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EE GATE Paper 2009
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Q. No. 1 – 20 Carry One Mark Each 1.
2.
The pressure coil of a dynamometer type wattmeter is (A) highly inductive
(B) highly resistive
(C) purely resistive
(D) purely inductive
The measurement system shown in the figure uses three sub-systems in cascade 1 whose gains are specified as G1 , G2 and . The relative small errors associated G3
G1 , G2 and G3 are ε1 , ε2 and ε3 .
with each respective subsystem associated with the output is: Input
G1
(A) ε1 + ε2 +
3.
1 ε3
1 G3
G2
(B)
ε1.ε2 ε3
The
error
Output
(C) ε1 + ε2 − ε3
(D) ε1 + ε2 + ε3
The following circuit has a source voltage Vs as shown in the graph. The current through the circuit is also shown. a
b
+ Vs −
R
10
1
Current (mA)
1.5
Vs (volts)
15
5 0
10k
0.5 0
−0.5
−5
−1
−10
−1.5
−15
0
100
200 Time (ms)
300
0
400
100
200 300 Time (ms)
400
The element connected between a and b could be (A)
a
(B)
b
b
a
(C)
(D) a
b a
b
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The two inputs of a CRO are fed with two stationary periodic signals. In the X-Y mode, the screen shows a figure which changes from ellipse to circle and back to ellipse with its major axis changing orientation slowly and repeatedly. The following inference can be made from this. (A) The signals are not sinusoidal (B) The amplitudes of the signals are very close but not equal (C) The signals are sinusoidal with their frequencies very close but not equal (D) There is a constant but small phase difference between the signals
5.
The increasing order of speed of data access for the following devices is (i) Cache Memory (ii) CDROM (iii) Dynamic RAM (iv) Processor Registers (v) Magnetic Tape
6.
(A)
( v ) , (ii) , (iii) , (iv ) , (i)
(B) ( v ) , (ii) , (iii) , (i) , (iv )
(C)
(ii) , (i) , (iii) , (iv ) , ( v )
(D) ( v ) , (ii) , (i) , (iii) , (iv )
A field excitation of 20 A in a certain alternator results in an armature current of 400A in short circuit and a terminal voltage of 2000V on open circuit. The magnitude of the internal voltage drop within the machine at a load current of 200A is (A) 1V
7.
(B) 10V
(C) 100V
(D) 1000V
The current through the 2 kΩ resistance in the circuit shown is 1kΩ
1kΩ
C
A
B 2kΩ 1kΩ
1kΩ
D
6V
(A) 0mA 8.
(B) 1mA
(C) 2mA
(D) 6mA
Out of the following plant categories (i) Nuclear
(ii) Run-of-river
(iii) Pump Storage (iv)Diesel
The base load power plants are (A) (i) and (ii) 9.
(B) (ii) and (iii)
(C) (i), (ii) and (iv) (D) (i), (iii) & (iv)
For a fixed value of complex power flow in a transmission line having a sending end voltage V, the real power loss will be proportional to (A) V
(B) V2
(C) 1/V2
(D) 1/V
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EE GATE Paper 2009 10.
How many 200W/220V incandescent lamps connected in series would consume the same total power as a single 100W/220V incandescent lamp? (A) not possible
11.
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(B) 4
(C) 3
(D) 2
A Linear Time Invariant system with an impulse response h(t) produces output y(t) when input x(t) is applied. When the input x ( t − τ ) is applied to a system with impulse response h ( t − τ ) , the output will be (A) y ( t )
12.
(B) y ( 2 ( t − τ ) )
(D) y ( t − 2τ )
(C) y ( t − τ )
The nature of feedback in the opamp circuit shown is (A) Current - Current feedback
+6V
1kΩ
2kΩ
− +
(B) Voltage - Voltage feedback
Vout
−6V
(C) Current - Voltage feedback
Vin
∼
(D) Voltage - Current feedback 13.
14.
The complete set of only those Logic Gates designated as Universal Gates is (A) NOT, OR and AND Gates
(B) XNOR, NOR and NAND Gate
(C) NOR and NAND Gates
(D) XOR, NOR and NAND Gates
The single phase, 50Hz, iron core transformer in the circuit has both the vertical arms of cross sectional area 20cm2 and both the horizontal arms of cross sectional area 10cm2. If the two windings shown were wound instead on opposite horizontal arms, the mutual inductance will (A) double (B) remain same (C) be halved (D) become one quarter
15.
A 3-phase squirrel cage induction motor supplied from a balanced 3-phase source drives a mechanical load. The torque-speed characteristics of the motor (solid curve) and of the load (dotted curve) are shown. Of the two equilibrium points A and B, which of the following options correctly describes the stability of A and B? (A) A is stable B is unstable
(C) Both are stable
Torque
(B) A is unstable B is stable
•B A
•
(D) Both are unstable 0
1.0
N Nsync
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EE GATE Paper 2009 16.
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An SCR is considered to be a semi-controlled device because (A) it can be turned OFF but not ON with a gate pulse (B) it conducts only during one half-cycle of an alternating current wave (C) it can be turned ON but not OFF with a gate pulse (D) it can be turned ON only during one half-cycle of an alternating voltage wave The polar plot of an open loop stable system is shown below. The closed loop system is (A) Always stable
ω=∞
Imaginary
17.
−1.42
(B) Marginally stable
Real
(C) Unstable with one pole on the RH s-plane ω=0
(D) Unstable with two poles on the RH s-plane 18.
The first two rows of Routh's tabulation of a third order equation are as follows. s3 s2
2 2 4 4
. This means there are
(A) two roots at s = ± j and one root in right half s-plane (B) two roots at s = ± j2 and one root in left half s-plane (C) two roots at s = ± j2 and one root in right half s-plane (D) two roots at s = ± j and one root in left half s-plane 19.
The asymptotic approximation of the log-magnitude vs frequency plot of a system containing only real poles and zeros is shown. Its transfer function is −40dB / dec
80dB
−60dB / dec
0.1
(A)
10 ( s + 5 )
s ( s + 2 ) ( s + 25 )
(B)
2 5
1000 ( s + 5 ) 2
s
( s + 2 ) ( s + 25)
2.5
(C)
ω rad / s
100 ( s + 5)
s ( s + 2) ( s + 25)
(D)
80( s + 5)
s (s + 2) ( s + 25) 2
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EE GATE Paper 2009 20.
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The trace and determinant of a 2×2 matrix are known to be -2 and -35 respectively. Its eigen values are (A) -30 and -5
(B) -37 and -1
(C) -7 and 5
(D)
17.5 and -2
Q. No. 21 – 56 Carry Two Marks Each
21.
The following circuit has R = 10 kΩ, C = 10 µF R. The input voltage is a sinusoid at 50Hz with an rms value of 10V. Under ideal conditions, the current is from the source is R 10kΩ Vs=10V rms, 50Hz
(A) 10π mA leading by 900 (B) 20π mA leading by 900 (C) 10 mA leading by 900
is ∼
+ OPAMP −
10kΩ R
C 10 µF
(D) 10π mA lagging by 900
22.
In the figure shown, all elements used are ideal. For time t<0, S1 remained closed and S2 open. At t=0, S1 is opened and S2 is closed. If the voltage Vc2 across the capacitor C2 at t=0 is zero, the voltage across the capacitor combination at t=0+ will be S1
3V
(A) 1V
23.
