GATE 2018 [Afternoon Session]

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3 GATE 8 [Afternoon Session].General Aptitude. Since 4 Question Ans. Q. to Q.5 carry one mark each ( k ) For what values of k given below is k 3 an integer? (A) 4, 8, 8 (B) 4,, 6 (C) 4, 8, 8 (D) 8, 6, 8 (C) ( k ) Sol. Given : f( k) k 3 Going by options when k 4, f( k) 36, which is integer when k 8, f( k), which is integer when k 8, f( k) 36, which is integer Hence, the correct option is (C). Question 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 Ans. (A) Question 3 A common misconception among writers is that sentence structure mirrors thought, the more the structure, the more complicated the ideas. The words that best fill the blanks in the above sentence are (A) detailed (B) simple (C) clear (D) convoluted Ans. (D) Question 4 Functions Fab (, ) and Ga (, b ) are defined as follows : Fab (, ) ( a b) and Ga (, b) a b, where x represents the absolute value of x. What would be the value of GF ( (, 3), G (, 3))? (A) (B) 4 (C) 6 (D) 36 Ans. (A) Sol. Given : f ( ab, ) ( a b) gab (, ) a b So, f (, 3) ( 3) 4 g(, 3) 3 g[ f(, 3), g(, 3)] g(4, ) 4 g[ f(, 3), g(, 3)] Hence, the correct option is (A). More questions and video solutions of GATE - 8 are available in the Facebook group GATE ACADEMY (EE/EC/N). To join click here

4 Question 5 GATE 8 [Afternoon Session] Since 4 The three roots of the equation f( x) are x {,, 3}. What are the three values of x for which f( x3)? (A) 5, 3, (B),, 3 (C), 6, 8 (D), 3, 6 Ans. (D) Sol. Given : f( x) for x (,,3) f( x) ( x) x( x3) f( x3) ( x3)( x3)( x33) x, 3, 6 Hence, the correct option is (D). Question 6 Q.6 to Q. carry two marks each An password must contain three characters. The password has to contain one numeral from to 9, one upper and one lower case character from the English alphabet. How many distinct passwords are possible? (A) 6,76 (B) 3,5 (C) 4,56 (D),5,456 Ans. (C) Sol. Number of different passwords 6C6C3! 456. Hence, the correct option is (C). Question 7 n a certain code, AMCF is written as EQGJ and NKUF is written as ROYJ. How will DHLP be written in that code? (A) RSTN (B) TLPH (C) HLPT (D) XSVR Ans. (C) Sol. BCD A NOP M DEF C GH F E Q G J OPQ N LMN K VWX U GH F R O Y J EFG D JK H MNO L QRS P H L P T Hence, the correct option is (C). Copyright Head Office : A/4-5, Smriti Nagar, Bhilai (C.G.), Contact : , Branch Office : Raipur, : , Bhopal, : 9-476, ndore, :

5 3 GATE 8 [Afternoon Session] Since 4 Question 8 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) 35 (B) (C) (D) Ans. (*) 4/ Sol. Given : Total children = Here boys are two more than girls hence, Boy (B) = 7 and Girl (G) = 5 Now three children are randomly picked. Hence, Total combination for that, N C 3 Now in the group, if girls are more than boys, then we have possible combinations : (i) Girls, Boy N 5 7 C C (ii) 3 Girls, Boy N C C N N 7 4 Hence, the probability ( P) N Hence, the given options are not correct. The right answer for this question is 4. Question 9 P, Q, R 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) 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 R. (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) R (D) S Ans. (C) Sol. P has to row on return journey as he can not row with anyone else. On next trip P will be leaving R and R will row. After this R will return back to pickup Q so on the last trip Q and R will go together in which Q will be rowing the boat. Hence R will be rowing the boat twice. Hence, the correct option is (C). More questions and video solutions of GATE - 8 are available in the Facebook group GATE ACADEMY (EE/EC/N). To join click here 5

6 4 GATE 8 [Afternoon Session] Since 4 Question 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 Red Yellow 4% 5% % 5% (A) (B) 4.5 (C) (D) 46.5 Ans. (C) Sol. Assume a total of marbles each of cost unit Blue paper = 4 Black paper = 5 Red paper = Yellow paper = 5 Let x marbles are reduced from each type & cost of each marble is.5 Total cost should remain (4 x).5 (5 x).5 ( x).5 (5 x).5 5 5x x 5 Blue = 35 Black = Red = 5 Yellow = 35 % Blue 43.75% 8 Hence, the correct option is (C)..Technical Part. Q. to Q.5 carry one mark each Question (Digital Electronics) n the logic circuit shown in the figure, Y is given by A (A) Y B C D ABCD (B) Y ( AB)( C D) (C) Y ABC D (D) Y AB CD Y Copyright Head Office : A/4-5, Smriti Nagar, Bhilai (C.G.), Contact : , Branch Office : Raipur, : , Bhopal, : 9-476, ndore, :

7 5 GATE 8 [Afternoon Session] Since 4 Ans. (D) Sol. Given : A A B B Y ( AB) ( CD) C D C D f AB CD AB CD AB CD [By De-Morgan theorem AB A B and A B AB] Hence, the correct option is (D). Question (Power Electronics) The waveform of the current drawn by a semi-converter from a sinusoidal AC voltage source is shown in the figure. f A, the rms value of fundamental component of the current is A (up to decimal places). Voltage and current Vm sin( t) 3 Ans Sol. Given : Assuming single phase symmetrical semi-converter (we can also opt single phase asymmetrical semi-converter as the output voltage and current waveform of load and source will not be affected) = A t 3 The given circuit is a symmetrical single phase semi-converter, + i S,,3 T T i R V S v L NHC D PHC D More questions and video solutions of GATE - 8 are available in the Facebook group GATE ACADEMY (EE/EC/N). To join click here

8 6 GATE 8 [Afternoon Session] Since 4 The Fourier series representation of supply current is given by, 4 n n is() t cos sinnt n, 3, 5... n From the given diagram Voltage and current Vm sin( t) 3 The value of firing angle ( ) = 3 Hence, the RMS value of supply fundamental current in a single phase semi-converter is given by, 4 cos 4 cos RMS Hence, the RMS value of fundamental component of the supply current is 7.39 A. t Copyright Head Office : A/4-5, Smriti Nagar, Bhilai (C.G.), Contact : , Branch Office : Raipur, : , Bhopal, : 9-476, ndore, :

