G F 1,3, G 1,3 G 4,2 F 1,3 4&G 1,3 2

Transcription

1 EE-GATE 08

2 EE-GATE 08 Section-I: General Ability. Functions F(a,b) and G(a,b) are defined as follows: F(a,b) = (a-b) and G(a,b) = a-b, where x represents the absolute value of x. What would be the value of G(F(,3), G(,3))? (A) (B) 4 (C) 6 (D) 36 Key: (A) Exp: Given, F,3 3 ; G,3 3 F,3 4 ; G,3 4 Ga,b a b GF,3, G,3 F a,b a b and G a,b a b G F,3, G,3 G 4, F,3 4&G,3. Since you have gone off the, the sand is likely to damage the car. The words that best fill the blanks in the above sentence are (A) course, coarse (B) course, course (C) coarse, course (D) coarse, coarse Key: (A) 3. A common misconception among writers is that sentence structure mirrors thought; the more the structure, the more complicated the ideas. The word that best fills the blank in the above sentence is (A) detailed (B) simple (C) clear (D) convoluted Key: (D) 4. For what values of k given below is k an integer? k 3 (A) 4,8,8 (B) 4, 0, 6 (C) 4,8,8 (D) 8, 6, 8 Key: (C) k Exp: Actually we can find an infinite no. of values of k, for which k 3 From the given choices; k 4 If k 4, then 36 integer. k to be an integer

3 EE-GATE 08 k 8 00 If k 8, then 0 integer k If k=8, then 36 integer The three roots of the equation f(x) = 0 are x = {-, 0, 3}. What are the three values of x for which Key: (D) f(x-3) = 0? (A) -5, -3, 0 (B) -, 0, 3 (C) 0, 6, 8 (D), 3, 6 Exp: Given that; X= -, 0, 3 are the 3 roots of f x 0. i.e., f 0, f 0 0 and f 3 0. Now we should find the values of x for which f x 3 0 x 3 ; x 3 0; x 3 3 f 0; f 0 0; f 3 0 x 3; x 3; x 6 x, 3, 6 6. A class of twelve children has two more boys than girls. A group of three children are randomly picked from this class to accompany the teacher on a field trip. What is the probability that the group accompanying the teacher contains more girls than boys? (A) 0 Key: (B) (B) (C) Exp: Given, a class of children has no two more boys than girls. Let, the no.of girls x Then no.of boys x x x x 0 x5 No.of Grils 5 No.of Boys 7 Total no. of ways of selection of 3 children = C n s Let A be the event that no. of girls are more than boys 3 (D) 5 3

4 EE-GATE 08 i.e. Case I or Case II Two cases possible 3G & 0B G & B Favorable no. of ways of A C 3 C 0 C C 3 n A ! Prob A n S C C C eq Prob An password must contain three characters. The password has to contain one numeral from 0 Key: (C) to 9, one upper case and one lower case character from the English alphabet. How many distinct passwords are possible? (A) 6,760 (B) 3,50 (C) 40,560 (D),05,456 Exp: We know that; Total no. of digits 0,,,...9 Total no. of upper case English alphabets 6A,B,C,...,Z Total no. of lower case English alphabets 6a,b,c,...,z No. of ways of selecting a digit out of 0 0C 0 No. of ways of selecting an upper case alphabet 6 C 6 No. of ways of selecting an lower case alphabet 6 C 6 Permutation = selection + arrangement Total no.of distinct passwords ! A a arrangement of B b 3things 3... arrangement 9 Z Z Selection ! 6 40,560 4

5 EE-GATE In a certain code, AMCF is written as EQGJ and NKUF is written as OYJ. How will DHLP be Key: (C) Exp: Given written in that code? (A) STN (B) TLPH (C) HLPT (D) XSV A M C F E Q G J & N K U FO Y J DH L P H LP T A designer uses marbles of four different colours for his designs. The cost of each marble is the same, irrespective of the colour. The table below shows the percentage of marbles of each colour used in the current design. The cost of each marble increased by 5%. Therefore, the designer decided to reduce equal numbers of marbles of each colour to keep the total cost unchanged. What is the percentage of blue marbles in the new design? Blue Black ed Yellow 40% 5% 0% 5% (A) (B) 40.5 (C) (D) 46.5 Key: (C) Exp: Assume that, Total no. of marbles = 00 C.P of marble =s 00 Total C.P of 00 marbles New cost price of marble = s 5 [Given] No. of marbles with new price No.of reduced marbles Blue Black ed Yellow Now; the no. of blue marbles = 40-5=35 Total no. of marbles =0-0=80 5

