Page 1 ?EE-Paper Code-A? GATE 2011 Q. No. 1 â€“ 25 Carry One Mark Each 1. Roots of the algebraic equation 3 2 x x x 1 0 + + + = are (A) ( ) 1, j, j + + - (B) ( ) 1, 1, 1 + - + (C) (0,0,0) (D) ( ) 1, j, j - + - Answer: - (D) Exp: - ( )( ) 3 2 2 x x x 1 0 x 1 x 1 0 + + + = ? + + = 2 x 1 0; x 1 0 ? + = + = x 1 x j ? = - = ± 2. With K as a constant, the possible solution for the first order differential equation 3x dy e dx - = is (A) 3x 1 e K 3 - - + (B) 3x 1 e K 3 - + (C) 3x 1 e K 3 - - + (D) x 3e K - - + Answer: - (A) Exp: - 3x 3x dy e dy e dx dx - - = ? = Integrate on both sides 3x 3x e 1 y K e K 3 3 - - = + = - + - 3. The r.m.s value of the current i(t) in the circuit shown below is (A) 1 A 2 (B) 1 A 2 (C) 1A (D) 2A Answer: - (B) Exp: - 1 rad /sec ? = L C X 1 ; X 1 = O = O ( ) Sin t I t Sin t; 1 = = O rms 1 I A 2 = 1F 1H ~ 1O ( ) i t ( ) 1.0sint V + 1O 1Sint 1O + - Page 2 ?EE-Paper Code-A? GATE 2011 Q. No. 1 â€“ 25 Carry One Mark Each 1. Roots of the algebraic equation 3 2 x x x 1 0 + + + = are (A) ( ) 1, j, j + + - (B) ( ) 1, 1, 1 + - + (C) (0,0,0) (D) ( ) 1, j, j - + - Answer: - (D) Exp: - ( )( ) 3 2 2 x x x 1 0 x 1 x 1 0 + + + = ? + + = 2 x 1 0; x 1 0 ? + = + = x 1 x j ? = - = ± 2. With K as a constant, the possible solution for the first order differential equation 3x dy e dx - = is (A) 3x 1 e K 3 - - + (B) 3x 1 e K 3 - + (C) 3x 1 e K 3 - - + (D) x 3e K - - + Answer: - (A) Exp: - 3x 3x dy e dy e dx dx - - = ? = Integrate on both sides 3x 3x e 1 y K e K 3 3 - - = + = - + - 3. The r.m.s value of the current i(t) in the circuit shown below is (A) 1 A 2 (B) 1 A 2 (C) 1A (D) 2A Answer: - (B) Exp: - 1 rad /sec ? = L C X 1 ; X 1 = O = O ( ) Sin t I t Sin t; 1 = = O rms 1 I A 2 = 1F 1H ~ 1O ( ) i t ( ) 1.0sint V + 1O 1Sint 1O + - ?EE-Paper Code-A? GATE 2011 4. The fourier series expansion ( ) 0 n n n 1 f t a a cosn t b sin n t 8 = = + ? + ? ? of the periodic signal shown below will contain the following nonzero terms ( ) 0 n A a and b , n 1,3,5,... = 8 ( ) 0 n B a and a , n 1,2,3,... = 8 ( ) 0 n n C a a and b , n 1,2,3,... = 8 ( ) 0 n D a anda n 1,3,5,... = 8 Answer: - (D) Exp: - ? it satisfies the half wave symmetry, so that it contains only odd harmonics. ? It satisfies the even symmetry. So n b 0 = 5. A 4 â€“ point starter is used to start and control the speed of a (A) dc shunt motor with armature resistance control (B) dc shunt motor with field weakening control (C) dc series motor (D) dc compound motor Answer: - (A) 6. A three-phase, salient pole synchronous motor is connected to an infinite bus. Ig is operated at no load a normal excitation. The field excitation of the motor is first reduced to zero and then increased in reverse direction gradually. Then the armature current (A) Increases continuously (B) First increases and then decreases steeply (C) First decreases and then increases steeply (D) Remains constant Answer: - (B) 7. A nuclear power station of 500 MW capacity is located at 300 km away from a load center. Select the most suitable power evacuation transmission configuration among the following options (A) (B) (C) (D) Answer: - (A) ( ) f t 0 t ~ 132kV,300km doublecircuit Load center ~ 132kv,300 km single circuitwith40%seriescapacitor compensation Load center ~ 400kV,300kmsinglecircuit Load center ~ 400kV,300km doublecircuit Load center Page 3 ?EE-Paper Code-A? GATE 2011 Q. No. 1 â€“ 25 