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1 GATE-2016 Index 1. Question Paper Analysis 2. Question Paper & Answer keys : , info@thegateacademy.com Copyright reserved. Web:
2 ANALYSIS OF GATE 2016 Electrical Engineering Mathematics 15% GA 15% Network Theory Network 8% Theory 8% Control System Control 10% System 8% CN Mathematics 11% 15% Power Electronics Power 3% Electronics Power System 8% Power 9% System 11% Electrical Machine 10% Signal & System Signal 8% & System 6% Digital Electronics Digital 6% Electronics 5% Analog Circuits EMT Analog 5% Circuits 5% EMT 3% 6% Measurement Measurement 3% 5% GATE EE 7 Feb -Afternoon Session-8 SUBJECT NO OF QUESTION Topics Asked in Paper Level of Toughness Total Marks Network Theory Control System Signal & System 1 M: 4 2 M:2 1 M: 2 2 M: 3 1 M: 2 2 M: 2 Reduction of networks with controlled voltage and current sources; Transient; Resonance; AC circuit analysis Routh Stability; Lag-lead system; Bode plot and transfer function LTI system; Fourier transform; Laplace transform Moderate 8 Tough 8 Tough 6 : , info@thegateacademy.com Copyright reserved. Web: 1
3 Digital Electronics Analog Circuits EMT Measurement Electrical Machines Power System Power Electronics Mathematics GA 1 M: 1 2 M: 2 1 M: 1 2 M:1 1 M: 2 2 M: 2 1 M: 1 2 M: 2 1 M: 2 2 M:4 1 M: 3 2 M: 4 1 M: 2 2 M: 3 1 M: 5 2 M: 5 1 M: 5 2 M: 5 Boolean algebra K-Map Easy 5 Op.AMP analysis;amplifier Easy 3 *Electrostatic field - Properties Moderate 6 Power Factor measurement;energy meter Moderate 5 DC Machine-Shunt motor, efficiency; Transformer-Losses, Autotransformer; Induction machine, Synchronous Machine;Sinle phase & 3 Phase induction Motor Stability; LLG fault; Line parameters Transmission Line VSI; Rectifier - full wave - input power factor, efficiency; Chopper;Convertor Complex integration; Differential Equation; infinite series; Probability-basic. Time & Work ;Paragraph; English fill in Blank;Number theory; Venn Diagram ; Mensuration& Area. Easy 10 Tough 11 Tough 8 Moderate 15 Easy 15 Total 65 Moderate 100 * Indicates Questions from New Syllabus Faculty Feedback: The question paper was bit tough and was little offbit than conventional GATE papers. Plenty of Numerical Answer Type (NAT) questions asked. Online Calculator was difficult to handle without Prectice. Numerical & Verbal ability was relatively easy. Practice previous Year Questions & Online Test Series will be beneficial. : , info@thegateacademy.com Copyright reserved. Web: 2
4 GATE-2016 Question Paper & Answer Keys
5 GATE 2016 Examination Electrical Engineering Test Date: 07/02/2016 Test Time: 9:00 AM to 12:00 PM Subject Name: ELECTRICAL ENGINEERING Q NO. 1. Section: General Aptitude [Ans. C] Q NO. 2. [Ans. B] Q NO. 3. [Ans. B] 13, 23, 43, 53 are all prime numbers; Only 33 is composite 33, (11 3); Odd one out is 33. Q NO. 4. [Ans. B] : , info@thegateacademy.com Copyright reserved. Web: 3
