GATE 2017 EE - Session 1
Kreatryx GATE 2017 EE – Session 1 Answer Key and Solutions
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GATE 2017 EE - Session 1
01. A closed loop system has the characteristic equation given by s3 Ks2 K 2 s 3 0 .
For this system to be stable, which one of the following conditions should be satisfied? (a) 0 K 0.5 (b) 0.5 K 1 (c) 0 K 1 (d) K 1 01. Ans: (d) Solution: Characteristic equation s3 ks2 k 2 s 3 0
Routh array s3
1
k 2
s2
k
3
s1 0
s
k k 2 3 k 3
For system to be stable, all elements in first column must be positive k 2 2k 3 0 k 3 k 1 0 Root are k=-3 & k=1 So equations will be positive k 3 and1 and 1 k
n
02. Consider the system with following input-output relation y n 1 1 x n
Where x [n] is the input and y [n] is the output. The system is (a) Invertible and time invariant (b) invertible and time varying (c) non-invertible and time invariant (d) non-invertible and time varying
GATE 2017 EE - Session 1
01. A closed loop system has the characteristic equation given by s3 Ks2 K 2 s 3 0 .
For this system to be stable, which one of the following conditions should be satisfied? (a) 0 K 0.5 (b) 0.5 K 1 (c) 0 K 1 (d) K 1 01. Ans: (d) Solution: Characteristic equation s3 ks2 k 2 s 3 0
Routh array s3
1
k 2
s2
k
3
s1 0
s
k k 2 3 k 3
For system to be stable, all elements in first column must be positive k 2 2k 3 0 k 3 k 1 0 Root are k=-3 & k=1 So equations will be positive k 3 and1 and 1 k
n
02. Consider the system with following input-output relation y n 1 1 x n
Where x [n] is the input and y [n] is the output. The system is (a) Invertible and time invariant (b) invertible and time varying (c) non-invertible and time invariant (d) non-invertible and time varying 02. Ans: (d) Solution: n y n 1 1 x n
For n= Odd, y n 0 n= Even y n 2x n Hence, odd samples of x n are removed & cannot be recovered so it is not invertible. Since y n depends on x n as f n x n , the system is time variant 03. A solid iron cylinder is place in region containing a uniform magnetic field such that the cylinder axis is parallel to the magnetic field direction. The magnetic field lines inside the cylinder will
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GATE 2017 EE - Session 1
(a) bend closer to the cylinder axis (c) remain uniform as before
(b) bend farther away from the axis (d) cease to exist inside the cylinder
03. Ans: (a) Solution: Iron is a ferromagnetic material so magnetic field lines are attracted & they closer to cylinder axis
04. The power supplied by the 25 V source in the figure f igure shown below is ____________ W.
04. Ans: 250 Solution: By KCL, current in R2 1.4I
And 1.4I 14 I 10A Power supplied by 25V source= 25 10 250W 05. A 4 pole induction machine is working as an induction generator. The generator supply frequency is 60 Hz. The rotor current frequency is 5 Hz. The mechanical speed of the rotor in RPM is (a) 1350 (b) 1650
GATE 2017 EE - Session 1
01. A closed loop system has the characteristic equation given by s3 Ks2 K 2 s 3 0 .
For this system to be stable, which one of the following conditions should be satisfied? (a) 0 K 0.5 (b) 0.5 K 1 (c) 0 K 1 (d) K 1 01. Ans: (d) Solution: Characteristic equation s3 ks2 k 2 s 3 0
Routh array s3
1
k 2
s2
k
3
s1 0
s
k k 2 3 k 3
For system to be stable, all elements in first column must be positive k 2 2k 3 0 k 3 k 1 0 Root are k=-3 & k=1 So equations will be positive k 3 and1 and 1 k
n
02. Consider the system with following input-output relation y n 1 1 x n
Where x [n] is the input and y [n] is the output. The system is (a) Invertible and time invariant (b) invertible and time varying (c) non-invertible and time invariant (d) non-invertible and time varying 02. Ans: (d) Solution: n y n 1 1 x n
For n= Odd, y n 0 n= Even y n 2x n Hence, odd samples of x n are removed & cannot be recovered so it is not invertible. Since y n depends on x n as f n x n , the system is time variant 03. A solid iron cylinder is place in region containing a uniform magnetic field such that the cylinder axis is parallel to the magnetic field direction. The magnetic field lines inside the cylinder will
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GATE 2017 EE - Session 1
(a) bend closer to the cylinder axis (c) remain uniform as before
(b) bend farther away from the axis (d) cease to exist inside the cylinder
03. Ans: (a) Solution: Iron is a ferromagnetic material so magnetic field lines are attracted & they closer to cylinder axis
04. The power supplied by the 25 V source in the figure f igure shown below is ____________ W.
04. Ans: 250 Solution: By KCL, current in R2 1.4I
And 1.4I 14 I 10A Power supplied by 25V source= 25 10 250W 05. A 4 pole induction machine is working as an induction generator. The generator supply frequency is 60 Hz. The rotor current frequency is 5 Hz. The mechanical speed of the rotor in RPM is (a) 1350 (b) 1650
GATE 2017 EE - Session 1
(a) bend closer to the cylinder axis (c) remain uniform as before
(b) bend farther away from the axis (d) cease to exist inside the cylinder
03. Ans: (a) Solution: Iron is a ferromagnetic material so magnetic field lines are attracted & they closer to cylinder axis
04. The power supplied by the 25 V source in the figure f igure shown below is ____________ W.
04. Ans: 250 Solution: By KCL, current in R2 1.4I
And 1.4I 14 I 10A Power supplied by 25V source= 25 10 250W 05. A 4 pole induction machine is working as an induction generator. The generator supply frequency is 60 Hz. The rotor current frequency is 5 Hz. The mechanical speed of the rotor in RPM is (a) 1350 (b) 1650 (c) 1950 (d) 2250 05. Ans: (c) Solution: Slip = 5 1 60 12 For induction generator, slip is negative Hence, s 1 12 Rotor speed, Nr Ns 1 s Ns 1 1 12 120 60 1800rpm Synchronous speed, Ns 4 Nr 1800 1 1 1800 13 1950rpm 12 12
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GATE 2017 EE - Session 1
06. The slope and level detector circuit in a CRO has a delay of 100 ns. The start-stop sweep generator has a response time of 50 ns. In order to display correctly, a delay line of (a) 150 ns has to be inserted into the y-channel
(b) 150 ns has to be inserted into the x-channel (c) 150 ns has to be inserted into both x and y channels (d) 100 ns has to be inserted into both x and y channels 06. Ans: (a) Solution: The y-channel signal to be delayed w.r.t. x-channel signal for correct display Delay line=100+50=150 s
150 s delay line needs to be inserted in y-channel 07. A three-phase, 50 Hz, star-connected cylindrical-rotor synchronous machine is running as a motor. The machine is operated from a 6.6 kV grid and draws current at unity power factor (UPF). The synchronous reactance of the motor is 30 per phase. The load angle is 30
degrees. The power delivered to the motor in kW is __________. (Give the answer up to one decimal place). 07. Ans: 838.3 Solution: For unity pf, motor must be normally excited Hence, Ef cos Vt Vt 6.6kV cos cos 30 30 Since, armature resistance is negative
Excitation emf Ef
6.6kV cos30 6.6kV 6.6kV 6.6kV 0
2
GATE 2017 EE - Session 1
06. The slope and level detector circuit in a CRO has a delay of 100 ns. The start-stop sweep generator has a response time of 50 ns. In order to display correctly, a delay line of (a) 150 ns has to be inserted into the y-channel
(b) 150 ns has to be inserted into the x-channel (c) 150 ns has to be inserted into both x and y channels (d) 100 ns has to be inserted into both x and y channels 06. Ans: (a) Solution: The y-channel signal to be delayed w.r.t. x-channel signal for correct display Delay line=100+50=150 s
150 s delay line needs to be inserted in y-channel 07. A three-phase, 50 Hz, star-connected cylindrical-rotor synchronous machine is running as a motor. The machine is operated from a 6.6 kV grid and draws current at unity power factor (UPF). The synchronous reactance of the motor is 30 per phase. The load angle is 30
degrees. The power delivered to the motor in kW is __________. (Give the answer up to one decimal place). 07. Ans: 838.3 Solution: For unity pf, motor must be normally excited Hence, Ef cos Vt Vt 6.6kV cos cos 30 30 Since, armature resistance is negative
Excitation emf Ef
P
Ef Vt Xs
2
6.6kV 6.6kV cos30 6.6kV 6.6kV 6.6 sin sin 30 0
0
30
08. Let I c
R
30
838.31kW tan 300 MW 0.8383125MW
104 . The value xy 2 dxdy , where R is the region shown in the figure and c 6 10
of I equals __________. (Given the answer up to two decimal places).
