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 2 of 4.
Vinothkumar said:
1 decade ago
Thanks sandeep.
Mustafa said:
1 decade ago
If this is a single phase generator , how are we considering the angle to be 120 degrees?
Ribka A said:
1 decade ago
In three phase generators we will consider the angle between three coils should be 120 degrees.
AMARENDRA DASH said:
1 decade ago
We know N = 120F/P.
SO F = N*P/120.
Here n = 400rps(revolution per second).
n = 400*60rpm(rps=Hz=60*rpm).
Thus,
F = 400*60*2/120.
F = 400Hz answer.
SO F = N*P/120.
Here n = 400rps(revolution per second).
n = 400*60rpm(rps=Hz=60*rpm).
Thus,
F = 400*60*2/120.
F = 400Hz answer.
Ahmed said:
1 decade ago
n= 120*f/p.
n : rapid per minute.
f: frequency.
p:no of poles.
n=400*60(rps) =24000(rpm).
f=n(rpm)*p/120.
f=24000*2/120.
f=400 HZ.
n : rapid per minute.
f: frequency.
p:no of poles.
n=400*60(rps) =24000(rpm).
f=n(rpm)*p/120.
f=24000*2/120.
f=400 HZ.
Thiru punk said:
1 decade ago
n(rps) = 400 rps.
p(pole) = 2.
f = (pn/2).
f = (2*400/2) => 400Hz.
p(pole) = 2.
f = (pn/2).
f = (2*400/2) => 400Hz.
Ch Nikhil Chakravarthy said:
1 decade ago
Since f=1/T.
Here f is frequency.
T is time in seconds.
Here f is frequency.
T is time in seconds.
Girishgowda said:
1 decade ago
p = 2.
n = 400.
rps its converted in to rpm its value is 6.66 rpm.
Then f = pn/120.
= 2*6.66/120.
= 0.111 hz.
n = 400.
rps its converted in to rpm its value is 6.66 rpm.
Then f = pn/120.
= 2*6.66/120.
= 0.111 hz.
NAvya said:
1 decade ago
Where 1sec=1/60min. But how you multiplied 60 in the speed.
Mrityunjay said:
1 decade ago
Please any one tell from where 2 is came?
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