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Sagot :
Certainly! Let's tackle each part of the question involving the given AC current expression: [tex]\( i(t) = 120 \sin(100\pi t + 0.36) \)[/tex] in a detailed, step-by-step manner.
### Part (a): Peak Value, Periodic Time, Frequency, and Phase Angle Relative to [tex]\(120 \sin 100\pi t\)[/tex]
1. Peak Value (Amplitude):
The peak value of the current is the maximum value of the sine function.
[tex]\[ \text{Peak Value} = 120 \text{ amperes} \][/tex]
2. Periodic Time:
The periodic time [tex]\( T \)[/tex] is the time it takes for the function to complete one full cycle. It is given by the inverse of the frequency.
[tex]\[ T = \frac{1}{f} \][/tex]
Given the frequency [tex]\( f = 100 \text{ Hz} \)[/tex],
[tex]\[ T = \frac{1}{100} = 0.01 \text{ seconds} \][/tex]
3. Frequency:
The frequency is given directly in the problem.
[tex]\[ f = 100 \text{ Hz} \][/tex]
4. Phase Angle Relative to } 120 \sin 100\pi t \):
To find the phase angle relative to the standard sine wave [tex]\( 120 \sin 100\pi t \)[/tex], we look at the phase shift in our given expression.
[tex]\[ \text{Phase Angle} = 0.36 \text{ radians} \][/tex]
### Part (b): Value of the Current when [tex]\( t = 0 \)[/tex]
Using the given equation [tex]\( i(t) = 120 \sin(100\pi t + 0.36) \)[/tex],
[tex]\[ i(0) = 120 \sin(100\pi \cdot 0 + 0.36) = 120 \sin(0.36) \][/tex]
Evaluating the sine function,
[tex]\[ i(0) \approx 42.27 \text{ amperes} \][/tex]
### Part (c): Value of the Current when [tex]\( t = 8 \)[/tex] ms
First, convert 8 milliseconds to seconds:
[tex]\[ t = 8 \text{ ms} = 0.008 \text{ s} \][/tex]
Now, use the equation:
[tex]\[ i(0.008) = 120 \sin(100\pi \cdot 0.008 + 0.36) \][/tex]
Evaluating the sine function,
[tex]\[ i(0.008) \approx 31.81 \text{ amperes} \][/tex]
### Part (d): Time when the Current First Reaches 60A
We need to solve for [tex]\( t \)[/tex] when the current [tex]\( i(t) = 60 \)[/tex] amperes.
Given:
[tex]\[ 60 = 120 \sin(100\pi t + 0.36) \][/tex]
Divide both sides by 120:
[tex]\[ 0.5 = \sin(100\pi t + 0.36) \][/tex]
Take the inverse sine:
[tex]\[ 100\pi t + 0.36 = \arcsin(0.5) \][/tex]
[tex]\[ 100\pi t + 0.36 = \frac{\pi}{6} \][/tex]
Now solve for [tex]\( t \)[/tex]:
[tex]\[ 100\pi t = \frac{\pi}{6} - 0.36 \][/tex]
[tex]\[ t = \frac{\frac{\pi}{6} - 0.36}{100\pi} \][/tex]
Evaluating the above expression:
[tex]\[ t \approx 0.00052075 \text{ seconds} \][/tex]
### Part (e): Time when the Current is First a Maximum
The current reaches its maximum when the sine term is equal to 1.
[tex]\[ \sin(100\pi t + 0.36) = 1 \][/tex]
Therefore:
[tex]\[ 100\pi t + 0.36 = \frac{\pi}{2} \][/tex]
Solving for [tex]\( t \)[/tex]:
[tex]\[ 100\pi t = \frac{\pi}{2} - 0.36 \][/tex]
[tex]\[ t = \frac{\frac{\pi}{2} - 0.36}{100\pi} \][/tex]
Evaluating the expression:
[tex]\[ t \approx 0.003854 \text{ seconds} \][/tex]
### Summary of Results:
(a)
- Peak Value: 120 amperes
- Periodic Time: 0.01 seconds
- Frequency: 100 Hz
- Phase Angle: 0.36 radians
(b) Current at [tex]\( t = 0 \)[/tex]: [tex]\( \approx 42.27 \)[/tex] amperes
(c) Current at [tex]\( t = 8 \)[/tex] ms: [tex]\( \approx 31.81 \)[/tex] amperes
(d) Time when current first reaches 60A: [tex]\( \approx 0.00052075 \)[/tex] seconds
(e) Time when current is first a maximum: [tex]\( \approx 0.003854 \)[/tex] seconds
### Part (a): Peak Value, Periodic Time, Frequency, and Phase Angle Relative to [tex]\(120 \sin 100\pi t\)[/tex]
1. Peak Value (Amplitude):
The peak value of the current is the maximum value of the sine function.