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GATE 2017 EE - Session 1
08. Ans: 0.99 Solution: The equation of boundary of region R is y 2x 5 2x
Ic
xy dxdy c xy dydx 2
R
2
1 0
5
2x 5 5 y3 8x3 dx c 8x 6 104 8 55 15 I c x dx c x 1 3 0 3 15 1 15 1 I 0.99968 5
09. The equivalent resistance between the terminals A and B is __________ .
09. Ans: 3 Solution: Circuit can be redrawn as
GATE 2017 EE - Session 1
08. Ans: 0.99 Solution: The equation of boundary of region R is y 2x 5 2x
Ic
xy dxdy c xy dydx 2
2
R
1 0
5
2x 5 5 y3 8x3 dx c 8x 6 104 8 55 15 I c x dx c x 1 3 0 3 15 1 15 1 I 0.99968 5
09. The equivalent resistance between the terminals A and B is __________ .
09. Ans: 3 Solution: Circuit can be redrawn as
6 || 2 1.5 1.5 || 3 1
1 2 3 3 || 6 2 3 || 2 1.2 Req 1 1.2 0.8 3
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GATE 2017 EE - Session 1
10. The transfer function of a system is given by,
V0 s Vi s
1 s . Let the output of the system 1 s
be v0 t Vm sin t for the input, vi t Vm sin t . Then the minimum and maximum values of (in radians) are respectively (a)
2
and
2
(c) 0and 2
(b)
2
and0
(d) and0
10. Ans: (d) Solution: Transfer function 1s H s 1s 1 j H j 1 j
tan1 tan1 2tan1 As varies from 0 to At 0 , 0 At , 2tan1 11. A 3-phase voltage source inverter is supplied from a 600V DC source as shown in the figure below. For a star connected resistive load of 20 per phase, the load power for 120°
device conduction, in kW, is ___________.
GATE 2017 EE - Session 1
10. The transfer function of a system is given by,
V0 s Vi s
1 s . Let the output of the system 1 s
be v0 t Vm sin t for the input, vi t Vm sin t . Then the minimum and maximum values of (in radians) are respectively (a)
2
and
2
(b)
2
and0
(d) and0
(c) 0and 2 10. Ans: (d) Solution: Transfer function 1s H s 1s 1 j H j 1 j
tan1 tan1 2tan1 As varies from 0 to At 0 , 0 At , 2tan1 11. A 3-phase voltage source inverter is supplied from a 600V DC source as shown in the figure below. For a star connected resistive load of 20 per phase, the load power for 120°
device conduction, in kW, is ___________.
11. Ans: 9 Solution:
In 1200 conduction mode, Vphrms For Y connected resistive load P
Vdc
6 2 3Vph rms R
6002 6 6002 3 9kW 20 40
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GATE 2017 EE - Session 1
12. Let z(t) x(t) y(t) , where " " denotes convolution, Let c be a positive real-valued
constant. Choose the correct expression for z (ct). (a) c x (ct) y(ct)
(b) x(ct) y(ct)
(c) c x (t) y(ct)
(d) c x (ct) y(t)
12. Ans: (a) Solution: By convolution property of Fourier transform Z j X j Y j FT Fourier transform of Z ct
j c c
1
Z
1 j 1 j 1 j 1 j j FT z ct Z X Y c X Y c c c c c c c c c
Z ct c x t * y ct
t t , t 0 13. Consider g t , wheret t t , otherwise
.
Here, t represents the largest integer less than or equal to t and t denotes the smallest integer greater than or equal to t. The coefficient of the second harmonic component of the Fourier series representing g t is ___________ 13. Ans: 0.07957 Solution:
GATE 2017 EE - Session 1
12. Let z(t) x(t) y(t) , where " " denotes convolution, Let c be a positive real-valued
constant. Choose the correct expression for z (ct). (a) c x (ct) y(ct)
(b) x(ct) y(ct)
(c) c x (t) y(ct)
(d) c x (ct) y(t)
12. Ans: (a) Solution: By convolution property of Fourier transform Z j X j Y j FT Fourier transform of Z ct
j c c
1
Z
1 j 1 j 1 j 1 j j FT z ct Z X Y c X Y c c c c c c c c c
Z ct c x t * y ct
t t , t 0 13. Consider g t , wheret t t , otherwise
.
Here, t represents the largest integer less than or equal to t and t denotes the smallest integer greater than or equal to t. The coefficient of the second harmonic component of the Fourier series representing g t is ___________ 13. Ans: 0.07957 Solution:
t t , t 0 gt , wheret t t , otherwise Based on this definition, the graph of g t vs t is shown below, It is periodic with a period of 1 Second harmonic component T
1 jk t ck g t e 0 dt T0
1
1 j2 t c2 te 0 dt 10
1
j2 t j2 t j2 j2 e e 1 e e j2 t c2 t e dt t 2 2 2 I II j2 j2 4 4 0 0 0 0 0 j20 1
0
0
0
0
0
0
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GATE 2017 EE - Session 1
0 2 2 T
j4 e e j4 1 j 1 1 j c2 j4 4 22 4 22 4 162 162 4 Magnitude
1 0.07957 4
14. For the circuit shown in the figure below, assume that diodes D1 ,D2 and D3 are ideal.
The DC components of voltages v1and v2 , respectively are (a) 0 V and 1 V (c) 1 V and 0.5 V 14. Ans: (b) Solution: During positive half cycle, D1 is forward biased
V1 V2
2
sin100t
(b) – 0.5 V and 0.5 V (d) 1 V and 1 V
GATE 2017 EE - Session 1
0 2 2 T
j4 e e j4 1 j 1 1 j c2 j4 4 22 4 22 4 162 162 4 Magnitude
1 0.07957 4
14. For the circuit shown in the figure below, assume that diodes D1 ,D2 and D3 are ideal.
The DC components of voltages v1and v2 , respectively are (a) 0 V and 1 V (c) 1 V and 0.5 V
(b) – 0.5 V and 0.5 V (d) 1 V and 1 V
14. Ans: (b) Solution: During positive half cycle, D1 is forward biased
V1 V2
2
sin100t
During negative half cycle, D2 and D3 are forward biased V1 sin100t V2 0 DC component means average value 1 V1avg sin t d t 2 0 2
1 1 1 sin td t 2 2 2 1 2 2 V 2 1 1 1 1 sin t d t 0sin td t 2 V 2 0 2 2 2 2
V2avg
2
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GATE 2017 EE - Session 1
15. A 10-bus power system consists of four generator buses indexed as G1, G2, G3, G4 and six load buses indexed as L1, L2, L3, L4, L5, L6. The generator-bus G1 is considered as slack bus, and the load buses L3 and L4 are voltage controlled buses. The generator at bus G2 cannot supply the required reactive power demand, and hence it is operating at its maximum reactive power limit. The number of non-linear equations required for solving the load flow problem using Newton-Raphson method in polar form is _____________. 15. Ans: 14 Solution: G1: Slack bus L3: L4: Voltage controlled bus (PV bus) G2: Voltage reactive power limit (PQ bus) PV bus= 4-2+2=4 PQ bus= 6-2+1=5 Number of non-linear equations 2 PQ PV 2 5 4 15 16. A 3-bus power system is shown in the figure below, where the diagonal elements of Ybus matrix are: Y11 j12pu, Y22 j15puandY33 j7pu
The per unit values of the line reactance’s p, q and shown in the figure are (a) (b) 02 01 0 0.2 01 0
GATE 2017 EE - Session 1
15. A 10-bus power system consists of four generator buses indexed as G1, G2, G3, G4 and six load buses indexed as L1, L2, L3, L4, L5, L6. The generator-bus G1 is considered as slack bus, and the load buses L3 and L4 are voltage controlled buses. The generator at bus G2 cannot supply the required reactive power demand, and hence it is operating at its maximum reactive power limit. The number of non-linear equations required for solving the load flow problem using Newton-Raphson method in polar form is _____________. 15. Ans: 14 Solution: G1: Slack bus L3: L4: Voltage controlled bus (PV bus) G2: Voltage reactive power limit (PQ bus) PV bus= 4-2+2=4 PQ bus= 6-2+1=5 Number of non-linear equations 2 PQ PV 2 5 4 15 16. A 3-bus power system is shown in the figure below, where the diagonal elements of Ybus matrix are: Y11 j12pu, Y22 j15puandY33 j7pu
The per unit values of the line reactance’s p, q and shown in the figure are (a) p 0.2, q 0.1,r 0.5 (b) p 0.2,q 0.1,r 0.5 (c) p 5,q 10,r 2
(d) p 5, q 10, r 2
16. Ans: (b) Solution: j j Y11 j12 1 1 12 i q r q r
Y22
j
Y33
j
q
j j15 1 1 15 ii
p
j j7 1 1 7 iii
p
q
r
p
p
r
Add (i), (ii) & (iii)
2 1 1 1 34 q r p © Kreatryx. All Rights Reserved.
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GATE 2017 EE - Session 1
1 1 1 17 iv p q r
Subtract one by one (i), (ii) & (iii) from (iv) 1 5 1 10 1 2 q p r Hence p 0.2,q 0.1,r 0.5 17 In the converter circuit shown below, the switches are controlled such that the load voltage v o t is a 400 Hz square wave.