S2
C1
(B) 2 V
1F
C2
2F
(C) 1.5 V
(D) 3 V
Transformer and emitter follower can both be used for impedance matching at the output of an audio amplifier. The basic relationship between the input power Pin and output power Pout in both the cases is (A) Pin = Pout for both transformer and emitter follower (B) Pin > Pout for both transformer and emitter follower (C) Pin < Pout for transformer and Pin = Pout for emitter follower (D) Pin = Pout for transformer and Pin < Pout for emitter follower
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EE GATE Paper 2009 24.
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The equivalent capacitance of the input loop of the circuit shown is 1kΩ
i1
1kΩ
100 µF
1kΩ 49i1
input loop 100 µF
(A) 2 µF 25.
(B) 100 µF
(B) 0F H
For the Y-bus matrix of a 4-bus 2 −5 2 −10 elements are YBUS = j 2.5 2.5 4 0 (A) 3 and 4
27.
(D) 4 µF
In an 8085 microprocessor, the contents of the Accumulator, after the following XRA A instructions are executed will become MVIB F0H SUB B (A) 01 H
26.
(C) 200 µF
(C) F0 H
(D) 10 H
system given in per unit, the buses having shunt 2.5 0 2.5 4 −9 4 4 −8
(B) 2 and 3
(C) 1 and 2
(D) 1,2 and 4
The unit-step response of a unity feedback system with open loop transfer function G(s) = K/ ((s + l) (s + 2)) is shown in the figure. The value of K is 1
response
0.75 0.5
0.25 0 0
(A) 0.5 28.
1
(B) 2
2 times ( s )
3
4
(C) 4
(D) 6
The open loop transfer function of a unity feedback system is given by G ( s ) = e−0.1s / s . The gain margin of this system is
(
(A) 11.95dB
)
(B) 17.67dB
(C) 21.33dB
(D 23.9dB
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EE GATE Paper 2009 29.
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Match the items in List-I with the items in List-II and select the correct answer using the codes given below the lists. List I
List II
To
30.
Use
a.
improve power factor
1.
shunt reactor
b.
reduce the current ripples
2.
shunt capacitor
c.
increase the power flow in line
3.
series capacitor
d.
reduce the Ferranti effect
4.
series reactor
(A) a → 2 b → 3 c → 4 d → 1
(B) a → 2 b → 4 c → 3 d → 1
(C) a → 4 b → 3 c → 1 d → 2
(D) a → 4 b → 1 c → 3 d → 2
Match the items in List-I with the items in List-II and select the correct answer using the codes given below the lists. List I
31.
List II
Type of transmission line
Type of distance relay preferred
a.
Short Line
1.
Ohm Relay
b.
Medium Line
2.
Reactance Relay
c.
Long Line
3.
Mho Relay
(A) a → 2 b → 1 c → 3
(B) a → 3 b → 2 c → 1
(C) a → 1 b → 2 c → 3
(D) a → 1 b → 3 c → 2
Three generators are feeding a load of 100MW. The details of the generators are Rating(MW)
Efficiency (%)
Regulation (p.u.) on 100 MVA base
Generator-1
100
20
0.02
Generator-2
100
30
0.04
Generator-3
100
40
0.03
In the event of increased load power demand, which of the following will happen? (A) All the generators will share equal power (B) Generator-3 will share more power compared to Generator-1 (C) Generator-1 will share more power compared to Generator-2 (D) Generator-2 will share more power compared to Generator-3 32.
A 500MW, 21kV,, 50Hz, 3-phase, 2-pole synchronous generator having a rated p.f=0.9, has a moment of inertia of 27.5 x 103 kg-m2. The inertia constant (H) will be (A) 2.44s
33.
(B) 2.71s
f(x,y) is a continuous function defined over 2
(C) 4.88s
(D) 5.42s
( x, y ) ∈ 0,1 × 0,1 .
Given the two
2
constraints, x>y and y>x , the volume under f(x,y) is © All rights reserved by GATE Forum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission. Discuss this questions paper at www.gateforum.com. Page 7 of 16
EE GATE Paper 2009 y =1 x = y
(A)
∫ ∫
f ( x, y ) dxdy
www.gateforum.com y =1 x =1
(B)
y = 0 x = y2
y= x x= y
y =1 x =1
(C)
∫ ∫ f ( x, y ) dxdy
(D)
y =0 x =0
34.
∫
y =0
∫
f ( x, y ) dxdy
x =0
Assume for simplicity that N people, all born in April (a month of 30 days), are collected in a room. Consider the event of at least two people in the room being born on the same date of the month, even if in different years, e.g. 1980 and 1985. What is the smallest N so that the probability of this event exceeds 0.5? (A) 20
35.
∫ f ( x, y ) dxdy
∫
y = x2 x = y2
(B) 7
(C) 15
(D) 16
A cascade of 3 Linear Time Invariant systems is causal and unstable. From this, we conclude that (A) Each system in the cascade is individually causal and unstable (B) At least one system is unstable and at least one system is causal (C) At least one system is causal and all systems are unstable (D) The majority are unstable and the majority are causal
36.
The Fourier Series coefficient, of a periodic signal x(t), expressed as x (t) =
∞
∑k = −∞ ak ej2πkt / T are given by
a−2 = 2 − j1; a−1 = 0.5 + j 0.2; a0 = j2; a1 = 0.5 − j 0.2; a2 = 2 + j1; and ak = 0; for k > 2 . Which of the following is true?
(A) x(t) has finite energy because only finitely many coefficients are non-zero (B) x(t) has zero average value because it is periodic (C) The imaginary part of x(t) is constant (D) The real part of x(t) is even 37.
The z-transform of a signal x n is given by 4z−3 + 3z−1 + 2 − 6z2 + 2z3 . It is applied to a system, with a transfer function H ( z ) = 3z−1 − 2 . Let the output be y(n). Which of the following is true? (A) y(n) is non causal with finite support (B) y(n) is causal with infinite support (C) y(n) = 0;|n|>3 (D) Re Y ( z ) z = ej θ = − Re Y ( z ) z = e− j θ ; Im Y ( z ) z = e j θ = Im Y ( z ) z = e− j θ ; − π ≤ θ < π
38.
A cubic polynomial with real coefficients (A) can possibly have no extrema and no zero crossings (B) may have up to three extrema and up to 2 zero crossings
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(C) cannot have more than two extrema and more than three zero crossings (D) will always have an equal number of extrema and zero crossings Let x2 -117 = 0. The iterative steps for the solution using Newton-Raphson's method is given by
39.
(A) xk +1 =
1 117 xk + xk 2
(B) xk +1 = xk −
117 xk
xk 117
(D) xk +1 = xk −
1 117 xk + xk 2
(C) xk +1 = xk −
(
)
(
)
F ( x, y ) = x2 + xy ˆ ax + y2 + xy ˆ ay . It's line integral over the straight line from
40.
(x,y)= (0,2) to (x,y) = (2,0) evaluates to (A) -8 (B) 4 41.
(C) 8
(D) 0
An ideal opamp circuit and its input waveform are shown in the figures. The output waveform of this circuit will be Vin
1kΩ
− +
3 2 1
V 0 t 1 −1
t4 t3
t2
t5
6V Vout
−3V
t6
2kΩ
t
1kΩ
−2 −3
(A)
(B)
6 ↑V
6 ↑V
t3
0
t6
t3
0
t→
−3
t→
t6
−3
(C)
(D) 6
6
↑V
↑V
0 −3
t6 t2
t4
t→
0
t2
t 4 t6 t→
−3
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EE GATE Paper 2009 42.
