The voltage of an a.c. source varies with time according to relation
E=120 sin `100pitcos100 pi t` then

Options

A. the peak voltage of source is 120v
B. the peak voltage = 60 v
C. the peak voltage = `(120//sqrt2)v`
D. frequency of source voltage is 100 Hz

Answer

To solve the problem, we need to analyze the given voltage equation of the AC source, which is:

\[ E(t) = 120 \sin(100 \pi t) \cos(100 \pi t) \]

### Step 1: Convert the voltage equation into standard form
We can use the trigonometric identity for the product of sine and cosine:

\[ \sin A \cos A = \frac{1}{2} \sin(2A) \]

In our case, \( A = 100 \pi t \). Therefore, we can rewrite the equation as:

\[ E(t) = 120 \cdot \frac{1}{2} \sin(2 \cdot 100 \pi t) = 60 \sin(200 \pi t) \]

### Step 2: Identify the peak voltage
From the standard form of the voltage equation \( E(t) = V_0 \sin(\omega t) \), we can identify the peak voltage \( V_0 \):

\[ V_0 = 60 \, \text{V} \]

### Step 3: Calculate the angular frequency
The angular frequency \( \omega \) is given by the coefficient of \( t \) in the sine function:

\[ \omega = 200 \pi \, \text{rad/s} \]

### Step 4: Calculate the frequency
The frequency \( f \) can be calculated using the relationship between angular frequency and frequency:

\[ f = \frac{\omega}{2 \pi} \]

Substituting the value of \( \omega \):

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

### Final Results
- **Peak Voltage**: \( 60 \, \text{V} \)
- **Frequency**: \( 100 \, \text{Hz} \)

### Summary
The peak voltage of the AC source is \( 60 \, \text{V} \) and the frequency is \( 100 \, \text{Hz} \).

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