The RMS value of the fundamental component of v o t in volts is ______________. 17. Ans: 198.06 Solution: The Fourier series of output voltage is 4V dc V0 sinnt n1,3,5 n
RMS fundamental voltage
3 2
0
2 2
1 2
Vdc
2 2
220 198.06V
GATE 2017 EE - Session 1
1 1 1 17 iv p q r
Subtract one by one (i), (ii) & (iii) from (iv) 1 5 1 10 1 2 q p r Hence p 0.2,q 0.1,r 0.5 17 In the converter circuit shown below, the switches are controlled such that the load voltage v o t is a 400 Hz square wave.
The RMS value of the fundamental component of v o t in volts is ______________. 17. Ans: 198.06 Solution: The Fourier series of output voltage is 4V dc V0 sinnt n1,3,5 n
RMS fundamental voltage
2 2
Vdc
2 2
220 198.06V
3 1 2 0 2 18. The matrix A 0 1 0 has three distinct eigenvalues and one of its eigenvectors is 1 3 0 2 2 1 0 . Which one of the following can be another eigenvector of A? 1 0 1 (a) 0 (b) 0 1 0 1 1 (c) 0 (d) 1 1 1 © Kreatryx. All Rights Reserved.
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GATE 2017 EE - Session 1
18. Ans: (c) Solution: I A 0
3 2
0
1
1 2
1 2 0 0
0
0
3 2
2
3 11 1 1 0 2 2 2 2 3 1 1 2 4 0 1 2 3 2 0
1 2 1 0 1,1, 2 For given Eigen vector 3 1 0 1 1 2 2 2 0 1 0 0 0 2 0
1 2
0
2
3 2 1
1
Hence, it corresponds to Eigen value 2 Checking given options 1 3 0 0 1 0 2
GATE 2017 EE - Session 1
18. Ans: (c) Solution: I A 0
3 2
0
1
1 2
1 2 0 0
0
0
3 2
2
3 11 1 1 0 2 2 2 2 3 1 1 2 4 0 1 2 3 2 0
1 2 1 0 1,1, 2 For given Eigen vector 3 1 0 1 1 2 2 2 0 1 0 0 0 2 0
1 2
2
3 2 1
0
1
Hence, it corresponds to Eigen value 2 Checking given options 1 3 0 0 1 0 2 2 2 0 0 1 0 0 0 A:
1 2
3 2 B: 0 1 2 3 2 C: 0 1 2
0 1 3 2 3 1 0 1 1 2 2 1 0 0 0 0 0 0 3 0 1 2 2 1 0 1 1 1 2 1 0 0 0 1 0 3 1 1 0 1 2 3 2 1
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GATE 2017 EE - Session 1
19. A source is supplying a load through a 2-phase, 3-wire transmission system as shown in figure below. The instantaneous voltage and current in phase-a are van 220sin 100t V
and ia 10sin 100t A, respectively. Similarly for phase-b, the instantaneous voltage and current are vbn 220cos 100t Vand ib 10cos 100 t A, respectively.
The total instantaneous power flowing from the source to the load is (a) 2200 W (b) 2200sin2 100t W (d) 2200sin 100 t cos 100 t W
(c) 4400 W 19. Ans: (a) Solution: Total instantaneous power Vania Vbnib
P 220sin 100t .10sin 100t 220cos 100 t .10cos 100 t P 2200 sin2 100t cos2 100t 2200W 20. For a complex number z,lim z i
(a) –2i (c)i
z2 1 z3 2z i z2 2
is
(b) –i (d) 2i
GATE 2017 EE - Session 1
19. A source is supplying a load through a 2-phase, 3-wire transmission system as shown in figure below. The instantaneous voltage and current in phase-a are van 220sin 100t V
and ia 10sin 100t A, respectively. Similarly for phase-b, the instantaneous voltage and current are vbn 220cos 100t Vand ib 10cos 100 t A, respectively.
The total instantaneous power flowing from the source to the load is (a) 2200 W (b) 2200sin2 100t W (d) 2200sin 100 t cos 100 t W
(c) 4400 W 19. Ans: (a) Solution: Total instantaneous power Vania Vbnib
P 220sin 100t .10sin 100t 220cos 100 t .10cos 100 t P 2200 sin2 100t cos2 100t 2200W 20. For a complex number z,lim z i
z2 1 z3 2z i z2 2
is
(a) –2i (c)i
(b) –i (d) 2i
20. Ans: (d)Solution: z2 1 lim 3 z i z 2z i z2 2
As z i , the expression becomes
0 0
Applying L’Hospital rule 2z 2i lim 2i 2 z i 1 2 3z 2 2iz
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21. The Boolean expression AB AC BC simplifies to (a) BC AC (b) AB AC B (c) AB AC (d) AB BC 21. Ans: (a) Solution: If C=0, expression, AB A 0 AB A A C=1, expression, AB A 0 B B AB B Hence, it can expressed as, AC BC 22. Consider an electron, a neutron and a proton initially at rest and placed along a straight line such that the neutron is exactly at the center of the line joining the electron and proton. At t = 0, the particles are released but are constrained to move along the same straight line. Which of these will collide first? (a) the particles will never collide (b) all will collide together (c) proton and neutron (d) electron and neutron 22. Ans: (d) Solution: The arrangement is shown Only proton & electron experience columbic force but since mass of e mass of proton e moves faster towards proton and collides with neutron in this process 23. The following measurements are obtained on a single phase load: V 220V 1%,
I 5.0 A 1% and
W 555 W 2% . If the power factor is calculated using these
measurements, the worst case error in the calculated power factor in percent is ________. (Given answer up to one decimal place).
GATE 2017 EE - Session 1
21. The Boolean expression AB AC BC simplifies to (a) BC AC (b) AB AC B (c) AB AC (d) AB BC 21. Ans: (a) Solution: If C=0, expression, AB A 0 AB A A C=1, expression, AB A 0 B B AB B Hence, it can expressed as, AC BC 22. Consider an electron, a neutron and a proton initially at rest and placed along a straight line such that the neutron is exactly at the center of the line joining the electron and proton. At t = 0, the particles are released but are constrained to move along the same straight line. Which of these will collide first? (a) the particles will never collide (b) all will collide together (c) proton and neutron (d) electron and neutron 22. Ans: (d) Solution: The arrangement is shown Only proton & electron experience columbic force but since mass of e mass of proton e moves faster towards proton and collides with neutron in this process 23. The following measurements are obtained on a single phase load: V 220V 1%,
I 5.0 A 1% and
W 555 W 2% . If the power factor is calculated using these
measurements, the worst case error in the calculated power factor in percent is ________. (Given answer up to one decimal place). 23. Ans: 4 Solution:
The expression for pf is pf
W VI
The maximum error in pf is %pf %w %v %I 2% 1% 1% 4% 24. Consider the unity feedback control system shown. The value of K that results in a phase margin of the system to be 30° is _________. (Give the answer up to two decimal places).