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A 220V, 50Hz, single-phase induction motor has the following connection diagram and winding orientations shown. MM' is the axis of the main stator winding M1 (M1M2) and AA' is that of the auxiliary rm = 0.1 Ω winding (A1A2). Directions of the L m = 0.1 / πH winding axes indicate direction of flux ra = 1 Ω M2 when currents in the windings are in L a = 10 / πH the directions shown. Parameters of M A1 A2 each winding are indicated. When S switch S is closed, the motor (B) rotates anticlockwise
A′
A
220V 50Hz
(A) rotates clockwise (C) does not rotate
M′
(D) rotates momentarily and comes to a halt 43.
The circuit shows an ideal diode connected to a pure inductor and is connected to a purely sinusoidal 50Hz voltage source. Under ideal conditions the current waveform through the inductor will look like D +
vs = 10sin100πt
−
+ ∼
L = (0.1 / π ) H
−
(A)
current
1.5 1 0.5 0
0
10
20 30 time ( ms )
40
50
(B)
current
1.5 1 0.5 0
0
10
20 30 time ( ms )
40
50
0
10
20 30 time ( ms )
40
50
current
1.5
(C)
1 0.5 0
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EE GATE Paper 2009 1.5 current
(D)
1 0.5 0
44.
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0
10
20 30 time ( ms )
40
50
The Current Source Inverter shown in figure, is operated by alternately turning on thyristor pairs (T1, T2) and (T3, T4). If the load is purely resistive, the theoretical maximum output frequency obtainable will be
T1
T3 0.1 µF −
+
D1
D3 10Ω
10A
D4
D2 −
0.1 µF
T4
(A) 125kHz 45.
+
T2
(B) 250kHz
(C) 500kHz
(D) 50kHz
In the chopper circuit shown, the main thyristor (TM) is operated at a duty ratio of 0.8, which is much larger the commutation interval. If the maximum allowable reapplied dv/dt on TM is 50 V/µs, what should be the theoretical minimum value of C1? Assume current ripple through Lo to be negligible. + TM L1
100V −
C1
TA
+
L0 D0
−
(A) 0.2 µF 46.
D1
(B) 0.02 µF
(C) 2 µF
C0
8Ω
(D) 20 µF
Match the switch arrangements on the top row to the steady-state V-I characteristics on the lower row. The steady state operating points are shown by large black dots
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( A) −
+
is
−
+
is
(I)
47.
−
+
is
(II)
vs
(D)
(C )
(B )
is
(III)
vs
−
+
(IV )
vs
vs
(A) A − I B − II C − III D − IV
(B) A − II B − IV C − I D − III
(C) A − IV B − III C − I D − II
(D) A − IV B − III C − II D − I
For the circuit shown, find out the current flowing through the 2Ω resistance. Also identify the changes to be made to double the current through the 2Ω resistance (A)
(5A; Put Vs
= 20V )
(B) ( 2A; Put Vs = 8V ) (C)
(5A; Put Is
VS = 4V
iS = 5A
2Ω
±
= 10A )
(D) ( 7A; Put Is = 12A ) 48.
The figure shows a three-phase delta connected load supplied from a 400V, 50 Hz, 3-phase balanced source. The pressure coil (PC) and current coil (CC) of a wattmeter are connected to the load as shown, with the coil polarities suitably selected to ensure a positive deflection. The wattmeter reading will be a
3-Phase Balanced Supply 400Volts 50Hz
Z2 = (100 + j0 ) Ω
Z1 = (100 + j0 ) Ω
Z1
Z2
CC c
b PC
(A) 0 49.
(B) 1600 Watt
(C) 800 Watt
(D) 400 Watt
An average-reading digital multimeter reads 10V when fed with a triangular wave, symmetric about the time-axis. For the same input an rms-reading meter will read. (A) 20 / 3
(B) 10 / 3
(C) 20 3
(D) 10 3
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Figure shows the extended view of a 2 pole dc machine with 10 armature conductors. Normal brush positions are shown by A and B, placed at the interpolar axis. If the brushes are now shifted, in the direction of rotation, to A' and B' as shown, the voltage waveform VA′B′ will resemble S
N
B −
B' −
1
A +
3
2
4
5
A' +
1'
2'
3'
4'
5'
rotation at speed ω rad / sec
(A)
VA 'B '
ωt 0
0.2π
0.4π
0.6π
0.8π
π
VA 'B '
(B) ωt 0
(C)
0.2π
0.4π
0.6π
0.8π
π
0.2π
0.4π
0.6π
0.8π
π
0.2π
0.4π
0.6π
0.8π
π
VA 'B '
ωt 0 VA 'B '
(D) ωt 0
Common Data Questions: 51 & 52 A
a
B
b
C
c
N
S1
S2
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The star-delta transformer shown above is excited on the star side with a balanced, 4-wire, 3-phase, sinusoidal voltage supply of rated magnitude. The transformer is under no load condition. 51.
52.
With both S1 and S2 open, the core flux waveform will be (A) A sinusoid at fundamental frequency
(B) Flat topped with third harmonic
(C) Peaky with third harmonic
(D) None of these
With S2 closed and S1 open, the current waveform in the delta winding will be (A) a sinusoid at fundamental frequency
(B) flat topped with third harmonic
(C) only third harmonic
(D) none of these
Common Data Questions: 53 & 54
The circuit diagram shows a two winding, lossless transformer with no leakage flux, excited from a current source, i(t), whose waveform is also shown. The transformer has a magnetizing inductance of 400/πmH. A •
1 :1
i (t )
S i (t )
30Ω
10A 0
5ms
10ms
15ms
20ms
25ms 10A
• B
53.
The peak voltage across A and B, with S open is (A) 400/πV
54.
30ms t
(B) 800V
(C) 4000/πV
(D) 800/πV
If the waveform of i(t) is changed to i ( t ) = 10 sin (100πt ) A, the peak voltage across A and B with S closed is (A) 400V
(B) 240V
(C) 320V
(D) 160V
Common Data Questions: 55 & 56
A system is described by the following state and output equations dx1 ( t ) dt dx2 ( t )
= −3x1 ( t ) + x2 ( t ) + 2u ( t )
= −2x2 ( t ) + u ( t ) dt y ( t ) = x1 ( t )
where u ( t ) is the input and y ( t ) is the output
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EE GATE Paper 2009 55.
The system transfer function is (A)
56.
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s+2 s + 5s − 6
(B)
2
s+3 s + 5s + 6
(C)
2
s+5 s + 5s + 6 2
(D)
2s − 5 s + 5s − 6 2
The state transition matrix of the above system is e−3t (A) −2t − 3t e + e e−3t (C) 0
e−3t (B) 0
0 e−2t
e−2t + e−3t e−2t
e3t (D) 0
e−2t − e−3t e−2t e−2t − e−3t e−2t
Linked Answer Questions: Q.57 to Q.60 Carry Two Marks Each
Statement for Linked Answer Questions: 57 & 58 C
A
•
•
Coil 1
Coil 2
B
D
The figure above shows coils 1 and 2, with dot markings as shown, having 4000 and 6000 turns respectively. Both coils have a rated current of 25A. Coil 1 is excited with single phase, 400V, 50Hz supply 57.
58.
