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.

Surya said:   8 years ago
N= 120f/p.
F= 400*60*2/120.
F= 400.

Waheed ullah khattak said:   9 years ago
If we change RPS to RPM then we should have to divide by 60.

Bhanu said:   9 years ago
Great job @Jansi.

Fatima said:   9 years ago
@Hassan.

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

Hassan said:   9 years ago
How did you guys get that 120?

Rakshith said:   9 years ago
As we all know,

Ns = ( (120 * f) /p); Here Ns is in rpm. Therefore,

Ns = Ns/60 rps.

So,

Ns = ((2 * f)/p),
And f = ((p * ns)/2),

f = 2 * 400/2,

f = 400Hz.

Jansi said:   10 years ago
Here given data is:

No. of poles = 2;

Speed (N) = 400 rps;

Usually speed is calculated in rpm. So here we convert 400 rps (revolution per second) in terms of rpm (revolutions per minute).

As 1 min = 60 sec; we multiply 400x60 rpm;

As we know N = 120f/p.

f = np/120;

f = 400x60x2/120.

f = 400 hz.

So hope no one has any doubts regarding this explanation.
(1)

Kalai sasi said:   10 years ago
Its a number of poles.

Mrityunjay said:   1 decade ago
Please any one tell from where 2 is came?

NAvya said:   1 decade ago
Where 1sec=1/60min. But how you multiplied 60 in the speed.


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