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GATE 2017 EE - Session 1
24. Ans: 1.05 Solution: ke j G j j
For gain crossover frequency G j 1 k
1 or k 2
At k, k 2
PM 300 PM 1800 k 2 6 k 2 1.0479 2 6 6 3 25. For the power semiconductor devices IGBT, MOSFET, Diode and Thyristor, which one of the following statements is TRUE? (a) All the four are majority carrier devices (b) All the four are minority carrier devices (c) IGBT and MOSFET are majority carrier devices, whereas Diode and Thyristor are minority carrier devices. (d) MOSFET is majority carrier device, whereas IGBT, Diode, Thyristor are minority carrier devices. 25. Ans: (d) Solution: IGBT, Diode, Thyristor, Minority carrier device MOSFET: Majority carrier device
GATE 2017 EE - Session 1
24. Ans: 1.05 Solution: ke j G j j
For gain crossover frequency G j 1 k
1 or k 2
At k, k 2
PM 300 PM 1800 k 2 6 k 2 1.0479 2 6 6 3 25. For the power semiconductor devices IGBT, MOSFET, Diode and Thyristor, which one of the following statements is TRUE? (a) All the four are majority carrier devices (b) All the four are minority carrier devices (c) IGBT and MOSFET are majority carrier devices, whereas Diode and Thyristor are minority carrier devices. (d) MOSFET is majority carrier device, whereas IGBT, Diode, Thyristor are minority carrier devices. 25. Ans: (d) Solution: IGBT, Diode, Thyristor, Minority carrier device MOSFET: Majority carrier device 26. The load shown in the figure is supplied by a 400 V (line-to-line), 3-phase source (RYB sequence). The load is balanced and inductive, drawing 3464 V A. When the switch S is in position N, the three watt-meters W1 , W2 and W3 read 577.35 W each. If the switch is moved
to position Y, the readings of the watt-meters in watts will be: (a) W1 1732andW2 W3 0 (b) W1 0, W2 1732 and W3 0 (c) W1 866, W2 0, W3 866 (d) W1 W2 0 and W3 1732
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GATE 2017 EE - Session 1
26. Ans: (d) Solution: Total real power 3 577.35 1732.05W Apparent power 3467VA 1732 0.5 pf P S 3464 When switch is move to Y, the potential coil of second wattmeter gets shorted Hence W2 0
W1 : VPC VRY ; ICC IR W1 VRY IR cos 30 60 0 W3 : VPC VBY ; ICC IB W3 VL IL cos30
S
cos30 S 1732W 2 3
27. A load is supplied by a 230 V, 50 Hz source. The active power P and the reactive power Q consumed by the load are such that 1 kW P 2kWand 1kVAR Q 2kVAR . A capacitor
connected across the load for power factor correction generates 1 kVAR reactive power. The worst case power factor after power factor correction is (a) 0.447 lag (b) 0.707 lag (c) 0.894 lag (d)1 27. Ans: (b) Solution: For worst case pf, Q= Reactive power must be maximum and P= Real power must be minimum
GATE 2017 EE - Session 1
26. Ans: (d) Solution: Total real power 3 577.35 1732.05W Apparent power 3467VA 1732 0.5 pf P S 3464 When switch is move to Y, the potential coil of second wattmeter gets shorted Hence W2 0
W1 : VPC VRY ; ICC IR W1 VRY IR cos 30 60 0 W3 : VPC VBY ; ICC IB W3 VL IL cos30
S
cos30 S 1732W 2 3
27. A load is supplied by a 230 V, 50 Hz source. The active power P and the reactive power Q consumed by the load are such that 1 kW P 2kWand 1kVAR Q 2kVAR . A capacitor
connected across the load for power factor correction generates 1 kVAR reactive power. The worst case power factor after power factor correction is (a) 0.447 lag (b) 0.707 lag (c) 0.894 lag (d)1 27. Ans: (b) Solution: For worst case pf, Q= Reactive power must be maximum and P= Real power must be minimum Q QL QC 2 1 1KVAR Q 1kW
tan Q
P
1 => 450
pf cos 0.707lag 28. The magnitude of magnetic flux density (B) in micro Teslas (µT), at the center of a loop of wire wound as a regular hexagon of side length 1 m carrying a current (I = 1 A) and placed in vacuum as shown in the figure is _________. (Give the answer up to two decimal places).
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GATE 2017 EE - Session 1
28. Ans: 0.6928 Solution: By Pythagoras theorem 2
r l2 l
4
3 l 2
nI sin n 2r nI B 0 sin n 2r Number of sides, n=6 H
4 10 6 1 B sin 7
2 3
2
1
6
12 2 3
1 107 2
12 3
101 T 0.6928T
29. Consider a causal and stable LTI system with rational transfer function H (z), whose 5 corresponding impulse response begins at n = 0. Furthermore, H(1) . The poles of H (z) 4 2k 1 1 j for k = 1, 2, 3, 4. The zeros of H (z) are all at z = 0. Let p exp are k 4 2
g n jnh n . The value of g[8] equals ___________________. (Given the answer up to three decimal places). 29. Ans: 0.097 Solution:
GATE 2017 EE - Session 1
28. Ans: 0.6928 Solution: By Pythagoras theorem 2
r l2 l
3 l 2
4
nI sin n 2r nI B 0 sin n 2r Number of sides, n=6 H
4 10 6 1 B sin 7
2 3
1
2
6
12 2
1 107
12
2
3
3
101 T 0.6928T
29. Consider a causal and stable LTI system with rational transfer function H (z), whose 5 corresponding impulse response begins at n = 0. Furthermore, H(1) . The poles of H (z) 4 1
are pk
2
2k 1 for k = 1, 2, 3, 4. The zeros of H (z) are all at z = 0. Let 4
exp j
g n jnh n . The value of g[8] equals ___________________. (Given the answer up to three decimal places). 29. Ans: 0.097 Solution: Since, the impulse response begins at n=0, the degree of numerator & denominator must be same Since, there are 4 poles, there must be 4 zeros kz4 H z 4 1 exp j 2k 1 z 4 k 1 2 H1 5
4
k
5 4
1 1 cos jsin 1 cos 3 jsin 3 1 4 4 4 4 2 2
1
1 1 cos 5 jsin5 1 cos 7 jsin 7 1 4 4 4 4 2 2
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GATE 2017 EE - Session 1
k
5 4 1
j j j j 2 2 32 2 32 2 1 2 2
k k 5 4 5 9 1 1 1 4 4 4 4 4 k 25 16 g n jnh n
G z H z j Gz
kz4
z 4
j
exp 2k 1 4 2
k 1
Gz
Gz
g n
kz4 2 2 1 1 z 1 1 z 2 4 2 4
kz4
z
4
1
k
4
1
i 0 n 4i
1 1 4z 4
i
z 4
n
25
16
g 8 , n=8 and i=2
g 8 25
16
1
2
4
25
256
0.0976
kz4
z2 1
2
2
z2
GATE 2017 EE - Session 1
k
5 4 1
j j j j 2 2 32 2 32 2 1 2 2
k k 5 4 5 9 1 1 1 4 4 4 4 4 k 25 16 g n jnh n
G z H z j Gz
kz4
z 4
j
exp 2k 1 4 2
k 1
Gz
Gz
g n
kz4 2 2 1 1 z 1 1 z 2 4 2 4
kz4
z
4
1
1
i 0 n 4i
kz4
z2 1
2
2
z2
k
4
1 1 4z 4
i
z 4
n
25
16
g 8 , n=8 and i=2
g 8 25
16
1
2
4
25
256
0.0976
30. For a system having transfer function G s
s 1 , a unit step input is applied at time t= s 1
0. The value of the response of the system at t = 1.5 sec. (rounded off to three decimal places) is ___________ 30. Ans: 0.554 Solution:
1 s 1 1 s 1 2 1 s s s 1 s s s 1 y t 1 2et u t
Y s G s X s
At t 1.5sec
y 1 2e1.5 0.5537
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GATE 2017 EE - Session 1
31. A three-phase, three winding / / Y (1.1 kV/6.6 kV/400 V) transformer is energized from AC mains at the 1.1 kV side. It supplies 900 kVA load at 0.8 power factor lag from the 6.6 kV winding and 300 kVA load at 0.6 power factor lag from the 400 V winding. The RMS line current in ampere drawn by the 1.1 kV winding from the mains is ____________. (Given the answer up to one decimal place). 31. Ans: 625.1 Solution:
S1 90036.86kVA S2 30053.13kVA Total power drawn from supply S1 S2 900 0.8 j0.6 300 0.6 j0.8 900 j78 1190.95 40.90 kVA S 3VL IL IL
1190.955 3 1.1
625.089A
32. The figure shows the single line diagram of a power system with a double circuit transmission line. The expression for electrical power is 1.5 sin , where is the rotor angle. The system is operating at the stable equilibrium point with mechanical power equal to 1 pu. If one of the transmission line circuits is removed, the maximum value of , as the rotor swings, is 1.221 radian. If the expression for electrical power with one transmission line circuit removed is Pmax sin , the value of Pmax , in pu is ____________. (Give the answer up to three
decimal places)
GATE 2017 EE - Session 1
31. A three-phase, three winding / / Y (1.1 kV/6.6 kV/400 V) transformer is energized from AC mains at the 1.1 kV side. It supplies 900 kVA load at 0.8 power factor lag from the 6.6 kV winding and 300 kVA load at 0.6 power factor lag from the 400 V winding. The RMS line current in ampere drawn by the 1.1 kV winding from the mains is ____________. (Given the answer up to one decimal place). 31. Ans: 625.1 Solution:
S1 90036.86kVA S2 30053.13kVA Total power drawn from supply S1 S2 900 0.8 j0.6 300 0.6 j0.8 900 j78 1190.95 40.90 kVA S 3VL IL IL
1190.955 3 1.1
625.089A
32. The figure shows the single line diagram of a power system with a double circuit transmission line. The expression for electrical power is 1.5 sin , where is the rotor angle. The system is operating at the stable equilibrium point with mechanical power equal to 1 pu. If one of the transmission line circuits is removed, the maximum value of , as the rotor swings, is 1.221 radian. If the expression for electrical power with one transmission line circuit removed is Pmax sin , the value of Pmax , in pu is ____________. (Give the answer up to three
decimal places)
32. Ans: 1.220 Solution: With one transmission line removal, the effective reactance increases so Pmax decreases, the
power angle by equal area criterion 2
1
1 P
max
sin d
P
max
sin 1 d
1
0
1 0 Pmax cos 0 cos 1 Pmax cos 1 cos 2 2 1 2 0 Pmax cos 0 cos 2 Pmax
2 0 cos 0 cos 2
0 sin1 1
1.5
0.72972
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GATE 2017 EE - Session 1
Pmax
1.221 0.72972 1.22pu cos 0.7297 cos 1.221
33. Consider the line integral I
x C
2
iy2 dz, where z x iy. The line c is shown in the
figure below.
