Question

6 pole lap wound



d.C generator has 600 conductor on its armature the flux per pole is 0.02 wb.The speed at which generator must be run to generate 300v is

Answer 1

Answer:

by using EMF equation of DC generator,1500 rpm is the right answer

Answer 2

Answer:

The speed of the generator, N, calculated is $1500rpm$.

Explanation:

Given data,

The number of poles of lap wound generator, P = $6$

The number of conductors in the generator, Z = $600$

The flux pole, Ф = $0.02wb$

The emf generated by the generator, E = $300V$

The speed of the generator, N =?

As we know, for a lap wound generator;

• The number of parallel paths ( A ) =  number of poles ( P ) = $6$

From the formula given below, we can find out the speed of the generator:

• E = $\frac{\phi ZNP}{60A}$

Thus,

• N = $\frac{60 A E}{\phi ZP}$
• N = $\frac{60 E}{\phi Z}$    ( A = P )
• N = $\frac{60\times 300}{0.02\times600 }$
• N = $1500rpm$

Hence, the speed of the generator, N = $1500rpm$.