1.Armature reaction is magnetizing in nature due to a purely resistive load in the synchronous generator.
a) True
b) False
Explanation: Due to a purely resistive load, armature current is in quadrature with the field magneto-motive force. Armature magneto-motive force produced due to this current will be in quadrature with the field flux. It will try to increase the net magnetic field
2. The slope of the V-I curve is 4.9°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.
a) .03 Ω
b) .08 Ω
c) .04 Ω
d) .07 Ω
Explanation: The slope of the V-I curve is resistance. The slope given is 4.9° so R=tan(4.9°)=.08 Ω. The slope of the I-V curve is reciprocal of resistance.
3. Calculate the velocity of the satellite if the angular speed is 87 rad/s and radius is 7.4 m.
a) 643.8 m/s
b) 642.4 m/s
c) 641.9 m/s
d) 643.2 m/s
Explanation: The velocity of the satellite can be calculated using the relation V=Ω×r. The velocity is the vector product of angular speed and radius. V=Ω×r = 87×7.4 = 643.8 m/s.
4. A 3-phase induction motor runs at almost 70 rpm at no load and 50 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) 20 revolution per minute
b) 30 revolution per minute
c) 40 revolution per minute
d) 50 revolution per minute
Explanation: Supply frequency=50 Hz. No-load speed of motor = 70 rpm. The full load speed of motor=50 rpm. Since the no-load speed of the motor is almost 70 rpm, hence synchronous speed near to 70 rpm. Speed of rotor field=70 rpm. Speed of rotor field with respect to rotor=70-50= 20 rpm.
5. Calculate the active power in a 9.854 H inductor.
a) 4.98 W
b) 0 W
c) 8.59 W
d) 1 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.
6. Calculate the reactive power in a 45 Ω resistor with 1.78 A current flowing through it.
a) 28.8 VAR
b) 23.4 VAR
c) 25.82 VAR
d) 0 VAR
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 0°. Q=VIsin(0°)=0 VAR.
7. Calculate the value of the frequency if the inductive reactance is 72 Ω and the value of the inductor is 7 H.
a) 1.63 Hz
b) 1.54 Hz
c) 1.78 Hz
d) 1.32 Hz
Explanation: The frequency is defined as the number of oscillations per second. The frequency can be calculated using the relation XL = 2×3.14×f×L. F = XL÷2×3.14×L = 72÷2×3.14×7 = 1.63 Hz.
8. Calculate the active power in a 2 Ω resistor with 8 A current flowing through it.
a) 125 W
b) 128 W
c) 123 W
d) 126 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 0°. P=I2R=8×8×2=128 W.
9. In the chopper circuit, commutation times are _______
a) Current commutation is equal to voltage commutation
b) Current commutation is less as compared to that of voltage commutation
c) Current commutation is more as compared to that of voltage commutation
d) Both commutation techniques are not comparable
Explanation: In current commutation, commutation time=CVr÷Io. In voltage commutation, commutation time= CVs÷Io. Hence, the commutation time of the current commutation is less as compared to voltage commutation.
10. The generated e.m.f from 4-pole armature having 1 conductors driven at 1 rev/sec having flux per pole as 10 Wb, with wave winding is ___________
a) 30 V
b) 40 V
c) 70 V
d) 20 V
Explanation: The generated e.m.f can be calculated using the formula Eb = Φ×Z×N×P÷60×A, Φ represent flux per pole, Z represents the total number of conductors, P represents the number of poles, A represents the number of parallel paths, N represents speed in rpm. In wave winding number of parallel paths are 2. Eb = 10×4×1×60÷60×2 = 20 V.