The value of I is 1 (a) i 2 3 (c) i 4
2 i 3 4 (d) i 5 (b)
33.Ans: (b) Solution: z x iy
dz dx idy The equation of line is y=x dz 1 i dx 1
1
3 I 1 i x 1 i dx 1 i x 3 0 0 2
2
I 1 i 1 3 2i
2
GATE 2017 EE - Session 1
Pmax
1.221 0.72972 1.22pu cos 0.7297 cos 1.221
33. Consider the line integral I
x C
2
iy2 dz, where z x iy. The line c is shown in the
figure below.
The value of I is 1 (a) i 2 3 (c) i 4
2 i 3 4 (d) i 5 (b)
33.Ans: (b) Solution: z x iy
dz dx idy The equation of line is y=x dz 1 i dx 1
1
3 I 1 i x 1 i dx 1 i x 3 0 0 2
2
2
I 1 i 1 3 2i
3
34. The transfer function of the system Y(s)/U(s) whose state-space equations are given below is:
x1 t 1 2 x1 t 1 u t x2 t 2 0 x2 t 2 x1 t . y t 1 0 x2 t (s 2) (a) 2 (s 2s 2) (s 4) (c) 2 (s s 4)
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(s 2) (s2 s 4) (s 4) (d) 2 (s s 4)
(b)
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GATE 2017 EE - Session 1
34. Ans: (d) Solution:
1 2 A 2 0
s 1 2 s
sI A 2 1
sI A
s 2 s s 1 4 2 s 1 1
Transfer function, C sI A
1
1 0 s 4 1 0 s 2 1 s4 B 2 s s 4 2 s 1 2 s2 s 4 2s s2 s 4
35. The approximate transfer characteristic for the circuit shown below with an ideal operational amplifier and diode will be
(a)
(b)
GATE 2017 EE - Session 1
34. Ans: (d) Solution:
1 2 A 2 0
s 1 2 s
sI A 2 1
sI A
s 2 s s 1 4 2 s 1 1
Transfer function, C sI A
1
1 0 s 4 1 0 s 2 1 s4 B 2 s s 4 2 s 1 2 s2 s 4 2s s2 s 4
35. The approximate transfer characteristic for the circuit shown below with an ideal operational amplifier and diode will be
(a)
(b)
(c)
(d)
35. Ans: (a) Solution: For positive input, V0 0 and diode becomes forward biased
The circuit then acts as voltage follower V0 Vin For negative input, the output is negative and diode is reverse biased so no current flows through resistor & V0 0 © Kreatryx. All Rights Reserved.
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GATE 2017 EE - Session 1
36. Two parallel connected, three-phase, 50 Hz, 11 kV, star-connected synchronous machines A and B, are operating as synchronous condensers. They together supply 50 MVAR to a 11 kV grid. Current supplied by both the machines are equal. Synchronous reactance’s of machine A and machine B are 1 and 3 , respectively. Assuming the magnetic circuit to be
linear, the ratio of excitation current of machine A to that of machine B is ____________. (Given the answer up to two decimal places). 36. Ans: 0.744 Solution:
Load current
S 3VL
50 106 3 11 103
2.6243kA
Current supplied by each machine= 1.3121kA Excitation emf of each machine 11 11 E1 1.3121 90 j1 1.3121 7.66290 0 3 3 11 E2 1.3121 90 j3 10.287kV 3 I E Since, magnetic circuit is linear i.e. f1 1 If2 E2 If1 7.669 0.745 If2 10.287 37. A separately excited DC generator supplies 150 A to a 145 V DC grid. The generator is running at 800 RPM. The armature resistance of the generator is 0.1 . If the speed of the
generator is increased to 1000 RPM, the current in amperes supplied by the generator to the DC gird is ___________. (Give the answer up to one decimal place).
GATE 2017 EE - Session 1
36. Two parallel connected, three-phase, 50 Hz, 11 kV, star-connected synchronous machines A and B, are operating as synchronous condensers. They together supply 50 MVAR to a 11 kV grid. Current supplied by both the machines are equal. Synchronous reactance’s of machine A and machine B are 1 and 3 , respectively. Assuming the magnetic circuit to be
linear, the ratio of excitation current of machine A to that of machine B is ____________. (Given the answer up to two decimal places). 36. Ans: 0.744 Solution:
Load current
S
50 106
3VL
3 11 103
2.6243kA
Current supplied by each machine= 1.3121kA Excitation emf of each machine 11 11 E1 1.3121 90 j1 1.3121 7.66290 0 3 3 11 E2 1.3121 90 j3 10.287kV 3 I E Since, magnetic circuit is linear i.e. f1 1 If2 E2 If1 7.669 0.745 If2 10.287 37. A separately excited DC generator supplies 150 A to a 145 V DC grid. The generator is running at 800 RPM. The armature resistance of the generator is 0.1 . If the speed of the
generator is increased to 1000 RPM, the current in amperes supplied by the generator to the DC gird is ___________. (Give the answer up to one decimal place). 37. Ans: 550 Solution: By KVL, E1 Vt IaRa
E1 145 150 0.1 160V In separately excited machine Constant E N E2 N2 E1
N1
1000 200V 800 E V 200 145 550A Ia 2 Ra 0.1
E2 160
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GATE 2017 EE - Session 1
38. In the system whose signal flow graph is shown in the figure, U1 s and U2 s are inputs.
The transfer function
(a) (c)
k1 JLs2 JRs k1k 2
Y s U1 s
is
(b)
k1 U2 R sL JLs2 JR U2L s k1k 2 U2R
38. Ans: (a) Solution:
For transfer function Forward path, G1
Y s U1 s
;U2 s 0
k1 JLs2
kk R ; L2 1 22 sL JLs By Mason’s Gain formula k1 G1 1 JLs2 T k1k 2 1 L1 L2 1 R
Loops, L1
(d)
k1 JLs2 JRs k 1k 2 k1 U2 sL R JLs2 JR U2L s k1k 2 U2R
GATE 2017 EE - Session 1
38. In the system whose signal flow graph is shown in the figure, U1 s and U2 s are inputs. Y s
The transfer function
k1
(a) (c)
JLs2 JRs k1k 2
U1 s
is
(b)
k1 U2 R sL
(d)
JLs2 JR U2L s k1k 2 U2R
k1 JLs2 JRs k 1k 2 k1 U2 sL R JLs2 JR U2L s k1k 2 U2R
38. Ans: (a) Solution:
For transfer function Forward path, G1
Y s U1 s
;U2 s 0
k1 JLs2
kk R ; L2 1 22 sL JLs By Mason’s Gain formula k1 G1 1 JLs2 T kk 1 L1 L2 1 R 1 22 sL JLs
Loops, L1
T
k1 s JL s JR k1k 2 2
39. A 220 V DC series motor runs drawing a current of 30 A from the supply. Armature and field circuit resistances are 0.4 and 0.1 , respectively. The load torque varies as the square
of the speed. The flux in the motor may be taken as being proportional to armature current. To reduce the speed of the motor by 50%, the resistance in ohms that should be added in series with the armature is _________. (Give the answer up to one decimal place). 39. Ans: 10.75 Solution: E Vt Ia R a Rse 220 30 0.5 205V
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GATE 2017 EE - Session 1
If speed is reduced by 50%,
N2
N1
0.5
Since, T N2 T2
2
N 2 0.25 T1 N1
In case of series motor 2
Ia2 I 0.25 a1 Ia2
Ia1
0.5
Ia2 15A EMF, E N IaN E2 E1
Ia2 Ia1
N2 N1
E2 205 0.5 0.5 51.25V R2
V E2 I2
220 51.25 11.25 15
Rext 11.25 0.5 10.75 40. The input voltage VDC of the buck-boost converter shown below varies from 32 V to 72
V. Assume that all components are ideal, inductor current is continuous, and output voltage is ripple free. The range of duty ratio D of the converter for which the magnitude of the steady-state output voltage remains constant at 48 V is
GATE 2017 EE - Session 1
If speed is reduced by 50%,
N2
N1
0.5
Since, T N2 T2
2
N 2 0.25 T1 N1
In case of series motor 2
Ia2 I 0.25 a1 Ia2
Ia1
0.5
Ia2 15A EMF, E N IaN E2
E1
Ia2 Ia1
N2 N1
E2 205 0.5 0.5 51.25V R2
V E2 I2
220 51.25 11.25 15
Rext 11.25 0.5 10.75 40. The input voltage VDC of the buck-boost converter shown below varies from 32 V to 72
V. Assume that all components are ideal, inductor current is continuous, and output voltage is ripple free. The range of duty ratio D of the converter for which the magnitude of the steady-state output voltage remains constant at 48 V is
(a)
2 3 D 5 5
2 3 D 3 4 1 2 (d) D 3 3
(b)
(c) 0 D 1 40. Ans: (a) Solution: D V0 Vin 1 D
For Vin 32V and V0=48 © Kreatryx. All Rights Reserved.
