GATE-2020 Electrical Engineering (EE) Past Paper PDF

Branch: Electrical Engineering Time: 3 Hours PRE- GATE-2020 Marks: 100 GATE-2020 General Aptitude (GA) Q. 1 – Q. 5 carry ONE mark each. Ans: (C)...

Branch: Electrical Engineering Time: 3 Hours PRE- GATE-2020 Marks: 100 GATE-2020 General Aptitude (GA) Q. 1 – Q. 5 carry ONE mark each. Ans: (C) Sol: Troop consists of monkeys just as a colony 01. Fill in the blank with an appropriate consists of bacteria. phrase Jobs are hard to _____ these days 03. Choose the most appropriate word from (A) Come by Shared on www.ErForum.Net (B) Come down the options given below to complete the (C) Come of (D) Come from following sentence: Ans: (A) If you had gone to see him, he _______ Sol: ‘Come by’ means to manage to get delighted. something. (A) Would have been (B) Will have been 02. The question below consists of a pair of (C) Had been related words followed by four pairs of (D) Would be words. Select the pair that best expresses Ans: (A) the relation in the original pair. Ans: ‘A” conditional tense type 3 grammatical MONKEY : TROOP: code is (A) sheep : hard If +had+V3, would +have+V3 (B) elephant : Parliament (C) bacteria : Colony (D) wolves : School ACE Engineering Academy GATE-2020 Electrical Engineering (EE) ACE Engineering Publications 04. Which of the following options is closest Q. 6 – Q. 10 carry TWO marks each. in meaning to the underlined word? European intellectuals have long debated 06. Critical reading is a demanding process. To the consequences of the hegemony of read critically, you must slow down your American popular culture around the world. reading and, with pencil in hand, perform (A) regimen (B) vastness specific operations on the text mark up the (C) dominance (D) popularity text with your reactions, conclusions, and Ans: (C) questions, then you read, become an active Sol: Dominance means influence or control over participant. another country, a group of people etc. This passage best supports the statement that (A) Critical reading is a slow, dull but 05. How many one-rupee coins, 50 paise coins essential process. 25 paise coins in total of which the numbers (B) The best critical reading happens at are proportional to 5, 7 and 12 are together critical times in a person’s life. work ₹115? (C) Readers should get in the habit of questioning the truth of what they read. (A) 50, 70, 120 (B) 60, 70, 110 Shared on www.ErForum.Net (D) Critical reading requires thoughtful and (C) 70, 80, 90 (D) None of these careful attention. Ans: (A) Ans: (D) Sol: (51+70.5+120.25) x = 115 Sol: Choice (A) is incorrect because the author (5+3.5+3)x = 115 never says that reading is dull. 11.5x = 115 Choice (B) and (C) are not support by the x = 10 paragraph.  Number of one rupee coin = 5x = 510 Choice (D) is correct as it is implied in the = 50 entire passage. Number of 5-paise coin = 7x = 710 = 70 Number of 25-paise coin = 12x = 1210 07. Anil’s house faces east. From the back-side = 120 of the house, he walks straight 50 metres, then turns to the right and walks 50m again finally, he turns towards left and stops after walking 25 m Now Anil is in which direction from the starting point? 2 ACE Engineering Academy Shared on www.ErForum.Net GATE-2020 Electrical Engineering (EE) ACE Engineering Publications (A) South-east (B) South-west (D) None of these (C) North-east (D) North- west Ans: (B) Ans: (D) Sol: A’s share : B’s share : C’s share Sol: The movement of Anil are shown in the (504+258) : (455+22.57) : (707) adjoining figure 400 : 382.5 : 490 800 : 765 : 980 25m D C 160 : 153 : 196 Total profit = ₹ 254 50m 160 Profit of A =  254 B A 160  153  196 160 He starts walking from back of his house   254 = ₹80 509 (i.e) towards west now, the final position is 153 D, which is to the north west of his starting Profit of B =  254 = ₹76 509 point A. 196 Profit of C =  254 = ₹98 509 Shared on www.ErForum.Net 08. A and B enter into a partnership, A puts in ₹ ∵Hence option ‘B’ is correct. 