[tex]\[ \text{Peak Value} = 120 \text{ amperes} \][/tex]
2. Periodic Time:
The periodic time [tex]\( T \)[/tex] is the time it takes for the function to complete one full cycle. It is given by the inverse of the frequency.
[tex]\[ T = \frac{1}{f} \][/tex]
Given the frequency [tex]\( f = 100 \text{ Hz} \)[/tex],
[tex]\[ T = \frac{1}{100} = 0.01 \text{ seconds} \][/tex]
3. Frequency:
The frequency is given directly in the problem.
[tex]\[ f = 100 \text{ Hz} \][/tex]
4. Phase Angle Relative to } 120 \sin 100\pi t \):
To find the phase angle relative to the standard sine wave [tex]\( 120 \sin 100\pi t \)[/tex], we look at the phase shift in our given expression.
[tex]\[ \text{Phase Angle} = 0.36 \text{ radians} \][/tex]
### Part (b): Value of the Current when [tex]\( t = 0 \)[/tex]
Using the given equation [tex]\( i(t) = 120 \sin(100\pi t + 0.36) \)[/tex],
[tex]\[ i(0) = 120 \sin(100\pi \cdot 0 + 0.36) = 120 \sin(0.36) \][/tex]
Evaluating the sine function,
[tex]\[ i(0) \approx 42.27 \text{ amperes} \][/tex]
### Part (c): Value of the Current when [tex]\( t = 8 \)[/tex] ms
First, convert 8 milliseconds to seconds:
[tex]\[ t = 8 \text{ ms} = 0.008 \text{ s} \][/tex]
Now, use the equation:
[tex]\[ i(0.008) = 120 \sin(100\pi \cdot 0.008 + 0.36) \][/tex]
Evaluating the sine function,
[tex]\[ i(0.008) \approx 31.81 \text{ amperes} \][/tex]
### Part (d): Time when the Current First Reaches 60A
We need to solve for [tex]\( t \)[/tex] when the current [tex]\( i(t) = 60 \)[/tex] amperes.
Given:
[tex]\[ 60 = 120 \sin(100\pi t + 0.36) \][/tex]
Divide both sides by 120:
[tex]\[ 0.5 = \sin(100\pi t + 0.36) \][/tex]
Take the inverse sine:
[tex]\[ 100\pi t + 0.36 = \arcsin(0.5) \][/tex]
[tex]\[ 100\pi t + 0.36 = \frac{\pi}{6} \][/tex]
Now solve for [tex]\( t \)[/tex]:
[tex]\[ 100\pi t = \frac{\pi}{6} - 0.36 \][/tex]
[tex]\[ t = \frac{\frac{\pi}{6} - 0.36}{100\pi} \][/tex]
Evaluating the above expression:
[tex]\[ t \approx 0.00052075 \text{ seconds} \][/tex]
### Part (e): Time when the Current is First a Maximum
The current reaches its maximum when the sine term is equal to 1.
[tex]\[ \sin(100\pi t + 0.36) = 1 \][/tex]
Therefore:
[tex]\[ 100\pi t + 0.36 = \frac{\pi}{2} \][/tex]
Solving for [tex]\( t \)[/tex]:
[tex]\[ 100\pi t = \frac{\pi}{2} - 0.36 \][/tex]
[tex]\[ t = \frac{\frac{\pi}{2} - 0.36}{100\pi} \][/tex]
Evaluating the expression:
[tex]\[ t \approx 0.003854 \text{ seconds} \][/tex]
### Summary of Results:
(a)
- Peak Value: 120 amperes
- Periodic Time: 0.01 seconds
- Frequency: 100 Hz
- Phase Angle: 0.36 radians
(b) Current at [tex]\( t = 0 \)[/tex]: [tex]\( \approx 42.27 \)[/tex] amperes
(c) Current at [tex]\( t = 8 \)[/tex] ms: [tex]\( \approx 31.81 \)[/tex] amperes
(d) Time when current first reaches 60A: [tex]\( \approx 0.00052075 \)[/tex] seconds
(e) Time when current is first a maximum: [tex]\( \approx 0.003854 \)[/tex] seconds
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