Answer / santosh kawade

If slip rings are shorted it will run at 1500 rpm.

santosh kawade

Answer / suraj patel

In Indaction motor 120f/p wher consider P is number of pole
for indaction motor this pole cosider only stater coze we
give supply on stater. thus Ans is 1500rpm

Answer / nagaachi

hey fools

in SIM no:P IN ROTOR = no: poles in staor

what could be the speed i dont know

but i dont answer foolishness

Answer / mohit

If stator poles and rotor poles are not equal then resultant
torque will be zero and motor will not run.

Answer / a.sakthi

answer b) 1000 rpm is correct. because 120f/p =N.here N is
the speed of the motor. hence f and p is the input frequency
and no of poles on the stator or rotor respectively
(depending upon which one is the rotating portion of the
motor) so and so, in general rotor is rotating so the answer
b) is correct. if the stator is rotating, answer b) is correct.


thank you....

Answer / bhautik patel

its bcoz Ns=120f/p
where Ns is the stator sychronous speed
f= supply frequency
p= no of supply poles i.e. stator poles
substituting values f=50 hz , p=4 ,
we obtain Ns=1500 r.p.m

Answer / xxxxxxxxxxx

zero rpm

Answer / wong

why the input current is higher even at no-load.

Answer / angel

how do you know the answer because the number of turn
changed with change in the windings on the coil from the
poles if you only know the freguency it's the same aas not
being given any thing because even the value of the
standard voltage has changed so the answer can not be
determine reason being we don't know the flux windings.

Answer / sayem

1000

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