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Electronics and Electrical Engineering interview Question and Answers for fresh engineers (Part 2)

Electronics and Electrical Engineering interview Question and Answers for fresh engineers (Part 2)

In this topic we will find some question about transmission voltage, regulator difference, motor type of fans, unit of electrical energy

Question 6: What are the advantages of high transmission voltage? What are the limitations of the high transmission voltage?

Answer 6:  Advantage of high transmission voltage:
1. It reduces the volume of conductor material.
                   2. It increases the transmission efficiency.
                   3. It decreases the line drop percentage.

       Limitations of high transmission voltage:

1.The insulation cost of the conductor increased.
2.The other terminal operators, switchgear and transformers cost is increasing.

Question 7:    Which is better between electrical rheostat regulator and electronic regulator for fans? Which is the basic component difference for them?

Answer 7: The electronic regulator is better then electrical rheostat regulator. In electronic regulator the power loss is very low. The electronic regulator only transfers the necessary power for the necessary speed of the fans. But in electrical rheostat regulator the power loss is same for every speed, so no power is saved.

The electronic regulator use TRIAC for speed control by varying the firing angle speed.
 The electrical rheostat regulator use control resistance, which is increased or decreased by the variation of speed.

Question 8: In our normal household fans (such as exhaust fan, pedestal fan, ceiling fan, table fan etc) which types of motor is used? What is the rotor type and starting type normally?

Answer 8:  In our normal household fans single phase induction motor is normally used.
The rotors are mostly squirrel cage rotor. The fans are normally capacitor start and run.

Question 9:  Which is the unit of electrical energy?

Answer 9: Answer: the unit of electrical energy is joule or watt-second  

Electrical energy in joule or watt-second  
= voltage in volts* current in amperes* time in seconds.

Joule or watt-second is very small for the measuring of electrical energy in the practical field purpose.

1 watt-hour = 1watt * 1 hour
                   = 1watt * 3600 seconds
                   = 3600 watt-sec
1 kilowatt-hour = 1kW * 1hr
                =1000watt* 3600 sec


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