6 % of blue marbles in new design EE-GATE % % % 0. P,Q, and S crossed a lake in a boat that can hold a maximum of two persons, with only one set of oars. The following additional facts are available. (i) Key: (C) The boat held two persons on each of the three forward trips across the lake and one person on each of the two return trips. (ii) P is unable to row when someone else is in the boat. (iii) Q is unable to row with anyone else except. (iv) Each person rowed for at least one trip. (v) Only one person can row during a trip. Who rowed twice? (A) P (B) Q (C) (D) S Exp: Section of persons out of 4 (P, Q,, S), i.e., 4 C = 6ways PQ P PS Q QS S 3 forward trips Two person can travel From the fact (i), there are returned trips only person travel From the facts (ii) and (iii); P and Q should not travel. To satisfy (iv) fact; only Q and should travel in st trip. First trip : Q owed in forward trip,q Q owed in return trip nd Second trip : In trip only & P should travel. owed in forward trip P, P owedin return trip P 6

7 EE-GATE 08 rd Third trip : In 3 triponly P & Swill travel S must rowed in forward trip rowed twice. P,S From ii fact S Section-II: Electrical Engineering. Four power semiconductor devices are shown in the figure along with their relevant terminals. The devices (s) that can carry dc current continuously in the direction shown when gated appropriately is (are) A MT A D G K I Thyristor G I MT Triac G K I GTO G I S MOSFET (A) Triac only (C) Triac and GTO Key: (B) Exp: SC: conducts only from anode to cathode Triac: A (B) Triac and MOSFET (D) Thyristor and Triac G I K () Triac is combination of two anti parallel SCs Triac MT G MT 7

8 EE-GATE 08 () When MTis ve & MT is ve I (3) When MTis ve & MT is ve G I GTO: GTO conducts only from anode to cathode A G K MOSFET D D G G S S () When D is + ve & S is ve, diode is OFF () When D is ve & S is +ve, diode is ON D ve I S ve S D I 8

9 EE-GATE 08 vin. The op-amp shown in the figure is ideal. The input impedance t in is given by i in Z V 0 V in (A) Z (B) Z (C) Z (D) Z Key: (B) Exp: Given op- Amp is ideal Z i in V 0 V in V By virtual ground concept V in 0 Vin V V0 Vin in V Vin V0 iin Z By voltage divider principle i V i V Vin Vin Vin Vin Z Z Z in in 9

10 EE-GATE Two wattmeter method is used for measurement of power in a balanced three phase load supplied from a balanced three phase system. If one of the wattmeters reads half of the other (both positive), then the power factor of the load is (A) 0.53 (B) 0.63 (C) (D) Key: (D) Exp: In two Wattmeter of power measurement we know well known standard formula 3 tan It is given that one meter reads half of the other or wecan take tan tan tan 30 3 power factor cos 3 cos The graph of a network has 8 nodes and 5 independent loops. The number of branches of the graph is (A) (B) (C) 3 (D) 4 Key: (B) Exp: we know in a network b n Where : number of loops = 5 (given) b:number of branches (to be found) h:number of nodes = 8 (given) b n 5 b 8 b 0

11 EE-GATE A continuous time input signal x(t) is an eigenfunction of an LTI system, if the output is (A) kx(t), where k is an eigenvalue (B) Key: (A) ke (C) x(t) jt xt, where k is an eigenvalue jt e, where j t e j t e is a complex exponential signal is a complex exponential signal (D) kh, where k is an eigenvalue and H is a frequency response of the system x t e Exp: Consider for example eigen function, st st e ht yt Hse st Where H(s) is Laplace transform of h(t). At specific S S 0, Thus yt K.e st or K.x t Where K is eigen value. Thus option (A) is correct. HS becomes a constant quantity. 0 In option (B) extra frequency forms one getting generated at output which is not possible for LTI system. Option (C) and Option (D) one also not the correct answer due to extra frequency term getting generated in output. 6. In the logic circuit shown in the figure, Y is given by A B C D Y Key: (D) (A) Y=ABCD (B) Y = (A+B) (C+D) (C) Y=A+B+C+D (D) Y=AB+CD Exp: y MN y AB. CD y AB CD A B C D M AB N CD Y

