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 1 of 4.

Jek Culabres said:   4 years ago
Frequency = 1 cycles/second.
1 cycles/second = 1 Hertz.

Cycles is the same revolution or rotation.
(1)

Aurangzaib said:   5 years ago
Actually, rps is rev/sec so to convert it to rpm =rev/sec/60 that's why we multiply with 60 directly.

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)

Shoon said:   6 years ago
1min = 60sec.

If 400rps then 400/60 rpm.

As from small quantities to bigger quantities we do division, from bigger quantities to small we do multiplication.

Kindly explain it.

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

Asim said:   7 years ago
Good explanation @Rakshith.

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

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)

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.

Sunnyleon said:   8 years ago
Thank you @Sandeep.


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