50 and B puts in ₹ 45. At the end of 4 months, A withdraws half his capital and at 09. A sum of ₹25400 was lent out in two parts, 1 the end of 5 months B withdraws of his, 1 2 one of 12% and the other at 12 % of the 2 C then enters with a capital of ₹ 70. At the total annual income is ₹3124.2, the money end of 12 months, the profit of concern is ₹ lent at 12% is ________. 254, how can the profit be divided among A, (A) ₹15240 (B) ₹25400 B and C? (C) ₹10160 (D) ₹31242 (A) ₹ 76, ₹ 80 and ₹ 98 Ans: (C) (B) ₹ 80, ₹ 76 and ₹ 98 Sol: Overall rate of interest 3124.2 (C) ₹ 76, ₹ 98 and ₹ 80 100  12.3% 25400 ACE Engineering Academy 3 GATE-2020 Electrical Engineering (EE) ACE Engineering Publications (A) 4 (B) 3 2nd part 1st part (C) 2 (D) 1 1 12% 12 % 2 Ans: (A) Sol: Sector angle for operators in the year 1990 12.3% 18   360  81 80 0.3% Sector angle for operator in the year 1995 0.2% 32   360 = 85.33 ≃85%  The sum will be divided in the ratio 135 0.2:0.3 (or) 2:3 ∴Required difference = 85– 81= 4 2  The sum lent at 12% = 25400  5 Q. 1 – Q. 25 carry ONE mark each. = ₹10160. 01. A 2 winding transformer supplies a leading 10. The following question is to be answered on power factor load at rated secondary the basis of the table given below. voltage. For a given load current, if Shared on www.ErForum.Net magnitude of the load power factor varies, Category of Number of Number of the core losses, as compared with the no personnel staff in the staff in the load core losses, will year-1990 year-1995 (A) remain unchanged. Data preparation 18 25 (B) increase. Data control 5 8 (C) decrease. Operators 18 32 (D) either increase, remain constant, or Programmers 21 26 decrease. Analysts 15 31 Ans: (D) Managers 3 3 Sol: Assume (for simplicity) that the Total 80 135 approximate equivalent circuit shown in fig.1 correctly represents the transformer. What is the increase in the sector angle for operators in the year 1995 over the sector angle for operators in the year 1990? 4 ACE Engineering Academy  GATE-2020 Electrical Engineering (EE) ACE Engineering Publications is opened. If the field flux dies away I2 , req jxeq  +ve completely in 0.075 sec, average field + + + induced emf during the 0.075 sec period is Supply V2  V2 0 Leading (Not rated V1Rc jxm V (V2 = rated (capacitive __________ V. 1 Value) Value). Load) Ans: (1600) – – – PZ 400 200 Equivalent circuit ref secondary. Sol: 1. K   . Fig.1 2A 2  An approximate but widely used expression 2. Initial operation: for (V1 –V2) is 2 2.1.   300   10 r / sec(mech) 60 (V1  V2) = I2(req cos  – xeq sin ). 2.2.Flux/pole =  (unknown). On no load, I2 = 0. 200  Drop across the impedance (req +jxeq) is 2.3. E   (10 )  200  V22 zero and V1 =V2. No load core losses =. (from given data). Rc  = 0.1 Wb. If cos is large, sin  will be small. 3. The field flux decreases from  = 0.1 (Assume 0    90). (V1 –V2) is positive. Wb to zero Shared on www.ErForum.Net in 0.075 sec. V 2 V1 >V2. core losses 1 > no load core d 0.1 4 Rc   Wb/sec. dt ave 0.075 3 losses. (Ignore the negative sign). If cos is small, sin  will be large. (V1 – 4. With 1200 field turns, average field 2 V induced emf during the period V2) is negative. V1 < V2. core losses 1 < no Rc 4 = 1200   1600V. load core losses. 3 There will be some  for which V1 – V2 = 0. V12 03. A 3- star connected 400 V (line to line), 8 Then core loss equals no load core loss. Rc kW (output) synchronous motor with full load efficiency of 88% operating with 02. The armature of a 2-pole, 200 V dc minimum possible current. The separately excited generator has 400 synchronous impedance per phase is 8  conductors and runs at 300 rpm. The number with negligible resistance. The induced of field turns are 1200. Now the field circuit ACE Engineering Academy 5 Shared on www.ErForum.Net GATE-2020 Electrical Engineering (EE) ACE Engineering Publications emf/ph is Sol: P + jQ (A) 233.7 V (B) 238.7 V (C) 248.7 V (D) 253.7 V 20 kVAR  Qc  80 kVAR load Qc Ans: (D) S = 200 kVA cos   = 0.8 lag 400 Sol: VL = 400 V  Vph = = 231 V 3 Motor output = 8 kW Load real and reactive power consumptions, Pin = Pout = 8 kW = 9091 W P  S. cos  = 200  0.8 = 160 kW  0.88 Q   S. sin   = 200  0.6 = 120 kVAR Pin = 3 VLILcos ; current will be Net real power drawn, P = P = 160 kW