Electrical Engineering (EE)

In the given logic diagram the propagation delays of both the gates are 20 nsec each. The output y is at logic ‘1’ from a long time and at t = 0, the input x change from 0 to 1, so the output goes momentary to logic 0, determine the time required by the output to again maintain its level at logic 1.

• a)
  0 nsec
• b)
  20 nsec
• c)
  40 nsec
• d)
  60 nsec

• a)
  0
• b)
  30
• c)
  45
• d)
  90

• a)
  2A
• b)
  5A
• c)
  6A
• d)
  3A

• a)
  1
• b)
  0
• c)
  2
• d)
  4

• a)
  Alternators.
• b)
  Switchgear.
• c)
  Transformers.
• d)
  Underground cable.

• a)
  Linear circuits only

   
• b)
  Both the linear and nonlinear circuits only
• c)
  Passive time invariant circuits
• d)
  Nonlinear circuits only

• a)
  for high start-up torque applications
• b)
  as a frequency converter
• c)
  any of the mentioned
• d)
  none of the mentioned

• a)
  a material with low work function is preferable.
• b)
  a material with high work function is preferable.
• c)
  the work function of the material has no importance.
• d)
  none of the above

For Bulk-Boost regulator, calculate critical capacitance and source-current. Duty cycle = 0.6 and = 10 sec.

• a)
  1/2 F, 15A
  
• b)
  15F, 15A
• c)
  1/3F, 15A
• d)
  16F, 16A

If cross-sectional area of base-emitter junction of transistor is 2 times of transistor then determine the value of Assume of both the transistors are very large

• a)
  4.65 mA
  
• b)
  9.3 mA
• c)
  - 4.65 mA
• d)
  - 9.3 mA

• a)
  
• b)
  
• c)
  
• d)
  

• a)
  directly proportional to applied voltage
• b)
  directly proportional to the square of the applied voltage
• c)
  inversely proportional to applied voltage
• d)
  none of the above

• a)
  10-20 lumen/m²
• b)
  30-50 lumen/m²
• c)
  40-75 lumen/m²
• d)
  100-140 lumen/m²

• a)
  High input impedance and low output impedance
• b)
  Low input impedance and high output impedance
• c)
  High voltage gain
• d)
  High current gain

• a)
  Decrease
• b)
  Increase
• c)
  Be unaffected
• d)
  Exponentially increase

• a)
  reduces the x-section of the conductor by n times
• b)
  increases the x-section of the conductor by n times
• c)
  reduces the x-section of the conductor by n² times
• d)
  increases the x-section of the conductor by n² times.

• a)
  INR
• b)
  DAA
• c)
  DAD
• d)
  INX

• a)
  L-L
• b)
  LLG
• c)
  LG
• d)
  3φ

• a)
  1
• b)
  2
• c)
  1 or 2
• d)
  4

• a)
  0.6 V
• b)
  0.1 V
• c)
  0.2 V
• d)
  0.5 V

• a)
  the angle between voltage & current will increase
• b)
  the pf will increase
• c)
  the pf will decrease
• d)
  the power drawn will decrease

What is the load current I_L in the circuit shown below:

• a)
  -5 mA
• b)
  -10 mA
• c)
  25 mA
• d)
  50 mA

 Match List-I (Logic gates) with List-ll (Applications) and select the correct answer using the codes given below the lists:

List-i
A. Exclusive OR gate
B. NOT gate
C. Exclusive NOR gate
D. NOR gate

List-ll
1. As a controlled inverter
2. As an universal gate
3. Complementation
4. Even-odd parity check

• a)
  a
• b)
  b
• c)
  c
• d)
  d

• a)
  very fine gauge copper wire
• b)
  heavy gauge copper wire
• c)
  insulation material
• d)
  interpoles do not need winding

• a)
  more for shunt capacitor
• b)
  more for series capacitor
• c)
  it is the same for both series and shunt
• d)
  uncertain

• a)
  the same
• b)
  1.5 times
• c)
  2 times
• d)
  3 times

• a)
  200 km
• b)
  160 km
• c)
  100 km
• d)
  80 km

• a)
  160.2 C
• b)
  -160.2 C
• c)
  16.02 C
• d)
  -16.02 C

• a)
  1
• b)
  0
• c)
  2
• d)
  4

• a)
  50%
• b)
  75%
• c)
  90%
• d)
  86%

• a)
  hysteresis losses only
• b)
  hysteresis & leakage losses only
• c)
  hysteresis, leakage & copper losses only
• d)
  hysteresis, leakage, copper & friction losses

• a)
  0.2 H
• b)
  0.24 H
• c)
  0.8 H
• d)
  0.42 H

• a)
  emitter, base, collector
• b)
  emitter, base, drain
• c)
  source, base, drain
• d)
  source, base, collector

• a)
  18.36 kV
• b)
  17.96 kV
• c)
  15.96 kV
• d)
  12.65 kV

• a)
  62.75W
• b)
  4.5275W
• c)
  5.275W
• d)
  0.5275W

• a)
  822.56 N-m    
• b)
  954.93 N-m
• c)
  525.76 N-m    
• d)
  730 N-m

• a)
  250 VA, 217 VAR
• b)
  188VA, 206 VAR
• c)
  125VA, 105VAR
• d)
  None of these

v1 = ?

• a)
  -11 V
• b)
  5 V
• c)
  8 V
• d)
  18 V

• A:

  -30 V

• B:

  25 V

• C:

  -20 V

• D:

  15 V

The answer is d.

• a)
  Both A and R are correct and R is correct explanation of A
• b)
  Both A and R are correct but R is not correct explanation of A
• c)
  A is true, R is false
• d)
  A is false, R is true

• a)
  zero
• b)
  low
• c)
  high
• d)
  infinite

• a)
  100%
• b)
  50%
• c)
  25%
• d)
  16.7%

• a) 
  current controlled current source
• b) 
  current controlled voltage source
• c) 
  voltage controlled voltage source
• d) 
  voltage controlled current source

The resistance of series RLC circuit is doubled and inductance is halved, then BW becomes

• a)
  half
• b)
  double
• c)
  one-fourth
• d)
  four times

• a)
  Phase constant β = ωμε
• b)
  Phase constant ω = με
• c)
  Phase constant β = ω√(με)
• d)
  Phase constant β = 1/ωμε

• a)
   Remains same
• b)
   Twice of the former
• c)
   Four times of former emf
• d)
   Half of the former emf

• a)
  Reduce weight of copper
• b)
  Improve stability
• c)
  Reduce corona
• d)
  Increase power transmission capacity

• a)
  0.5
• b)
  2
• c)
  2.5
• d)
  50

• a)
  72 AT/m
• b)
  144 AT/m
• c)
  196 AT/m
• d)
  222 AT/m

• a)
  3-f induction motor squirrel cage type
• b)
  3-f induction motor slip ring type
• c)
  synchronous motor working at leading power factor
• d)
  synchronous motor working at lagging power factor

• a)
  Lead to a null indication and setting will correspond to a frequency of 200 Hz
• b)
  Lead to a null indication and setting will correspond to 260 Hz
• c)
  Lead to a null indication and setting will correspond to 400 Hz
• d)
  Not lead to null indication

• a)
  
• b)
  
• c)
  
• d)
  

• a)
  1, 3, 2
• b)
  1, 2, 3
• c)
  3, 1, 2
• d)
  2, 1, 3