6 Q NO. 5. [Ans. C] 9y 6 = 3 Either (9y 6 = 3)or (9y 6 = 3); [y = 1]or [y = 1 3 ] [y 2 4y 3 ] Put y = ; (1 3 ) = 1 3 or y 2 4y 3 Put y = 1, = 1 3 Q NO. 6. [Ans. *]Range: 120 to 120 : , info@thegateacademy.com Copyright reserved. Web: 4
7 According to information in question Large plants are 1, 4, 8, 9 which are having installed capacity of at least 200 tonnes. Total production of large plant [ ] =770 Remaining plant number 2, 3, 5, 6, 7 all are small plants with capacity less than 200 tonnes. Total production of small plants = = 650 Difference = = z120 Q NO. 7. [Ans. C] Q NO. 8. [Ans. C] : , info@thegateacademy.com Copyright reserved. Web: 5
8 Q NO. 9. [Ans. A] Probability of free throw = = 0.6 Probability of NOT free throw = = 0.4 So required probability of exactly 6 throws in 10 attempts will be given by 10 C 6 (0.6) 6 (0.4) 4 = Q NO. 10. [Ans. *] Range: 7 to 7 Unit digit of is = 7 : , info@thegateacademy.com Copyright reserved. Web: 6
9 Q NO. 1. Section: Technical [Ans. B] BC A F = A + C Q NO [Ans. A] R 2 V in R 1 + C V out V R out 2 = [ V in 1 R jωc 2 + 1/jωc ] R 1 : , info@thegateacademy.com Copyright reserved. Web: 7
10 V out R 2 = [ V in R 1 (R 2 jωc + 1) ] So the system is a low pass filter Q NO. 3. [Ans. *] Range: 9.5 to 10.5 The below circuit can be drawn by transferring secondary circuit to primary side. 10 jω 100 V ~ 100V I = (8 + 10j 4)Ω = 100V (8 + 6j)Ω So the rms value of I will be 10 A. I (100) 2 Ω (100) 2 jω Q NO. 4. [Ans. D] The differential equation dy(t) + 1 y(t) = 3x(t) dt 6 So, sy(s) + 1 Y(s) = 3X(s) 6 : , info@thegateacademy.com Copyright reserved. Web: 8
11 Y(s) = 3X(s) (s + 1 ) ; x(s) = 9 (s + 1 ) So, Y(s) = (s + 1 ) (s + 1 ) = 54 (s + 1 ) 54 (s + 1 ) So, y(t) = (54e 1 6t 54e 1 3t) u(t) Q NO. 5. [Ans. *] Range: 6 to 6 Since x(t) is band limited to 5 khz then maximum frequency in x(t) cos (20007πt) is 6 khz. Q NO. 6. [Ans. B] f(z) = z + z f(z) = 2x is continuous (polynomial) u = 2x v = 0 u x = 2 u y = 0 v x = 0 u y = 0 C.R. equation not satisfied. Nowhere analytic. Q NO. 7. [Ans. D] By Cayley Hamilton theorem λ 3 = λ λ = 0, 1, 1 : , info@thegateacademy.com Copyright reserved. Web: 9
12 Q NO. 8. [Ans. A] The differential equation is y (t) + 2y (t) + y(t) = 0 So, (s 2 Y(s) sy(0) v (0)) + 2 [sy(s) y(0)] + Y(s) = 0 So, Y(s) = sy(0) + y (0) + 2y(0) (s 2 + 2s + 1) Given that y (0) = 1, y(0) = 0 1 So, Y(s) = (s + 1) 2 So, y(t) = te t u(t) Q NO. 9. [Ans. B] F d r C Where F = xy 2 i + 2x 2 yj + k F = 0 (F is irrotational F is conservative) F = ϕ (ϕ is scalar potational function ) ϕ x = 2xy 2 ϕ y = 2x 2 y ϕ z = 1 ϕ = x 2 y 2 + z + C Where, F is conservative F d r c (1,1,1) = dϕ = (0,0,0) [x 2 y 2 (1,1,1) + z] (0,0,0) = 2 : , info@thegateacademy.com Copyright reserved. Web: 10