9 7 GATE 8 [Afternoon Session] Since 4 Question 3 (Signals & Systems) Let f be the real valued function of a real variable defined f ( x) x [ x], where [ x ] denotes the largest.5 integer less than or equal to x. The value of f ( xdx ) is (up to decimal plces)..5 Ans..5 Sol. Given : f ( x) x x Where, x represents the largest integer less than or equal to x. So, For x f( x) x x x x For x f( x) x x x x.5 x f( x) x x.5 f( x) x.5 f ( xdx ) Area or triangle + Area of Rectangle + Area of triangle Hence, the correct answer is.5. Question 4 (Network Theory) V n the two-port network shown, the h parameter (where, h when V ) in ohms is (up to decimal places). V V More questions and video solutions of GATE - 8 are available in the Facebook group GATE ACADEMY (EE/EC/N). To join click here

10 8 GATE 8 [Afternoon Session] Ans..5 Sol. Given : The two port network is shown below, Since 4 V V To find the value of h V V, the equivalent circuit is shown below, V () () V V Copyright Applying KVL in loop (), V ( ) V (i) Applying KVL in loop (), ( ) ( ) 3 (ii) From equation (i) and (ii), 3 V Hence, V h 3.5 V V.5 Hence, the correct answer is.5 Head Office : A/4-5, Smriti Nagar, Bhilai (C.G.), Contact : , Branch Office : Raipur, : , Bhopal, : 9-476, ndore, :

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12 9 GATE 8 [Afternoon Session] Since 4 Question 5 (Power System) A bus admittance matrix for an electric power system has 8 non zero elements. The minimum number of branches (transmission lines and transformers) in this power system are (up to decimal places). Ans. 35 Sol. Given : A bus system of contains 8 non zero elements. Method : The total number of elements 8 99 % of sparcity (% x) 99.% n ( x) n Hence the number of transmission lines (.99) 8 35 Hence, the minimum number of branches are 35. Method : Number of non-zero element = 8 Number of diagonal element = Number of non-zero off diagonal element = 7 Number of non-zero off diagonal element 7 Hence, Number of transmission lines 35 Hence, the minimum number of branches are 35. Question 6 (Signals & Systems) A continuous time input signal x() t is an eigen function of an LT system, if the output is (A) kx() t, where k is an eigen value jt j t (B) ke x() t, where k is an eigen value and e is a complex exponential signal (C) x() te j t j t, where e is a complex exponential signal (D) kh( ), where k is an eigen value and H ( ) is a frequency response of the system. Ans. (A) Sol. f the output signal is a scalar multiple of input signal, the signal is refereed as an eigen function (or characteristic function) and the multiplier is referred as an eigen value (or characteristic value). f x(t) is the eigen function and is the eigen value, then output, yt () xt (). Hence, the correct option is (A). Question 7 (Engineering Mathematics) Let f be a real valued function of a real variable defined as f ( x) x for x, and f ( x) x for x. Which of the following statement is true? (A) f ( x ) is discontinuous at x (B) f ( x ) is continuous but not differentiable at x (C) f ( x) is differentiable but its first derivative is not continuous at x (D) f ( x ) is differentiable but its first derivative is not differentiable at x. More questions and video solutions of GATE - 8 are available in the Facebook group GATE ACADEMY (EE/EC/N). To join click here

13 Ans. (D) Sol. Given : Copyright GATE 8 [Afternoon Session] x f( x) x t s first derivative, for x for x x x f '( x) x x Let x t x t xt Since 4 t t f '( x) t t f '( t) t The first derivative is mod function according to property, it is continuous but not differentiable at t x. x Also, f "( x) x First derivative is not differentiable. Hence, the correct option is (D). Question 8 (Control System) Consider a unity feedback system with forward transfer function is given by, Gs () ( s)( s) The steady state error in the output of the system for a unit-step input is (up to decimal places). Ans..67 Sol. Given : The open loop transfer function for given unity feedback system is G() s H() s H() s ( s)( s) Steady state error for unit step input ( e ss ) K p Where, K (Position error coefficient) lim G( s) H( s) K p Steady state error p lim s ( s )( s ) s ( ess ).67.5 Hence, steady state error in the output of the system for a unit-step input is.67. Head Office : A/4-5, Smriti Nagar, Bhilai (C.G.), Contact : , Branch Office : Raipur, : , Bhopal, : 9-476, ndore, :

14 GATE 8 [Afternoon Session] Since 4 Question 9 (Power System) n the figure, the voltages are V( t) cos( t), V( t) cos( t /8) and V 3 ( t ) cos( t / 36). The circuit is in sinusoidal steady state, and R L, P, P and P 3 are the average power outputs. Which one of the following statements is true? R L R L P v () t v () t v () t P P 3 3 (A) P P P 3 (B) P, P, P 3 P, P, P (D) P, P, P 3 (C) 3 Ans. Sol. (C) Given : The circuit is shown below, R L R L P v () t v () t v () t P P 3 3 v ( t) cos t ( ) cos v t t 8 ( ) cos v3 t t 36 From the above expression, it is clear that v is leading v by and v 3 is leading v by 5. Since, active power ( P ) is always flow from leading angle to lagging angle. Hence, the current will flow from v () t towards v () t and v () t. Hence 3 P and P and P3. Hence, the correct option is (C). More questions and video solutions of GATE - 8 are available in the Facebook group GATE ACADEMY (EE/EC/N). To join click here

15 GATE 8 [Afternoon Session] Since 4 Question (Engineering Mathematics) z The value of the integral dz in counter clockwise direction around a circle C of radius with C z 4 center at the point z is i (A) (B) i (C) i (D) i Ans. (A) z Sol. Given : dz C z 4 Counter Cz This counter only encircles z = pole and hence we need residue at z =. z ( z) i i dz i Re s f ( z) i lim( z ) C z z 4 z ( z )( z ) 4 Hence, the correct option is (A). Question (Analog Electronics) vin The op-amp shown in the figure is ideal. The input impedance is given by i Z in V in ~ in R V R Ans. Sol. R R R (A) Z (B) Z (C) Z (D) Z R R R R (B) Given : deal Op-Amp circuit, Z in V f V in ~ V f A A R R V Copyright Head Office : A/4-5, Smriti Nagar, Bhilai (C.G.), Contact : , Branch Office : Raipur, : , Bhopal, : 9-476, ndore, :

16 3 GATE 8 [Afternoon Session] Since 4 V is acting on series combination of R and R Hence, V R V f [By VDR] R R VN V Vf [Due to virtual ground concept] Also, Vf Vin Applying KCL at inverting terminal, Vin V in Z R Vin Vin R in Z R Vin R in Z Vin R Z in R Hence, the correct option is (B). Question (Electrical & Electronics Measurement) Two wattmeter method is used for measurement of power in a balanced three-phase load supplied from a balanced three-phase system. f one of the wattmeters reads half of the other (both positive), then the power factor of the load is (A).53 (B).63 (C).77 (D).866 Ans. (D) Sol. Given : Two wattmeter method measuring power in a balanced three phase load supplied from three phase balanced three phase system. Hence, W W The power factor angle is given by, tan 3 W W W W W W tan 3 W W Power factor tan 3 3 cos cos Hence, the correct option is (D). More questions and video solutions of GATE - 8 are available in the Facebook group GATE ACADEMY (EE/EC/N). To join click here