The coils are to be connected to obtain a single phase, 400/1000V, auto transformer to drive a load of 10kVA. Which of the options given should be exercised to realize the required auto transformer? (A) Connect A and D; Common B
(B) Connect B and D; Common C
(C) Connect A and C; Common B
(D) Connect A and C; Common D
In the autotransformer obtained in Question 57, the current in each coil is (A) Coil-1 is 25 A and Coil-2 is 10 A
(B) Coil-1 is 10 A and Coil-2 is 25 A
(C) Coil-1 is 10 A and Coil-2 is 15 A
(D) Coil-1 is 15 A and Coil-2 is 10 A
Statement for Linked Answer Questions: 59 & 60 2kΩ
+ −
5V
+ −
2kΩ
3VAB
A
1kΩ
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EE GATE Paper 2009 59.
For the circuit given above, the Thevenin’s resistance across the terminals A and B is (A) 0.5kΩ
60.
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(B) 0.2kΩ
(C) 1kΩ
(D) 0.11kΩ
For the circuit given above, the Thevenin’s voltage across the terminals A and B is (A) 1.25V
(B) 0.25V
(C) 1V
(D) 0.5V
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⏐EE⏐ GATE Paper 2010
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Q. No. 1 – 25 Carry One Mark Each 1
1.
The value of the quantity P, where P =
∫ xe dx,is equal to x
0
(A) 0 2.
(B) 1
(
ˆ (C) ˆi + ˆj + k
(B) 1/r
ˆ (D) 3 ˆi + ˆj + k
)
π⎞ ⎛ The period of the signal x ( t ) = 8 sin ⎜ 0.8πt + ⎟ is 4 ⎝ ⎠ (A) 0.4πs
4.
(D) 1/e
Divergence of the three-dimensional radial vector field r is (A) 3
3.
(C) e
(B) 0.8πs
(C) 1.25s
(D) 2.5s
The system represented by the input-output relationship y ( t ) =
5t
∫ x ( τ )dτ, t > 0
is
−∞
5.
(A) Linear and causal
(B) Linear but not causal
(C) Causal but not linear
(D) Neither linear nor causal
The switch in the circuit has been closed for a long time. It is opened at t = 0. At t = 0+, the current through the 1μF capacitor is 1Ω
t =0
(A) 0A 6.
4Ω
1μF
5V
(B) 1A
(C) 1.25A
(D) 5A
The second harmonic component of the periodic waveform given in the figure has an amplitude of +1 (A) 0
(B) 1
7.
(C) 2 / π
0
(D)
−1
5
T /2
T
t
As shown in the figure, a 1Ω resistance is connected across a source that has a load line v + i = 100. The current through the resistance is (A) 25A (C) 100A
(B) 50A (D) 200A
i
Source
+ v
1Ω
−
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A wattmeter is connected as shown in the figure. The wattmeter reads (A) Zero always
Current coil
(B) Total power consumed by Z1 and Z2
Potential coil Wattmeter
(C) Power consumed by Z1
Z2
(D) Power consumed by Z2 9.
10.
An ammeter has a current range of 0 - 5 A, and its internal resistance is 0.2Ω . In order to change the range to 0 - 25 A, we need to add a resistance of (A) 0.8Ω in series with the meter
(B) 1.0Ω in series with the meter
(C) 0.04Ω in parallel with the meter
(D) 0.05Ω in parallel with the meter
As shown in the figure, a negative feedback system has an amplifier of gain 100 with ±10% tolerance in the forward path, and an attenuator of value 9/100 in the feedback path. The overall system gain is approximately: +
(A) 10±1% (B) 10 ±2%
100 ± 10% −
(C) 10 ±5%
9 100
(D) 10 ±10% 11.
Z1
For the system 2
( s + 1)
, the approximate time taken for a step response to reach 98%
of its final value is (A) 1s 12.
(B) 2s
(C) 4s
(D) 8s
If the electrical circuit of figure (b) is an equivalent of the coupled tank system of figure (a), then
D
B
h1
h2
( a) Coupled tank
A
C
(b ) Electrical equivalent
(A) A, B are resistances and C, D capacitances (B) A, C are resistances and B, D capacitances (C) A, B are capacitances and C, D resistances (D) A, C are capacitances and B, D resistances © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission. Discuss this question paper at www.gatementor.com. 2
⏐EE⏐ GATE Paper 2010 13.
A single-phase transformer has a turns ratio of 1:2, and is connected to a purely resistive load as shown in the figure. The magnetizing current drawn is 1A, and the secondary current is 1A. If core losses and leakage reactance’s are neglected, the primary current is 1A
1.2
(A) 1.41A 14.
(B) 2A
(C) 2.24 A
Power Flow ⎯⎯⎯⎯⎯⎯ →
AC
C
l VAB
System A
AC
VCD
System B D
B Re ctifier
Inverter
(A) VAB<0, VCD<0, VAB>VCD
(B) VAB>0, VCD>0, VAB>VCD
(C) VAB>0, VCD>0, VAB
(D) VAB>0, VCD<0
A balanced three-phase voltage is applied to a star-connected induction motor, the phase to neutral voltage being V. The stator resistance, rotor resistance referred to the stator, stator leakage reactance, rotor leakage reactance referred to the stator, and the magnetizing reactance are denoted by rs ,rr , xs , xr and Xm , respectively. The magnitude of the starting current of the motor is given by (A)
(C)
16.
(D) 3 A
Power is transferred from system A to system B by an HVDC link as shown in the figure. If the voltages VAB and VCD are as indicated in the figure, and I > 0, then A
15
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V
(rs
(B)
+ rr ) + ( xs + xr ) 2
2
V
(rs + rr )
2
(D)
+ ( Xm + xr )
2
V rs + ( xs + Xm )
2
2
V rs + ( Xm + xr ) 2
2
Consider a step voltage wave of magnitude 1pu travelling along a lossless transmission line that terminates in a reactor. The voltage magnitude across the reactor at the instant the travelling wave reaches the reactor is A
Re actor
(A) –1pu
(B) 1pu
(C) 2pu
(D) 3pu
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18.
Consider two buses connected by an impedance of (0+j5)Ω. The bus 1 voltage is 100∠30o V , and bus 2 voltage is 100∠0o V. The real and reactive power supplied by bus 1, respectively, are (A) 1000W, 268VAr
(B) –1000W, –134Var
(C) 276.9W, –56.7Var
(D) –276.9W, 56.7Var
A three-phase, 33kV oil circuit breaker is rated 1200A, 2000MVA, 3s. The symmetrical breaking current is (A) 1200 A
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(B) 3600 A
(C) 35 kA
(D) 104.8 kA
Consider a stator winding of an alternator with an internal high-resistance ground fault. The currents under the fault condition are as shown in the figure. The winding is protected using a differential current scheme with current transformers of ratio 400/5 A as shown. The current through the operating coil is CT ratio 400 / 5
CT ratio 400 / 5
(220 + j0 ) A
(250 + j0 ) A
Operating coil
(A) 0.17875 A 20.
(B) 0.2A
(C) 0.375A
(D) 60 kA
The zero-sequence circuit of the three phase transformer shown in the figure is r
R
b Y y
B
(A)
(B)
r
R
r
R
G
G
(C)
r
R
G
(D)
r
R
G
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Given that the op-amp is ideal, the output voltage V0 is (A) 4V
2R
(B) 6V
R
− +
(C) 7.5V
+2V
+10V V0
−10V
(D) 12.12V 22.
Assuming that the diodes in the given circuit are ideal, the voltage V0 is 10kΩ
(A) 4V (B) 5V
10kΩ
10V
15V V0
10kΩ
(C) 7.5V (D) 12.12V 23.
The power electronic converter shown in the figure has a single-pole double-throw switch. The pole P of the switch is connected alternately to throws A and B. The converter shown is a A (B) half-wave rectifier
+
B
VIN
(C) step-up chopper (boost converter)
L
P
(A) step-down chopper (buck converter)
VOUT
−
(D) full-wave rectifier 24.
