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
40 Hz
100 Hz
400 Hz
indeterminable
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
38 comments Page 2 of 4.

Feroz khan said:   7 years ago
1rpm=60rps.
so 1rps 1/60 rpm and,
400rps=400 * 1/60 rps.

I think the answer is wrong @Kiran.
(1)

Trupti said:   5 years ago
rps=400*60=24000=Ns.

P = 2
Ns = 120f/p.
120*f/Ns*p.
f = Ns*p/120,
f = 24000*2/120,
f = 400.
(3)

D.Jyothi said:   7 years ago
As per the definition, frequency means that the number of cycles completed by one rotation.

Mustafa said:   1 decade ago
If this is a single phase generator , how are we considering the angle to be 120 degrees?

Kiran said:   8 years ago
rps=400*60=24000=Ns.

P=2
Ns=120f/p
120*f/Ns*p
f=Ns*p/120
f=24000*2/120
f=400.

Mehr said:   6 years ago
For rpm; Simply convert this to the minute, we multiply 60 with rps 400*60.

Thiru punk said:   1 decade ago
n(rps) = 400 rps.
p(pole) = 2.

f = (pn/2).
f = (2*400/2) => 400Hz.

Fatima said:   9 years ago
@Hassan.

Formula contains Ns syncronous speed ,Ns = (( 120 * f ) /p).

Vipul Parmar said:   1 decade ago
F = PN/120
= 400*60*2/120
= 400Hz Because 400*60 rps= 400 rpm.

Yanusha A said:   1 decade ago
N=120F/P
N=400rps=400*60rpm
p=2
F=NP/120
=400*60*2/120
=400HZ


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