11. 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
Answer: C
Torque produced by the fan is directly proportional to the square of speed throughout the range of usable fan speeds. This type of load exhibits non-linearly rising torque characteristics.
12. 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
Answer: A
Only motoring mode is available in the case of a step-down chopper (Type-A chopper).
The value of output voltage (Vo) is less than the input voltage (Vin) in the case of the step-down chopper.
13. Calculate the value of the angular acceleration of the 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
Answer: B
Using the dynamic equation of motor J×(angular acceleration)
= Motor torque – Load torque: 20×(angular acceleration)
= 60-20=40
angular acceleration=2 rad/s2.
14. 230V, 10A, 1500rpm DC separately excited motor having a resistance of .2 ohm excited from an 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
Answer: B
Back emf developed in the motor during full load can be calculated using the equation
Eb = Vt- I×Ra = 228 V
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.
15. 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
Answer: B
The output voltage of the boost converter is Vo = Vin / 1 – D.
The value of the duty cycle is less than 1 which makes the Vo > Vin as the denominator value decrease and become less than 1.
The boost converter is used to step up the voltage.
16. Reverse motoring mode is available in the fourth quadrant.
A. True
B. False
Answer: A
In reverse motoring, the mode motor rotates in opposite to the original direction as the direction of motor torque changes which makes the motor run in the opposite direction and load torque tries to oppose the motion of the motor.
17. Calculate the power developed by the 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
Answer: B
Power developed by the motor can be calculated using the formula
P = Eb × I = 20 × 10 = 200 W.
If rotational losses are neglected, the power developed becomes equal to the shaft power of the motor.
18. Which one is an example of variable loss?
A. Windage loss
B. Hysteresis loss
C. Armature copper loss
D. Friction loss
Answer: C
Armature copper losses are variable losses as they depend on armature current which further depends on the load. As load changes armature current changes hence armature copper losses (I2 × r) also change.
19. What is the empirical formula for the tractive force required to overcome curve resistance? (W-the weight of the body, R – radius of curvature)
A. 710×W÷R
B. 700×W÷R
C. 720×W÷R
D. 750×W÷R
Answer: B
Fc= 700×W÷R is the tractive force required to overcome curve resistance where W is the weight of the body in Kg and R is the radius of curvature in meters.
20. Force resisting the upward motion of a body on an inclined plane is given by (alpha – the angle of inclination, W- the weight of the body).
A. F = W×sin(alphA.
B. F = W×cosec(alphA.
C. F = W×sec(alphA.
D. F = W×cos(alphA.
Answer: A
When a body is moving upward on an inclined plane its weight can be resolved into two perpendicular components that are
W×sin(αA)
W×cos(αA)
W×cos(αA). is the component that is opposite to the normal of the inclined plane
W×sin(αA) is the component that opposes the upward motion of the body.