17 4 GATE 8 [Afternoon Session] Question 3 (Electrical Machine) Ans. Sol. Since 4 n a salient pole synchronous motor, the developed reluctance torque attains the maximum value when the load angle in electrical degrees is (A) (B) 45 (C) 6 (D) 9 (B) Given : A salient pole synchronous motor. The power flow in a silent pole synchronous machine is given by, EV f T Vt P sin sin Xd Xq X d P 6 EV f T Vt The torque produced, T sin Ns Xd Xq X d EV f T Where, sin Electromagnetic power X The reluctance power is given by, d Vt Preluctance sin Xq X d To get maximum reluctance torque, the reluctance power must be maximum. Hence the condition for maximum reluctance torque is sin sin () Hence, the correct option is (B). Question 4 (Power Electronics) Four power semiconductor devices are shown in the figure along with their relevant terminals. The device (s) that can carry dc current continuously in the direction shown when gated appropriately is (are) G A k G MT A MT G k G D s Copyright Thyristor Triac GTO MOSFET (A) Triac only (B) Triac and MOSFET (C) Triac and GTO (D) Thyristor and triac Head Office : A/4-5, Smriti Nagar, Bhilai (C.G.), Contact : , Branch Office : Raipur, : , Bhopal, : 9-476, ndore, :

18 5 GATE 8 [Afternoon Session] Since 4 Ans. Sol. (A) (i) A G k Thyristor An SCR allows only anode to cathode current. Hence the given current can not flow through SCR. (ii) MT G MT Triac A triac is a bidirectional current flow device. Hence the given current can flow from terminal MT to terminal MT. (iii) A G k GTO A GTO is a gate turn off thyristor. t is a unidirectional current flow device. t allows only anode to cathode current. Hence the given current can not flow through GTO. (iv) G D s MOSFET Note : A MOSFET with body diode is a by directional current conduction device. But here the body diode is not mentioned. The given MOSFET is a D-MOSFET. t allows only drain to source current. Hence the given current can not flow through MOSFET. Hence, the correct option is (A). More questions and video solutions of GATE - 8 are available in the Facebook group GATE ACADEMY (EE/EC/N). To join click here

19 6 GATE 8 [Afternoon Session] Since 4 Question 5 (Network Theory) 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 Ans. (B) Sol. Given : (i) Number of nodes = 8 (ii) Number of independent loop = 5 n graph theory, number of independent loop represents number of links bn Hence, bn 5 b 8 5 b Number of branch = Hence, the correct option is (B). Question 6 (Power System) The series impedance matrix of a short three-phase transmission line in phase coordinates is Zs Zm Zm Zm Zs Z m. f the positive sequence impedance is ( j) and the zero sequence is (4 j 3) Zm Zm Z s, then the imaginary part of Z m (in ) is (up to decimal places). Ans. 7 Sol. The positive sequence impedance ( Z) Zs Zm j (i) The zero second sequence impedance ( Z) Zs Zm 4 j3 (ii) From equation (i) and (ii), Z Z 3Zm 3 j Zm j7 Hence, imaginary part of Z m is 7. Question 7 (Power Electronics) A single phase fully controlled rectifier is supplying a load with an anti-parallel diode as shown in the figure. All switches and diodes are ideal. Which one of the following is true for instantaneous load voltage and current? L o a d i + v (A) v and i (B) v and i (C) v and i (D) v and i Copyright Head Office : A/4-5, Smriti Nagar, Bhilai (C.G.), Contact : , Branch Office : Raipur, : , Bhopal, : 9-476, ndore, :

20 7 GATE 8 [Afternoon Session] Since 4 Ans. Sol. (C) Given : A single phase fully controlled rectifier with freewheeling diode. As the converter has freewheeling diode, hence the instantaneous output voltage will never be negative whether the converter is working under continuous mode or discontinuous mode. The instantaneous value of output current is always unidirectional as it is a AC to DC converter. Here we can think about two cases : Case : Discontinuous conduction mode Here the output current, i. Case : Continuous conduction mode Here the output current, i. Hence, the correct option is (C). Question 8 (Engineering Mathematics) Ans. Sol. The value of the directional derivative of the function in the direction of the vector p ij k is ( x, y, z) xy yz zx at the point (,, ) (A) (B).95 (C).93 (D).9 (A) xy yz zx Hence, the correct option is (A). Question 9 (Power System) ˆ ˆ ˆ x y z ˆ ˆ (,, ) 5iˆ 3ˆj kˆ p ( ˆ ˆ ˆ ˆ ˆ ˆ i j k) 564 pˆ (5i 3 j k) p 3 3 i j k ( y xz) i ( xy z ) j ( yz x ) k The positive negative and zero sequence impedances of a 5 MVA, three-phase, 5.5 kv, star-grounded, 5 Hz generator are j.pu, j.5 pu and j. pu respectively on the machine rating base. The machine is unloaded and working at the rated terminal voltage. f the grounding impedance of the generator is j. pu, then the magnitude of fault current for a b-phase to ground fault (in ka) is (up to decimal places). Ans Sol. Given : (i) Positive sequence impedance Z j. pu (ii) Negative sequence impedance Z j.5 pu (iii)zero sequence impedance Z j. pu (iv) Grounding impedance Zg j. pu More questions and video solutions of GATE - 8 are available in the Facebook group GATE ACADEMY (EE/EC/N). To join click here ˆ

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22 8 GATE 8 [Afternoon Session] Since 4 The magnitude of fault current in LG fault, 3VTH 3 3 f (pu) ZZ Z 3Zn Hence, the value of fault current in actual ampere can be calculated as f f f (pu) base VA 3V f VA f (pu) ka 3V Hence, the magnitude of fault current for a b-phase to ground fault is 73.5 ka. Question (Control System) Ans. Sol. 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 5s5. Over damped Q. 5 s s5. Critically damped R. 35 s 8s35. Underdamped (A) P-, Q-, R- (B) P-, Q-, R- (C) P-, Q-, R- (D) P-, Q-, R- (C) Option Characteristic equation Damping ratio ( ) Damping P. s 5s Under damping Q. s s 5 Critical damped R. s Over damped Hence, the correct option is (C). Question (Engineering Mathematics) Consider a non-singular square matrix A. f trace (A) = 4 and trace ( A ) = 5. The determinant of the matrix A is (up to decimal place). Ans. 5.5 Sol. Given : Trace ( A) Eigen value[ A], 4 Trace Copyright ( A ) Eigen value Trace, Det. = Prod. ( ) [ A ], 5 Head Office : A/4-5, Smriti Nagar, Bhilai (C.G.), Contact : , Branch Office : Raipur, : , Bhopal, : 9-476, ndore, :