minimum at upf Net reactive power drawn, Q = Q   Q C Pin 9091  IL = = = 13.12 A 3VL 3  400 = 120 – QC E= (V cos   I a R a ) 2  (V sin   I a X s ) 2 Net apparent power, S = P2  Q2 = 160 2  120  Q C  2 = (231  1  0) 2  (231  0  13.12  8) 2 = 253.7 V Shared on www.ErForum.Net Minimum net apparent power (S min) occurs for lowest value of ‘Q’ or highest value of 04. A variable shunt capacitor bank of reactive ‘QC’ power rating ‘Qc’ connected in parallel with Smin = 1602  120  802 a load. = 165 kVA The load consumes an apparent power of 200 kVA at 0.8 power factor lagging. The 05. A 150 bus power system network consists reactive power ‘Qc’ varies as 20 kVAR  Qc 25 generator buses, 5 buses having fixed  80 kVAR. The combination of capacitor shunt capacitor banks, 3 buses having bank and load will draw the lowest apparent SVC’s, 2 buses having STATCOM’s and power from connected bus bar for any value remaining buses are treated as load buses. of ‘Qc’ is How many number of equations to be (A) 165 kVA (B) 188 kVA solved in load flow analysis of this system (C) 200 kVA (D) 212 kVA by Gauss Seidel method and Newton Ans: (A) Raphson method (rectangular form) respectively 6 ACE Engineering Academy GATE-2020 Electrical Engineering (EE) ACE Engineering Publications (A) 150, 264 (B) 150, 298 100   20  0.8 = 1131.4 W (C) 149, 269 (D) 149, 298 2 Ans: (D) Sol: Number of equations to be solved in Gauss 07. A single phase one pulse converter with Seidel method = n – 1 = 149 RLE load has the following parameter: Number of equations to be solved in Supply voltage: 230 V at 50 Hz Newton Raphson method (rectangular form) Load: R = 2, E = 120 V = 2n – 2 = 300 – 2 = 298 Extinction angle,  = 275 Firing angle,  = 25 06. Find the PIV (Peak Inverse Voltage) for the M L SCR. C V L (A) 325.27 volt (B) 445.27 volt Diode O A Z = 4 + j3  100V, 50Hz, (C) 444.03 volt (D) 650.54 volt AC Supply D Ans: (C) Sol: PIV for  > 270 100V, 50Hz, PIV =|Vm sin   E| AC Supply Shared on www.ErForum.Net PIV = 230 2 sin 275  120 The reading of wattmeter is (A) 1600 W (B) 800 W PIV = 444.03 volt (C) 1414 W (D) 1131 W Ans: (D) 08. The IGBT used in the circuit has the V following data: Sol: Load current I L  Z t on  3 μs, t off  1 μs, Duty cycle (D) = 0.7, 100 100 VCE ( sat )  1.5 V and f s  1 kHz.    20A 4 3 2 2 5 100  2 Vp.c  RL 10 2 + V IC VCE = 200 V [∵Half wave sinusoidal, rms value is m ]  2 G + E VGE R 4 cos     0.8  Z 5 Pavg = Vrms × Irms × cos ACE Engineering Academy 7  GATE-2020 Electrical Engineering (EE) ACE Engineering Publications The switching energy loss (in mJ) during Sol: RMS value of phase voltage for 120° turn on is ________. (Give upto one conduction mode is decimal place) Vdc 600 VL  VPh   V Ans: 1.985 (Range: 1.9 to 2.1) 2 2 VCC  VCE sat  Power delivered to load Sol: I C  RC Po  3 V ph2 = 3 600 / 2 2 = 18 kW 200  1.5 R 30   19.85 A 10 VCC I C 10. Resonance will occur in the circuit shown Turn ON energy loss =  Ton 6 only when 200  19.85   3  10  6 6 IS R = 1.985 mJ 0.025 F VS ~ 25 mH 09. A three-phase voltage source inverter (VSI) as shown in the figure is feeding a delta (A) R > 1000  (B) R < 1000  connected resistive load of 30/phase. If it (C) R > 2000  (D) R < 2000  Shared on www.ErForum.Net Ans: (B) is fed from a 600 V battery, with 120 Sol: As resonance frequency conduction of solid-state device, the power consumed by the load, in kW, is 1 R 2C fr  1 2 LC L Resonance will occur in circuit only when + R 2C R 2C L 1 > 0, 1  , R2  L L C 600 V 30  L 30 R 30  C 25  10 3 – R< = 1000  0.025  10 6 (A) 12 (B) 18 (C) 24 (D) 8 11. The resistance between terminals A, B is Ans: (B) __________ . 8 ACE Engineering Academy GATE-2020 Electrical Engineering (EE) ACE Engineering Publications 8 A 4 4 1 RAB = 2 // = 3 = = 0.8  3 10 5 1 1 3 1 1 1 1 12. An electric field (x2ax + yay + 2z3az) N/C 1 1 exists in free space. Corresponding charge 1 density at the origin is _________ p C/m3. 