13 Q NO. 10. [Ans. B] Q NO. 11. [Ans. C] Q NO. 12. [Ans. *] Range: 249 to 251 Method 1: The average power consumed by the load P = V I I 1 cos ϕ 1 Method 2: V(t) = 100 sin(ωt) i(t) = 10 sin(ωt 60 ) + 2 sin(3ωt) + 5 sin(5ωt) P = V(t)i(t) = 1000 sin ωt sin(ωt 60 ) sin ωt sin 3ωt sin ωt sin 5ωt = 500 [cos(ωt ωt + 60 ) cos(ω t + ωt 60 ] + 100[cos(ωt 3ωt) cos(ωt + 3ωt)] [cos(ωt 4ωt) cos(6ωt)] = 500 cos 60 = 250 W Average value of cos (2ωt 60 ), cos (2ωt), cos (4ωt), cos (6 ωt) will be zero. : , info@thegateacademy.com Copyright reserved. Web: 11
14 Q NO. 13. [Ans. D] Q NO. 14. [Ans. A] Size of the Jacobian matrix is, 2m m 1 2n m 1 Given that 10 generator buses, we need to assume with in the 10 buses one bus as slack bus then ( ) ( ) Q NO. 15. [Ans. B] Z n = L C = = 400Ω SIL = = 400 MW 400 In the second case SIL decreases means Z n increases. Z n increases with increase in inductance L So, it is inductive; Load < SIL means, line behaves capacitive to compensate it inductor to be placed. : , info@thegateacademy.com Copyright reserved. Web: 12
15 Q NO. 16. [Ans. C] Q NO. 17. [Ans. *] Range: 4.9 to 5.1 I = E g X ; I = E g d X ; I I = X X = 1.0 = 5.0 p. u d 0.2 Q NO. 18. [Ans. *] Range: 0.69 to H(s) = (s + 1) Put s = jω, H(jω) = 1 jω H(jω) = ω Input x(t) = cos(t) Here ω = 1 rad/sec and x(t) = 1; Hence, steady state output y(t) = x(t) H(jω) ω=1 cos[t + H(jω)] A = [ x(t) H(jω) ] ω=1 A = 1 2 = : , info@thegateacademy.com Copyright reserved. Web: 13
16 Q NO. 19. [Ans. *] Range: 57 to 58 In the 3-ϕ diode bridge rectifier each diode conducts for 120 for one complete cycle I D rms = 1 2π/3 2π I 0 2 dωt = I o 3 = = 57.7 A 0 = 2π 2π 3 Q NO. 20. [Ans. *] Range: 59.5 to 60.5 When switch S is OFF, diode D is ON then + 24 V 36 V + Peak voltage across switch = = 60 V Q NO. 21. [Ans. *] Range: 0.30 to 0.33 : , info@thegateacademy.com Copyright reserved. Web: 14
17 Assuming R 1 = 9 Ω; R 2 = 1 Ω We can write V o (s) V in (s) = R sc R 1 + R sc 1 + R 2 Cs = 1 + (R 1 + R 2 )Cs 1 + R 2 Cs = 1 + ( R 1+R 2 ) R R 2 Cs 2 Let R 2 C = T R 1 + R 2 = β R 2 Hence V o(s) 1 + Ts (1 + βts) V in (s) = Which represent a lag compensator Here T = R 2 C = 1.1 = 1 sec β = = 10 1 Maximum phase lag occurs at frequency ω n = 1 T β = = rad/sec Q NO. 22. [Ans. C] Inter changing the terminals of the auxiliary winding Q NO. 23. [Ans. D] : , info@thegateacademy.com Copyright reserved. Web: 15
