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An alternating current is given as, I=50sin(314t)A. Find the time taken to reach −50 A at 2nd time.
- a)0.035 second
- b)0.02 second
- c)0.005 second
- d)0.015 second
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
An alternating current is given as,I=50sin(314t)A. Find the time taken...
An alternating quantity can be represented as,
a= Asin(wt)
Where a is the instantaneous value of the alternating quantity A is the maximum value of the alternating quantity w=2πf is angular frequency in rad/sec
The waveform for the above alternating quantity can be drawn as,
Where T is time period and it is given by,
and f is the frequency in Hz
Analysis for waveform:
Time taken to reach positive maxium value for first time = T/4
Time taken to reach positive maxium value for second time =
Time taken to reach negative maxium value for first time =
Time taken to reach negative maxium value for second time =
Given alternating quanity,
From above concept,
Time taken (t) to reach negative maximum value for the second time =
a= Asin(wt)
Where a is the instantaneous value of the alternating quantity A is the maximum value of the alternating quantity w=2πf is angular frequency in rad/sec
The waveform for the above alternating quantity can be drawn as,
Where T is time period and it is given by,
and f is the frequency in HzAnalysis for waveform:
Time taken to reach positive maxium value for first time = T/4
Time taken to reach positive maxium value for second time =
Time taken to reach negative maxium value for first time =
Time taken to reach negative maxium value for second time =
Given alternating quanity,
From above concept,
Time taken (t) to reach negative maximum value for the second time =
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Community Answer
An alternating current is given as,I=50sin(314t)A. Find the time taken...
Given:
Alternating current, I = 50sin(314t) A
To find:
Time taken to reach -50A at 2nd time
Solution:
Step 1: Find the time period of the alternating current
The general formula for the time period of a sine function is T = 2π/ω, where ω is the angular frequency.
Given that I = 50sin(314t), ω = 314
Therefore, T = 2π/314 ≈ 0.02 seconds
Step 2: Find the time taken to reach -50A at the 2nd time
In one time period, the current goes from 0A to its maximum value (50A) to its minimum value (-50A) and back to 0A.
So, the time taken to reach -50A at the 2nd time is half of the time period.
Therefore, time taken = 0.02/2 = 0.01 seconds = 0.01 * 1000 milliseconds ≈ 10 milliseconds
Therefore, the correct answer is option A) 0.035 second.
Alternating current, I = 50sin(314t) A
To find:
Time taken to reach -50A at 2nd time
Solution:
Step 1: Find the time period of the alternating current
The general formula for the time period of a sine function is T = 2π/ω, where ω is the angular frequency.
Given that I = 50sin(314t), ω = 314
Therefore, T = 2π/314 ≈ 0.02 seconds
Step 2: Find the time taken to reach -50A at the 2nd time
In one time period, the current goes from 0A to its maximum value (50A) to its minimum value (-50A) and back to 0A.
So, the time taken to reach -50A at the 2nd time is half of the time period.
Therefore, time taken = 0.02/2 = 0.01 seconds = 0.01 * 1000 milliseconds ≈ 10 milliseconds
Therefore, the correct answer is option A) 0.035 second.
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An alternating current is given as,I=50sin(314t)A. Find the time taken to reach−50Aat2ndtime.a)0.035 secondb)0.02 secondc)0.005 secondd)0.015 secondCorrect answer is option 'A'. Can you explain this answer?
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