1. In a synchronous machine, the phase sequence can be reversed by reversing the _________
a) Rotor direction
b) Field polarities
c) Armature terminal
d) Rotor direction and armature terminal
Explanation: In synchronous generator, the phase sequence is governed by the direction of rotation of the rotor and in a synchronous motor, the phase sequence governs the direction of rotation of the rotor
2. The slope of the V-I curve is 7°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.
a) .122 Ω
b) .360 Ω
c) .377 Ω
d) .578 Ω
Explanation: The slope of the V-I curve is resistance. The slope given is 7° so R=tan(7°)=.122 Ω. The slope of the I-V curve is reciprocal of resistance.
3. In an induction motor, when the number of stator slots is not equal to an integral number of rotor slots _________
a) There may be a discontinuity in torque slip characteristics
b) A high starting torque will be available
c) The machine performs better
d) The machine may fail to start
Explanation:When the number of stator slots is not an integral multiple of a number of rotor slots the machine will not fail to start. It does not cause the cogging phenomenon.
4. A 3-phase induction motor runs at almost 1500 rpm at no load and 900 rpm at full load when supplied with power from a 50 Hz, 3-phase supply. What is the corresponding speed of the rotor field with respect to the rotor?
a) 300 revolution per minute
b) 400 revolution per minute
c) 600 revolution per minute
d) 500 revolution per minute
Explanation: Supply frequency=50 Hz. No-load speed of motor = 1500 rpm. The full load speed of motor=900 rpm. Since the no-load speed of the motor is almost 1500 rpm, hence synchronous speed near to 1500 rpm. Speed of rotor field=1500 rpm. Speed of rotor field with respect to rotor = 1500-900 = 600 rpm
5. For a practical synchronous motor, the pull-out torque will occur when the torque angle is nearly equal to ________
a) 0°
b) 30°
c) 45°
d) 75°
Explanation:In a practical synchronous motor, the armature resistance cannot be neglected and hence the pull-out occurs at delta=beta which is the impedance angle and is practically 75°
6. Calculate the active power in an 8 Ω resistor with 8 A current flowing through it.
a) 512 W
b) 514 W
c) 512 W
d) 518 W
Explanation: The resistor is a linear element. It only absorbs real power and dissipates it in the form of heat. The voltage and current are in the same phase in case of the resistor so the angle between V & I is 90°. P = I2R = 8×8×8 = 512 W.
7. Calculate the active power in a .45 H inductor.
a) 0.11 W
b) 0.14 W
c) 0.15 W
d) 0 W
Explanation: The inductor is a linear element. It only absorbs reactive power and stores it in the form of oscillating energy. The voltage and current are 90° in phase in case of the inductor so the angle between V & I is 90°. P=VIcos90° = 0 W.
8. A 10-pole, 3-phase, 60 Hz induction motor is operating at a speed of 100 rpm. The frequency of the rotor current of the motor in Hz is __________
a) 52.4
b) 54.8
c) 51.66
d) 51.77
Explanation: Given a number of poles = 10. Supply frequency is 60 Hz. Rotor speed is 100 rpm. Ns = 120×f÷P=120×60÷10 = 720 rpm. S=Ns-Nr÷Ns = 720-100÷720=.86. F2=sf=.86×60=51.66 Hz.
9. Calculate the phase angle of the sinusoidal waveform z(t)=.99sin(4578πt+78π÷78).
a) π÷3
b) 2π
c) π÷7
d) π
Explanation: Sinusoidal waveform is generally expressed in the form of V=Vmsin(ωt+α) where Vm represents peak value, Ω represents angular frequency, α represents a phase difference.
10. Calculate the moment of inertia of the disc having a mass of 4 kg and diameter of 1458 cm.
a) 106.288 kgm2
b) 104.589 kgm2
c) 105.487 kgm2
d) 107.018 kgm2
Explanation: The moment of inertia of the disc can be calculated using the formula I=mr2×.5. The mass of the disc and diameter is given. I=(4)×.5×(7.29)2=106.288 kgm2. It depends upon the orientation of the rotational axis.