1 1 (rounded off to two decimal places). B Ans: 8.85 (Range: 8.84 to 8.86) Ans: 0.8 (Range: 0.75 to 0.85) Sol: It is useful to check that the given vector Sol: Joints nodes which one of same potential by field E is indeed a static electric field (static mirror image symmetry because t does not appear in the field A 1 expression). Find E. Is it zero? Then E 1 is indeed a static electric field. (Otherwise 1 1 the problem is wrongly framed). 1 A 1 Shared1 on www.ErForum.Net 1/2 1 1 1 1 1/2 1 1 B 1/2 1/2 Mirror image 1 Delta-to star 1/2 B A Balanced bridge 1/2 A ax ay az 1    E  1/6 x y z 1/6  4/3 2 x2 y 2z 3 1/6 1 1/2 B B ACE Engineering Academy 9 GATE-2020 Electrical Engineering (EE) ACE Engineering Publications       Sol:  a x  (2z 3 )  ( y)  a y  ( x 2 )  2z 3  input + G (s) output  y z   z x  –     a z  ( y)  ( x 2 )  H(s)  x y  =0 G (s) CLTF  E is a static electric field. 1  G (s)H(s)  Negative feedback reduces gain of Then, . E = (Gauss;s law) 0 system But, 1 SGM  is less than unity      1  G (s)H(s) .E   a x  a y  a z .x 2 a x  ya y  2z 3 a z   x y z  i.e., less sensitive to the forward path parameter variations. = 2x + 1 + 6z2 Bandwidth increases hence rise time At origin (0, 0, 0), .E = 1. decreases, speed increases, time constant ( 0 , 0 , 0 ) decreases.  =1 0  (0, 0, 0) = 0 = 8.854  1012 C/m3 Shared on www.ErForum.Net 14. State space representation of a system is given as 13. Negative feedback is employed in a control  2 0  1 x ( t )    x ( t )    u ( t ), y( t )  1 1x ( t ) system then which one of the following  0  4  1 statement (s) is/are true.. Where y(t) is the output and u(t) is the 1. Gain increases input. Then the undamped natural frequency 2. Bandwidth increases of the system is _____ rad/sec. (round up to 3. Sensitivity of the output with respect to two decimal places). parameter changes in the forward path Ans: 2.83 (Range 2.8 to 2.9) decreases. Sol: Characteristic equation (s+2) (s+4) = 0 4. Time constant of the system decreases s2 + 6s + 8 = 0 (A) 1, 2, 3, 4 (B) Only 2, 3, 4 2n  8 (C) Only 3, 4 (D) Only 2,4 n = 2 2 rad/sec = 2.83 rad/sec Ans: (B) 10 ACE Engineering Academy GATE-2020 Electrical Engineering (EE) ACE Engineering Publications 15. The circuit shown used to provide regulated 16. A, B, and C are Input bits and ‘Y’ is the output voltage of 6V across 1k resistor. output bit in the circuit shown below. If the Assume Vz = 6V, Iz (knee) = 4mA. The output ‘Y’ is set to logic ‘1’, which of input voltage may vary by 10% from normal the following option can satisfies the input value of 10V. The required value of R for binary combinations. the satisfactory operation of the circuit is A ______  B R C + Vi Y 6V 1k V0 – (A) Two (or) more number of input’s A,B, and C are ‘0’ Ans: 300 (B) Two (or) more number of input’s A,B, Sol: Given that Vz = 6V= V0 and C, are ‘1’ Iz knee = Iz min = 4 mA (C) A = B  C (or) A = C  B Vi = (10  10%) V Shared on www.ErForum.Net (D) A = B =C Vi range is 9V to 11V Ans: (B) V0 Sol: IL   6 mA RL A As Vi varies, I varies B A B Imin = Iz(min) + IL = 10 mA C C(A  B) Vi (min)  Vz  10  10 3 R AB Y 96 R = 300 10  10 3 Y  CA  B.AB R   Y  C AB  A B  AB I Iz IL Vi  ABC  A B C  AB(C  C) 6V 1k ABC A BC ABC ABC         m3 m5 m6 m7 ACE Engineering Academy 11 Shared on www.ErForum.Net GATE-2020 Electrical Engineering (EE) ACE Engineering Publications 18. The below program is stored from B C B C BC B C Address 0101H A 1 LXI H, 7788 H  Y=AB+BC+CA A 1 1 1 MOV A, L ANA H If two more number of input’s are ‘1’ output JPO SKIP (Y) = 1 ADD L Because Y = 1+(anything) = 1 SKIP: MOV H,A DAD H 17. For which of the following Boolean function SHLD 1234H the dual and complement values are same? PCHL (A) f = A B From which address next instruction will be (B) f = AB fetched? (C) f  A B (A) 0110H (B) 1111H (C) 1110H (D) 1234H (D) All of the above Ans: (C) 17. Ans: (A) Shared on www.ErForum.Net Sol:  (HL) = 7788H Sol: Given AB  A B  (A)  (L) = 88H   f compt  A  B. A  B   (A) = 88H = A A  AB  B A  B B  (A)  88H  1000 1000 = 0  AB  A B +0  (H)  77 H  0111 0111 (A)  00H  0000 0000 fcompt = AB  A B CY  O, P  1, AC  1, Z  1, S  0 Given f = AB  A B  JPO means Jump if parity odd   f dual  A  B A  B  i.e., test for P = 0 = 0  A B  BA  0 This test fails as P = 1 because of ANA H f dual  AB  A B Therefore, p does not jump but continues with next