18 2 Ω + 10 V 5 A + 5 A V out So V out = (5 2) + 10 = 20 V Q NO. 24. [Ans. D] Number of KCL equation = n 1 = 5 1 = 4 Number of KVL equation = b (n 1) = 7 (5 1) = 3 Q NO. 25. [Ans. A] Q NO. 26. [Ans. D] F = (a + b + c + d ) + (b + c ) = (a + b + c + d ) (b + c ) = a b c d b c = 0 : , info@thegateacademy.com Copyright reserved. Web: 16
19 Q NO. 27. [Ans. D] 9 Ω V 3 V 1 1 Ω 1 Ω V B V A + 4 Ω V out V 2 V A = ( 4 5 V V 2) ; V out = 9V V A = 9V 3 + 8V 1 + 2V 2 Q NO. 28. : , info@thegateacademy.com Copyright reserved. Web: 17
20 [Ans. B] Given that Bandwidth of X 1 (ω) = B 1 Bandwidth of X 2 (ω) = B 2 System has h(t) = e 2 t and input to the system is x 1 (t) x 2 (t) The bandwidth of x 1 (t)is B 1 + B 2 The bandwidth of output B 1 + B 2 So sampling rate will be 2(B 1 + B 2 ) Q NO. 29. [Ans. D] The Fourier transform of 2 sin(tτ 2) 2πrect ( ω t τ ) sin(2πt) rect ( ω πt 4π ) So, sin(2πt) e jωt dt = rect ( ω πt 4π ) Putting ω = 0in above equation sin(2πt) dt = 1 πt 2 sin(2π) dt = 2 πt Q NO. 30. [Ans. *] Range: 7.0 to 7.5 A. E m 2 4m + 4 = 0 m = 2, 2 y = (C 1 + C 2 x)e 2x y(0) = 0 C 1 = 0 y = C 2 xe 2x y = C 2 e 2x + 2C 2 xe 2x C 2 = 1 y = xe 2x ; y(1) = e 2 = 7.38 : , info@thegateacademy.com Copyright reserved. Web: 18
21 Q NO. 31. [Ans. *] Range: 4.40 to 4.45 E = 5xzi + (3x 2 + 3y)j + x 2 zk = F. d r c = 5xzdx + (3x 2 + 2y) dy + x 2 zdz c x = t, y = t 2, z = t, t = 0 to 1 dx = dt; dy = 2t dy, dz = dt 1 = 5t 2 dt + (3t 2 + 2t 2 )2tdt + t 3 dt 0 1 = (5t t 3 )dt 0 1 = [ 5t t4 4 ] 0 = = = 4.41 Q NO. 32. [Ans. D] Q NO. 33. [Ans. A] ae 4x x < 0 f x (x) = { 3 2 e 3x x 0 f x (x) = 1 : , info@thegateacademy.com Copyright reserved. Web: 19
22 ae 4x dx + [ ae4x 0 4 ] a = 1 a = 2 + [ P(x < 0) = 3 2 e 3x e 3x 3 ] 0 0 = [ e4x 0 2 ] = 1 2 2e 4x dx = 1 dx = 1 Q NO. 34. [Ans. *] Range: 19.5 to 20.5 To find impedance seen by V s V + 1 A I s 2 Ω 2 Ω + V s 3 Ω 4 V 1 4 Ω Z s = V s I s V 1 = 2I s Applying KCL at node A I s + 4V 1 = V A 3 + V A 6 V A = V s V 1 and V 1 = 2I s So, I s + 8I s = V s 2I s + V s 2I s I s = 2V s 4I s + V s 2I S 3V s = 60 I s V s = 20 Ω I s : , info@thegateacademy.com Copyright reserved. Web: 20
23 Q NO. 35. [Ans. *] Range: 34 to Ω I 1 I V V 1 [Z] V 2 20 Ω Given Z 11 = 40Ω ; Z 12 = 60 W Z 21 = 80 Ω; Z 22 = 100Ω From the figure V 2 = 20I 2. 1 And V 1 = 40 I I 2. 2 V 2 = 80I I 2. 3 From equation 1 and 3 we get So, 20I 2 = 80I I 2 I 2 = 2 3 I 1. 4 Using equation 2 and 4, we get So, V 1 = 40 I I 2 = 40I ( 2 3 ) I 1 V 1 = 0 From the figure, 20 = 10I 1 + V 1 Since V 1 = 0 So, I 1 = 2A So, I 2 = 4 3 A Power dissipated in R L = I 2 2 R L = ( 4 3 ) 2 20 = = W 9 : , info@thegateacademy.com Copyright reserved. Web: 21
24 Q NO. 36. [Ans. *] Range: 2.9 to 3.1 Using star to delta conversion L L L Line current will be zero when the parallel pair of induction-capacitor is resonant at f = 50 Hz So, 50 2π = 1 LC 3 100π = 1 LC 3 Since, L = 10 mh C will be 3.03 mf Q NO. 37. C C L L C/3 C/3 C/3 L [Ans. *] Range: 6.8 to 7.2 : , info@thegateacademy.com Copyright reserved. Web: 22