12 EE-GATE 08 z 7. The value of the integral dz in counter clockwise direction around a circle C of radius z 4 with center at the point z = - is (A) Key: (A) i z Exp: dz z 4 C z 4 0 z C (B) i z lies inside & z lies outside 'C' By Cauchy's integralformula, z z z z dz dz dz z 4 z z z C C C if ( ) i i (C) i (D) i C y 43 0 x 8. Let f be a real valued function of a real variable defined as f(x) = x-[x], where [x] denotes the Key: (0.5) largest integer less than or equal to x. The value of Exp: We know that x x x where x integral partof x, x x x where x fractional part of 'x '.5.5 y f x dx x dx.5 x dx x dx 0.5 x x x f xdx is (up to decimal places). 0 3 x

13 EE-GATE The value of the directional derivative of the function x, y, z xy yz zx at the point (,-, Key: (A) ) in the direction of the vector p = i + j + k is (A) (B) 0.95 (C) 0.93 (D) 0.9 Exp: i j K x y z i y zx j xy z k yz x 5i 3 j k,, i j k equired directional deivative 5 i 3 j k Match the transfer functions of the second order systems with the nature of the systems given below. Transfer functions Nature of system P. 5 s 5s 5 I. Overdamped Q. 5 s 0s 5 II. Critically damped. 35 s 8s 35 III. Underdamped (A) P-I, Q-II, -III (B) P-II, Q-I, -III (C) P-III, Q-II, -I (D) P-III, Q-I, -II Key: (C) Exp: 5 n P s 5s 5 s s n n n n underdamped 5 5 Q s 0s 5 s s n n n n critically damped 3

14 EE-GATE s 8s 35 s s n n n n n over damped 35 P III Q II I. A bus admittance matrix for an electric power system has 8000 non zero elements. The minimum number of branches (transmission lines and transformers) in this system are (up to decimal places). Key: (3500) Exp: Given bus matrix N 000 Total root Buses given no of non zero elements 8000 Sparsity s 0 No of zero elements Total noof elements 0.99 No of transmission lines and transformers N s N A single phase 00kVA, 000 V/00V. 50Hz transformer has a voltage drop of 5% across its series Key: A Exp: impedance at full load. Of this, 3% is due to resistance. The percentage regulation of the transformer at full load with 0.8 lagging power factor is (A) 4.8 (B) 6.8 (C) 8.8 (D) 0.8 Impedance 5% 0.05pu Zpu esistance 3% 0.03pu eactance, X z pu pu pu pu pu V 0.8 p.f lag cos X sin pu pu pu 4.8% 4

15 EE-GATE Let f be a real valued function of a real variable defined as f(x) = x for x 0, and f x x for x 0. Key: (D) Exp: Given Which one of the following statements is true? (A) f(x) is discontinuous at x=0. (B) f(x) is continuous but not differentiable at x=0 (C) f(x) is differentiable but its first derivative is not continuous at x=0. (D) f(x) is differentiable but its first derivative is not differentiable at x=0. f x x x f x in continuals&differentiable xx f ' x x x x fx is differentiable but its first derivative is not differentiable at x=0 4. A positive charge of nc is placed at (0, 0, 0.) where all dimensions are in metres. Consider the x-y plane to be a conducting ground plane. Take F / m. The Z component of the E field at (0, 0, 0.) is closest to (A) 899.8V/m (B) V/m (C) V/m (D) V/m Key: (D) Exp: E E Q aˆ z a Z a a z a zv m z 5. A separately excited dc motor has an armature resistance a =0.05. The field excitation is kept constant. At an armature voltage of 00V, the motor produces a torque of 500Nm at zero speed. Neglecting all mechanical losses, the no-load speed of the motor (in radian/s) for an armature voltage of 50 V is (up to decimal places). Key: (600) Exp: Torque Produced by motor = 500N-m 5

16 At zero speed, such that E K 0 L b I b V E Also,T ki b a a nl b 000A EE-GATE k T Ia For armature voltage of 50V, no load speed in At no load I 0 V E 50V E k Field excitation or flux is constant, k constant E 50 K rad sec 00V 0.05 nl 6. In a salient pole synchronous motor, the developed reluctance torque attains the maximum value when Key: (B) the load angle in electrical degrees is (A) 0 (B) 45 (C) 60 (D) 90 Exp: eluctance torque sin Torque ix maximum when Consider a lossy transmission line with V and V as the sending and receiving end voltages, Key: (B) respectively. Z and X are the series impedance and reactance of the line, respectively. Z and X are the series impedance and reactance of the line, respectively. The steady-state stability limit for the transmission line will be VV (A) greater than X VV (C) equal to X Exp: Power with impudence Z, V V V P cos cos Z Z P max V V V cosfor Z Z Power with reactence X, VV (B) less than X VV (D) equal to Z 6