instruction. ∴ fcompt = fdual  (A)  (A) + (L) 12 ACE Engineering Academy GATE-2020 Electrical Engineering (EE) ACE Engineering Publications  (A)  00H  0000 0000   (L)   88H  1000 1000 For the stability  | h (t ) | dt    (A)  88H  1000 1000  1  (H)  (A) = 88H  t2 dt   ∴ Stability 1  (H) = 88H  The system is stable & NC (HL)  8888H  1000 1000 1000 1000  (HL)  8888H  1000 1000 1000 1000 20. An FIR system with input x(n) and output (HL) 1110H  0001 0001 0001 0000 y(n) related as y(n) = 0.2x(n) 0.5x(n 2) +  (HL) = 1110 H stored into 2 locations 0.4x(n3). If the input x(n) = {–1, 1, 0, 1} 1234H is applied then the output at n = 2 is And ________ 1235H Ans: 0.5  (HL) = 1110 H copied into P.C Sol: y(n) = 0.2x(n) – 0.5 x(n–2) + 0.4 x(n–3)  (P.C) = 1110H y(2) = 0.2x(2) – 0.5 x(0) + 0.4 x(–1) …..(1) ∴ 8085 P fetches next instruction from x(n) = {–1,1,0,1} x(0) = –1 x(1) = 1; x(2) = 0 Address 1110H Shared on www.ErForum.Net ; x(3) = 1 Substituting in equation (1) 19. An LTT system with impulse response y(2) = 0.2(0) – 0.5 (–1) + 0.4 (0)=0.5 1 h(t) = u ( t  1) is y(2) = 0.5 t2 (A) Causal & Stable (B) Causal & Unstable 21. Two LTI systems with impulse response (C) Unstable & Non causal h1(n) = (n) and h2(n) =  (n) –  (n2) are (D) Non causal & Stable connected in cascade. If the input x(n) = Ans: (D) u(n) is applied then the output is 1 (A)  (n) (B)  (n1) Sol: h ( t )  u ( t  1) t2 (C)  (n) + (n1) (D) (n1) +  Because the signal (IR) starts at t = –1  (n2) Non causal. Ans: (C) Sol: Two systems are connected in cascade the ACE Engineering Academy 13 GATE-2020 Electrical Engineering (EE) ACE Engineering Publications overall IR is 23. If directional derivative of  = 2xz – y2 , at h(n) = h1(n) h2(n) the point (1, 3, 2) becomes maximum in the h(n) = (n) [(n) –(n–2)] direction of a , then magnitude of a is h(n) =(n) – (n–2) __________. (Give upto two decimal place) y(n) = x(n)  h(n) Ans: 7.48 (Range 7.45 to 7.50) y(n) = u(n)  [(n) – (n–2)] Sol: Given  = 2xz – y2 y(n) = u(n) – u(n–2)      i j k x y z = 1, 1   2z i  2 yj  2x k y(n) = (n) + (n–1)  Required direction vector = a = () at (1, 3, 2) = (4i  6 j  2k ) 22. For the function f(x, y) = x 2  y 2 , the point Magnitude of a = 16  36  4 = 56 (0, 0) is = 7.48 (A) a local minimum (B) a saddle point 24. A continuous random variable X has a (C) a local maximum Shared on www.ErForum.Net (D) not a stationary point probability density function Ans: (B) f(x) = ex , 0 < x < . Then P(X > 2) is (A) 0.1353 (B) 0.2354 Sol: Given f x, y   x 2  y 2 (C) 0.2343 (D) 1.1353  f x  2x, f y  2y and Ans: (A) f xx  2, f xy  0, f yy  2  Consider fx = 0 and fy = 0 Sol: P(X > 2) =  f ( x)  dx 2  2x = 0 and –2y = 0  =  e  x dx   0, 0  is a stationary point 2 At (0, 0) , f xx f yy  f xy   4  0 2  ex  1 2  f(x, y) has neither a maximum nor minimum at (0, 0). = e 2  0.1353 14 ACE Engineering Academy Shared on www.ErForum.Net GATE-2020 Electrical Engineering (EE) ACE Engineering Publications 25. The solution to x2y11 + xy1 – y = 0 is Q. 26 – Q. 55 carry TWO marks each. (A) y = C1x2 + C2 x–3 (B) y = C1 + C2x–2 26. Three identical 2-winding single-phase C2 transformers are connected as shown in fig. (C) y = C1 x + x 1. 4 Supply (D) y = C1 x + C2 x Ans: (C) R Y B V0 V120 t + – + – Sol: Put ln x = t so that x = e and N-turns N-turns N-turns dy d2y Let x  Dy , x 2 2 = D(D –1) y (1) (2) (3) dx dx N-turns N-turns N-turns 1 2 d Where D = Vs dt + – Fig.1 Given differential equation is x2y11 + xy1 – y = 0 The two possible values for Vs are, (A) 3V, V (B) 0, 2 V  D(D–1)y+ Dy – y = 0 (C) 0, V (D) 0, 2V  60.  (D2 – 1) y = 0 Shared on www.ErForum.Net 26. Ans: (B) Consider Auxiliary equation f(D) = 0 Sol: Fig. 1 of the question does not give any  D2 – 1 = 0 information about the relative polarities of  D = 1, –1 are different real roots the primary and secondary voltages. (This information is usually given by placing dots  The general solution of given equation is at the two windings of each of the y = c1et + c2e–t transformers as per the dot convention. Such c = c1 x + 2 dots are not placed any where). x One meaning of dots (which is needed for this problem) is as follows: + + +  V1 V2 V1 V2    + V1 and V2 are in phase V1 and V2 are 180 out of phase. Fig. 2 ACE Engineering Academy 15  GATE-2020 Electrical Engineering (EE) ACE Engineering Publications (A) Let the dots be located as shown in There are several other ways of connecting fig.3. the three secondaries. Everyone of them R Y B leads to either Vs = 0 or Vs = 2V. V0 V120 V120 + – + – + – Thus there are only two possible values for N-turns N-turns N-turns Vs: Zero, and 2V. (1) (2) (3) N-turns N-turns N-turns 1 2 27. A 100 kW, belt-driven dc shunt generator is V0 V120 V120 + – + – + – running at 300 rpm in the clockwise + Vs direction, delivering power to a 200 V bus. – Vs = V0 +V120 + V120 = 0 Now the belt breaks, but the machine The terminals 1 & 2 can now be safely joined continues to run, drawing 4 kW from the Together to make a  –  transformer Fig. 3 supply. Neglect armature reaction. Armature and field resistances are 0.1  and 100  (B) Let the dots be located as shown in fig.4. respectively. Its speed and direction of rotation after the belt breaks are, R Y B V0 V–120 V120 + – + – + – respectively. N-turns Shared on N-turns www.ErForum.Net N-turns (A) 238, clockwise (1) (2) (3) (B) 238, anti clockwise N-turns N-turns N-turns 1 2 (C) 242, clockwise V0 V–120 V120 +  +   + (D) 242, anti clockwise V0+ V–120 – V120 =Vs Ans: (A) + – Sol: 1. Initial operation: Fig.4 200 100  10 3 Vs is determined in the phasor diagram of  2A  500 A 100 200 fig.4. F1 A1 V120 – 50.2 V 0.1  100  + 502 A = Ia + V0 – 200 Volt bus + 60 E Prime – V–120 F2 mover A2 Vs =2V–60 E – 50.2 = 200  E = 250.2 V. Fig. 1 –V120 Terminals 1 and 2 of fig. 4 should not be joined Fig. 5. 16 ACE Engineering Academy GATE-2020 Electrical Engineering (EE) ACE Engineering Publications For a field current of 2A, let the flux/pole be reaction is neglected, change in armature  Wb. current from 502 A to 18 A does not affect  2  25.02 the flux/pole in any way). k   300    250.2  k  .  60    25.02   2    N    198.2 The armature is given to be rotating in the     60  cw direction.  N  237.65rpm  238 rpm. 2. Direction of developed torque during When belt breaks, there is no prime mover the initial operation: anymore. However, the stored kinetic With the armature and field currents directed energy in the rotor keeps the rotor running in as shown in fig.1, the developed torque in the clock wise direction, but with decreasing the machine must be in the anti clock wise speed. The machine is in regenerative direction. This is because the prime mover, braking mode, which continues as long as E in driving the armature in the clock wise in fig.1 is greater than 200V, and Ia of fig. 1 direction, has to do mechanical work against is positive (developed torque will be in acw the developed torque, which work is direction, it opposes motion and causes converted into electrical energy and losses speed to fall). When E becomes less than by the machine. Shared on www.ErForum.Net 200 V, as in fig. 2, Ia reverses. With If 3. Operation after the belt breaks: continuing to be in the original direction, Td Steady state conditions after the belt breaks is now in the clock wise direction, and are specified the problem. Corresponding drives the machine in the clock wise circuit is as shown in fig. 2. direction as a motor. 4000 2A 200  20A 238 rpm, clock wise; is the correct F1 18A A1 answer. + 1.8 V 0.1  – + 200 Volt bus 28. A squirrel cage induction motor has slip of – + E = 198.2 V 4% at full load. Its starting current is five F2 – A2 times the full load current. The stator Fig. 2 impedance and magnetizing current may be With field current unchanged, k remains neglected, the rotor resistance is assumed 25.02 constant. The percentage of slip at which unchanged at. (Since armature  maximum torque occurs is _______. ACE Engineering Academy 17 GATE-2020 Electrical Engineering (EE) ACE Engineering Publications Ans: 20 (No range) bus of 400 kV voltage. The ratings of Sol: We have apparatus are given as R2 Generator (G): 11 kV, 200 MVA, Xd = 0.5 Tem = KI 22 ………….. (1) s p.u, From eq. (1) Transformer (T): 11 kV/400 kV, 200 MVA, R2 Xt = 