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liberstina [14]
3 years ago
6

Maximum voltage produced in an AC generator completing 60 cycles in 30 sec is 250V. (a) What is period of armature? (b) How many

cycles are completed in T/2 sec? (c) What is the maximum emf produced when the armature completes 180 0 rotation?
Physics
1 answer:
Vlada [557]3 years ago
8 0

Answer:

a. 2 Hz b. 0.5 cycles c . 0 V

Explanation:

a. What is period of armature?

Since it takes the armature 30 seconds to complete 60 cycles, and frequency f = number of cycles/ time = 60 cycles/ 30 s = 2 cycles/ s = 2 Hz

b. How many cycles are completed in T/2 sec?

The period, T = 1/f = 1/2 Hz = 0.5 s.

So, it takes 0.5 s to complete 1 cycles. At t = T/2 = 0.5/2 = 0.25 s,

Since it takes 0.5 s to complete 1 cycle, then the number of cycles it completes in 0.25 s is 0.25/0.5 = 0.5 cycles.

c. What is the maximum emf produced when the armature completes 180° rotation?

Since the emf E = E₀sinθ and when θ = 180°, sinθ = sin180° = 0

E = E₀ × 0 = 0

E = 0

So, at 180° rotation, the maximum emf produced is 0 V.

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Answer:

Output power = 500 KW

Explanation:

Given the following data;

Efficiency = 20%

Input power = 2500 KW

To find the output power;

Efficiency = \frac {Out-put \; power}{In-put \; power} * 100

Substituting into the equation, we have;

LET Output power = OP

20 = \frac {OP}{2500} * 100

Cross-multiplying, we have;

20 * 2500 = OP * 100

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