17 VV P sin X VV Pmax for 90 X i.e P P max max EE-GATE 08 VV Steady state stability limit for transmission line Pmax X 8. A single phase fully controlled rectifier is supplying a load with an anti-parallel diode as shown in the Key: (C) figure. All switches and diodes are ideal. Which one of the following is true for instantaneous load voltage and current? (A) 0 0 &i0 0 (B) 0 0 &i0 0 (C) 0 0 &i0 0 (D) 0 0 &i0 0 Exp: During forward bias i 0 L O A D V 0 ON off off ON i 0 L o a d V o Free wheeling diode is off Let V t sin t, V t sin tpositive in m 0 n During reverse bias i 0 off ON ON Off L o a d o Free wheeling diode isoff V t V sin t positive m 7

18 EE-GATE 08 Vin t V m 0 V m V m 0 Key concept: SC allows current only in one direction, Anode to cathode i Consider a unity feedback system with forward transfer function given by Gs s s The steady state error in the output of the system for a unit step input is (upto decimal places). Key: (0.67) Exp: unit stepinput, Kp limgs s0 A for unit step input, ess K p V 0. In the two port network shown, the h parameter Where, h,when V 0 I (up to decimal places). in ohms is I V I V Key: (0.5) Exp: when V 0 the output port is short circuited 8

19 EE-GATE 08 I I I I V I I I I V 0 Writing KVL on the input loop we have V I I I 0 V I I... Writing KVL on the output loop we have I I I I 0 I 3I 0 3 I I...() Using equation in equation, we have 3 V I I V 0.5I V h 0.5 I v 0 h 0.5. The waveform of the current drawn by a semi-converter from a sinusoidal AC voltage source is shown in the figure. If I 0 =0 A, the rms value of fundamental component of the current is A (up decimal places). voltage and current Vm sint 0 30 I 0 80 I0 t 0 9

20 EE-GATE 08 Key: (7.39) Exp: i0 t I 0 0 t I 0 Fourier series representation of i o (t) is 4I0 n n i0 t cos sin nt n,3,5 n 4I0 n n MS value of nth harmonic is n cos I0 cos n n MS value of fundamental is IS I0 cos 30 0cos 7.39A. In the figure, the voltages are t 00cos t, t 00cos t / 8and 3 t 00cos t / 36. The circuit is in sinusoidal steady state, and << L. P, P and P 3 are the average power outputs. Which one of the following statements is true? L L v t P v t P P3 v3 t Key: (C) (A) P = P = P 3 = 0 (B) P < 0, P >0, P 3 > 0 (C) P < 0, P > 0, P 3 < 0 (D) P >0, P < 0, P 3 >0 Exp: V t 00 cos t V t 00 cost 8 0

21 EE-GATE 08 V3 t 00 cos t 36 V t V t V t 00 3 V t 0 80 V t V3 t 5 36 V t is leading 3 V t andv t by0and 5. V t is source & delivering power P 0 V tandv t are sinks & absorbing power P 0, P V t 0 V t 5 3 V t 0 3. Consider a non singular square matrix A. If trace (A) = 4 and trace (A ) = 5, the determinant of the matrix A is (up to decimal place). Key: (5.5) a b Exp: Let A Trace of A a d c d &A a b a b a bc ab bd c d c d ac cd bc d Trace of A a bc bc d a bc d Given a d 4... & a bc d 5... a d 6 a d ad 6 5 bc ad 6 from ad bc ad bc The series impedance matrix of a short three phase transmission line in phase coordinates is Zs Zm Zm Zm Zs Z m. If the positive sequence impedance is (+j0), and the zero sequence is (4+j3). Z m Zm Z s Then the imaginary part of Z m (in ) is (up to decimal places).

22 EE-GATE 08 Key: (7) Exp: s m Z Z Z j s m 0 m m m Z Z Z 4 j3... Z Z 3Z 3 j Z j7 Im Z 7 5. The positive, negative and zero sequence impedances of a 5 MVA, three phase, 5.5kV, stargrounded, 50Hz generator are j0. pu, j0.05 pu and j0.0 pu respectively on the machine rating base. The machine is unloaded and working at the rated terminal voltage. If the grounding impedance of the generator is j0.0 pu, then the magnitude of fault current for a b-phase to ground fault (in ka) is (up to decimal places). Key: (73.5) Exp: I pu f 3 Z Z Z 3Z 0 n 3 3 j0. j0.05 j I actual I I f f pu b base MVA 3 S b ka 0.9 3V b KV kA The signal energy of the continuous time signal xt t ut t u t t 3 u t 3 t 4 u t 4 is (A) /3 (B) 7/3 (C) /3 (D) 5/3 Key: (D) Exp: xt t ut t u t t 3u t 3 t 4u t 4 a b c 0 3 4