Figure shows a composite switch consisting of a power transistor (BJT) in series with a diode. Assuming that the transistor switch and the diode are ideal, the I-V characteristic of the composite switch is +
V
−
I I
(A)
I V
(B)
I
(C)
V
I V
(D)
V
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The fully controlled thyristor converter in the figure is fed from a single-phase source. When the firing angle is 0°, the dc output voltage of the converter is 300 V. What will be the output voltage for a firing angle of 60°, assuming continuous conduction? +
(A) 150V (B) 210V
Vdc
(C) 300V (D) 100πV
−
Q. No. 26 – 51 Carry Two Marks Each
26.
27.
At t = 0, the function f ( t ) = (A) a minimum
(B) a discontinuity
(C) a point of inflection
(D) a maximum
A box contains 4 white balls and 3 red balls. In succession, two balls are randomly selected and removed from the box. Given that the first removed ball is white, the probability that the second removed ball is red is (A) 1/3
28.
(B) 3/7
⎛1 ⎜ An eigenvector of P = ⎜ 0 ⎜0 ⎝ T
For
the
30.
2 0
0⎞ ⎟ 2 ⎟ is 3 ⎟⎠ T
(B) ⎣⎡1 2 1⎦⎤
differential
x ( 0 ) = 1 and
1
(D) 4/7
(C) 1/2
T
(A) ⎣⎡−1 1 1⎦⎤
29.
sin t has t
equation
(C) ⎣⎡1 − 1 2⎦⎤ d2 x dx +6 + 8x = 0 dt dt2
with
T
(D) ⎣⎡2 1 − 1⎦⎤
initial
conditions
dx = 0 , the solution is dt t =0
(A) x ( t ) = 2e−6t − e−2t
(B) x ( t ) = 2e−2t − e−4t
(C) x ( t ) = −e−6t + 2e−4t
(D) x ( t ) = e−2t + 2e−4t
For the set of equations, x1 + 2x2 + x3 + 4x4 = 2 and 3x1 + 6x2 + 3x2 + 12x4 = 6 . The following statement is true (A) Only the trivial solution x1 = x2 = x3 = x 4 = 0 exists (B) There are no solutions (C) A unique non-trivial solution exists (D) Multiple non-trivial solutions exist
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x(t) is a positive rectangular pulse from t = -1 to t = +1 with unit height as shown ∞
in the figure. The value of
∫ | X ( ω) |
dω {where X ( ω) is the Fourier transform of
2
−∞
x(t)} is
x (t)
(A) 2
1
(B) 2π (C) 4
−1
(D) 4π 32.
1
0
Given the finite length input x[n] and the corresponding finite length output y[n] of an LTI system as shown below, the impulse response h[n] of the system is
x [n] = {1, −1}
h [n]
x [n] = {1, 0, 0,0, −1}
↑
(A) h ⎡⎣n⎦⎤ = {1, 0, 0,1} ↑ 33.
↑
(B) h ⎡⎣n⎦⎤ = {1, 0,1} ↑
(C) h ⎡⎣n⎦⎤ = {1,1,1,1} (D) h ⎣⎡n⎦⎤ = {1,1,1} ↑ ↑
If the 12Ω resistor draws a current of 1A as shown in the figure, the value of resistance R is 1Ω
(A) 4Ω
R
12Ω
1A
2A
34.
t
(B) 6Ω
6V
(C) 8Ω
(D) 18Ω
The two-port network P shown in the figure has ports 1 and 2, denoted by terminals (a, b) and (c, d), respectively. It has an impedance matrix Z with parameters denoted by zij. A 1Ω resistor is connected in series with the network at port 1 as shown in the figure. The impedance matrix of the modified two-port network (shown as a dashed box) is e
1Ω
a
c P d
b
⎛ z11 + 1 z12 + 1 ⎞ (A) ⎜ ⎟ z22 + 1 ⎠ ⎝ z21
z12 ⎞ ⎛ z11 + 1 (B) ⎜ ⎟ z z 22 + 1 ⎠ ⎝ 21
⎛ z11 + 1 (C) ⎜ ⎝ z21
z12 ⎞ ⎛ z11 + 1 ⎟ (D) ⎜ z22 ⎠ ⎝ z21 + 1
z12 ⎞ ⎟ z22 ⎠
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The Maxwell's bridge shown in the figure is at balance. The parameters of the inductive coil are R + jωL
R3 R4
R2
36.
− j / ( ωC 4 )
(A) R = R 2R 3 / R 4 , L = C4R 2R 3
(B) L = R 2R 3 / R 4 , R = C4R 2R 3
(C) R = R 4 / R 2R 3 , L = 1 ( C4R 2R 3 )
(D) L = R 4 / R 2R 3 ,R = 1 / ( C4R 2R 3 )
The frequency response of G ( s ) = 1 / ⎡⎣s ( s + 1) ( s + 2 ) ⎤⎦ plotted in the complex G ( jω) plane (for 0 < ω < ∞) is Im
(A)
ω=0
−3 / 4
Im
(B) Re
Re
−3 / 4 ω=0
(C)
ω=0
Im
Im
(D) Re
Re
ω=0 −1 / 6
−1 / 6
37.
38.
⎡ −1 The system x = Ax + Bu with A = ⎢ ⎣0
2⎤ ⎡0⎤ ⎥ , B = ⎢ ⎥ is 2⎦ ⎣1 ⎦
(A) stable and controllable
(B) stable but uncontrollable
(C) unstable but controllable
(D) unstable and uncontrollable
The characteristic equation of a closed-loop system s(s+1)(s+3)+k(s+2)=0, k>0 . Which of the following statements is true?
is
(A) Its roots are always real (B) It cannot have a breakaway point in the range −1 < Re ⎣⎡s⎦⎤ < 0 (C) Two of its roots tend to infinity along the asymptotes Re[s] = -1 (D) It may have complex roots in the right half plane © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission. Discuss this question paper at www.gatementor.com. 8
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A 50 Hz synchronous generator is i n i t i a l l y connected to a long lossless transmission l i n e which is open circuited at the receiving end. With the f i e l d voltage held constant, the generator is disconnected from the transmission l i n e . Which of the following may be s a i d about the steady state terminal voltage and field current of the generator?
Long Transmission Line receiving end
(A) The magnitude of terminal voltage decreases, and the field current does not change (B) The magnitude of terminal voltage increases, and the field current does not change (C) The magnitude of terminal voltage increases, and the field current increases (D) The magnitude of terminal voltage does not change, and the field current decreases 40.
A separately excited dc machine is coupled to a 50Hz, three-phase, 4-pole induction machine as shown in the figure. The dc machine is energized first and the machines rotate at 1600 rpm. Subsequently the induction machine is also connected to a 50Hz, three-phase source, the phase sequence being consistent with the direction of rotation. In steady state, DC machine
Induction machine 4 pole, 50Hz 50Hz, balanced three − phase sup ply
(A) Both machines act as generators (B) The dc machine acts as a generator, and the induction machine acts as a motor (C) The dc machine acts as a motor, and the induction machine acts as a generator (D) Both machines act as motors
41.
A balanced star-connected and purely resistive load is connected at the secondary of a star-delta transformer as shown in the figure. The line-to-line voltage rating of the transformer is 110V/220V. Neglecting the non-idealities of the transformer, the impedance 'Z' of the equivalent star-connected load, referred to the primary side of the transformer, is
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r
R
4Ω 4Ω
b Y y
B
R Z Z
Z
Y
B
(A) (3 + j0 ) Ω 42.