Carry One Mark Each 1. Roots of the algebraic equation 3 2 x x x 1 0 + + + = are (A) ( ) 1, j, j + + - (B) ( ) 1, 1, 1 + - + (C) (0,0,0) (D) ( ) 1, j, j - + - Answer: - (D) Exp: - ( )( ) 3 2 2 x x x 1 0 x 1 x 1 0 + + + = ? + + = 2 x 1 0; x 1 0 ? + = + = x 1 x j ? = - = ± 2. With K as a constant, the possible solution for the first order differential equation 3x dy e dx - = is (A) 3x 1 e K 3 - - + (B) 3x 1 e K 3 - + (C) 3x 1 e K 3 - - + (D) x 3e K - - + Answer: - (A) Exp: - 3x 3x dy e dy e dx dx - - = ? = Integrate on both sides 3x 3x e 1 y K e K 3 3 - - = + = - + - 3. The r.m.s value of the current i(t) in the circuit shown below is (A) 1 A 2 (B) 1 A 2 (C) 1A (D) 2A Answer: - (B) Exp: - 1 rad /sec ? = L C X 1 ; X 1 = O = O ( ) Sin t I t Sin t; 1 = = O rms 1 I A 2 = 1F 1H ~ 1O ( ) i t ( ) 1.0sint V + 1O 1Sint 1O + - ?EE-Paper Code-A? GATE 2011 4. The fourier series expansion ( ) 0 n n n 1 f t a a cosn t b sin n t 8 = = + ? + ? ? of the periodic signal shown below will contain the following nonzero terms ( ) 0 n A a and b , n 1,3,5,... = 8 ( ) 0 n B a and a , n 1,2,3,... = 8 ( ) 0 n n C a a and b , n 1,2,3,... = 8 ( ) 0 n D a anda n 1,3,5,... = 8 Answer: - (D) Exp: - ? it satisfies the half wave symmetry, so that it contains only odd harmonics. ? It satisfies the even symmetry. So n b 0 = 5. A 4 â€“ point starter is used to start and control the speed of a (A) dc shunt motor with armature resistance control (B) dc shunt motor with field weakening control (C) dc series motor (D) dc compound motor Answer: - (A) 6. A three-phase, salient pole synchronous motor is connected to an infinite bus. Ig is operated at no load a normal excitation. The field excitation of the motor is first reduced to zero and then increased in reverse direction gradually. Then the armature current (A) Increases continuously (B) First increases and then decreases steeply (C) First decreases and then increases steeply (D) Remains constant Answer: - (B) 7. A nuclear power station of 500 MW capacity is located at 300 km away from a load center. Select the most suitable power evacuation transmission configuration among the following options (A) (B) (C) (D) Answer: - (A) ( ) f t 0 t ~ 132kV,300km doublecircuit Load center ~ 132kv,300 km single circuitwith40%seriescapacitor compensation Load center ~ 400kV,300kmsinglecircuit Load center ~ 400kV,300km doublecircuit Load center ?EE-Paper Code-A? GATE 2011 8. The frequency response of a linear system ( ) G j? is provided in the tubular form below ( ) G j? 1.3 1.2 1.0 0.8 0.5 0.3 ( ) G j ? ? O 130 - O 140 - O 150 - O 160 - O 180 - O 200 - ( ) O A 6 dB and 30 ( ) O B 6 dB and 30 - ( ) O C 6 dB and 30 - ( ) O D 6 dB and 30 - - Answer: - (A) Exp: - At ( ) G jw 180 ? = - magnitude M=0.5 So G.M = 20 log 1 6dB 0.5 ? ? = ? ? ? ? At ( ) G jw 1 = phase angle ( ) G jw 150 ? = - So P.M = ( ) 0 180 150 30 + - = 9. The steady state error of a unity feedback linear system for a unit step input is 0.1. The steady state error of the same system, for a pulse input r(t) having a magnitude of 10 and a duration of one second, as shown in the figure is (A) 0 (B) 0.1 (C) 1 (D) 10 Answer: - (A) Exp: - For step input ss 1 e 0.1 k 9 1 k = = ? = + ( ) 9 G S S 1 = + Now the input is pulse ( ) ( ) ( ) r t 10 u t u t 1 ? ? = - - ? ? ( ) s 1 e r s 10 s - ? ? - = ? ? ? ? ( ) ( ) ( ) s ss S 0 S 0 S 10 1 e S.R s S e Lt Lt S 10 1 