0.15 p.u, Tst = KIst2 1 Transmission line: X = 40 , R = 0, 400 kV R2 Tfl = KIfl2 s fl G T Line 2 Tst  Ist     s fl …………… (2) -bus Tfl  I fl  Also we have The stable angle made by rotor of alternator Tst s2  s2 with respect to infinite bus is ______  2T max fl …………. (3) Tfl (s T max  1)s fl electrical degrees From (2) & (3) Ans: 22.64 (Range: 21.0 to 24.0) 2  Ist  s T2 max  s fl2 Sol:   s fl  200 MVV  I fl  Shared on www.ErForum.Net s fl (s T2 max  1) 11 kV 11 kV/400 kV 200 MVA Given Ist = 5Ifl G Base 2 Base  Ist  s s2 2 11 kV 400 kV    T max fl  I fl  s s  1 2 2 fl T max 200 MVA 200 MVA 2  5I  s 2  0.04 2   fl   T max Let us choose common base as 11 kV, 200  I fl  0.04 2 s T2 max  1   MVA at ‘G’ location. s  0.04 2 2 Line: X   40   25 = T max 0.04 s T2 max  1 2 2 Solving for Z  400 k  = 800  base 200 M STmax = 0.2 or 20% 40 X  = 0.05 p.u 29. Consider the following single line diagram  800 ( p.u ) in which an alternator with emf of 15 kV 15 supplying 150 MW real power to infinite G: Ep.u = = 1.3636 p.u 11 18 ACE Engineering Academy Shared on www.ErForum.Net  GATE-2020 Electrical Engineering (EE) ACE Engineering Publications 400 Ans: 300 (Range: 295 km to 305 km) Infinite bus: Vp.u = = 1 p.u 400 Sol: Equivalent- model, The per phase equivalent circuit is given as, Z jXd jXt jX  y' y' 2 2 + + E  E  V  V 0o – – z ' y' Parameter A = 1  2 Real power flow, P = 150 MW For transmission line, A = cos  P 150 z ' y' P   = 0.75 p.u  cos  = 1  (p.u ) S 200 base 2 E V = 1+ (j87.5  j574  10–6) P.sin X eq = 0.9498  = cos–1 (0.9498) Where Xeq = Xd + Xt + X  = 0.7 p.u  = 0.31828 radians 1.3636  1 0.75 = sin   = 22.64 Where  = 1.06  10–3 rad/km 0.7 Shared on www.ErForum.Net 0.31828   km = 300 km 1.06  10 3 30. A 3- loss less transmission line has the propagation constant  = 0 + j1.06  10–3 31. A transmission line of length 200 km with radians per km. The transmission line is series reactance per km as 0.4  is supplied represented in its equivalent  model as by a source of 220 kV (LL) with source shown in the figure. reactance of 20 . The line is protected by j87.5  a over current relay (R) with relay setting 80% and associated with a CT of ratio j574 ℧ j574 ℧ 2000/5A. The relay will provide protection for The approximate length of transmission line ________ length of transmission line for is ________ km. three phase fault on transmission line. ACE Engineering Academy 19 GATE-2020 Electrical Engineering (EE) ACE Engineering Publications = 148.47 km VS =220 kV(L-L) Line ~ XS=20 200 km 32. A solidly grounded neutral Y-connected Relay (R) X=0.4/km alternator rated for 11 kV, 80 MVA has the (A) 175 km (B) 198.5 km resistance and reactances as Ra = 0.1 pu, (C) 148.5 km (D) 108.77 km X d = 0.2 pu, X d = 0.4 pu, Xd = 0.8 pu, X2 = Ans: (C) 0.2 pu, X0 = 0.05 pu. The open circuit Sol: CT ratio : 2000A/5A voltage of alternator is 14 kV(LL). A single Relay setting = 0.8 phase load of impedance 1.5 + j0  is Pickup current on primary side of CT connected at the terminals of alternator. The = 0.8  2000 load current is = 1600 A (A) 4.23 kA (B) 4.66 kA The per phase model onto a 3- fault on (C) 4.85 kA (D) 5.39 kA transmission line Ans: (B) XS=20 Sol: Ia XRF F ~ 11kV, 80 MVA VSph R Shared on www.ErForum.Net ~ Zl If Let XRF is the portion of line reactance upto Ra = 0.1 pu ; Z1 = 0.1 + j0.8 which the relay can identify the fault. X1 = Xd = 0.8 pu ; Z2 = 0.1 + j0.2 For verge of operation, If = 1600 A X2 = 0.2 pu ; Z0 = 0.1 + j0.05 VS( ph ) X0 = 0.05 pu = 1600 X S  X RF Zbase = 11 2 = 1.5125  220 80  10 3 3 = 1600 Z  () 20  X RF Load impedance in per unit, Zl pu = Z base XRF = 59.388  1.5  = 0.9917 pu Line reactance, Xl = 0.4 /km 1.5125 The length of line protected by relay 59.388 Open circuit voltage (or) internal unit of = km 0.4 alternator, Ea1 (LL) = 14 kV 20 ACE Engineering Academy GATE-2020 Electrical Engineering (EE) ACE Engineering Publications 14 Sol: Ea1(pu) = = 1.2727 pu 11 10 10 3E a1 Load current , Ia = Z1  Z 2  Z 0  3Z  I1 G 3  1.2727 = I2 V 0.3  j1.05  2.975 20 10 I 3.8181 I2 = 3.275  j1.05 10 V |Ia| = 1.11 pu I2  20 = 5 Ia (kA) = Ia(p.u)  Ibase 5 I2   0.25A 80 20 = 1.11  