Slip of an induction motor lies between_______ - Three-Phase Systems in Power Applications

Electrical and Electronic Engineering (EEE) : Three-Phase Systems in Power Applications - Quiz(MCQ)

Slip of an induction motor lies between_______

0 to 5 %

0 to 10 %

0 to 400 RPM

0 to 4000 RPM

Correct Answer : Option (A) : 0 to 5 %

Explanation : Slip is the difference between the speed of rotating field and actual speed of rotor. The speed of the rotor must be less than the speed of the rotating field, else there would be no relative motion and the rotary motion would cease to exist.

The speed of the rotor should be such that the magnitude of the rotor current is enough to exert the necessary torque. The slip speed denotes the speed of rotor relative to that of the field. Though it is measured in revolutions per minute, more commonly it is denoted in percentage and ranges from 0 to 5%.

Recently Updated in Three-Phase Systems in Power Applications Questions

In a three phase, delta connection

Line current is equal to phase current

Line voltage and line current is zero

Line voltage is equal to phase voltage

None of the above

Correct Answer : Line voltage is equal to phase voltage

Explanation : A delta connected AC circuit is achieved by connecting the start end of a winding to the finish end of another winding such that all three windings form a mesh. Since each end of the windings forms the line connection, voltage across each winding is equal to the potential difference between the corresponding lines taken from that winding. Hence the phase voltage is equal to the line voltage.

For a three-phase delta connected load, fed from a star connected network, the power transferred to the load is_

3 KW

4.7 KW

5 KW

7KW

Correct Answer : 4.7 KW

Explaination :

For a balanced three phase, three wire system with input power of 10kW, at 0.9 power factor, the readings on both wattmeter are ________ respectively

7kW, 3kW

5000W, 5000W

6350W, 3650W

7600W, 1200W

Correct Answer : 6350W, 3650W

Explanation :

Let reading of one Wattmeter = W1

Reading of second Wattmeter = W2

Input Power, P = W1+W2 = √3V_LI_Lcosφ = 10000………………1

Power Factor, cosφ = 0.9

Phase angle, φ = 25.8 degrees

Therefore, W1 = V_LI_Lcos(30-φ) = 0.99V_LI_L = 6350W