23 9 GATE 8 [Afternoon Session] We have to find det (A) that means. det( A)? By squaring trace ( A ), we get ( ) (4) 6 Put the value of, from trace ( A ), then Since Hence, the determinant of the matrix A is 5.5. Question (Electrical Machine) A separately excited dc motor has an armature resistance Ra.5. The field excitation is kept constant. At an armature voltage of V, the motor produces a torque of 5 Nm at zero speed. Neglecting all mechanical losses, the no-load speed of the motor (in radians) for an armature voltage of 5 V is (up to decimal places.) Ans. 6 Sol. Given :. Separately excited DC motor. Field excitation ( ) = Constant Case : At zero speed (N = ), torque ( T ) 5 Nm, induced voltage ( E) Hence, T K a T K' a [as Constant] E V a R a V E a A Ra.5 T 5 K ' 4 a Case : No load condition, At no load, load torque ( TL ), E a T, and a nduced voltage, E K K' 5 V, and K ' Constant 4 E 5 6 rad/sec K ' 4 Hence, at no load motor will operate at 6 rad/sec. More questions and video solutions of GATE - 8 are available in the Facebook group GATE ACADEMY (EE/EC/N). To join click here

24 GATE 8 [Afternoon Session] Since 4 Question 3 (Power System) Consider a lossy transmission line with V and V as the sending and receiving end voltages, 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 Ans. Sol. VV (A) greater than X (B) Given : A lossy transmission line VV (B) less than X Case : Power with impedance (Z) Z V V VV (C) equal to X VV (D) equal to Z VV V P cos( ) cos z z Where, mpedance angle and Load angle To get maximum power, Hence, P max VV Vcos z z Case : Power with reactance, V V X VV P ' sin X To get maximum power, 9 Hence, ' VV Pmax X From above analysis, it is clear that Pmax ' Pmax Hence, the maximum power with resistance is less than the maximum power without resistance. Hence, the correct option is (B). Question 4 (Electromagnetic Field) A positive charge of nc is placed at (,,.) where all dimensions are in metres. Consider the x-y plane to be a conducting ground plane. Take 8.85 F/m. The Z component of the E field at (,,.) is closest to (A) V/m (B) V/m (C) V/m (D) V/m Copyright Head Office : A/4-5, Smriti Nagar, Bhilai (C.G.), Contact : , Branch Office : Raipur, : , Bhopal, : 9-476, ndore, :

25 GATE 8 [Afternoon Session] Since 4 Ans. (D) Sol. Given : Charge nc at (,,.). Since the charge is placed above conducting grounded plane there will be an image charge below the grounded conducting plane as per method of image concept. z = plane z nc (,,.) Conducting grounded plane z nc (,,.) P (,,.) nc (,,.) Electric field at any point is given by, QR E 3 4 R R displacement vector between charge to point of interest. Electric field at point P due to point charge nc, 9 ( ) ax ( ) ay (..) a z E 3 4 (.) E V/m Electric field point P due to point charge nc, 9 ( ) ax ( ) ay (..) a z E 3 4 (.3) Total electric field at point P due to both charges, E EE E V/m V/m Hence, the correct option is (D). Question 5 (Electrical Machine) A single-phase kva. V/ V. 5 Hz transformer has a voltage drop of 5% across its series impedance at full load. Of this, 3% is due to resistance. The percentage regulation of the transformer at full load with.8 lagging power factor is (A) 4.8 (B) 6.8 (C) 8.8 (D).8 Ans. (A) Sol. Given : (i) % R 3%, % Z 5% (ii) Power factor (pf) =.8 lagging Also, % X Z R 5 3 4% As at full load, x Fraction of loading =. Hence, voltage regulation (V.R.) x(% Rcos % X sin ) V.R. ( ) % Hence, the correct option is (A). More questions and video solutions of GATE - 8 are available in the Facebook group GATE ACADEMY (EE/EC/N). To join click here z = plane V = Equipotential surface

26 GATE 8 [Afternoon Session] Since 4 Q.6 to Q.55 carry two marks each Question 6 (Power System) The positive, negative and zero sequence impedances of a three phase generator are Z, Z and Z Ans. Sol. respectively. For a line-to-line fault with fault impedance the fault current with zero fault impedance. The relation between Z f, the fault current is f Z f and k is ( ZZ)( k) ( ZZ)( k) (A) Z f (B) Z f k k ( Z Z) k ( Z Z) k (C) Z f (D) Z f k k (A) Given : Line to line fault Case : Fault current with Z f k f, where f is f Z Z f E g Z 3E g f ZZ Z f Case : Fault current without Z f f Z Z E g As, Copyright f Z k f f 3E g Z 3Eg k 3E Z Z Z Z Z f ZZ k( ZZ) kz f ( ZZ)( k) Z f k Hence, the correct option is (A). g Head Office : A/4-5, Smriti Nagar, Bhilai (C.G.), Contact : , Branch Office : Raipur, : , Bhopal, : 9-476, ndore, :

27 3 GATE 8 [Afternoon Session] Since 4 Question 7 (Signals & Systems) Consider the two continuous-time signals defined below : t, t t, t x() t, x() t, otherwise, otherwise These signals are sampled with a sampling period of T =.5 seconds to obtain discrete-time signals x [ n] Ans. Sol. 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. (A) x () t x ( n) Sampling T.5Sec s n x () t x ( n) Sampling T.5Sec s n x( n) (i) n E x( n) (ii) n E From equation (i) and (ii) E E Hence, the correct option is (A). More questions and video solutions of GATE - 8 are available in the Facebook group GATE ACADEMY (EE/EC/N). To join click here

28 4 GATE 8 [Afternoon Session] Since 4 Question 8 (Power Electronics) A phase controlled single phase rectifier, supplied by an AC source, feeds power to an R-L-E load as shown in the figure. The rectifier output voltage has an average value given V (3 cos ), where Vm 8 volts and is the firing angle. f the power delivered to the lossless battery is 6 W, in degree is (up to decimal places). V m Vm sin( t) V mh 8 V Battery Ans. 9 Sol. Given : Power transfer to the 8 V battery, E 6W 6 Hence, A 8 V 8 V (3cos ) (3cos ) V 4(3 cos ) (i) The relation between V and E is given by, V RE 8 V 4 8 V From equation (i), V 4(3 cos ) cos 3 4 cos () 9 Hence, the firing angle ( ) will be 9. Question 9 (Electrical & Electronics Measurement) A - Ampere moving iron ammeter has an internal resistance of 5 m and inductance of. mh. A shunt coil is connected to extend its range to - Ampere for all operating frequencies. The time constant in milliseconds and resistance in m of the shunt coil respectively are (A), 5.55 (B), (C).8,.55 (D)., Copyright Head Office : A/4-5, Smriti Nagar, Bhilai (C.G.), Contact : , Branch Office : Raipur, : , Bhopal, : 9-476, ndore, :

