In a squirrel cage induction motor, the rotor slots are_______ - Three-Phase Systems in Power Applications

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

In a squirrel cage induction motor, the rotor slots are_______

Parallel to the rotor shaft

Skewed at a certain angle to the rotor shaft

Placed in line with the rotor shaft

None of the above

Correct Answer : Option (B) : Skewed at a certain angle to the rotor shaft

Explanation : A squirrel cage rotor consists of a laminated steel cylinder with slots in which either Aluminium conductors are die-cast of copper bars are used. These bars are shorted at both ends by heavy end rings of same materials. To reduce the magnetic noise and hence produce a more uniform torque and to prevent possible magnetic locking of the rotor with the stator, the rotor slots are skewed at a certain angle to the rotor shaft.

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