In an AC circuit `I=100sin 200pit`. The time required for the current to achieve its peak value of will be

Options

A. `1/100 sec`
B. `1/200 sec`
C. `1/300 sec`
D. `1/400 sec`

Explanation

To solve the problem, we need to determine the time required for the current in the AC circuit, given by the equation \( I = 100 \sin(200 \pi t) \), to reach its peak value.

### Step-by-Step Solution:

1. **Identify the Maximum Current**:
The maximum current, \( I_{\text{max}} \), occurs when the sine function equals 1. Therefore, we need to find when:
\[ \sin(200 \pi t) = 1 \]

2. **Determine the Condition for Maximum**:
The sine function reaches its maximum value of 1 at:
\[ 200 \pi t = \frac{\pi}{2} + n \cdot 2\pi \quad (n \in \mathbb{Z}) \]
This is because the sine function is periodic with a period of \( 2\pi \).

3. **Solve for Time \( t \)**:
To find \( t \), we can rearrange the equation:
\[ t = \frac{\frac{\pi}{2} + n \cdot 2\pi}{200 \pi} \]
Simplifying this gives:
\[ t = \frac{1}{400} + \frac{n}{100} \]

4. **Find the First Occurrence of Maximum**:
For the first occurrence of the peak value, we set \( n = 0 \):
\[ t = \frac{1}{400} \text{ seconds} \]

5. **Conclusion**:
Therefore, the time required for the current to achieve its peak value is:
\[ t = 0.0025 \text{ seconds} \text{ or } 2.5 \text{ ms} \]

### Final Answer:
The time required for the current to achieve its peak value is \( 0.0025 \) seconds or \( 2.5 \) milliseconds.

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