29 5 GATE 8 [Afternoon Session] Since 4 Ans. (A) Sol. Given : m A (meter current) A (extended current) R m L m R 5 m (meter resistance) m L. mh (meter inductance) m Z R ( L ) Z R L m sh sh sh sh m m ( m) m A A sh 9A R sh L sh m sh m sh R R sh R R m sh m L R sh sh L Rm m sh m For the meter to be independent of frequency time constant of meter should be equal to time constant of shunt. i.e. sh m m Rsh R 9 sh m Rm 5 m Rsh 5.55 mω 9 9. mh Now, m sec and as system is independent of frequency m sh sec 5 m Hence, the correct option is (A) Question 3 (Power System) Consider the two bus power system network with given loads as shown in the figure. 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, QG and line reactive power loss ( Q loss ) respectively are G jqg. 5 j5 j. Q loss. j. (A) 5.,.68,.68 (B) 6.34,.,.34 (C) 6.34,.34,.68 (D) 5.,.34,.34 More questions and video solutions of GATE - 8 are available in the Facebook group GATE ACADEMY (EE/EC/N). To join click here G 5 jqg

30 6 Ans. (C) Sol. Given : GATE 8 [Afternoon Session] P S j. Since 4 G 5 j5 QS Q loss QR j. G At G load demand is pu, G supplies only 5. Remaining supplied by G through transmission line. At : P s VV S X L R sin 5 sin. 3 VS VSVR Q S cos X X Q S G G load L L cos3.34 p.u... VV V QR cos X X S R R L.. Q R cos3 L.34 p.u. Qloss QS QR Qloss.34 (.34).68p.u. Q Q Q QG p.u. At G : Q Qload ( Q ) G S R QG (.34).34 p.u. Hence, the correct option is (C). Question 3 (Power Electronics) The figure shows two buck converters connected in parallel. The common input dc voltage for the converters has value of V. The converters have inductors of identical value. The load resistance is. The capacitor voltage has negligible rippel. 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. Assume 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). Copyright Head Office : A/4-5, Smriti Nagar, Bhilai (C.G.), Contact : , Branch Office : Raipur, : , Bhopal, : 9-476, ndore, :

31

32 7 GATE 8 [Afternoon Session] Since 4 i s L S V + C Switch control signals S L S t S.5 ms ms t Ans. 5 Sol. Given : (i) V V s (ii) nterleaved converters. (iii)load resistance (iv) Switching frequency = khz Case : When S ON, S Open V V (As the average voltage across inductor is zero at steady state) S Case : When S ON, S Open V V (As the average voltage across inductor is zero steady state) S Hence, V is always equal to V S V A R Since the load current is equal divided in two inductive element. Hence, L( avg ) L( avg ) 5A Now D and s( avg ) DL( avg ) s( avg ) L( avg ) T.5 Where, D Duty ratio ON.5 T Hence, ( ) D ( ) A s avg L avg ( ) D ( ).55 5 A s avg L avg Hence, the average value of switch ( S ) current is 5 A. More questions and video solutions of GATE - 8 are available in the Facebook group GATE ACADEMY (EE/EC/N). To join click here

33 8 GATE 8 [Afternoon Session] Question 3 (Electrical Machine) Ans. 85 Sol. Since 4 A V DC series motor, when operating from rated voltage while driving a certain load, draws A current and runs at rpm. 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 rpm (rounded to the nearest integer) is. Given : DC series motor (i) Terminal voltage, VT V (ii) a A N rpm a (iii) R R R T a se R T R se (iv) (Linear magnetization) R a V T (v) T.44T A DC series motor circuit is shown below, ZPa For motor, T ka A For, DC series motor a E T a T T.44 a a. A a a Also, E N E E N N E E N N N a E Ea (i) E VT a RT 9 V E VT art 88 V 88 N rpm 9. Hence, the speed of the motor (rounded to the nearest integer) is 85 rpm. Copyright Head Office : A/4-5, Smriti Nagar, Bhilai (C.G.), Contact : , Branch Office : Raipur, : , Bhopal, : 9-476, ndore, :

34 9 GATE 8 [Afternoon Session] Since 4 Question 33 (Analog Electronics) n the circuit shown in the figure, the bipolar junction transistor (BJT) has a current gain. The base-emitter voltage drop is a constant, VBE.7 V. The value of the Thevenin equivalent resistance R Th (in ) as shown in the figure is (up to decimal places). 5 V k k RTh.7 V Ans..76 Sol. Given : Common collector amplifier RTH R k Where, R is output impedance as seen in the figure equal to output impedance of common collector amplifier R gm C gm VT Applying KVL in B-E loop,.7.7 ( ) B B B B C B 5 V C k.7 V E B R k RTh B ma g C m B ma C 4 A/V V T Rs hie gm R hfe g RTH R.86 Hence, the correct answer is.86. m More questions and video solutions of GATE - 8 are available in the Facebook group GATE ACADEMY (EE/EC/N). To join click here

35 3 GATE 8 [Afternoon Session] Since 4 Question 34 (Network Theory) The voltage v(t) across the terminals a and b as shown in the figure, is a sinusoidal voltage having a frequency = 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). vt () a it () L Z F Ans. 5 Sol. Given : a b L Z F b j j j C Z jl C C jl j C C Since, the power factor is unity, unity power factor represents resonance. So, imaginary part of Z is zero only real part exist. Re( Z) C 6 ( C) ( ) C Re( Z) 5 Hence, the magnitude of the impedance Z seen between the terminals a and b is 5 Ω. Question 35 (Digital Electronics) Which one of the following statements is true about the digital circuit shown in the figure. D Q D Q D Q f out C C C Copyright f in (A) t can be used for dividing the input frequency by 3. (B) t can be used for dividing the input frequency by 5. (C) t can be used for dividing the input frequency by 7. (D) t cannot be reliably used as a frequency divider due to disjoint internal cycles. Head Office : A/4-5, Smriti Nagar, Bhilai (C.G.), Contact : , Branch Office : Raipur, : , Bhopal, : 9-476, ndore, :

36 3 GATE 8 [Afternoon Session] Since 4 Ans. 5 Sol. Given : DA QA DB QB DC QC f out C C C f in From the above sequential circuit, Present state FF inputs Next state A B C DA B D C B A DC B A B State diagram : C unique states MOD 5 counter divide by 5 counter t can be used for dividing the input frequency by 5 T Q A Q B Q C 5T More questions and video solutions of GATE - 8 are available in the Facebook group GATE ACADEMY (EE/EC/N). To join click here

