The alternating current in a circuit is given by `I = 50sin314t`. The peak value and frequency of the current are

Options

A. `I_(0)`=25 A and f=100Hz
B. `I_(0)`=50 A and f=50 Hz
C. `I_(0)` = 50 A and f=100 Hz
D. `I_(0)`=25 A and f=50 Hz

Explanation

To solve the problem, we need to determine the peak value and frequency of the alternating current given by the equation \( I = 50 \sin(314t) \).

### Step-by-Step Solution:

1. **Identify the given equation**:
The alternating current is given by:
\[ I = 50 \sin(314t) \]

2. **Compare with the standard form**:
The standard form of an alternating current is:
\[ I = I_0 \sin(\omega t) \]
where \( I_0 \) is the peak current (amplitude) and \( \omega \) is the angular frequency.

3. **Determine the peak current \( I_0 \)**:
From the equation \( I = 50 \sin(314t) \), we can see that:
\[ I_0 = 50 \, \text{A} \]
Thus, the peak value of the current is \( 50 \, \text{A} \).

4. **Find the angular frequency \( \omega \)**:
From the equation, we have:
\[ \omega = 314 \, \text{rad/s} \]

5. **Calculate the frequency \( f \)**:
The relationship between angular frequency \( \omega \) and frequency \( f \) is given by:
\[ \omega = 2\pi f \]
Rearranging this gives:
\[ f = \frac{\omega}{2\pi} \]
Substituting the value of \( \omega \):
\[ f = \frac{314}{2\pi} \]

6. **Calculate \( f \)**:
Using \( \pi \approx 3.14 \):
\[ f = \frac{314}{2 \times 3.14} = \frac{314}{6.28} \approx 50 \, \text{Hz} \]

### Final Results:
- The peak value \( I_0 \) is \( 50 \, \text{A} \).
- The frequency \( f \) is \( 50 \, \text{Hz} \).

### Summary:
The peak value of the current is \( 50 \, \text{A} \) and the frequency is \( 50 \, \text{Hz} \).