The voltage of an `AC` source varies with time according to the equation,`V=100 sin 100 pi I cos 100 pi t`.Where `t` is in second and `V` is in volt.Then:

Options

A. the peak voltage of the source is `100` volt
B. the peak voltage of the source is `(100//2)` volt
C. the peak voltage of the source is `50` volt
D. the peak voltage of the source is `50 Hz`

Explanation

To solve the problem, we will analyze the given equation for the AC voltage and determine the peak voltage and frequency.

### Step-by-Step Solution:

1. **Identify the given equation**:
The voltage of the AC source is given by:
\[ V = 100 \sin(100\pi t) \cos(100\pi t) \]

2. **Use the trigonometric identity**:
We can use the trigonometric identity:
\[ \sin A \cos A = \frac{1}{2} \sin(2A) \]
Here, let \( A = 100\pi t \). Thus, we can rewrite the voltage equation as:
\[ V = 100 \cdot \frac{1}{2} \sin(2 \cdot 100\pi t) = 50 \sin(200\pi t) \]

3. **Determine the peak voltage**:
The peak voltage (also known as the amplitude) of the sine function \( V = A \sin(\omega t) \) is given by \( A \). From our equation:
\[ V = 50 \sin(200\pi t) \]
Therefore, the peak voltage \( V_0 \) is:
\[ V_0 = 50 \text{ volts} \]

4. **Calculate the frequency**:
The angular frequency \( \omega \) is given by:
\[ \omega = 200\pi \text{ rad/s} \]
The frequency \( f \) can be calculated using the formula:
\[ f = \frac{\omega}{2\pi} = \frac{200\pi}{2\pi} = 100 \text{ Hz} \]

5. **Conclusion**:
The peak voltage of the source is \( 50 \) volts and the frequency is \( 100 \) Hz.

### Final Answer:
- Peak voltage \( V_0 = 50 \) volts
- Frequency \( f = 100 \) Hz