37 3 GATE 8 [Afternoon Session] Question 36 (Network Theory) Since 4 The voltages across the circuit in the figure, and the current through it are given that the following expressions : V( t) 5 cos( t 6) and i () t 5 X cos( t) i i Where rad/sec. f the average power delivered to the circuit is zero then the value of X (in ohm) is (upto two decimal places) it i( ) + Vt i( ) Electric Circuit Ans. Sol. Given : (i) V( t) 5 cos( t 6) i (ii) i () t 5 X cos t i The average power is given by, P V V cos avg Where, V cos Fundamental power. Hence, P avg X (55) cos 6 X 5 cos6 X 5 X Hence, the value of X is. Question 37 (Engineering Mathematics) Let A 3 and B A A 4A 5 where, is the 33identity matrix. The determinant of B is.(upto one decimal places) Copyright Head Office : A/4-5, Smriti Nagar, Bhilai (C.G.), Contact : , Branch Office : Raipur, : , Bhopal, : 9-476, ndore, :

38 33 GATE 8 [Afternoon Session] Since 4 Ans. Sol. To find the eigen value of A, A ( )( )( ),, 3 B A A 4A 5 The eigen value of matrix B is given by, (4) ( ) ( ) [4 ( )] 5 B 3 Hence, the determinant of B is. Question 38 (Signals & Systems) The signal energy of the continuous time signal xt ( ) ( t) ut ( ) ( t) ut ( ) ( t3) ut ( 3) ( t4) ut ( 4) is (A) 3 (B) 7 3 (C) 3 (D) 5 3 Ans. (D) Sol. we know that, rt () tut () put t t r( t) ( t) u( t ) put t t r( t) ( t) u( t ) put t t3 r( t3) ( t3) u( t 3) put t t4 r( t4) ( t4) u( t 4) Given : f( t) ( t) u( t) ( t) u( t) ( t3) u( t3) ( t4) u( t 4) f( t) rt ( ) rt ( ) rt ( 3) rt ( 4) f() t 3 4 t f() t ( t) f() t ( t3) 4 3 f () t t f() t t3 f () t t 4 More questions and video solutions of GATE - 8 are available in the Facebook group GATE ACADEMY (EE/EC/N). To join click here

39 34 GATE 8 [Afternoon Session] Since E f() t dt ( t) dt (4 t) dt E ( t t) dt (6t 8 t) dt t ( ) t E tt t t t E 4 (3 ) E 3 Hence, the correct option is (D). Question 39 (Power System) The per unit power output of a salient pole generator which is connected to an infinite bus is given by the expression, P.4sin.5sin, where is the load angle. Netwton Raphson method is used to calculate the value of for P.8 pu. f the initial guess is 3, then its value (in degree) at the end of the first iteration is (A) 5 (B) 8.48 (C) 8.74 (D) 3. Ans. (C) Sol. Given : P.4sin.5sin, P.8, 3 f ( ).4sin.5sin.8 f ( ) n n f '( ).4sin.5sin P.4 cos.3cos.4sin 3.5sin rad 6.4 cos 3.3cos 6 (3) 8.74 Hence, the correct option is (C). Question 4 (Control System) The unit step response y(t) of a unity feedback system with open loop transfer function K GsHs () () is shown in the figure. The value of K is (up to decimal places). ( s) ( s) yt () Copyright. t (sec) Head Office : A/4-5, Smriti Nagar, Bhilai (C.G.), Contact : , Branch Office : Raipur, : , Bhopal, : 9-476, ndore, :

40 35 GATE 8 [Afternoon Session] Since 4 Ans. 8 K + Sol. Given : Gs () X() s Gs () Y() s ( s) ( s) CLTF is given by, Ys () GsHs () () Gs () [Unit feedback system] X () s G() s H() s G() s K Y() s ( s) ( s) X() s K ( ) s ( s ) Y() s K X () s ( s)( s) K K Y() s s ( s) ( s) K From final value theorem, y( ) lim sy( s).8 s [From time response shown in the figure steady state value in time domain is.8] K.8 K K.6.8 K K 8 Hence, the value of K is 8. Question 4 (Network Theory) The equivalent impedance Z eq for the infinite ladder circuit shown in the figure is a j9 j9 Ans. Sol. b Z eq j5 j (A) j (B) j (C) j3 (D) 3 (A) Given : The circuit is given below, j9 j9 a j5 j b Z eq j5 j j5 j More questions and video solutions of GATE - 8 are available in the Facebook group GATE ACADEMY (EE/EC/N). To join click here

41 36 GATE 8 [Afternoon Session] Since 4 The given network consists only reactive element, the equivalent figure can be represented as, a j9 j9 b Z eq j5 j j5 j The above circuit can be re-drawn as shown below, The equivalent impedance ( Z eq ) is given as j4 Zeq Zeq j9 j4 Zeq ( j9) Zeq ( j4) Zeq 36 Zeq Zeq j4 Zeq ( j9) Zeq 36 Zeq () j (3) j Zeq 36 ( Z j) ( Z j3) Zeq eq eq j [ Z j3 is neglected as the whole circuit is a inductive nature] eq a b Z eq Z eq j9 j5 j Hence, the correct option is (A). Question 4 (Control System) The number of roots of the polynomial, s s 7s 4s 3s 73s 5s, in the open left half of the complex plane is (A) 3 (B) 4 (C) 5 (D) 6 Ans. (A) Sol Given : Characteristic equation, s s 7s 4s 3s 73s 5s. The R-H table is given by, 7 s s s s s 3 96 s 4 s 7.67 s Z eq Copyright Head Office : A/4-5, Smriti Nagar, Bhilai (C.G.), Contact : , Branch Office : Raipur, : , Bhopal, : 9-476, ndore, :

42 37 GATE 8 [Afternoon Session] Since 4 From the above table number of sign change in the first column is 4. So, the number of left hand pole are s 48s s x 8x 48x x 6x5 x 3 j4 s 3 j4 s 3 j4 s 3 j4, s 3 j4 x jy 3 j4 ( x jy) 3 j4 x y jxy 3 j4 x y 3 xy 4 xy ( x y ) ( x y ) 4x y ( x y ) ( 3) x y 5 x y 3 x y 5 x x x x, x y, y xy ve s j s j Now, x jy 3 j4 x y jxy 3 j4 x y 3 xy 4xy ( x y ) ( x y ) 4 x y ( 3) 4( ) 5 x y 5 + j j j + j j More questions and video solutions of GATE - 8 are available in the Facebook group GATE ACADEMY (EE/EC/N). To join click here

