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maw [93]
3 years ago
6

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 th

e 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
Engineering
2 answers:
WITCHER [35]3 years ago
4 0

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

sergey [27]3 years ago
3 0

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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8 0
2 years ago
A company buys a machine for $12,000, which it agrees to pay for in five equal annual payments, beginning one year after the dat
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Answer:

$7,778.35

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First, calculate the uniform payments A:

A = 12000(A/P, 4%, 5)

= 12000(0.2246) = 2695.2  (from the calculator)

Then take the last three payments as its own cash flow.

To calculate the new P:

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3 years ago
During January, at a location in Alaska winds at -20°C can be observed, However, several meters below ground the temperature rem
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Answer:

a) \eta_{th} = 10.910\,\%, b) Yes.

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a) The maximum thermal efficiency is given by the Carnot's Cycle, whose formula is:

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b) The claim of the inventor is possible since real efficiency is lower than maximum thermal efficiency.

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