G S H S S 1 - ? ? ? ? - ? ? = = + + + 0 0 10 = = 10. Consider the following statement (i) The compensating coil of a low power factor wattmeter compensates the effect of the impedance of the current coil. (ii) The compensating coil of a low power factor wattmeter compensates the effect 0of the impedance of the voltage coil circuit. ( ) r t 10 1s t Page 4 ?EE-Paper Code-A? GATE 2011 Q. No. 1 â€“ 25 Carry One Mark Each 1. Roots of the algebraic equation 3 2 x x x 1 0 + + + = are (A) ( ) 1, j, j + + - (B) ( ) 1, 1, 1 + - + (C) (0,0,0) (D) ( ) 1, j, j - + - Answer: - (D) Exp: - ( )( ) 3 2 2 x x x 1 0 x 1 x 1 0 + + + = ? + + = 2 x 1 0; x 1 0 ? + = + = x 1 x j ? = - = ± 2. With K as a constant, the possible solution for the first order differential equation 3x dy e dx - = is (A) 3x 1 e K 3 - - + (B) 3x 1 e K 3 - + (C) 3x 1 e K 3 - - + (D) x 3e K - - + Answer: - (A) Exp: - 3x 3x dy e dy e dx dx - - = ? = Integrate on both sides 3x 3x e 1 y K e K 3 3 - - = + = - + - 3. The r.m.s value of the current i(t) in the circuit shown below is (A) 1 A 2 (B) 1 A 2 (C) 1A (D) 2A Answer: - (B) Exp: - 1 rad /sec ? = L C X 1 ; X 1 = O = O ( ) Sin t I t Sin t; 1 = = O rms 1 I A 2 = 1F 1H ~ 1O ( ) i t ( ) 1.0sint V + 1O 1Sint 1O + - ?EE-Paper Code-A? GATE 2011 4. The fourier series expansion ( ) 0 n n n 1 f t a a cosn t b sin n t 8 = = + ? + ? ? of the periodic signal shown below will contain the following nonzero terms ( ) 0 n A a and b , n 1,3,5,... = 8 ( ) 0 n B a and a , n 1,2,3,... = 8 ( ) 0 n n C a a and b , n 1,2,3,... = 8 ( ) 0 n D a anda n 1,3,5,... = 8 Answer: - (D) Exp: - ? it satisfies the half wave symmetry, so that it contains only odd harmonics. ? It satisfies the even symmetry. So n b 0 = 5. A 4 â€“ point starter is used to start and control the speed of a (A) dc shunt motor with armature resistance control (B) dc shunt motor with field weakening control (C) dc series motor (D) dc compound motor Answer: - (A) 6. A three-phase, salient pole synchronous motor is connected to an infinite bus. Ig is operated at no load a normal excitation. The field excitation of the motor is first reduced to zero and then increased in reverse direction gradually. Then the armature current (A) Increases continuously (B) First increases and then decreases steeply (C) First decreases and then increases steeply (D) Remains constant Answer: - (B) 7. A nuclear power station of 500 MW capacity is located at 300 km away from a load center. Select the most suitable power evacuation transmission configuration among the following options (A) (B) (C) (D) Answer: - (A) ( ) f t 0 t ~ 132kV,300km doublecircuit Load center ~ 132kv,300 km single circuitwith40%seriescapacitor compensation Load center ~ 400kV,300kmsinglecircuit Load center ~ 400kV,300km doublecircuit Load center ?EE-Paper Code-A? GATE 2011 8. The frequency response of a linear system ( ) G j? is provided in the tubular form below ( ) G j? 1.3 1.2 1.0 0.8 0.5 0.3 ( ) G j ? ? O 130 - O 140 - O 150 - O 160 - O 180 - O 200 - ( ) O A 6 dB and 30 ( ) O B 6 dB and 30 - ( ) O C 6 dB and 30 - ( ) O D 6 dB and 30 - - Answer: - (A) Exp: - At ( ) G jw 180 ? = - magnitude M=0.5 So G.M = 20 log 1 6dB 0.5 ? ? = ? ? ? ? At ( ) G jw 1 = phase angle ( ) G jw 150 ? = - So P.M = ( ) 0 180 150 30 + - = 9. The steady state error of a unity feedback linear system for a unit step input is 0.1. The steady