23 Energyof x t a dt b dt c.dt a dt c dt 3 3 Energy of x t a dt b dt 3 t dt dt 3 t EE-GATE A 0- Ampere moving iron ammeter has an internal resistance of 50 m and inductance of 0.mH. A Key: (A) Exp: shunt coil is connected to extend its range to 0-0 Ampere for all operating frequencies. The time constant in milliseconds and resistance in m of the shunt coil respectively are (A), 5.55 (C), (C).8,0.55 (D)., m Lm MI ammeter sn Lsn Shunt coil It is given that L m m 50m 0. mh Existing range 0 ampere eauired range 0 0 ampere equired 0 Scaling factor m 0 Existing The required value shunt coil resistance is given by 3 m 500 sn 5.55m. m 0 To make the meter independent of frequency, the time constant of both parallel branch should be same i.e. 3

24 EE-GATE 08 L sn sn L sn sn m m sn 0.00 m.sec m.sec 5.55 ma. 8. The voltage v(t) across the terminals a and b as shown in the figure, is a sinusoidal voltage having a frequency =00 radian/s. When the inductor current i(t) is in phase with the voltage v(t), the magnitude of the impedance Z (in ) seen between the terminals a and b is (up to decimal places). Key: (50) vt Exp: when the Source voltage and current are in phase, then the circuit is under resonance and the impedance is purely real. Z Transforming the given network into its equivalent phasor domain we have j00l it a b L 00 F 00 j00 00 at 00 Z jl j00l L j ZC j00 C Z 00 Zeq j00l 00 j00 4 j0 j00l 00 j00 j00l Z 00 rad sec, given eq 4 j0 00 j00 00 j00 00 j00 4

25 EE-GATE j 00L Since the circuit is under resonance, the imaginary part of Z eq Zeq 0, hence 9. Which one of the following statements is true about the digital circuit shown in the figure? D Q D Q D Q fout C C C f IN (A) It can be used for dividing the input frequency by 3. (B) It can be used for dividing the input frequency by 5. (C) It can be used for dividing the input frequency by 7. (D) It cannot be reliably used as a frequency divider due to disjoint internal cycles. Key: (B) Exp: D0 Q0 D Q D Q fout C C C f IN D Q Q 0 Clk Q Q Q Q Q Q Q 0 0 D0 D D epeated i.e., MOD 5 Hence, f o fi 5 5

26 EE-GATE The voltage across the circuit in the figure. And the current through it, are given by the following expressions: vt 5 0cos t 60V it 5 Xcos t A Where 00 radian/s. If the average power delivered to the circuit is zero, then the value of X (in Ampere) is (up to decimal places). it vt Electrical Circuit Key: (0) Exp: If the voltage and currents of electrical circuit is in the form of 0 v v 0 i i V t V V cos t V cos t... i t i i cos t i cos t... Then the average power is given by P avg T T 0 V t i t dt, V i V I V0i 0 cos v i cos v i... its a very well known standard result It is given that 5 0cos t 0 V t 5 0cos t 60 i t 5 x cos t 0 0 x then Pavg 55 cos0 0, 0x 0 5 cos 0 P 0, given 0 5 5x 5 x 5 x 0 avg 6

27 EE-GATE The equivalent impedance Z eq for the infinite ladder circuit shown in the figure is j9 j9 Z eq j5 j j5 j... (A) j (B) -j (C) j3 (D) 3 Key: (A) Exp: j9 j9 j9 j9 j5 j j5 j j4 j4 Z eq Z eq Z eq Z j9 j4 Z eq eq j4zeq j9 j4 Z eq j4 Z eq Zeq j9 j4 Zeq j4zeq j4zeq Zeq 36 j9zeq j4zeq eq Z j9 Z 36 0 eq j9 j4 Z eq Z eq j9 j Z eq j9 844 j9 5 j9 j5 j9 j5 jq j5 or j or j3 7

28 EE-GATE 08 Since the nature of the network is overally inductive the reactance must should be positive (for capacitive case the reactance could have ve). Hence we are discarding j3. Z eq j 3. A transformer with toroidal core of permeability is shown in the figure. Assuming uniform flux density across the circular core cross-section of radius r < <, and neglecting any leakage flux, the best estimate for the mean radius is vp ip I sin ax V cos t N P r is 0 NS vs (A) Key: (D) Vr N p I (B) Ir NpVs V (C) Vr N p I (D) Ir N p V Exp: The inductance of primary is Np L S Where S is reluctance of magnetic path S A r r Np Np r L r Np r XL L I..Np. r V IXL.I.r.Np. V 33. The number of roots of the polynomial, half of the complex plane is s s 7s 4s 3s 73s 5s 00 in the open left (A) 3 (B) 4 (C) 5 (D) 6 8