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GATE 2017 EE - Session 1
D 1 D 1.5 1.5D D D 0.6 For Vin 72V 48 32
D 1 D 2 2D 3D D 0.4 Hence, 0.4 D 0.6 2 D 3 48 72
5
5
41. Two passive two-port networks are connected in cascade as shown in figure. A voltage source is connected at port 1.
Given, V1 A1V2 B1 I2 I1 C1V2 D1 I2 V2 A2 V3 B2 I3 I2 C2V3 D2 I3
A1 ,B1 ,C1 ,D1 , A2 ,B2 , C2 , and D2 are the generalized circuit constants. If the Thevenin equivalent circuit at port 3 consists of a voltage source VT and impedance Z T , connected in series, then (a) V
V1
Z
A1B2 B1D2
(b) V
V1
Z
A1B2 B1D2
GATE 2017 EE - Session 1
D 1 D 1.5 1.5D D D 0.6 For Vin 72V 48 32
D 1 D 2 2D 3D D 0.4 Hence, 0.4 D 0.6 2 D 3 48 72
5
5
41. Two passive two-port networks are connected in cascade as shown in figure. A voltage source is connected at port 1.
Given, V1 A1V2 B1 I2 I1 C1V2 D1 I2 V2 A2 V3 B2 I3 I2 C2V3 D2 I3
A1 ,B1 ,C1 ,D1 , A2 ,B2 , C2 , and D2 are the generalized circuit constants. If the Thevenin equivalent circuit at port 3 consists of a voltage source VT and impedance Z T , connected in series, then V1 A B B1D2 , ZT 1 2 A1A2 A1A2 B1C2
(a) VT (c) VT
V1 A1 A2
, ZT
(b) VT
A1B2 B1D2
(d) VT
A1 A2
V1 A1A2 B1C2 V1 A1A2 B1C2
, ZT , ZT
A1B2 B1D2 A1A2 A1B2 B1D2 A1A2 B1C2
41. Ans: (d) Solution: The two networks are cascaded, so equivalent ABCD parameters can be obtained by multiplication of individual parameters A B A1 B1 A 2 B2 A1 A2 B1C2 A1B2 B1D2
C D C D C A D C C B D D C D 1 1 2 2 1 2 1 2 1 2 1 2 V1 A1 A2 B1C2 V3 A1B2 B1D2 I3
I1 C1A2 D1C2 V3 C1B2 D1D2 I3
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GATE 2017 EE - Session 1
For VT , I3 0 => VT V3 VT ZT
V1 A1 A2 B1C2 V3
I3
V1 0
A1B2 B1D2 A1 A2 B1C2
j39.9 j20 j20 42. The bus admittance matrix for a power system network is j20 j39.9 j20 pu. j20 j20 j39.9 There is a transmission line, connected between buses 1 and 3, which is represented by the circuit shown in figure.
If this transmission line is removed from service, what is the modified bus admittance matrix? j39.95 j20 j19.9 j20 0 0 j39.9 (a) j20 (b) j20 j20 pu j39.9 j20 pu
0 j20 j19.9 j19.95 j20 0 j39.9 (c) j20 j20 pu 0 j19.95 j20 42. Ans: (c)
0 j39.95 j20 j19.95 j20 j20 j39.9 (d) j20 j20 pu j20 j19.95 j20
GATE 2017 EE - Session 1
For VT , I3 0 => VT V3 VT ZT
V1 A1 A2 B1C2 V3
I3
V1 0
A1B2 B1D2 A1 A2 B1C2
j39.9 j20 j20 42. The bus admittance matrix for a power system network is j20 j39.9 j20 pu. j20 j20 j39.9 There is a transmission line, connected between buses 1 and 3, which is represented by the circuit shown in figure.
If this transmission line is removed from service, what is the modified bus admittance matrix? j39.95 j20 j19.9 j20 0 0 j39.9 (a) j20 (b) j20 j20 pu j39.9 j20 pu
0 j39.95 j20 j19.95 j20 j20 j39.9 (d) j20 j20 pu j20 j19.95 j20
0 j20 j19.9 j19.95 j20 0 j39.9 (c) j20 j20 pu 0 j19.95 j20
42. Ans: (c) Solution: For a transmission line connected between bus 1 and 3 1 y13 j20 j0.05
y10 j0.05 y30 j0.05 After removing the transmission line Y13new Y13 y13 j20 j20 0
Y11new Y11 y10 y13 j39.9 j0.05 j20 j19.95 Y33new Y33 y30 y13 j19.95 Rest all elements remain same
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GATE 2017 EE - Session 1
j19.95 j20 0 YBUS j20 j20 j39.9 0 j20 j19.95 43. The output expression for the Karnaugh map shown below is
(a) BD BCD (b) BD AB (c) BD ABC (d) BD ABC 43. Ans: (d) Solution: The given k-map is Y BD ABC
dy 5t y sin(t) with y(1) 2 . There exists a dt unique solution for this differential equation when t belongs to the interval 44. Consider the differential equation (t2 81)
GATE 2017 EE - Session 1
j19.95 j20 0 YBUS j20 j20 j39.9 0 j20 j19.95 43. The output expression for the Karnaugh map shown below is
(a) BD BCD (b) BD AB (c) BD ABC (d) BD ABC 43. Ans: (d) Solution: The given k-map is Y BD ABC
dy 5t y sin(t) with y(1) 2 . There exists a dt unique solution for this differential equation when t belongs to the interval (a) (–2, 2) (b) (–10, 10) (c) (–10, 2) (d) (0, 10) 44. Consider the differential equation (t2 81)
44. Ans: (a) Solution: The differential equation can be reframed as dy sint 5ty 2 dt t 81
f t,y
sint 5ty
t
2
81
5t df 2 dy t 81
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GATE 2017 EE - Session 1
The function f and
df are defined and continuous at all points expect t 9 & t 9 dy
So, solution is unique in an open interval around t=1 Hence, only option A doesn’t include either t 9 or t 9 45. The circuit shown in the figure uses matched transistors with a thermal voltage VT = 25 mV. The base currents of the transistors are negligible. The value of the resistance R in k that is required to provide 1 A bias current for the differential amplifier block shown
is __________ . (Give the answer up to one decimal place.)
45. Ans: 172.7 Solution: The current in second transistor is very small. So it can be assumed to be in saturation I 1 A
GATE 2017 EE - Session 1
The function f and
df are defined and continuous at all points expect t 9 & t 9 dy
So, solution is unique in an open interval around t=1 Hence, only option A doesn’t include either t 9 or t 9 45. The circuit shown in the figure uses matched transistors with a thermal voltage VT = 25 mV. The base currents of the transistors are negligible. The value of the resistance R in k that is required to provide 1 A bias current for the differential amplifier block shown
is __________ . (Give the answer up to one decimal place.)
45. Ans: 172.7 Solution: The current in second transistor is very small. So it can be assumed to be in saturation Is 1A
Since the transistor are matched Is1 1A
V Ic1 Is1 e
BE1
1
VT
V 1mA 1A e
BE1
0.025
1
VBE1
1001 e 0.025 VBE1 0.1727V
The voltage drop across both transistor must be same & since IB 0 , VBE2 0 , IE2 IC2 1 A IE2R VBE1 R
0.1727 172.7k 1A
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GATE 2017 EE - Session 1
ex , x 1 46. A function f(x) is defined as f(x) , where x . Which one of the 2 lnx ax bx, x 1 following statements is TRUE? (a) f(x) is NOT differentiable at x = 1 for any values of a and b. (b) f(x) is differentiable at x = 1 for the unique values of a and b. (c) f(x) is differentiable at x = 1 for all values of a and b such that a + b = e. (d) f(x) is differentiable at x = 1 for all values of a and b. 46. Ans: (b) Solution: Left hand derivative at x=1
f 1 ex
x 1
e
Right hand derivative at x=1 f 1 1 2ax b x 1 2a b 1 x For differentiability f 1 f 1
e 2a b 1 2a b e 1 For continuity Left hand limit, lim ex e x 1
Right hand limit, lim lnx ax2 bx a b x 1
RHL=LHL b
GATE 2017 EE - Session 1
ex , x 1 46. A function f(x) is defined as f(x) , where x . Which one of the 2 lnx ax bx, x 1 following statements is TRUE? (a) f(x) is NOT differentiable at x = 1 for any values of a and b. (b) f(x) is differentiable at x = 1 for the unique values of a and b. (c) f(x) is differentiable at x = 1 for all values of a and b such that a + b = e. (d) f(x) is differentiable at x = 1 for all values of a and b. 46. Ans: (b) Solution: Left hand derivative at x=1
f 1 ex
x 1
e
Right hand derivative at x=1 f 1 1 2ax b x 1 2a b 1 x For differentiability f 1 f 1
e 2a b 1 2a b e 1 For continuity Left hand limit, lim ex e x 1
Right hand limit, lim lnx ax2 bx a b x 1
RHL=LHL ab e Function is differentiable at x=1, if following two equation are satisfied 2a b e 1 ab e So function is differentiable at x=1 for unique values of a & b 47. A 375 W, 230 V, 50 Hz, capacitor start single-phase induction motor has the following constants for the main and auxiliary windings (at starting): Zm (12.50 j15.75) (main
winding), Za (24.50 j12.75) (auxiliary winding). Neglecting the magnetizing branch, the value of the capacitance (in F) to be added in series with the auxiliary winding to obtain maximum torque at starting is _____________.