state error of the same system, for a pulse input r(t) having a magnitude of 10 and a duration of one second, as shown in the figure is (A) 0 (B) 0.1 (C) 1 (D) 10 Answer: - (A) Exp: - For step input ss 1 e 0.1 k 9 1 k = = ? = + ( ) 9 G S S 1 = + Now the input is pulse ( ) ( ) ( ) r t 10 u t u t 1 ? ? = - - ? ? ( ) s 1 e r s 10 s - ? ? - = ? ? ? ? ( ) ( ) ( ) s ss S 0 S 0 S 10 1 e S.R s S e Lt Lt S 10 1 G S H S S 1 - ? ? ? ? - ? ? = = + + + 0 0 10 = = 10. Consider the following statement (i) The compensating coil of a low power factor wattmeter compensates the effect of the impedance of the current coil. (ii) The compensating coil of a low power factor wattmeter compensates the effect 0of the impedance of the voltage coil circuit. ( ) r t 10 1s t ?EE-Paper Code-A? GATE 2011 (A) (i) is true but (ii) is false (B) (i) is false but (ii) is true (C) both (i) and (ii) are true (D) both (i) and (ii) are false Answer: - (B) 11. A low â€“ pass filter with a cut-off frequency of 30Hz is cascaded with a high-pass filter with a cut-off frequency of 20Hz. The resultant system of filters will function as (A) an all-pass filter (B) an all-stop filter (B) an band stop (band-reject) filter (D) a band â€“ pass filter Answer: - (D) Exp: - So it is a band pass filter 12. The CORRECT transfer characteristic is (A) (B) (C) (D) R 12V + 12V + O V 12V - 12V - R R R R vi - + - + Vo 6v - 12v - Vi 12v + 12V + 6v - 12v - Vi O V 6v + Vo Vi 6v + 6v - 12v + Vo Vi 6v + 6v - 12v + 20 30 Page 5 ?EE-Paper Code-A? GATE 2011 Q. No. 1 â€“ 25 Carry One Mark Each 1. Roots of the algebraic equation 3 2 x x x 1 0 + + + = are (A) ( ) 1, j, j + + - (B) ( ) 1, 1, 1 + - + (C) (0,0,0) (D) ( ) 1, j, j - + - Answer: - (D) Exp: - ( )( ) 3 2 2 x x x 1 0 x 1 x 1 0 + + + = ? + + = 2 x 1 0; x 1 0 ? + = + = x 1 x j ? = - = ± 2. With K as a constant, the possible solution for the first order differential equation 3x dy e dx - = is (A) 3x 1 e K 3 - - + (B) 3x 1 e K 3 - + (C) 3x 1 e K 3 - - + (D) x 3e K - - + Answer: - (A) Exp: - 3x 3x dy e dy e dx dx - - = ? = Integrate on both sides 3x 3x e 1 y K e K 3 3 - - = + = - + - 3. The r.m.s value of the current i(t) in the circuit shown below is (A) 1 A 2 (B) 1 A 2 (C) 1A (D) 2A Answer: - (B) Exp: - 1 rad /sec ? = L C X 1 ; X 1 = O = O ( ) Sin t I t Sin t; 1 = = O rms 1 I A 2 = 1F 1H ~ 1O ( ) i t ( ) 1.0sint V + 1O 1Sint 1O + - ?EE-Paper Code-A? GATE 2011 4. The fourier series expansion ( ) 0 n n n 1 f t a a cosn t b sin n t 8 = = + ? + ? ? of the periodic signal shown below will contain the following nonzero terms ( ) 0 n A a and b , n 1,3,5,... = 8 ( ) 0 n B a and a , n 1,2,3,... = 8 ( ) 0 n n C a a and b , n 1,2,3,... = 8 ( ) 0 n D a anda n 1,3,5,... = 8 Answer: - (D) Exp: - ? it satisfies the half wave symmetry, so that it contains only odd harmonics. ? It satisfies the even symmetry. So n b 0 = 5. A 4 â€“ point starter is used to start and control the speed of a (A) dc shunt motor with armature resistance control (B) dc shunt motor with field weakening control (C) dc series motor (D) dc compound motor Answer: - (A) 6. A three-phase, salient pole synchronous motor is connected to an infinite bus. Ig is operated at no load a normal excitation. The field excitation of the motor is first reduced to zero and then increased in reverse direction gradually. Then the armature current (A) Increases continuously (B) First