29 EE-GATE 08 Key: (A) Exp: 7 s s sign s change s 3 s s 4 00 s sign s change 00 4 A s 8s 48 s 00 da s 3 3s 96 s ds No. of sign changes 4 ight hand roots 4 No. of imaginary roots order of auxillary eqn No.of sign changes below zero th row 4 0 left handroots The equivalent circuit of a single phase induction motor is shown in the figure, where the parameters are X l X, XM 40 and s is the slip. At no load, the motor speed can be approximated to be the synchronous speed. The no-load lagging power factor of the motor is (up to 3 decimal places). jx Key: (0.06) V 0 Exp: At no load, m s. S 0. S ' The modified equivalent Circuit is X M j X M j ' s X' j ' s X' j 9

30 j EE-GATE j Z eq 40 j 4 j ' ' s 4 eq Z j j0 j0 3 j6 3 j6 j0 j0 3 j6 j3 3 j6 4.7 j No load p.f cos cos lag. 35. A 3-phase 900kVA, 3kV/ 3 kv /Y, 50 Hz transformer has primary (high voltage side) resistance per phase of 0.3 and secondary (low voltage side) resistance per phase of 0.0,Iron loss of the transformer is 0kW. The full load % efficiency of the transformer operated at unity power factor is (up to decimal places). Key: (97.36) Exp: 3 a ph 3 ; IHv I 00A 33 a eq

31 EE-GATE 08 Cu loss 3 3I eq W 4.4kW Core loss 0kW 900 % 97.36% A DC voltage source is connected to a series L-C circuit by turning on the switch S at time t=0 as shown in the figure. Assume i(0)=0, v(0)=0. Which one of the following circular loci represents the plot of i(t) versus v(t)? S it t 0 L H 5V C F vt (A) it vt (B) it 0, 5 5,0 vt (C) it (D) it Key: (B) Exp: Since the network is of nd order, to obtain the expression of v(t)and i(t) we should use Laplace transform approach. Once we get the expression for v(t) and i(t) we can plot the loci by observing them at different time instant. Transforming the given time domain network into its equivalent s-domain form for t>0, have 0,5 vt 5,0 vt we 3

32 EE-GATE 08 5 s 5 Is i t s s s 5sin t s V s s 5 5 s 5 s Is s Vs s 5 s s s s j s j 5 5s s s v t 5 5 cos Finaly we have i t 5sint v t 5 5cos t t it vt Coordinate , , , , ,0 0,0 0,0 t 0 5,5 5, 5 t t 3 t 37. A three phase load is connected to a three phase balanced supply as shown in the figure. If Van 000 V, Vbn 00 0 V and V cn V (angles are considered positive in the anti clock wise direction). The value of for zero current in the neutral wire is (up to decimal places). a n c j0 j0 b 3

33 EE-GATE 08 Key: (5.77) Exp: a n c I C I a I N j0 X j0 b By KCL at node x, we have I I I I a b c N V V V j0 j0 an bn cn I N IN 0 given I b 38. The unit step response y(t) of a unity feedback system with open loop transfer function G(s) H(s) K is shown in the figure. The value of K is (up to decimal places). s s yt t sec 33

34 EE-GATE 08 Key: (8) Exp: Steady state output =0.8 Input is unit step input = Stead state error e ss = 0. For unit step input K limgshs p s0 ss s0 K lim K e s s A K p 0. K K 5 K 4; K The Fourier transform of a continuous- time signal x(t) is given by X,. 0 j Key: (0) where j and denotes frequency. Then the value of nx t at t (up to decimal place.) (ln denotes the logarithm to base e). Given X, Exp: 0 j Converting to s domain, 0t x t t.e u t Xs 0t we know e ut s 0 0t t.e s 0 0t n x t n t.e u t n t 0t n c s 0 n Consider a system governed by the following equations dx t x t x t dt dx t x t x t dt 34

35 Key: (C) Exp: EE-GATE 08 The initial conditions are such that x 0 x 0. Let x limx t and x limx t. f t Which one of the following is true? (A) xf xf (B) xf xf (C) xf xf dx t dt t (D) xf xf x t x t... dx x t x t... dt Differentiating w.r.t 't ' d x t dx t dx t dt dt dt dx t xt xt from dt dxt dxt xt xt from dt dt d x t dx t 0 dt dt D D 0 D D 0 D 0, Solution, x t C C e x lim x t C f t t Similarly, Differently () w.r.t t using () & () we get d x t dx t dt 0 dt solution, x t C C e t x lim x t C f t x x f f 4. Consider the two bus power system network with given loads as shown in the figure..0 j0..00 f t G 0 jq G Q loss 5 j5 0 j0 G 5 jqg 35

