For an 8-pole induction motor supplied with power by a 6 pole Alternator at 1200 revolutions per minute, the value of motor speed at slip of 3% is â€“ - Three-Phase Systems in Power Applications

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

For an 8-pole induction motor supplied with power by a 6 pole Alternator at 1200 revolutions per minute, the value of motor speed at slip of 3% is â€“

400 RPM

800 RPM

873 RPM

900 RPM

Correct Answer : Option (C) : 873 RPM

Explanation :

Speed of Alternator, N = 1200 RPM

Number of Alternator poles, P = 6

Supply frequency, f = PN/120 = 60 Hz

Hence, for number of 8 poles in Induction motor, the synchronous speed or speed of rotating field is: N_s = 120f/P = 900 RPM

Let N_ris actual speed of motor.

Thus, Percentage Slip, S % = ((N_s – N_r)/N_s) x 100

Substituting the values, we get: N_r = N_s (1-S/100) = 873 RPM.

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