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A DC generator turns at 2000 rpm and has an output of 200 V. The armature constant is 0.5 V-min/Wb, and the field constant of the machine is 0.02 H. What is most nearly the field current produced by the generator? a) 4.0 A b) 10 A c) 1000 A d) 4000 A

Respuesta :

Answer:

b. 10A

Explanation:

Using the formula, E= k × r×I

200= 0.5 ×2000×0.02×I

200=20×I

Dividing with 20

I = 200/20= 10A

batolisis

Answer:

10 A ( B )

Explanation:

output of the DC generator = 200 V

revolution = 2000 rpm

Armature constant = 0.5 V-min/Wb

field constant = 0.02 H

i = field current

output of a DC generator = ( rpm) * (armature constant) * (field constant) * i

200 = 2000 * 0.5 * 0.02 * i

200 = 20 i

therefore i = 200 / 20 =  10 A

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