1. Auto transformer rating is chosen so as to have best performance when the analogous two winding transformer has ___________
a) two voltage levels fairly close to each other
b) very high transformation ratio
c) any voltage levels
d) step down purpose
Explanation: The auto transformer is best suited with the voltage levels of same rating for a two winding transformer.
2. In the lab, two students X and Y conducted the experiments to verify the name plate ratings of the transformer.But student X conducted the same at higher frequency. The kVA reading of student X will be ___________
a) higher than Y’s reading
b) lower than Y’s reading
c) same as Y’s reading
d) none of the mentioned
Explanation: As rating is directly proportional to the frequency of operation.
3. Synchronous-induction coupled motors are required for ____________
a) high starting torque applications
b) high pull out torque applications
c) high running torques applications
d) none of the mentioned
Explanation: Induction motor provides high starting torque with proper values of slip ring terminal resistance. Coupled along with the synchronous motor it acts as an independent set and be employed in high torque application.
4. Which of the following can be used for braking purposes in electric trains?
a) Induction generator
b) Induction motor
c) Dc series motor
d) Dc differentially compounded generator
Explanation: Induction generator runs at a speed greater than the synchronous speed. To apply in braking purposes the induction generator can be run in the opposite direction to that of the operating machine direction and stopped.
5. A synchronous motor absorbing 60 kW is connected in parallel with an continuous load of 240 kW operating at 0.8 power factor lag.What is the kVAR supplied by the synchronous motor to obtain a total power factor of 0.9 lagging?
a) 35, leading
b) 35, lagging
c) 145, leading
d) None of the mentioned
Explanation: Total power = 240 + 60 = 300 kW
cos(phi) = 0.90. Total reactive power = P*tan(phi) = 300*tan(25.8) = 180 kVAR Load kVAR = 240*tan(36.87) = 180 kVAR
kVAR supplied by the motor = 145-180 = -35 or k3 kVAR leading.
6. The most widely used application of ac series motor is ____________
a) electric traction upto 1600 kW
b) paper mills
c) welding
d) none of the mentioned
Explanation: Ac series motor is dc series motor connected to ac supply having an unidirectional torque, which is not very high in magnitude.It is widely applicable for small torque requirements like portable drilling machines, mixers etc.
7. Universal motor is used in vacuum cleaners, table fans and portable drilling machine.
a) True
b) False
Explanation: The universal motor is dc series motor with ac supply with smaller torque. So it can be used for lower torque applications
8. _______ generator is used in arc welding purposes.
a) Differential compound dc
b) Dc series
c) Cumulative compounded dc
d) Shunt
Explanation: The external characteristics of the differentially compound generator have minimum voltage for the high current voltages. This is best harnessed feature for a high current requirement by the welding application.
9. The dc series generator is employed as ____________
a) a booster to maintain constant voltage at the end of the feeder
b) for supplying traction load
c) for battery bank charging
d) none of the mentioned
Explanation: Referring to the external characteristics of dc series generator shows that the terminal end voltage increases as the loading increases from no load to full load. In feeders due to variable requirements the voltage drops while transferring when loading is raised.
10. If it asked by the customer to provide tappings on the transformer, the most efficient and economical way to provide it at ____________
a) middle of hv winding
b) phase end of hv winding
c) neutral end of hv winding
d) middle of lv winding
Explanation: Tappings are provided to tap different levels of voltages at the high voltage end of the transformer. Providing it as low voltage side has no use and it is not economical.