25 S 1 4 V 5 μf 5 Ω i(t) = ( 4 5 e t τ ) ; τ = RC = sec Change lost by capacitor from t = 25 μs to 100 μs is 100 μs 25 μs Q NO. 38. i(t)dt = C [Ans. *] Range: 2.20 to 2.35 The ratio = P 1 P 2 = X 2 X 1 : , info@thegateacademy.com Copyright reserved. Web: 23
26 j0.1 j0.5 X 1 = j0.2 j0.1 j0.5 = j0.5 j0.6 X 2 = j j0.3 3 j0.3 j0.3 j0.6 = j j1.2 j0.3 j0.6 = j j1.2 j0.3 j0.3 = j j1.2 j0.1 j0.15 j0.15 j X 12 j0.35 j j0.75 X 12 = X 2 j0.35 j0.15 j j j0.075 P 1 = X 2 = j1.2 = 2. 4 P 2 x 1 j0.5 Q NO. 39. = j1.2 p. u. [Ans. C] : , info@thegateacademy.com Copyright reserved. Web: 24
27 Open loop transfer function K(s + 1) G(s) = s(1 + Ts)(1 + 2s) ; K > 0 and T > 0 For closed loop system stability, characteristic equation is 1 + G(s)H(s) = 0 K(s + 1) 1 + s(1 + Ts)(1 + 2s). 1 = 0 s(1 + Ts)(1 + 2s) + k(s + 1) = 0 2Ts 3 + (2 + T)s 2 + (1 + k)s + k = 0 Using Routh s criteria s 3 2T (1 + k) s 2 (2 + T) K s 1 (2 + T)(1 + k) 2Tk 0 (2 + T) s 0 K For stability, k > 0 And (2 + T)(1 + k) 2Tk > 0 k(2 + T 2T) + (2 + T) > 0 Or, (T 2)k + 2(2 + T) > 0 (T 2)k + 2(2 + T) > 0 (2 + T) k > (T 2) (2 + T) k > (T 2) Or, k < T + 2 (T 2) Hence for stability, 0 < k < T + 2 T 2 Q NO. 40. [Ans. *] Range: to At no load, V s = AV R 400 = A 420 A = = A = 1 + YZ = 1 + (r + jωl)(g + jωc) 2 2 For lossless line r = 0, g = 0 Then, A = 1 (ωc)(ωl) 2 βl = ωl ωc A = = 1 β2 l 2 βl = : , info@thegateacademy.com Copyright reserved. Web: 25
28 β = l V f = 2π β = 2π 50 l = km Q NO. 41. ( l ) [Ans. *] Range: 0.31 to 0.33 P 1 = = 400 kw Q 1 = = 300 kvar The power factor is to be raised to unity The motor has to supply 300 kvar The motor rating is 100 kw, 300 kvar ϕ m = tan 1 ( Q P ) ϕ m = tan 1 ( ) = Power factor of motor = cos ϕ m = cos = Q NO. 42. [Ans. *] Range: 186 to 190 Q NO. 43. [Ans. *] Range: 31.0 to V, 3-phase, Δ connected induction motor (I st ) line = 120 A at rated voltage. At, V = 110 V, i.e. reduced voltage I st = x(i st ) rated Where, x = V reduced V rated x = (I st ) at 110 V = ( ) 120 = A : , info@thegateacademy.com Copyright reserved. Web: 26
29 Q NO. 44. [Ans. *] Range: 5.9 to kva, V transformer, a 2winding = = 2 [(kva) auto ] max = (a 2winding + 1)(kVA) 2winding = (2 + 1) 2 = 6 Q NO. 45. [Ans. *] Range: 5.9 to 6.1 V o = 2 V m 200π cos α = 2 cos 120 π π V o = 200 V V o = 200 V Power balance equation, EI o = I 2 o R + V o I o 800 I o = I 2 o (20) I o I o = 30 A I o = I or Power fed to source = V o I o = = 6 kw : , info@thegateacademy.com Copyright reserved. Web: 27
30 Q NO. 46. [Ans. *] Range: 23 to 25 V L = V ph = 2 3 V s V ph = P = 3 V ph 2 R = = 24 kw 30 Q NO. 47. [Ans. *] Range: 3 to 4 i s I s T on T t : , info@thegateacademy.com Copyright reserved. Web: 28