increases and then decreases steeply (C) First decreases and then increases steeply (D) Remains constant Answer: - (B) 7. A nuclear power station of 500 MW capacity is located at 300 km away from a load center. Select the most suitable power evacuation transmission configuration among the following options (A) (B) (C) (D) Answer: - (A) ( ) f t 0 t ~ 132kV,300km doublecircuit Load center ~ 132kv,300 km single circuitwith40%seriescapacitor compensation Load center ~ 400kV,300kmsinglecircuit Load center ~ 400kV,300km doublecircuit Load center ?EE-Paper Code-A? GATE 2011 8. The frequency response of a linear system ( ) G j? is provided in the tubular form below ( ) G j? 1.3 1.2 1.0 0.8 0.5 0.3 ( ) G j ? ? O 130 - O 140 - O 150 - O 160 - O 180 - O 200 - ( ) O A 6 dB and 30 ( ) O B 6 dB and 30 - ( ) O C 6 dB and 30 - ( ) O D 6 dB and 30 - - Answer: - (A) Exp: - At ( ) G jw 180 ? = - magnitude M=0.5 So G.M = 20 log 1 6dB 0.5 ? ? = ? ? ? ? At ( ) G jw 1 = phase angle ( ) G jw 150 ? = - So P.M = ( ) 0 180 150 30 + - = 9. The steady state error of a unity feedback linear system for a unit step input is 0.1. The steady state error of the same system, for a pulse input r(t) having a magnitude of 10 and a duration of one second, as shown in the figure is (A) 0 (B) 0.1 (C) 1 (D) 10 Answer: - (A) Exp: - For step input ss 1 e 0.1 k 9 1 k = = ? = + ( ) 9 G S S 1 = + Now the input is pulse ( ) ( ) ( ) r t 10 u t u t 1 ? ? = - - ? ? ( ) s 1 e r s 10 s - ? ? - = ? ? ? ? ( ) ( ) ( ) s ss S 0 S 0 S 10 1 e S.R s S e Lt Lt S 10 1 G S H S S 1 - ? ? ? ? - ? ? = = + + + 0 0 10 = = 10. Consider the following statement (i) The compensating coil of a low power factor wattmeter compensates the effect of the impedance of the current coil. (ii) The compensating coil of a low power factor wattmeter compensates the effect 0of the impedance of the voltage coil circuit. ( ) r t 10 1s t ?EE-Paper Code-A? GATE 2011 (A) (i) is true but (ii) is false (B) (i) is false but (ii) is true (C) both (i) and (ii) are true (D) both (i) and (ii) are false Answer: - (B) 11. A low â€“ pass filter with a cut-off frequency of 30Hz is cascaded with a high-pass filter with a cut-off frequency of 20Hz. The resultant system of filters will function as (A) an all-pass filter (B) an all-stop filter (B) an band stop (band-reject) filter (D) a band â€“ pass filter Answer: - (D) Exp: - So it is a band pass filter 12. The CORRECT transfer characteristic is (A) (B) (C) (D) R 12V + 12V + O V 12V - 12V - R R R R vi - + - + Vo 6v - 12v - Vi 12v + 12V + 6v - 12v - Vi O V 6v + Vo Vi 6v + 6v - 12v + Vo Vi 6v + 6v - 12v + 20 30 ?EE-Paper Code-A? GATE 2011 Answer: - (D) Exp: - It is a Schmitt trigger and phase shift is zero. 13. A three-phase current source inverter used for the speed control of an induction motor is to be realized using MOSFET switches as shown below. Switches S 1 to S 6 are identical switches. The proper configuration for realizing switches S 1 to S 6 is (A) (B) (C) (D) Answer: - (C) 14. A point Z has been plotted in the complex plane, as shown in figure below. The plot of the complex number 1 y is z = d I A 1 S B 2 S 3 S A B 4 S 6 S 2 S l.M. A B A B A B A B Im unit circle o Z Re Im unit circle y â€¢ Re ( ) A Im unit circle y â€¢ Re ( ) B Im unit circle y â€¢ Re ( ) C Im unit circle y â€¢ Re ( ) DRead More

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