converter bridge The input power factor (PF).9 cosα IPF.9 cos3 IPF A single-phase SC based ac regulator is feeding power to a load consisting of 5Ω resistance and 6 mh inductance. The input supply is 3 V, 5 Hz ac. The maximum firing angle at which the voltage across the device becomes zero all throughout and the rms value of current through SC, under this operating condition, are Answer: (C) (A) 3 and 46 A (B) 3 and 3 A (C) 45 and 3 A (D) 45 and 3 A Vs 3V, 5Hz 5 Ω, L 6mH The maximum firing angle at which the volt across device becomes zero is the angle at which device trigger i.e. minimum firing angle to converter. India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India 3

32 EE-GATE-4 PAPE- XL α φ tan tan ωl 3 π 5 6 α φ tan The current flowing SC is max at their angle ie. when α φ, γ π π+ α Vm ITrms sin( ωt α).dωt π α I m Trms + Trms V 3 z I.9 3A 55. The SC in the circuit shown has a latching current of 4 ma. A gate pulse of 5 µs is applied to the SC. The maximum value of in Ω to ensure successful firing of the SC is. SC V + 5Ω mh Answer: 66Ω IL 4mt Width of gate pulse t 5µ sec I I + I When SC in ON with given pulse width of gate t ( τ ) V V I e + L τ 5 3 Time constant of L circuit, τ e Ω India s No. institute for GATE Training Lakh+ Students trained till date 65+ Centers across India 3