43 38 Copyright GATE 8 [Afternoon Session] Since 4 x y 3 x x x xy 4 xy x, x y, y s j j Hence, the correct option is (A). Question 43 (Engineering Mathematics) z f C is a circle z = 4 and f( z), then ( ) ( z 3z) f z dz is C (A) (B) (C) (D) Ans. (B) z Sol. Given : dz; z 4 iy ( z) ( z) Finding roots of denominator, ( z) ( z) O 4 z,,, [Multiple poles] Cauchy integral formula for multiple poles n f( z) i d dz f ( z) n n z ( za) ( n)! dz As both the poles are inside the closed curve, hence z z /( z) z /( z) dz dz C ( z) ( z) ( z) ( z) z i d z d z dz C ( z) ( z) ( z) dz ( z) dz ( z) z ( z) zz ( z) ( z) zz ( z) dz i C 4 4 ( z) ( z) ( z) ( z) z 48 dz i i ( 4 4) C ( z) ( z) Hence, the correct option is (B). Head Office : A/4-5, Smriti Nagar, Bhilai (C.G.), Contact : , Branch Office : Raipur, : , Bhopal, : 9-476, ndore, : C (circle) x

44 39 GATE 8 [Afternoon Session] Since 4 Question 44 (Electromagnetic Field) The capacitance of an air-filled parallel-plate capacitor is 6 pf. When a dielectric slab whose thickness is half the distance between the plates, is placed on one of the plates covering it entirely, the capacitance becomes 86 pf. Neglecting the fringing effects, the relative permittivity of the dielectric is (up to decimal places). Ans..53 Sol. Given : Capacitance of an air-filled parallel plate capacitor, A C 6 pf (i) d Overall capacitance, C 86pF d eq Plate area A Plate area A Air d / d / Fig. Fig. From fig., after filling with dielectric it has become series combination of two capacitances overall capacitance for series combination is given by, CC Ceq 86 pf (ii) C C A r C and C A d / d / From equation (ii), A r A d/ d/ A ( r ) Ceq A r A d / ( r ) d/ d/ r 86 r 86( r) r r r.59 Hence, the relative permittivity of the dielectric is.53. Question 45 (Power Electronics) A dc to dc converter shown in the figure is charging a battery bank, B whose voltage is constant at 5 V. B is another battery bank whose voltage is constant at 5 V. The value of the inductor. L is 5 mh and the ideal switch. S is operated with a switching frequency of 5 khz with a duty ratio of.4. Once the 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). Plate area A Plate area A More questions and video solutions of GATE - 8 are available in the Facebook group GATE ACADEMY (EE/EC/N). To join click here Air Dielectric ( r ) C C

45 4 GATE 8 [Afternoon Session] i L 5mH D Since 4 B 5 V B 5 V Ans. Sol. The given circuit is a step up chopper TON DT f 5.8msec T T T..8. msec OFF ON Device working status : Case : When switch is ON V L i i s 5 V 5 V di VL L 5 dt di dt L DT DT A Case : We have to check, whether we have continuous on discontinuous conduction. Hence we find T. OFF new During TOFF Switch is open and diode is ON. V L i i s 5 V 5 V Copyright Head Office : A/4-5, Smriti Nagar, Bhilai (C.G.), Contact : , Branch Office : Raipur, : , Bhopal, : 9-476, ndore, :

46 4 GATE 8 [Afternoon Session] V L di L ( ) dt T L OFF Since 4 T OFF T OFF LL msec As T( OFF )new T OFF. Hence conduction will be discontinuous. i L i s T ON T ( OFF) new T OFF t Now, L or T T T ON ( OFF )new s( avg) 3 3 s( avg).4. 5 (.4) A.8[ 3. f ] Hence, P V ( ) W avg s s avg Calculation of output power : i T ON T ( OFF) new t ( avg ) T T ( OFF )new ( avg ) ( avg ) A Hence, P V.85 W 3 3 More questions and video solutions of GATE - 8 are available in the Facebook group GATE ACADEMY (EE/EC/N). To join click here

47 4 GATE 8 [Afternoon Session] Question 46 (Digital Electronics) Ans. Sol. Since 4 Digital input signals A, B, C with A as the MSB and C as the LSB are used to realize the Boolean function F m m m3m5 m7, where m i denotes the i th minterm. n 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) ACBC AC (B) F m m m3m5 m7 Also, F has don t care at m, F( A, B, C) m(,,3,5,7) () BC A A A BC BC BC BC F m(,,,3) A F m(,3,5, 7) C F FF A C Hence, the correct option is (B). Question 47 (Electrical Machine) A transformer with toroidal core of permeability is shown in the figure. Assuming uniform flux density across the circular core cross - section of radius of r R, and neglecting any leakage flux, the best estimate for the mean radius R is i sin t p r i s ~ v Vcost NP p R NS vs (A) Vr N P (B) r NPNS V (C) Vr N P (D) r N P V Copyright Head Office : A/4-5, Smriti Nagar, Bhilai (C.G.), Contact : , Branch Office : Raipur, : , Bhopal, : 9-476, ndore, :

48 43 GATE 8 [Afternoon Session] Since 4 Ans. (D) Sol. Given : Transformer with toroidal core of permeability. From the given figure, r equivalent radius of core Ar cross sectional area of core R mean radius of core l R= mean length of core The applied input voltage at primary, d d Ni P P vp NP NP [ RL Reluctance] dt dt RL NP d NP vp ( sin t) cost V cos t RL dt RL Where, NP NP NPA NPr V R l L l R A Hence, NP r R V Hence, the correct option is (D). Question 48 (Engineering Mathematics) 3 Let f ( x) 3x 7x 5x 6. The maximum value of f ( x ) over the interval [, ] is (upto one decimal place) Ans. Sol. 3 Given : f ( x) 3x 7x 5x 6 To find the maximum value of the function f ( x ), f '( x) and f "( x) f '( x) 9x 4x5 5 x, 9 f "( x) 8x 4 f "( x) x 5 f "( x) x 9 9 x 5 9 f (x) f( x) max Hence, the function f ( x ) has the maximum value of. More questions and video solutions of GATE - 8 are available in the Facebook group GATE ACADEMY (EE/EC/N). To join click here

