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:

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

### Step 1: Use the trigonometric identity

We can use the trigonometric identity:
\[ 2 \sin \theta \cos \theta = \sin(2\theta) \]

In our case, we can rewrite the equation as:
\[ E = 120 \cdot \frac{1}{2} \cdot 2 \sin(100 \pi t) \cos(100 \pi t) \]

This simplifies to:
\[ E = 60 \sin(200 \pi t) \]

### Step 2: Identify the peak voltage

From the equation \( E = 60 \sin(200 \pi t) \), we can identify the peak voltage \( E_0 \):
\[ E_0 = 60 \, \text{volts} \]

### Step 3: Identify the angular frequency

The angular frequency \( \omega \) is given by:
\[ \omega = 200 \pi \]

### Step 4: Calculate the frequency

We know that the relationship between angular frequency and frequency \( f \) is given by:
\[ \omega = 2 \pi f \]

Substituting for \( \omega \):
\[ 200 \pi = 2 \pi f \]

Dividing both sides by \( 2 \pi \):
\[ f = \frac{200 \pi}{2 \pi} = 100 \, \text{Hz} \]

### Conclusion

From the analysis, we have determined:
- The peak voltage \( E_0 \) is **60 volts**.
- The frequency \( f \) is **100 Hz**.

### Final Answers:
- Peak Voltage: **60 volts**
- Frequency: **100 Hz**

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