Electrical Engineering - Alternating Current and Voltage - Discussion
Discussion Forum : Alternating Current and Voltage - General Questions (Q.No. 2)
2.
The conductive loop on the rotor of a simple two-pole, single-phase generator rotates at a rate of 400 rps. The frequency of the induced output voltage is
Discussion:
38 comments Page 4 of 4.
Mau chau said:
1 decade ago
Thanks raj.
Sandeep said:
1 decade ago
P=2
N= 400 r/s (rps )= 400 r/(m/60)=400*60 r/m
=400*60 rpm
we know f = N P /120= 400*60*2/120=400 Hz.
N= 400 r/s (rps )= 400 r/(m/60)=400*60 r/m
=400*60 rpm
we know f = N P /120= 400*60*2/120=400 Hz.
Yanusha A said:
1 decade ago
N=120F/P
N=400rps=400*60rpm
p=2
F=NP/120
=400*60*2/120
=400HZ
N=400rps=400*60rpm
p=2
F=NP/120
=400*60*2/120
=400HZ
Vipul Parmar said:
1 decade ago
F = PN/120
= 400*60*2/120
= 400Hz Because 400*60 rps= 400 rpm.
= 400*60*2/120
= 400Hz Because 400*60 rps= 400 rpm.
Talha said:
1 decade ago
Thanks for that answer.
Raj said:
1 decade ago
Because we have to change rps to rpm.
Mahendra said:
1 decade ago
From where did the 60 came please tell me vineeta verma.
Vineeta verma said:
1 decade ago
f = pn/120
f = 400*60*2/120
f = 400 HZ
f = 400*60*2/120
f = 400 HZ
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