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47. Ans: 98.87 Solution: For maximum starting torque, the phase shift between main & auxiliary winding current is 900 RR 24.5 12.5 12.75 32.1944 X c a m Xa Xm 15.75 C
1 1 98.87F Xc 2 50 32.194
48. Let a causal LTI system be characterized by the following differential equation, with initial d2y dy dx(t) rest condition 7 10y(t) 4 x(t) 5 2 dt dt dt
Where, x(t) and y(t) are the input and output respectively. The impulse response of the system is (u(t) is the unit step function) (a) 2e2tu(t) 7e5t u(t)
(b) 2e2tu(t) 7e5t u(t)
(c) 7e2tu(t) 2e5t u(t)
(d) 7e2tu(t) 2e5t u(t)
48. Ans: (b) Solution: Take Laplace transform both sides
s
2
7s 10 Y s 5s 4 X s
Transform function H s
Y s X s
5s 4 s 7s 10 2
GATE 2017 EE - Session 1
47. Ans: 98.87 Solution: For maximum starting torque, the phase shift between main & auxiliary winding current is 900 RR 24.5 12.5 12.75 32.1944 X c a m Xa Xm 15.75 C
1 1 98.87F Xc 2 50 32.194
48. Let a causal LTI system be characterized by the following differential equation, with initial d2y dy dx(t) rest condition 7 10y(t) 4 x(t) 5 2 dt dt dt
Where, x(t) and y(t) are the input and output respectively. The impulse response of the system is (u(t) is the unit step function) (a) 2e2tu(t) 7e5t u(t)
(b) 2e2tu(t) 7e5t u(t)
(c) 7e2tu(t) 2e5t u(t)
(d) 7e2tu(t) 2e5t u(t)
48. Ans: (b) Solution: Take Laplace transform both sides
s
2
7s 10 Y s 5s 4 X s
Transform function H s H s
Y s X s
5s 4 s 7s 10
2
5s 4 7 2 s 5 s 2 s 5 s 2
h t 7e5t 2e2t u t 49. Only one of the real roots of f(x) = x6 – x – 1 lies in the interval 1 x 2 and bisection
method is used to find its value. For achieving an accuracy of 0.001, the required minimum number of iterations is ___________ . 49. Ans: 10 Solution: The number of steps for achieving an accuracy of is
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GATE 2017 EE - Session 1
n
n n
loge
ba
loge 2 2 1
loge
0.001 loge 2
loge 1000 loge 2
9.965
Hence, n=10 50. Let the signal x(t)
k
(1) t 2000 be passed through an LTI system with frequency k
k
response H() , as given in the figure below The Fourier series representation of the output is given as (a) 4000 4000 cos(2000 t) 4000cos(4000 t) (b) 2000 2000cos(2000 t) 2000cos(4000 t) (c) 4000cos(2000 t) (d) 2000 cos(2000 t) 50. Ans: (c) Solution: x t
1 t k 2000 k
k
The period of signal x(t) is 2
1msec 2000
GATE 2017 EE - Session 1
n
n n
loge
ba
loge 2 2 1
loge
0.001 loge 2
loge 1000 loge 2
9.965
Hence, n=10 50. Let the signal x(t)
k
(1) t 2000 be passed through an LTI system with frequency k
k
response H() , as given in the figure below The Fourier series representation of the output is given as (a) 4000 4000 cos(2000 t) 4000cos(4000 t) (b) 2000 2000cos(2000 t) 2000cos(4000 t) (c) 4000cos(2000 t) (d) 2000 cos(2000 t) 50. Ans: (c) Solution: x t
1 t k 2000 k
k
The period of signal x(t) is 2
1msec 2000
The signal is shown below:
The signal is even
2 2 a1 x t cos 0 tdt t cos 0tdt t 0.5 cos 0tdt 1m 0.25 T0 0.25 0.25
T
0.75
a1 2000 1 cos 2
0.5 10 3 4000 10 3
Signal is half wave symmetric. So only odd harmonics are present & DC component is zero 2 Fundamental frequency 1 3 2000 10 © Kreatryx. All Rights Reserved.
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Second harmonic frequency 2
4 103
4000
6 6000 103 When the signal is passed through LPF only fundamental and second harmonic pass through 5000 but second harmonic is absent. filter c
Third harmonic frequency 3
y t 4000cos 2000 t 51. The positive, negative, and zero sequence reactances of a wye-connected synchronous generator are 0.2 pu, 0.2 pu and 0.1 pu, respectively. The generator is an open circuit with a terminal voltage of 1 pu. The minimum value of the inductive reactance, in pu, required to be connected between neutral and ground so that the fault current does not exceed 3.75 pu if a single line to ground fault occurs at the terminals is _________ (assume fault impedance to be zero). (Give the answer up to one decimal place.) 51. Ans: 0.1 Solution: Fault current for SLG fault 3 If X1 X2 X0 3Xn 3.75
3 0.2 0.2 0.1 3Xn
0.5 3Xn 0.8 Xn 0.1pu 52. The figure below shows an uncontrolled diode bridge rectifier supplied from a 220 V, 50
GATE 2017 EE - Session 1
Second harmonic frequency 2
4 4000 103
6 6000 103 When the signal is passed through LPF only fundamental and second harmonic pass through 5000 but second harmonic is absent. filter c
Third harmonic frequency 3
y t 4000cos 2000 t 51. The positive, negative, and zero sequence reactances of a wye-connected synchronous generator are 0.2 pu, 0.2 pu and 0.1 pu, respectively. The generator is an open circuit with a terminal voltage of 1 pu. The minimum value of the inductive reactance, in pu, required to be connected between neutral and ground so that the fault current does not exceed 3.75 pu if a single line to ground fault occurs at the terminals is _________ (assume fault impedance to be zero). (Give the answer up to one decimal place.) 51. Ans: 0.1 Solution: Fault current for SLG fault 3 If X1 X2 X0 3Xn 3.75
3 0.2 0.2 0.1 3Xn
0.5 3Xn 0.8 Xn 0.1pu 52. The figure below shows an uncontrolled diode bridge rectifier supplied from a 220 V, 50 Hz, 1-phase ac source. The load draws a constant current I o = 14 A. The conduction angle of the diode D1 in degrees (rounded off to two decimal places) is ____________ .
52. Ans: 210.836 Solution: For single phase full bridge controlled converter V I0 m cos cos Ls
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For diode bridge rectifier α=00 220 2 cos0 cos 100 0.01 14 cos 1 220 2
14
30.8360 Conduction angle of diode 180 210.8360 53. In the circuit shown below, the maximum power transferred to the resistor R is _____W.
53. Ans: 3.025 Solution: For maximum power transfer to R, we need to determine Thevenin equivalent across R. By KVL 5 5I 6 5I 10 0 10I 21 I 2.1A Vab 5 5I 5.5V
For R Th Short circuit the voltage sources
GATE 2017 EE - Session 1
For diode bridge rectifier α=00 220 2 cos0 cos 100 0.01 14 cos 1 220 2
14
30.8360 Conduction angle of diode 180 210.8360 53. In the circuit shown below, the maximum power transferred to the resistor R is _____W.
