EE GATE

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Harry Moore
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D 41 D 4 C 43 A 44 C 45 A 46 C 47 B 48 C 49 A 50 A 51 B 5 C 53 D 54 A 55 C 56 B 57 A 58 A 59 B 60 D Q.1 (B) Q. (C) Q.3 (A) Q.6 (D) Q.7 (A) Q.9 (B) Q.1 (B) Q.13 (A) Q.

1 EE GATE GATE Answer Keys 1 B C 3 A 4 D 5 B 6 D 7 A 8 A 9 B 10 D 11 D 1 B 13 A 14 C 15 B 16 C 17 C 18 D 19 B 0 C 1 D C 3 A 4 A 5 D 6 C 7 D 8 D 9 B 30 C 31 C 3 A 33 A 34 B 35 B 36 A 37 A 38 C 39 A 40 D 41 D 4 C 43 A 44 C 45 A 46 C 47 B 48 C 49 A 50 A 51 B 5 C 53 D 54 A 55 C 56 B 57 A 58 A 59 B 60 D Q.1 (B) Q. (C) Q.3 (A) Q.6 (D) Q.7 (A) Q.9 (B) Q.1 (B) Q.13 (A) Q.14 (C) Since all the applied voltage appears across it. u Let y=, % error in y = % error in u - % error in v v Here u = G G, v = G 1 3 When diode is ON, i=1ma. When diode is OFF, i is zero after small reverse recovery time. 000V Internal resistance = = 5Ω 400A Internal voltage drop = 00A 5 Ω=1000V It is a balanced Wheatstone bridge. EV V P = sin sin X δ X δ All other XNOR, NOR, NAND gates can be derived from NOT, OR, AND gates. written permission. Discuss this question paper at 1

2 EE GATE N φ M = L L ; L =, φ = B A i1 M1 0 0 M A1A = = 4 = M M M = 1 Q.15 (B) B is stable as slip 1 A is unstable as slip 0 Q.17 (C) a = 1.4; GM = -0 log a Since a>1, GM will be negative and unstable 1 jωt1 Now, G ( jω) H( jω ) = j ω C 1 ω ( j T ) (Assuming one pole in RHS plane) G jω H jω = 90 tan ωt tan ωt At ω=0, G jω H jω = 90 At ω=, G jω H jω = 70 Q.18 (D) From Routh table, The equation is 3 s 4s s 4 = 0 s 4 s s 4 = 0 s 1 s 4 = 0 s = ± j = 0.5 LHS Q.19 (B) written permission. Discuss this question paper at

3 EE GATE Since at ω= and ω=5, slope changes from 40dB/dec to -60dB/dec at both the value. So there are poles at ω= and ω = 5. Also at ω=0.1 slope is -40dB/dec. It means there are two poles at origin. Q.0 (C) Q.1 (D) Hence the transfer function of the form, K ( s5) ( ) ( ) T(s) = s s s 5 5k 80 = 0 log K = Trace = Sum of principal diagonal elements R = 10k ~ i s V V R = 10k V out I I 10µ F X 1 1 j100 = = = jωc jπ π c 6 V 10 jπ I = = = X X 100 c c 10,000 jπ Vout = 10 = 10 j100 π 100 V Vout j100 π is = = = j10 π ma = 10πmA 90ºs R written permission. Discuss this question paper at 3

4 EE GATE Q. (C) When S is closed, S is open 1 1 C is full charged to 3V 3V S1 S When S is open 1 S is closed Q=C V = C V 1 1 1F 3V = F V ; V = 1.5V C1 C C1 C Q.4 (A) Apply KVL for input loop V = kω I 49I I / 100µ F s in I1 I1 = ( KI1 ) = ki µ F µ F S C off = µ F Q.5 (D) XRA A Accumulator is cleared, A 00H ( A) FOH MVI B B FOH SUB B (A) A - B FFFFFF10H Q.6 (C) written permission. Discuss this question paper at 4

5 EE GATE y Bus here y11 y1 y13 y14 y1 y y3 y4 = y31 y3 y33 y34 y41 y4 y43 y44 y = y y y y = 5j [1] y = y y y y = 10j [] y = y y y y = 9j [3] y = y y y y y = y = y = j y = y = y =.5j y = y = y = 0j y = y = y =.5j y = y = y = 4j From equation 1, y = y y y y = 8j [4] y = 5j j.5j 0j = 5j 4.5j = 0.5j From Eqn. y = y y y y = 10j j.5j 4j = 1.5j From Eqn. 3 y = y y y y = 9j.5j.5j j4 = 0 From Eqn. 4, y = y y y y = 8j 0 4j j4 = S y 1 10 y S 0 y 1 y 4 y S 4 y 43 S 3 written permission. Discuss this question paper at 5