36 EE-GATE 08 Key: (C) Exp: All the values shown in the figure are in per unit. The reactive power supplied by generator G and G are Q G and Q G respectively. The per unit values of Q G,Q G, and line reactive power loss (Q loss ) respectively are (A) 5.00,.68,.68 (B) 6.34, 0.00,.34 (C) 6.34,.34,.68 (D) 5.00,.34,.34 VV S P x sin 5 sin 0. sin VS Qs VS V cos cos pu X 0. V Q VS cos V cos30.34pu x 0. Q Q Q loss S pu Q Q Q G load S pu Q Q Q G load QG Q Qload pu 4. A phase controlled single phase rectifier, supplied by an AC source, feeds power to an -L-E load as Vm shown in the figure. The rectifier output voltage has an average value given by V 0 3 cos, where V m = 80 volts and is the firing angle. If the power delivered to the lossless battery is 600W, in degree is (up to decimal places). V sin m t V 0 0mH 80V Battery 36

37 Key: (90) EE-GATE 08 Exp: Power delivered to lossless battery EI0 =600W 80 I 600 I 0 0 0A From given rectifier, V0 E I0 V m 3 cos E I cos cos 0 3 cos 3 cos The positive, negative and zero sequence impedances of a three phase generator are Z, Z and Z 0 respectively. For a line to line fault with fault impedance Z f, the fault current is If kif, where I f is the fault current with zero fault impedance. The relation between Z f and k is Z Z k Z Z k (A) Zf (B) Zf k k (C) Key: (A) Z f Exp: I L L fault f I L L fault f f f Z Z k k 3 Z Z with fault impedance, Z I ki given Z Z Z Z Z f Z Z k Z k Z Z f Z k Z Z k Z f f Z Z k k f (D) 3 Z Z Z f Z f Z Z k k 37

38 EE-GATE As shown in the figure, C is the arc from the point (3, 0) to the point (0, 3) and the circle x + y =9. The y yx dx xy x dy is (up to decimal places). C value of the integral y 0,3 C 3,0 x Key: (0) Exp: Given x y 9 y 9 x y 9 x x dy xdx dx 9 x 9 x 0 y yx dx xy x dy 9 x x 9 x dx x 9 x x C x3 xdx 9 x x 9 x x x 9x dx x 0 3 a x 9 t dt Consider, dx, put 9 x t x t0 t from, Digital input signals A,B,C with A as the MSB and C as the LSB are used to realize the Boolean Key: (B) function F = m 0 + m + m 3 + m 5 + m 7, where m i denotes the i th minterm. In addition, F has a don t care for m. The simplified expression for F is given by (A) AC BC AC (B) A C (C) C A (D) AC BC AC Exp: F m0,,3,5,7 d A AC A AA CDistributive law AC A A BC BC BC BC 0 3 x

39 EE-GATE Let f x 3x 7x 5x 6. The maximum value of f(x) over the interval [0,] is (up to Key: () decimal place). 3 Exp: f x 3x 7x 5x 6 f ' x 9x 4x 5 0 x, 5 9 f '' x 8x 4 f '' 0 & f '' At x 5 9, f x has local maximum & f f 0 6 f The maximum value of f x in 0, is"" 47. The per unit power output of salient-pole generator which is connected to an infinite bus, is given by the Key: (C) expression, P=.4sin +0.5sin, where is the load angle. Newton aphson method is used to calculate the value of for P=0.8pu. If the initial guess is 30, then its value (in degree) at the end of the first iteration is (A) 5 (B) 8.48 (C) 8.74 (D) 3.0 Exp: P sin 0.5 sin f.4sin 0.5sin 0.8 By, Newton aphson method f 0 f ' 0 0.4sin sin cos cos rad Consider the two continuous time signals defined below: t, t t, t x t, x t 0, otherwise 0, otherwise 39