(B) ( 0.866 − j0.5) Ω
(C)
(0.866 + j0.5) Ω
(D) (1 + j0 ) Ω
Consider a three-phase, 50Hz, 11kV distribution system. Each of the conductors is suspended by an insulator string having two identical porcelain insulators. The self capacitance of the insulator is 5 times the shunt capacitance between the link and the ground, as shown in the figure. The voltage across the two insulators is (A) e1 = 3.74kV, e2 = 2.61kV (B) e1 = 3.46kV, e2 = 2.89kV C
(C) e1 = 6.0kV, e2 = 4.23kV (D) e1 = 5.5kV, e2 = 5.5kV 43.
5C
e2
5C
e1
Conductor
Consider a three-core, three-phase, 50Hz, 11kV cable whose conductors are denoted as R, Y and B in the figure. The inter-phase capacitance (C1) between each pair of conductors is 0.2μF and the capacitance between each l i n e conductor and the sheath is 0.4μF . The per-phase charging current is C2 C1
R
Y
B C2
C1
C1
C2
Outer Sheath
(A) 2.0A
(B) 2.4A
(C) 2.7A
(D) 3.5A
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For the power system shown in the figure below, the specifications of the components are the following: G 1 : 25 kV, 100 MVA, X=9% G2: 25'kV, 100MVA, X=9% T1: 25 kV/220 kV, 90 MVA, X=12% T2: 220kV/ 25 kV, 90 MVA, X=12% Line1: 220 kV, X= 150 ohms T2
T1
Line 1 G1
G2 Bus 1
Bus 2
Choose 25 kV as the base voltage at the generator G1, and 200 MVA as the MVA base. The impedance diagram is j0.27
(A)
j0.42
j0.27
j0.18
j0.18
G1
G2
j0.27
(B)
j0.62
j0.27
j0.18
j0.18
G1
G2
j0.27
(C)
j0.42
j0.27 j0.21
j0.21 G1
G2
j0.3
(D)
j0.21 G1
j0.42
j0.3 j0.21
G2
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The transistor circuit shown uses a s i l i c o n transistor VBE = 0.7 V, IC ≈ IE and a dc current gain of 100. The value of V0 is
with
+10V
(A) 4.65A 50kΩ
10kΩ
(B) 5V (C) 6.3V
V0 100Ω
(D) 7.23V 46.
The TTL circuit shown in the figure is fed with the waveform X (also shown). All gates have equal propagation delay of 10 ns. The output Y of the circuit is X
X
100 ns
1 0
Y
t
Y
(A)
Y
1 0
(B)
0
t
Y
(C)
47.
t
Y
1 0
1
(D) t
1 0
t
When a "CALL Addr" instruction is executed, the CPU carries out the following sequential operations internally: Note: (R) means content of register R ((R)) means content of memory location pointed to by R PC means Program Counter SP means Stack Pointer (A) (SP) incremented
(B)
(PC ) ← Addr ( (SP ) ) ← (PC ) (C) (PC ) ← Addr
(SP ) incremented ( (SP ) ) ← (PC )
(D)
(PC ) ← Addr ( (SP ) ) ← (PC ) (SP ) incremented ( (SP ) ) ← (PC )
(SP ) incremented (PC ) ← Addr
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Common Data Questions: 48 & 49
A separately excited DC motor runs at 1500 rpm under no-load with 200 V applied to the armature. The field voltage is maintained at its rated value. The speed of the motor, when it delivers a torque of 5 Nm, is 1400 rpm as shown in the figure. The rotational losses and armature reaction are neglected. speed (rpm ) 1500 1400
0
48.
torque (Nm )
The armature resistance of the motor is, (B) 3.4Ω
(A) 2Ω 49.
5
(C) 4.4Ω
(D) 7.7Ω
For the motor to deliver a torque of 2.5 Nm at 1400 rpm the armature voltage to be applied is (A) 125.5V
(B) 193.3V
(C) 200V
(D) 241.7V
Common Data Questions: 50 & 51
Given f(t) and g(t) as shown below: g (t )
f (t) 1 0
50.
51.
1 1
t
0
t
5
3
g(t)can be expressed as
(A) g ( t ) = f (2t − 3)
⎛t ⎞ (B) g ( t ) = f ⎜ − 3 ⎟ ⎝2 ⎠
3⎞ ⎛ (C) g ( t ) = f ⎜ 2t − ⎟ 2⎠ ⎝
⎛ t 3⎞ (D) g ( t ) = f ⎜ − ⎟ ⎝2 2⎠
The Laplace transform of g(t) is (A)
1 3s e − e5s s
(C)
e−3s 1 − e−2s s
(
(
) )
(B)
1 −5s e − e−3s s
(D)
1 5s e − e3s s
(
(
)
)
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Linked Answer Questions: Q.52 to Q.55 Carry Two Marks Each Statement for Linked Answer Questions: 52 & 53
The following Karnaugh map represents a function F. F X
52.
11 10
0
1
1
1
0
1
0
0
1
0
A minimized form of the function F is (A) F = XY + YZ
53.
00 01
(B) F = XY + YZ
(C)
F = XY + YZ
(D) F = XY + YZ
Which of the following circuits is a realization of the above function F? (A)
X F
Y Z
(B)
X F
Y Z
(C)
X F
Y Z
(D)
X F Y
Z
Statement for Linked Answer Questions: 54 & 55 The L-C circuit shown capacitance C = 10μF .
in
the
figure
has
an
inductance
L=1mH
and
a
L i C t =0
− 100V
+
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The initial current through the inductor is zero, while the initial capacitor voltage is 100 V. The switch is closed at t = 0. The current i through the circuit is:
(
)
(
(A) 5 cos 5 × 103 t A
(
)
(B) 5 sin 104 t A
)
(
(C) 10 cos 5 × 103 t A
55.
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)
(D) 10 sin 104 t A
The L-C circuit of Q54 is used to commutate a thyristor, which is initially carrying a current of 5A as shown in the figure below. The values and initial conditions of L and C are the same as in Q54. The switch is closed at t = 0. If the forward drop is negligible, the time taken for the device to turn off is L i C t =0
100V
(A)
52μs
− 100V
+
5A
(B) 156μs
20Ω
(C) 312μs
(D) 26μs
General Aptitude (GA) Questions Q.No. 56-60 Carry One Mark Each
56.
25 persons are in a room. 15 of them play hockey, 17 of them play football and 10 of them play both hockey and football. Then the number of persons playing neither hockey nor football is (A) 2
57.
(B) 17
(C)13
(D) 3
The question below consists of a pair of related words followed by four pairs of words. Select the pair that best expresses the relation in the original pair. Unemployed: Worker
(A) fallow: land 58.
(B) unaware: sleeper (C) wit: jester
(D) renovated: house
Choose the most appropriate word from the options given below to complete the following sentence If we manage to ____________ our natural resources, we would leave a better planet for our children. (A) uphold
59.
(B) restrain
(C) cherish
(D) conserve
Which of the following options is closest in meaning to the word: Circuitous? (A) cyclic
(B) indirect
(C) confusing
(D) crooked
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Choose the most appropriate word from the options given below to the complete the following sentence: His rather casual remarks on politics ___________ his lack of seriousness about the subject. (A) masked (B) belied (C) betrayed (D)suppressed Q.No. 61 - 65 Carry Two Marks Each
61.