kA = 4.66 kA 3  11 I2(10) –V = 5 0.25 10 – V = 5 33. The resistance values of the bridge circuit 2.5 – V = 5 shown in the figure are R1 = R2 = R3 =10  V = – 2.5 V and R4 is 20. The bridge is balanced by Shared on www.ErForum.Net introducing a voltage source of ‘V’ as 34. A time varying voltage signal V(t) = X + shown in figure. Ysint is measured by a single channel Analog CRO (operated with coupling mode 10 10 set to DC) and also by Dual slope integrating DMM (operated with voltage G Range set to AC). After measurement, V DMM and CRO will display respectively are 20 10 2 2  X   Y  (A)     & X+Y sint  2  2 10 V (B) X & X+Y sint 2 The value of voltage source is ______ volts.  Y  (C) X2    & Ysint Ans: – 2.5 (No range)  2 (D) X & Y sint Ans: (B) ACE Engineering Academy 21 GATE-2020 Electrical Engineering (EE) ACE Engineering Publications Sol: DMM measures average value. Therefore di L KVL : R on i L  L  Vdc …… (1) displays X dt In DC coupling, the sensed signal as it is di L KVL : L   V0 …… (1) dt reaches to Y-input of CRO and hence d 0 0 displayed as X + Y sint. KCL : C   0 ….. (2) dt R d 0 0 35. A Buck-Boost converter is shown in figure. KCL : C   i L ……. (2) dt R Assume that inductor and capacitor are large Flux balance equation from KVL enough to treat iL and V0 are ripple free.  R on.I L.D  0  D.Vdc  V0 (1  D) ……. (1) MOSFET has ON resistance of 0.5 during Charge balance equation from its conduction. To maintain output voltage V0 of 400V at 10A to the load, the duty cycle KCL   I L 1  D  …… (2) R ratio of converter will be _______. (Give By substituting IL value from equation (2) upto two decimal place) into equation (1), we will get V0  R on. D  V0 (1  D)  D.Vdc R 1  D  – Vdc iL Shared L C on400V www.ErForum.Net 300V R D   +  V0  on   1  D   DVdc  R 1 D  V0 D   Ans: 0.58 (Range: 0.56 to 0.60) Vdc R on. D  1  D  R 1 D Sol: During MOSFET ON: Now, substitute the given data During MOSFET OFF: 400 D Ron 0A  300 0.5  D  1  D  – 40 1  D Vdc iL L C R V0  560D 2  876D  320  0 +  D = 0.983 (or) 0.5813 If D is near to unity, buck-boost converter – iL will be unstable. Hence choose, 0.5813. Vdc L C R V0 + 22 ACE Engineering Academy GATE-2020 Electrical Engineering (EE) ACE Engineering Publications 36. A single phase full wave half controlled the average output current 20.7 A, then the rectifier is supplying an inductive load and peak value of fundamental supply current is assume current is ripple free at 10 A. It has _________ A (Give upto two decimal been operated with firing angle delay of 45 places) then power factor on the AC supply lines is Ans: 32.52 (Range: 32 to 33) (A) 0.9238 (B) 0.8869 Sol: A single phase full wave rectifier consists of (C) 0.707 (D) 0.52 three diodes and one thyristor and supplying Ans: (B) a resistive load, then Sol: Single phase full wave half controlled Vm V0 = 3  cos  rectifier is a semi converter. 2 Vo V Power factor = C.D.F × D.F I0 = = m 3  cos  R 2R IS1 C.D.F = ISr 230  2  20.7 = 3  cos  2  10 2 2  IS1 = I o cos  cos = 1  2 =0 45 = 0.9 Io cos = 8.315 A 2Shared on www.ErForum.Net If  =0, then the shape of the supply current  is pure sinusoidal. ISr = Io  Peak value of the fundamental current 180  45 Vm 230  2 = 10 = 8.66 A component, IS1 = = 180 R 10 8.315 = 32.52 A C.D.F = = 0.96 8.66  45 38. Obtain Thevinin’s equivalent at terminals a D.F = cos = cos = 0.9238 2 2 –b Power factor = 0.96 × 0.9238 = 0.8869 12 a 37. A single phase 230V, 50Hz full wave + 0.05Vx rectifier consists of three diodes and one 24A VX 10  thyristor and supplying a resistive load of 10 . The firing angle delay is so selected that b ACE Engineering Academy 23  GATE-2020 Electrical Engineering (EE) ACE Engineering Publications VTh = 240  144 = 96 volts (A) 22  a RTh: 12 1/2 96V  a + 1/2 b 0.05Vx VX 10 VT 1A (B) 22  a b VX = 1 = 10 volts 96V  KVL: b 1 VT + 12 + 10 = 0 (C) 16 2 a VT = 16 volts 96V  VT So, RTh = = 16  1 b So, Thevenin’s equivalent (D) 16Shared on www.ErForum.Net 16 a a 96V  96V  b b Ans: (C) Sol: VTh: 39. Determine the RMS value of short circuit 12 current ISC in circuit shown a + + 1H VX 10 0.05Vx VTh 24A ISC 24cos4t V  1/2H 2H   b VX