40 EE-GATE 08 These signals are sampled with a sampling period of T=0.5 seconds to obtain discrete time signals x [n] and x [n], respectively. Which one of the following statements is true? (A) The energy of x [n] is greater than the energy of x [n]. (B) The energy of x [n] is greater than the energy of x [n]. (C) x [n] and x [n] have equal energies. (D) Neither x [n] nor x [n] is a finite energy signal. Key: (A) Exp: Given t, t x t 0, otherwise x t t, t x t 0, otherwise x t t t xn, 0.75, 0.5, 0.5, 0, 0.5, 0.5, 0.75, x n 0,0.5, 0.5,0.75,,0.75,0.5,0.5,0 We can see clearly, xn has greater energy than x n x n Energy in x n Energy in 49. The figure shows two buck converters connected in parallel. The common input dc voltage for the converters has a value of 00V. The converters have inductors of identical value. i S 00V S L C S Switch control signals t S L S 0 0.5ms ms t 40

41 Key: (.5) EE-GATE 08 The load resistance is. The capacitor voltage has negligible ripple. Both converters operate in the continuous conduction mode. The switching frequency is khz, and the switch control signals are as shown. The circuit operates in the steady state. Assuming that the converters share the load equally, the average value of i S, the current of switch S (in Ampere), is (up to decimal places). Exp: V0 DVs V 0 I0 50A. S V 50 Sence losser negligible, P V F V I S S out out 00 I I S 5A I s IS IS.5A in P out Let A Key: () Exp: is (up to decimal place). 0 Given A Characteristic equation A is A I By Cayley Hamilton theorem, and B = A 3 -A -4A+5I, where I is the 3 3 identity matrix. The determinant of B 3 A A 4A 4I 0 3 A A 4A 5I I B I According to the given question B I 4

42 EE-GATE A 00V DC series motor, when operating from rated voltage while driving a certain load, draws 0A Key: (85) current and runs at 000 r.p.m. The total series resistance is. The magnetic circuit is assumed to be linear. At the same supply voltage, the load torque is increased by 44%. The speed of the motor in r.pm (rounded to the nearest integer) is. Exp: 00V, 0A a f E V I b a a f V When torque becomes.44 times T.44T Magnetic circuit linear, I T I I T a a T Ia a a T b a a f b b T a a I.44 0 A, E V I E E N N N N I I b a b a Ia 00 88V E I 88 0 N N 000 E I rpm 85 rpm nearest integer 00V 5. The capacitance of an air filled parallel plate capacitor is 60pF. When a dielectric slab whose thickness is half the distance between the plates, is placed on one of the plates converting it entirely, the capacitance becomes 86pF. Neglecting the fringing effects, the relative permittivity of the dielectric is (up to decimal places). Key: (.53) A A Capacitance, C 60pF 0pF d / d 0 0 Exp: 4

43 r r EE-GATE 08 and 0 ra C 60rpF 0rpF d Now, CC 00r Ceq C C 0 86 r 0 86 or, r z If C is circle z 4 and f z. then f zdz is z 3z C (A) (B) 0 (C) - (D) - Key: (B) Exp: f z z z 3z z z z z & are poles of order, both lie inside z 4 esidue of f z at z d z es f z lim z z z dz z z z d z lim z dz z z z z.z 4 z lim 4 z esidue of f z at z d z es f z lim z z z! dz z z d z lim 4 z dz z z C By Cauchy esidue theorem, f z dz i A dc to dc converter shown in the figure is charging a battery bank. B whose voltage is constant at 50 V. B is another battery bank whose voltage is constant at 50V. The value of the inductor, L is 5mH and the ideal switch, S is operated with a switching frequency of 5kHz with a duty ratio of 0.4. Once the 43

44 EE-GATE 08 circuit has attained steady state and assuming the diode D to be ideal, the power transferred from B to B (in Watt) is (up to decimal places) i L 5mH L D 50 V B S B 50 V Key: () Exp: Given Vin 50V, Vout 50V V0 V D in D D But given duty cycle = 0.4 The boost converter is operating in discontinuous mode. T 00 sec f t DT 80sec. on V V D 0 s di I VS L L dt t on I 800 I I I I I 0.8A 6 II max min min max I max 6 Imax T I Lavg 0.4A 6 T 000 Power transfered from B to B V I W in Lavg I L 0 DT BT T t 44

45 EE-GATE In the circuit shown in the figure, the bipolar junction transistor (BJT) has a current gain 00. The base-emitter voltage drop is a constant, V BE =0.7V. The value of the Thevenin equivalent resistance th (in ) as shown in the figure is (up to decimal places). 0 a 5V 0 k 0.7V k b Th Key: (90.9) Exp: Form BJT amplifier, SC Th V I Th SC V I OC B S B Th can be written as it havedependent sources kI k I 0 0 0m IB 0k 0k VOC IE k 0mk IB m 0k 0k I I 0 m 0k Th B 45

46 EE-GATE 08 46