SSC JE Electrical Engineering 2010 Paper

Question : 23 of 50

Marks: +1, -0

For a 3-phase, 4-pole, 50 Hz synchronous motor the frequency, no. of poles, and the load torque are all halved. The motor speed will be

Solution:

N_1={120F_1}/P_1

F_1 =

50Hz 1

P_1 = 4

F_1 =

Initial frequency = 50Hz

P_1 =

Number of initial pole = 4
and

F_2=F_1/2

P_2 =P_1/2

given in question

N_2={120F_2}/P_2

={120×F_1}/{{P_2×2}/2}={120F_1}/P_1

={120×50}/4=

1500RPM

Go to Question:

Prev Question

Next Question