6 EE GATE Q.8 (D) Phase crossover frequency ω = 180 π ω=15.7 ( ω) ω= 15.7 a= G j ; GM=0 log a = 3.9dB Q.3 (A) P 500MW G = = = MVA cos φ πn 1 π 3000 KE stored = M = = MJ Q.40 10f N = = 3000rpm P KE H = = =.44MJ / MVA MVA ( 0,) (,0) 0 m = = 1 0 y = 1 x 0 = x x y = F x,y dxdy ( x,y) 0 = x x x dx y y y dy = 0 0 Q.41 (D) For V in R V out k 1k written permission. Discuss this question paper at 6

7 EE GATE Hysterisis Diagram is V out V th V in V tl 1k V = V = V = t th H 3k After t, Vout After t 4, Vout = 1k Vtl = ( 3) = 1V = t4 3k = 3VA 6VA Q.4 (C) Zauxiliary 10 = ra jω La = 1 j π 50 π = 1 j1000 Zmain 0.1 = rm jω Lm = 0.1 jπ 50 = 0.1 j π a m a m starting torque = KI I sin I,I sin 0 = 0 Q.43 (A) di 1 VL = L ; i= Vdt dt L π i = 10sin 100π t = 100π 0.1 i t cos100 π t = cos100πt 100πt i t = 0 when cos 100πt = πt=n π n t = 00 When n=1, t=5msec; n=, t=10msec; n=3, t=15msec Answer is (A) written permission. Discuss this question paper at 7

8 EE GATE Q.47 (B) By Superposition theorem, When 4V is shorted, i.1 =0, as all the current will flow through short. When 5A is open, i.11 =A Therefore i Ω =A Q.48 (C) Wattmeter reading = Current through the coil (I) x Voltage across Pressure coil (V) x Cos(phase angle between V & I) R 100Ω 100Ω RYB C Y B IBR = ICC = = V = V = YB PC W = cos(40) = 800W Q.49 (A) For triangular wave, Avg value rms value V V m 3 m 3 Vm = 10V, Vm = 30V 3 30 rms value = = Q.53 (D) Q.54 (A) written permission. Discuss this question paper at 8

9 EE GATE di' 400 dt V = L = m dt' π dt dt is max 0 5msec dt V= m V π 5msec π If i t max = 10 sin ( 100πt) mh 400 di t 400 V = = m 10 cos 100πt π dt π = 400 cos 100 πt V Q.55 (C) dx 1 dt = 400V ( t) ( t) = 3x t x t µ t 1 sx s = 3x s x s [1] dx dt 1 1 = x t µ t sx s = x s 1 1 s x = 1 ( s ) ( s 3) 1 1 x = ; From [1], ( s3) x1 = s s s x Q.56 (B) = s 5 y s s 5 s 5 y t = x1 t H(s)= = = X s s s 3 s 5s 6 At 1 1 e = L φ s ; φ s = si-a From given state equations = µ x& t 3x t x t t 1 1 = µ x& t x t t written permission. Discuss this question paper at 9

10 EE GATE s s 3 1 φ ( s) = = 0 s 0 0 s 1 s 1 = ( s ) ( s 3) 0 s s 3 ( s ) ( s 3 ) φ ( s) = 1 0 ( s ) 3t t 3t At 1 e e e e = L φ ( s) = t 0 e Q.57 (A) Q.58 (A) 10A C A B 400V D 5A 1000V 1000 = 5A 400V = 10A 1000V Q.59 (B) 5V kω kω 3V A 1kΩ.5V 1kΩ 3V I A 1kΩ B B written permission. Discuss this question paper at 10

11 EE GATE V OC sc = V.5 = 1kΩ I 3V V V = 1kΩ I.5 = V 3V V = 5V V = 0.5V Apply short across : V = 0, 3V = 0, I Q.60 (D).5V = =.5mA 1kΩ Voc 0.5V 1000 Rth = = = = 00Ω = 0.kΩ I.5mA 5 sc written permission. Discuss this question paper at 11

Basics of Network Theory (Part-I)

Basics of Network Theory (PartI). A square waveform as shown in figure is applied across mh ideal inductor. The current through the inductor is a. wave of peak amplitude. V 0 0.5 t (m sec) [Gate 987: Marks]