1. Fan type of loads exhibits which type of load torque characteristics?
a) Constant torque characteristics
b) Linearly rising torque characteristics
c) Non-Linearly rising torque characteristics
d) Non-Linearly decreasing torque characteristics
Explanation: Torque produced by the fan is directly proportional to square of speed throughout the range of usable fan speeds. This type of loads exhibits non-linearly rising torque characteristics.
2. Type-A chopper is used for obtaining which type of mode?
a) Motoring mode
b) Regenerative braking mode
c) Reverse motoring mode
d) Reverse regenerative braking mode
Explanation: Only motoring mode is available in case of step-down chopper (Type-A chopper). Value of output voltage (Vo) is less than the input voltage (Vin) in case of step-down chopper.
3. Calculate the value of angular acceleration of motor using the given data: J = 20 kg-m2, load torque = 20 N-m, motor torque = 60 N-m.
a) 5 rad/s2
b) 2 rad/s2
c) 3 rad/s2
d) 4 rad/s2
Explanation: Using the dynamic equation of motor J*(angular acceleration) = Motor torque – Load torque: 20*(angular acceleration) = 60-20=40, angular acceleration=2 rad/s2.
4. 230V, 10A, 1500rpm DC separately excited motor having resistance of .2 ohm excited from external dc voltage source of 50V. Calculate the torque developed by the motor on full load.
a) 13.89 N-m
b) 14.52 N-m
c) 13.37 N-m
d) 14.42 N-m
Explanation: Back emf developed in the motor during full load can be calculated using equation Eb = Vt- I*Ra = 228 V and machine constant Km = Eb / Wm which is equal to 1.452. Torque can be calculated by using the relation T = Km* I = 1.452*10 = 14.52 N-m.
5. Boost converter is used to _________
a) Step down the voltage
b) Step up the voltage
c) Equalize the voltage
d) Step up and step down the voltage
Explanation:Output voltage of boost converter is Vo = Vin / 1 – D. Value of duty cycle is less than 1 which makes the Vo > Vin as denominator value decreases and becomes less than 1. Boost converter is used to step up voltage
6. Reverse motoring mode is available in fourth quadrant.
a) True
b) False
Explanation: In reverse motoring mode motor rotates in opposite to original direction as direction of motor torque changes which makes the motor to run in opposite direction and load torque tries to oppose the motion of motor.
7. Calculate the power developed by motor using the given data: Eb = 20V and I = 10 A. (Assume rotational losses are neglected)
a) 400 W
b) 200 W
c) 300 W
d) 500 W
Explanation:Power developed by motor can be calculated using the formula P = Eb*I = 20*10 = 200 W. If rotational losses are neglected, power developed becomes equal to the shaft power of motor
8. Which one is an example of variable loss?
a) Windage loss
b) Hysteresis loss
c) Armature copper loss
d) Friction loss
Explanation: Armature copper losses are variable losses as they depend on armature current which further depends on load. As load changes armature current changes hence armature copper losses (I2 * r) also changes
9. Duty cycle (D) is _______
a) Ton÷Toff
b) Ton÷(Ton+ Toff)
c) Ton÷2×(Ton+ Toff)
d) Ton÷2×Toff
Explanation:Duty cycle (D) is defined as the ratio of time for which system is active to the total time period. It is also known as the power cycle. It has no unit
10. A 220 V, 1000 rpm, 60 A separately-excited dc motor has an armature resistance of .5 ω. It is fed from single-phase full converter with an ac source voltage of 230 V, 50Hz. Assuming continuous conduction, the firing angle for rated motor torque at (-400) rpm is _________
a) 122.4°
b) 117.6°
c) 130.1°
d) 102.8°
Explanation: During rated operating conditions of the motor, Eb = Vt-Ia×Ra = 220-60×.5=190 V. As Eb=Kmwm so Km=190×60÷(2×3.14×1000) = 1.8152 V-s/rad. Back emf at (-400 rpm) is Kmwm = 1.8152×(2×3.14×(-400))÷60 = -76 V. Now Vt = -76+60×.5 = -46 V. Average voltage of single-phase full converter is 2×Vm×cos(α)÷3.14. The output of the converter is connected to the input terminal of the motor so α = cos-1(-46×3.14÷2×230×1.414) = 102.8o.