Hari (H), Gita (G), Irfan (I) and Saira (S) are siblings (i.e. brothers and sisters). All were born on 1st January. The age difference between any two successive siblings (that is born one after another) is less than 3 years. Given the following facts: i. Hari’s age + Gita’s age > Irfan’s age + Saira’s age ii. The age difference between Gita and Saira is 1 year. However Gita is not the oldest and Saira is not the youngest. iii. There are no twins. In what order were they born (oldest first)? (A) HSIG (B) SGHI (C) IGSH (D) IHSG
62.
5 skilled workers can build a wall in 20days; 8 semi-skilled workers can build a wall in 25 days; 10 unskilled workers can build a wall in 30days. If a team has 2 skilled, 6 semi-skilled and 5 unskilled workers, how long will it take to build the wall? (A) 20 (B) 18 (C) 16 (D) 15
63.
Modern warfare has changed from large scale clashes of armies to suppression of civilian populations. Chemical agents that do their work silently appear to be suited to such warfare; and regretfully, there exist people in military establishments who think that chemical agents are useful tools for their cause. Which of the following statements best sums up the meaning of the above passage: (A) Modern warfare has resulted in civil strife. (B) Chemical agents are useful in modern warfare. (C) Use of chemical agents in warfare would be undesirable (D) People in military establishments like to use chemical agents in war.
64.
Given digits 2,2,3,3,4,4,4,4 how many distinct 4 digit numbers greater than 3000 can be formed? (A) 50 (B) 51 (C) 52 (D) 54
65.
If 137+276=435 how much is 731+672? (A) 534 (B) 1403
(C) 1623
(D)1513
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EE GATE Paper 2008
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Answer keys 1
A
2
B
3
C
4
C
5
D
6
A
7
D
8
C
9
A
10
A
11
D
12
B
13
D
14
C
15
D
16
B
17
A
18
B
19
D
20
D
21
C
22
C
23
A
24
B
25
B
26
B
27
28
B
29
C
30
B
31
D
32
A
33
D
34
D
35
A
36
D
37
C
38
39
B
40
B
41
C
42
A
44
B
45
B
46
B
47
B
48
C
49
B
C
52
C
53
54
C
55
B
56
B
43 50
C
51
57
B
58
64
A
65
71
59 A
72
78
B
85
A
79
C
66 73
C
60
C
61
C
62
B
63
C
67
C
68
C
69
B
70
A
76
C
77
83
C
84
74 C
80
75 B
81
D
82
D
Explanation:1.
(b − n + 1) b = No of branches n = No of nodes
2.
Ith =
=
Vth zth
−15.4
ref
−7.8
3.71 −15.9 2.4 −8.0
I
= 1.54 −7.8
V
I lead V so it is a combination of Resistance and capacitance
3.
When we apply e−dt sin ωt to LTI system O / P is of form Ke−dt sin ( ωt + φ ) , so, compare this equation with
Ke−βt sin ( υt + φ ) so β = α, υ=ω
x4 1 1= ∫ 4 0 4
4.
3 ∫ x dx =
5.
It should satisfy the characteristic equation λ3 + λ2 + 2λ + l = 0
(
)
⇒ P3 + P2 + 2P + 1 = 0 ⇒ P P2 + P + 2I + P −1 = 0
(
P ≠ 0 and P2 + P + 2I + P −1 = 0 ⇒ P −1 = − P2 + P + 2I
)
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7.
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Rank is 4, so Q will have four linearly independent columns and four linearly independent rows. SY ( s ) + y ( s ) = 1
Y ( s) = e − tu ( t )
1 s +1
8.
Up to 5.7V, the diode is in reverse biased so, V0=Vi and diode is forward biased after Vi > 5.7 and reaches peak value and comes to again 5.7V, in this output voltage is 5.7V, after that again diode is in reverse bias so V0=Vi
11.
Theory bit
12.
A 30º
YDI B
C
13.
It is a predefined statement
15.
l = 200km β = 0.00127radias per km λ=
2π 2π = = 4947.39km 0.00127 β
300 l = = 6.06% λ 4947.39 16.
ZP = Zs − Zm
Zo = Zs + 2Zm
48 = Zs − zm
15 = Zs + 2zm
Zs = 26Ω Zm = 11Ω
17.
X 50 φ = Tan− 2 = Tan−1 = 45º R 50 ∴ Below 45º of firing angle the out put voltage V0 is not controlable
18.
Line voltage is free from triplex harmonics
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Linear system holds superposition theorem Causal system means h(t) = 0 for t < 0
20.
In O.C. test pf is low, so we use low pf meters and In S.C. test pf is high, so we use high pf meters. S.C.Test 2kvA = 8.69A. so we use 10Ameter test 230v
21.
22.
R = 6, C =
2 ×1 2 6×2 = , τ = RC = =4 2 +1 3 3
(
Z = ( 0.1s ) + 1 / 1
s
0.15
) z⇒
equating imaginary pout to zero ω = 3rad / sec
23.
−vab + v2 − 5 = 0
1
s
2Ω
i = 1A v2 = 1 × 2 = 2v
1
+ 1A
vab
+ v2 vi
m 5V
−
vab = 2.5 = −3v
∈0 A ∈0 A 8.85 × 10−12 × 500 × 500 × 10−6 = = 1475PF ,C = −3 −3 d1 d2 d 4 × × 10 2 10 + + ∈1 ∈2 8 2
24.
C=
25.
(300) × µ0 × µr × 300 × 10−3 N2µ0µr A ⇒ L = l 300 × 10−3
26.
−5 + 2i1 = 0
(
2
)
2
Q l = 0.3m
1
+
i1 = 2.5 v = 2.5volt −2.5 + vab + i = 0 vab + i1 = 2.5 −4vab + 3i + i = 0
vab
3Ω
−
+
+ 5v −
1 i1
v −
1Ω
+ ∆v ab − i
i = vab i + 1 = 2.5 2.5 i= = 1.25 2
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28.
Hint: (i) Apply shifting, (ii) scaling and (iii) Time reversal
29.
Non-causal because the signal is defined for t<0 also time invariant since y(n-k) = [x(n-K)] where x (n) → y (n) 1
30.
sin cαt
−α / 2
sin c Bt
32.
α /2
−B / 2
convolution in time dmin is (
x (z) z z − (1 + j)
so it will be max ( α,B )
B /2
H (z)
(1 + j)n
) in frequency
y (z ) = x (z ) H (z ) y (n) =
{ z >11 + j }
z n Q O is zero for x (n) = (1 + j) and the z. Transform of x (n) = P z1 − (1 + j) 1 −1 The ROC of 1 − z H ( z ) 2 implies it has 1 pole and 1 zero
33.
Re sides of zP . × ( z ) =
n 1 dn ( z − a) P × z z ( ) (n − 1) ! dzn −1
( z − a)n at ( z = a) given problem
(
where z=a
n = order of pole
)
(
)
1 d n −1 × (z) / z = a z (2 − 1) ! dz 1 d n −1 z z = (2 − a)2 2 1! dz ( z − a) d n z z = a = nzn −1 = = nan −1 z=a dz
n = 2Re sides zn −1 × ( z ) =
(
)
( )
34.