31 V o = 1 V s 1 α = 1 1 α α = 0.1 V s I s = Power 360 I s = 4000 I s = 11.1 A Neglecting ripple in i s, I switch (rms) = I s ( T on = I s α = = 3.5 A 1 2 T ) Q NO. 48. [Ans. *] Range: 9.1 to 9.3 δ P = V s I s p. f = 220 I s 1 I s = A tan δ = I s X s V s δ = tan π ( ) 220 δ = 9.21 Q NO. 49. I s V C1 I s X 2 : , info@thegateacademy.com Copyright reserved. Web: 29
32 [Ans. A] x = Ax Eigen values are λ 1 and λ 2 We can write, ϕ(t) = [ eλ 1t 0 0 e λ 2t ] Response due to initial conditions, x(t) = ϕ(t). x(0) x(t) = [ eλ 1t 0 0 e λ 2t ] [α 0 ] = α eλ 1t Q NO. 50. [Ans. B] Damping ratio ξ = 0.5 Undamped natural frequency ω n = 10 rad sec Steady state output to a unit step input C ss = 1.02 Hence, steady state error = e ss = 0.02 Characteristics equation is, s 2 + 2ξω n s + ω 2 n = 0 s s = 0 s s = 0 From options, if we take option B Then, C ss = lim s. C(s) = lims 1 s 0 s 0 s 102 s s C ss = 1.02 Hence option B is correct answer Q NO. 51. [Ans. A] : , info@thegateacademy.com Copyright reserved. Web: 30
33 At 11 kv, load is 8 MW, 0.8 PF lagging (V ph) Δ (V ph ) Y = (I ph) Y (I ph ) Δ (I ph ) Δ = (I ph ) Y (V ph) Y (V ph ) Δ (I ph ) Δ (V ph ) Y = = 1: 10 3 Q NO. 52. [Ans. A] Ks OLTF G(s) = (s 1)(s 4) Now, characteristics equation 1 + G(s)H(s) = 0 Ks (s 1)(s 4) + 1 = 0 Ks + (s 2 5s + 4) = 0 For break away point: dk ds = 0 dk ds = [1 0 4 s 2] = 0 We get s = ±2 Therefore valid break away point is s = 2, now gain at s = 2 is Product of distances from all the poles to break away point K = Product of distance from all the zeros to break away point Gain, K = = 1 : , info@thegateacademy.com Copyright reserved. Web: 31
34 Q NO. 53. [Ans. *] Range: 5.5 to 6.5 j0.4 + ~ 1 0 ~ j j0.3 j0.3 I a 1 j0.15 j0.15 j0.2 + ~ 1 0 j0.15 I a 1 j0.15 I f = 3 I a1 = I f = 6 p. u : , info@thegateacademy.com Copyright reserved. Web: 32
35 Q NO. 54. [Ans. *] Range: 1.9 to rev kwh, 20 rev, 30 sec P loss = = 2 kw Q NO. 55. [Ans. *] Range: 9.4 to 9.7 Voltage induced = E m. dl 1 0 (Where E m is induced electric field) = E m. volts Since, voltage induced = 1 V So, E m = 1 V m As we know E m = V B Where V = (Radius of path) (Angular velocity) 1V = (V 1 Tesla) m v = 1 m sec v = r ω = 1 m sec Since, r = 1 m; So, ω = 1 rad sec Now from this we get ω = 2 π N 60 N = 30 = 9.55 revolutions per minute π = 1 rad sec : , info@thegateacademy.com Copyright reserved. Web: 33
GATE : , Copyright reserved. Web:www.thegateacademy.com
Index. Question Paper Analysis 2. Question Paper & Answer keys : 080-67 66 222, info@thegateacademy.com Copyright reserved. Web:www.thegateacademy.com ANALYSIS OF GATE 206 Electrical Engineering CN Mathematics
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