The voltage of an AC source varies with time according to the equation `V=100sin 100pitcos100pit` where `t` is in seconds and `V` is in volts. Then

Options

A. The peak voltage of the source is `100` volts
B. The peak voltage of the source is `50 `volts
C. The peak voltage of the source is `100sqrt(2)` volts
D. The frequency of the source is `50Hz`

Explanation

To solve the problem, we need to analyze the given voltage equation and extract the peak voltage and frequency from it. The given equation is:

\[ V = 100 \sin(100 \pi t) \cos(100 \pi t) \]

### Step 1: Simplify the Voltage Equation
We can use the trigonometric identity that relates sine and cosine:

\[ 2 \sin(a) \cos(a) = \sin(2a) \]

Here, let \( a = 100 \pi t \). Therefore, we can rewrite the equation as:

\[ V = 100 \cdot \frac{1}{2} \cdot 2 \sin(100 \pi t) \cos(100 \pi t) \]
\[ V = 100 \cdot \frac{1}{2} \sin(200 \pi t) \]
\[ V = 50 \sin(200 \pi t) \]

### Step 2: Identify the Peak Voltage
From the simplified equation \( V = 50 \sin(200 \pi t) \), we can identify the peak voltage \( V_0 \):

\[ V_0 = 50 \text{ volts} \]

### Step 3: Determine the Angular Frequency
In the equation \( V = V_0 \sin(\omega t) \), we can see that \( \omega = 200 \pi \).

### Step 4: Calculate the Frequency
The relationship between angular frequency \( \omega \) and frequency \( f \) is given by:

\[ \omega = 2 \pi f \]

We can rearrange this to find \( f \):

\[ f = \frac{\omega}{2 \pi} \]
\[ f = \frac{200 \pi}{2 \pi} \]
\[ f = 100 \text{ Hz} \]

### Summary of Results
- The peak voltage \( V_0 \) is 50 volts.
- The frequency \( f \) is 100 Hz.

### Final Answer
Thus, the correct options are:
- Peak Voltage: 50 volts
- Frequency: 100 Hz