11. Calculate the heat absorbed by a 20 Ω resistor when 40 V is applied across it.
A. 120 W
B. 100 W
C. 80 W
D. 110 W
Answer: C
The resistor always absorbs active power and dissipates it in the form of heat. Power absorbed by a 20 Ω resistor is
40×40÷20=80 W.
12. Full form of VFD is _____________
A. Variable frequency drive
B. Variable force drive
C. Value frequency drive
D. Variable force diverter
Answer: A
VFD stands for variable frequency drive. It is a controller for AC or DC motors to achieve the desired speed. It can perform the four-quadrant operation.
13. What is Load Equalisation?
A. Process of smoothing the fluctuating load
B. Process of removing the load
C. Process of changing the voltage level
D. Process of changing the speed
Answer: A
Load Equalisation is the process of smoothing the fluctuating load. Load fluctuation causes a large voltage dip in the system and can damage the device.
14. Which device is used to store the energy during load equalization?
A. Flywheel
B. Capacitor
C. Inductor
D. Resistor
Answer: A
The flywheel is used to store the energy during load equalization. It acts as a reservoir. It stores the energy during light load and utilizes the stored energy during peak load conditions.
15. Which test is used to determine the conditions where electric motors are being used?
A. Thermography
B. Geography
C. Seismography
D. Anthropology
Answer: A
Thermography is used to detect heat patterns. It uses an Infrared rays camera. This test is used to determine the conditions where electric motors are being used.
16. What is the range of size of the motor (kW) for efficiency nearly equals 90%?
A. 15-150
B. 0-2
C. 3-15
D. 150-250
Answer: A
As the size of the motor increases its speed decreases. The range of size of the motor (kW) for efficiency nearly equals 90 % is 15-150. The efficiency increases as the size increases.
17. Calculate the heat loss in the electric motor for time intervals of 0-5 sec using the following data: Ia=2 A, Ra=.8Ω.
A. 16 J-sec
B. 14 J-sec
C. 12 J-sec
D. 10 J-sec
Answer: A
The heat loss can be calculated using the Joule law of the heating effect. The value of the heat loss is
Ia2×Ra×t=4×4=16 J-sec.
18. 1 Btu/h is equal to ___________
A. .641
B. .334
C. .293
D. .417
Answer: C
One British thermal unit is equal to .293 Watt. 1 Btu/h is the amount of energy required to increase the temperature of one pound of water by 1℉.
19. Overloading can be prevented using _____________
A. Over-current protection
B. Speed protection
C. Over frequency protection
D. Oversize protection
Answer: A
Overloading can be prevented using over-current protection. It will detect the over-current and interrupt the supply. IDMT overcurrent relay can be used for protection purposes.
20. Full form of IDMT is _________
A. Inverse definite minimum time
B. Inverter definite minimum time
C. Inverter definite maximum time
D. Insert definite minimum time
Answer: A
IDMT stands for Inverse definite minimum time over-current relay. It has both the characteristics of inverse over-current and definite time over-current relay.