53. Ans: 3.025 Solution: For maximum power transfer to R, we need to determine Thevenin equivalent across R. By KVL 5 5I 6 5I 10 0 10I 21 I 2.1A Vab 5 5I 5.5V
For R Th Short circuit the voltage sources RTh 5 || 5 2.5 Pmax
2 Vab
4R th
2
5.5
4 2.5
2
5.5 10
3.025W
54. The logical gate implemented using the circuit shown below where, V 1 and V2 are inputs (with 0 V as digital 0 and 5 V as digital 1) and VOUT is the output, is
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GATE 2017 EE - Session 1
(a) NOT (c) NAND
(b) NOR (d) XOR
54. Ans: (b) Solution: If either V1 or V2 is logic 1; the transistor turns ON & Vout 0
If both V1 V2 0 , the output is large 1 Hence, the logic implements a NOR gate 55. The switch in the figure below was closed for a long time. It is opened at t = 0. The current in the inductor of 2 H for t 0 , is
(a) 2.5e–4t (c) 2.5e–0.25t 55. Ans: (a) Solution: The circuit enters steady state at t 0 . So inductor is shorted Req 6 8|| 8 10 I 50
10
5A
(b) 5e–4t (d) 5e–0.25t
GATE 2017 EE - Session 1
(a) NOT (c) NAND
(b) NOR (d) XOR
54. Ans: (b) Solution: If either V1 or V2 is logic 1; the transistor turns ON & Vout 0
If both V1 V2 0 , the output is large 1 Hence, the logic implements a NOR gate 55. The switch in the figure below was closed for a long time. It is opened at t = 0. The current in the inductor of 2 H for t 0 , is
(a) 2.5e–4t (c) 2.5e–0.25t
(b) 5e–4t (d) 5e–0.25t
55. Ans: (a) Solution: The circuit enters steady state at t 0 . So inductor is shorted Req 6 8|| 8 10 I 50
10
5A
Isc iL 0 5 2.5A 2 After opening the switch, Req across inductor Req 16 || 32 || 32 8
For source free RL circuit
iL iL 0 e
tReq L
8 t
2.5e
2
2.5e4t
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GATE 2017 EE - Session 1
GENERAL APTITUDE 01. Rahul, Murali, Srinivas and Arul are seated around a square table. Rahul is sitting to the left of Murali. Srinivas is sitting to the right of Arul. Which of the following pairs are seated opposite each other? (a) Rahul and Murali (b) Srinivas and Arul (c) Srinivas and Murali (d) Srinivas and Rahul 01. Ans: (c) Solution: The sitting arrangement looks like as shown Hence, pairs sitting opposite are Murali & Srinivas Rahul & Arul 02.The probability that a k-digit number does NOT contain the digits 0, 5, or 9 is (a) 0.3k (b) 0.6k (c) 0.7k (d) 0.9k 02. Ans: (c) Solution: There are 7 possible digits at each position of a k-digit number Hence, probability of each position= 7 0.7 10 k
P (Number is 0, 5, 9) 0.7
03. Find the smallest number y such that y × 162 is a perfect cube. (a) 24 (b) 27
GATE 2017 EE - Session 1
GENERAL APTITUDE 01. Rahul, Murali, Srinivas and Arul are seated around a square table. Rahul is sitting to the left of Murali. Srinivas is sitting to the right of Arul. Which of the following pairs are seated opposite each other? (a) Rahul and Murali (b) Srinivas and Arul (c) Srinivas and Murali (d) Srinivas and Rahul 01. Ans: (c) Solution: The sitting arrangement looks like as shown Hence, pairs sitting opposite are Murali & Srinivas Rahul & Arul 02.The probability that a k-digit number does NOT contain the digits 0, 5, or 9 is (a) 0.3k (b) 0.6k (c) 0.7k (d) 0.9k 02. Ans: (c) Solution: There are 7 possible digits at each position of a k-digit number Hence, probability of each position= 7 0.7 10 k
P (Number is 0, 5, 9) 0.7
03. Find the smallest number y such that y × 162 is a perfect cube. (a) 24 (b) 27 (c) 32 (d) 36 03. Ans: (d) Solution: 3
y 162 6y 27 6y 3
Hence, 6y must be a perfect cube which is possible if y 62 36 04. After Rajendra Chola returned from his voyage to Indonesia, he _______ to visit the temple in Thanjavur.
(a) was wishing (c) wished
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(b) is wishing (d) had wished
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GATE 2017 EE - Session 1
04. Ans: (c) Solution: After Rejendra chola returned from his voyage to Indonesia, he wished to visit the temple in Thanjavur 05. Research in the workplace reveals that people work for many reasons _____________ . (a) money beside
(b) beside money (c) money besides (d) besides money 05. Ans: (d) Solution: Research in the work place reveals that people work for many reasons besides money 06. Six people are seated around a circular table. There are at least two men and two women. There are at least three right-handed persons. Every woman has a left-handed person to her immediate right. None of the women are right-handed. The number of women at the table is (a) 2 (b) 3 (c) 4 (d) Cannot be determined 06. Ans: (a) Solution: For having a left handed person to immediate right the left handed people must sit consecutively Since, there are at least three right handed people & no women is right handed
GATE 2017 EE - Session 1
04. Ans: (c) Solution: After Rejendra chola returned from his voyage to Indonesia, he wished to visit the temple in Thanjavur 05. Research in the workplace reveals that people work for many reasons _____________ . (a) money beside
(b) beside money (c) money besides (d) besides money 05. Ans: (d) Solution: Research in the work place reveals that people work for many reasons besides money 06. Six people are seated around a circular table. There are at least two men and two women. There are at least three right-handed persons. Every woman has a left-handed person to her immediate right. None of the women are right-handed. The number of women at the table is (a) 2 (b) 3 (c) 4 (d) Cannot be determined 06. Ans: (a) Solution: For having a left handed person to immediate right the left handed people must sit consecutively Since, there are at least three right handed people & no women is right handed There are 2 possibilities 2 women left handed +1man left handed + 2 men right handed 3 women left handed+ 3 men right handed If there are 3 women, then 1 women will have a right man on his right which violates the condition given If there are 2 women are woman can have a left handed man to her right which satisfies all given conditions Hence, there can be only 2 women
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GATE 2017 EE - Session 1
07. A contour line joins locations having the same height above the mean sea level. The following is a contour plot of a geographical region. Contour lines are shown at 25 m intervals in this plot. If in a flood, the water level rises to 525 m, which of the villages P, Q, R, S, T get submerged?
(a) P, Q (c) R, S, T
(b) P, Q, T (d) Q, R, S
07. Ans: (c) Solution: Based on contour lines the approximate height of villages above sea level are P 575m Q 550m R 475m S 450m T 500m The villages having height< 525m get submerged So, R, S & T get submerged 08.“The hold of the nationalist imagination on our colonial past is such that anything inadequately or improperly nationalist is just not history.”
GATE 2017 EE - Session 1
07. A contour line joins locations having the same height above the mean sea level. The following is a contour plot of a geographical region. Contour lines are shown at 25 m intervals in this plot. If in a flood, the water level rises to 525 m, which of the villages P, Q, R, S, T get submerged?
(a) P, Q (c) R, S, T
(b) P, Q, T (d) Q, R, S
07. Ans: (c) Solution: Based on contour lines the approximate height of villages above sea level are P 575m Q 550m R 475m S 450m T 500m The villages having height< 525m get submerged So, R, S & T get submerged 08.“The hold of the nationalist imagination on our colonial past is such that anything inadequately or improperly nationalist is just not history.” Which of the following statements best reflects the author’s opinion? (a) Nationalists are highly imaginative. (b) History is viewed through the filter of nationalism. (c) Our colonial past never happened. (d) Nationalism has to be both adequately and properly imagined. 08. Ans: (b) Solution: The author wishes to state that historical events listed are filtered by the view of nationalism ( x y ) | x y | is equal to 2 (a) the maximum of x and y (c) 1 09. The expression
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(b) the minimum of x and y (d) none of the above 36
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GATE 2017 EE - Session 1
09. Ans: (b) Solution: x y x y f x,y 2 If x y
f x, y
x y x y 2
y
If y x f x, y
x y y x
2 So, f x, y min x, y
x
10. Arun, Gulab, Neel and Shweta must choose one shirt each from a pile of four shirts colored red, pink, blue and white respectively. Arun dislikes the color red and Shweta dislikes the colour white. Gulab and Neel like all the colors. In how many different ways can they choose the shirts so that no one has a shirt with a color he or she dislikes? (a) 21 (b) 18 (c) 16 (d) 14 10. Ans: (d) Solution: If Arun picks white shirt. Shweta can pick either of three remaining shirt & Gulab can get one of two remaining & Neel gets final shirt Number of ways = 3 2 1 6 If Arun picks Pink or Blue shirt, Shweta is left with 2 options as she cannot white shirt then again Gulab gets 1 of 2 remaining & Neel gets final shirt Total numbers of ways 2 2 2 8
GATE 2017 EE - Session 1
09. Ans: (b) Solution: x y x y f x,y 2 If x y
f x, y
x y x y 2
y
If y x f x, y
x y y x
2 So, f x, y min x, y
x
10. Arun, Gulab, Neel and Shweta must choose one shirt each from a pile of four shirts colored red, pink, blue and white respectively. Arun dislikes the color red and Shweta dislikes the colour white. Gulab and Neel like all the colors. In how many different ways can they choose the shirts so that no one has a shirt with a color he or she dislikes? (a) 21 (b) 18 (c) 16 (d) 14 10. Ans: (d) Solution: If Arun picks white shirt. Shweta can pick either of three remaining shirt & Gulab can get one of two remaining & Neel gets final shirt Number of ways = 3 2 1 6 If Arun picks Pink or Blue shirt, Shweta is left with 2 options as she cannot white shirt then again Gulab gets 1 of 2 remaining & Neel gets final shirt Total numbers of ways 2 2 2 8 Total numbers of ways =6+8=14
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GATE 2017 EE - Session 1
Kreatryx GATE 2017 Rank Predictor Form
GATE 2017 EE - Session 1
Kreatryx GATE 2017 Rank Predictor Form
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