49 44 GATE 8 [Afternoon Session] Question 49 (Electrical Machine) Since 4 A 3-phase 9 kva, 3 kv/ 3kV ( /Y). 5 Hz transformer has primary (high voltage side) resistance per phase of.3 and secondary (low voltage side) resistance per phase of.. ron loss of the transformer is kw. The full load % efficiency of the transformer operated at unity power factor is (up to decimal places). Ans Sol. Given : Three phase transformer (i) kva = 9 (ii) Primary line voltage V l ( ) 3 kv (iii)secondary line voltage V l (Y) 3 kv (iv) Primary resistance per phase R ph ( ).3 (v) Secondary resistance per phase R ph (Y). (vi) ron loss ( Pi ) kw The percentage efficiency of the transformer is given by, x(kva)cos % x(kva)cospi x Pcufl. (i) At full load, fraction of loading ( x) Where, P R R. (ii) cufl 3 ( ) 3 (Y) ph ph phy ph ph From equation (ii), cufl 9 9 A 3V 33 l and P phy kw Hence, percentage efficiency is given by, 9 % % Hence, the full load efficiency of the transformer is %. Question 5 (Signals & Systems) A 3V 3 3 The Fourier transform of a continuous-time signal xt () is given by X ( ),, where ( j ) j and denotes frequency. Then the value of ln xt ( ) at t is (up to decimal place). (ln denotes the logarithm to base e). ly Copyright Head Office : A/4-5, Smriti Nagar, Bhilai (C.G.), Contact : , Branch Office : Raipur, : , Bhopal, : 9-476, ndore, :

50 45 GATE 8 [Afternoon Session] Since 4 Ans. Sol. We know that Put a e at FT u() t a j e u() t j Apply multiplication by t property [frequency differentiation property] d i.e., tx() t j X ( ) d t d te u() t j d j t t te u() t j t ln xt ( ) ln te ut ( ) t t ln() ( t) ln u( t) ln( t) ln e t ln u( t) ln() t Question 5 (Electrical Machine) The equivalent circuit of a single phase induction motor is shown in the figure, where the parameters are R R' Xl Xl', X m 4 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). R jx l V X j M X j M R ' s X ' j l R ' ( s) X ' j l More questions and video solutions of GATE - 8 are available in the Facebook group GATE ACADEMY (EE/EC/N). To join click here

51 46 Ans..6 Sol. GATE 8 [Afternoon Session] Given : Single phase induction motor, A single phase induction motor having, R R Since 4 X 4 X m X As the single phase induction motor is running approximately at synchronous speed ( N ), hence Ns Nr Ns Ns Slip ( s) [as Nr Ns] N N s s The equivalent circuit under this condition is shown below, s a R X l j X M j R ' s X ' l j6 Z eq X M j R ' 3 ( s) 4 X ' l j6 b The equivalent impedance, (3 j6) j Zeq j j 3 j6 j Zeq j The input power factor is given by, (cos ) cos lagging. Hence, the no load lagging power factor of the motor is.6. Copyright Head Office : A/4-5, Smriti Nagar, Bhilai (C.G.), Contact : , Branch Office : Raipur, : , Bhopal, : 9-476, ndore, :

52 47 GATE 8 [Afternoon Session] Since 4 Question 5 (Network Theory) A DC voltage source is connected to a series L-C circuit by turning on the switch S at time t = as shown in the figure. Assume i(), v(). Which one of the following circular loci represents the plot of i(t) versus v(t)? S it () 5V t L H + C F vt () (A) it () (B) it () vt () 5 vt () 5 (C) it () (D) it () 5 vt () 5 vt () Ans. (B) Sol. Given : The circuit with i() v() S L H 5V t + it () F vt () The equivalent circuit is represented as, S 5 s + () s F s More questions and video solutions of GATE - 8 are available in the Facebook group GATE ACADEMY (EE/EC/N). To join click here

53 48 GATE 8 [Afternoon Session] 5 () s 5 s s s s it () 5sint t t V() t i() t dt 5sint C V() t 5( cos) t it () 5sint Since 4 5 B A 9 8 t 5cost 5 E vt () 5(cos) t C B 5 D B D t A 9 8 E t C The circular loci of it () is given by, it () 5 vt () Hence, the correct option is (B). Question 53 (Control System) Consider a system governed by the following equations dx () t x() t x() t dt dx() t x() t x() t dt The initial conditions are such that x() x(). Let xf lim x( t) and x f lim x( t). Which one t t of the following is true? (A) xf x f (B) x f xf (C) xf x f (D) xf x f Copyright Head Office : A/4-5, Smriti Nagar, Bhilai (C.G.), Contact : , Branch Office : Raipur, : , Bhopal, : 9-476, ndore, :

54 49 GATE 8 [Afternoon Session] Since 4 Ans. Sol. (C) Converting to state model x x x x A s ( s A) s s ( s A) s s s s ss ( ) ss ( ) s ( s) s ( s) ( s A) s ss ( ) ss ( ) s ( s) s ( s) t t e e () t t t e e t t t e e x() x() x() e [ x() x()] xt () () tx() t t t e e x() x() x() e [ x() x()] x() x() xf lim x( t) t x() x() Hence, xf x f Hence the correct option is (C). Question 54 (Engineering Mathematics) As shown in the figure, C is the arc from the point (3, ) to the point (, 3) on the circle x y 9. The value of the integral ( y xy) dx ( x xy) dy is (up to decimal places). C y (3, ) (3, ) x More questions and video solutions of GATE - 8 are available in the Facebook group GATE ACADEMY (EE/EC/N). To join click here

55 5 GATE 8 [Afternoon Session] Since 4 Ans. Sol. Given : x y 9 this equation is given for circle in first quadrant and from figure we know centre (, ) and radius of 3. For closed line integral we applies green s theorem. Green s theorem, dx dy dx dy x y C S..(i) Here, y xy (given) and x xy (given) So, x y and y x x y Put the above values in equation (i), we get dx dy x y (x y) dx dy C C dx dy S Hence the value of integral is. Question 55 (Network Theory) A three-phase load is connected to a three-phase balanced supply as shown in the figure. f Van V, Vbn V and Vcn 4 V (angles are considered positive in the anti-clockwise direction), the value of R for zero current in the neutral wire is (up to decimal places). a R n c j j Ans Sol. Given : V an V bn V 4 cn a b R n c b j j Copyright Head Office : A/4-5, Smriti Nagar, Bhilai (C.G.), Contact : , Branch Office : Raipur, : , Bhopal, : 9-476, ndore, :

56 5 GATE 8 [Afternoon Session] a b c n 4 R j j 5 R Since j j 7.3 j. (i) R Equating the real part of the equation (i), 7.3 R R Hence, the value of R is More questions and video solutions of GATE - 8 are available in the Facebook group GATE ACADEMY (EE/EC/N). To join click here

57 5 Answer Key : GATE 8 [Afternoon Session] General Aptitude Since 4. C. A 3. D 4. A 5. D 6. C 7. C 8. (*) 4/ 9. C. C Technical Section. D A 7. D C. A. B. D 3. B 4. A 5. B C 8. A C B 4. D 5. A 6. A 7. A A 3. C B D 39. C A 4. A 43. B B 47. D B 53. C Copyright Head Office : A/4-5, Smriti Nagar, Bhilai (C.G.), Contact : , Branch Office : Raipur, : , Bhopal, : 9-476, ndore, :

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