(
f ( x ) = x2 − 4
(
)
2
)
f1 ( x ) = 2 x2 − 4 .2x
(
)
f1 ( x ) = 0 ⇒ 4x x2 − 4 = 0, ⇒ x = 0, 2, −2
f11 ( x ) = 12x2 − 16 x = 0 f11 = −16 → max x = −z f11 = −64 → max x = z f11 = 0 f111 = 24x x = z f111 = 24 × 2 = 48 → max
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35.
xn +1 = xn −
f ( xn ) = exn − 1
f1 ( xn )
f ( x0 )
x1 = x0 −
36.
f ( xn )
f1 ( xn ) = exn = −1 −
f1 ( x0 )
(
A+ = AT A
)
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( ) (A ) = A
−1
A T = A −1 A T
−1
T
e −1 e −1
= 0.71828
−1
A by checking options AA + A = AA −1A = A ∴ AA + A ≠ A +
h y0 + 2y2 2
37.
x=
39.
F ( s) =
α
∫
f ( t ) eist dt
x (t ) 1
−α
=
1
0
1
∫ (1) e
ist
dt
f ( t ) = x ( t ) + x ( −t ) f (t )
−1
= 2 sin cs ω = 2 sin c 2π
40.
−1
1
The both transfer for a cont mirror
S0 I1 = I
t
t
1kΩ I1
ref
0 − 0.7 − 0.7 + 5 1 = 3.6mA
+5 Iref
I1 =
−5v
41.
( )
v1 0− = v1 ( 0t ) = 0v
+
at the half cycle after t = 0 at − vc half cycle vc1 = 5v
K5v ~
vc1
−
kvL
5v
−+ −5
+−
vc2
vc2 = 10v
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Combine two waveforms R vin
− +
vO1 ...(1)
(1) + (2) So
44.
R
vin
Q
− +
vO2 ...(2)
A is Answer
Here monoshot o/p’s Q1 and Q2 has pulse widths of TON1 and T0N2 . So, when triggered according to truth table given Q1 responds to T0N2 and Q2 for TON1 and waveform will be 1 f = T0N + T0N 1 2
45.
46.
DAD
T0N2 ,Duty cycle = T0N + T0N 1 2
TON1 TON1
SP
PCHL After first
instruction
HL − 2700H
PCHL so,
HL contes PC = 2700
exchanged with PC Sp = 2700
The first op-amp is a square wave generator and now after adjusting voltage at +ve terminal to 2.5v the second op-amp an follows C Now d (2.5 − v0 ) vi − 2.5 =e R at 1 vi − 2.5 = 2.5 − Vo RC ∫ so triangle wave shifts upward
R vi
2.5v
47.
1500 − 1425 7.8 7.8 S= = 0.05, effective rotor resis tan ce = Q 1500 2 − 0.05 2 − S
48.
Hint: core losses 1200w are also considered
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49.
Hint: Ers
50.
Direct
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t = 0 to 1 the o / p voltage is − ve and constant value t = 1 to 2 the o / p voltage is zero t = 2 to 2.5 the o / p voltage is + ve and constant
dθ = −N2 ; dt
v2
51.
L1 C1
SIL1 = SIL 2 v2
=
L2 C2
52.
L2 C2 L1 C1
(1 − 0.3) L C L C
=
= 0.7, SIL2 =
SIL1 0.7
dC1 dC2 = dPG1 dPG 2 ...... (1)
1 + 0.11 PG1 = 30.06 PG2
PG1 + PG2 = 250 ...... (2 )
solve (1) & (2 )
54.
Re move e4 then loss = (5) z + ( 3) z + (2 ) z = 25 + 9 + 4 = 36z 2
2
2
e3 then loss z 1112 z + (7 ) z + (2 ) z = 1 + 49 + 4 = 54z 2
2
So, if we remove C3 then system will operate with minimum loss. 2 2 ωεα π
55.
Hint : Pf =
57.
Eb1 = 220 − ( 2.5 × 20 ) = 170
Eb1 Eb2
=
1000 600
600 × 170 = 102 1000 V = 102 + 50 = 152
Eb2 =
duty cycle =
152 = 0.608 250
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Tst 2a = Tmax 1 + a2 1= R2 x2
2×a
⇔ a=1 1 + a2 1 = = 0.666 1.5
V = 0.66 × 400 = 266.6V frequency = sf = 0.666 × 50 = 33.3Hz
61.
2Vm cos α = −130 ⇔ α = 129º π
62.
TWD =
63.
V0 = Vs ×
64.
C (S)
R (S )
π2 2 × 10 × 3 − 9 = 0.31 = (31% ) ,Rms value = = 7.8 9 2×π 1 20 = = 40V, since poweris constant 20 × 4 = 40 × Id , Id = 2A 1 0.5 0.5 − ( )
= G (S)
R (S) = 1
Lt C ( t ) = Lt SC ( t )
x →∞
65.
C1 =
s →0
10 ( S + 1)
(S + 10 )
pole
zero
−10
−1
lead compensator
66.
C2 =
−10
S + 10 10 ( S + 1)
−1
lag compensator
At starting the stop is -40dB/decode ∴ At starting we have 2 poles and in next portion the stop is changes from -40dB/decode to -20dB/decode (i.e. we have one zero and next portion the stop is changes from (-20dB/decode to 0 dB/decode)(i.e. one zero) ∴ Total we have 2 poles and 2 zeros.
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K K = 2 C (S) S ( S + 3) ( S + 10 ) S + 3S ( S + 10 ) + K = K R (S ) K 1+ S ( S + 3 ) ( S + 10 ) = 3 2 S + 13S + 30S + K
(
C.E
)
S3 + 13S2 + 30S + K = 0 1 × K = 30 × 13 = 390
0 < K < 390
68.
C (S)
=
R (S )
ωn2
S2 + 2ξωns + ωn2
ωn = 10 2ξωn = 20 ξ = 1 so critical damped
69.
fy No. of horizontol tan gencies ω , fx = ω, fy = = fx No. of vertical tan gencies 2
70.
( z = 500 ) = ZBC × ZAD
76.
dq 1 1 q = ∫ idt Area upto 5sec i.e. A1 + A2 + A3 = × 2 × 4 + × 3 × 2 + (3 × 2) = 13GC, Qi = 2 2 dt
78.
T.F = C ( SI − A )
79.
ess =
80.
Hint :
B+D
1 K
V0 = Vi
ω→0
81.
−1
V0 = Vi
SC SC +
1 RA
ωC 2
put SC = Jω
1
=
1 2 2 +ω C R A
1+
1
( ω CR ) 2
A
V0 V = 0, ω → ∞ 0 = 1 so high pass filter. Vi Vi
From the given circuit
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I = Ish + Ia
240 + Ia 80 Ia = 12A
15 =
At plugging the net voltage across armature resistance is =V+Eb Eb = 2400 − 12 × 0.5 = 234 = 240 + 234 = 474v
83.
Ia2 = 1.25 × Ia = 1.25 × 12 = 15A,R =
84.
V = E + Ia xs
234 − 0.5 = 15.1Ω 15
v = 10 Ia = 0.6 0 xs = 1 90 1 0 = E + 0.6 01 90 Ef = 1 − j0.6 = 1.17 −30.96 = 1.17, 30.96 lag
85.
Ia = 1.2Ia1 = 1.2 × 0.6 = 0.72
(Ia2xs )2 = Ef2 + Ut2 − 2Ef Vt cos δ (0.72 × 1)2 = 1.172 + 12 − 2 × 1.17 × 1 × cos δ cos δ = 0.79 ⇒ δ = 37.73 sin δ = 0.6120 Ef vt sin δ = vt Ia2 cos θ ⇒ cos θ = 0.994 xs Ef cos δ = 1.17x0.79 = 0.924 Ef cos δ < V,hence PF is 0.994 lagging
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