The relation between an ac voltage source and time in SI units is `V=120 sin(100 pi t)cos (100 pi t) V`. The value of peak voltage and frequency will be respectively

Options

A. 120 V and 100 Hz
B. `(120)/(sqrt(2)) V` and 100 Hz
C. 60 and 200 Hz
D. 60 V and 100 Hz

Explanation

To solve the problem, we need to find the peak voltage and frequency from the given equation of the AC voltage source:

\[ V = 120 \sin(100 \pi t) \cos(100 \pi t) \, \text{V} \]

### Step 1: Simplify the Voltage Equation
We can use the trigonometric identity:
\[ 2 \sin \theta \cos \theta = \sin(2\theta) \]

In our case, we can rewrite the equation as follows:
\[ V = 120 \sin(100 \pi t) \cos(100 \pi t) = 60 \cdot 2 \sin(100 \pi t) \cos(100 \pi t) = 60 \sin(200 \pi t) \]

### Step 2: Identify the Peak Voltage
From the simplified equation:
\[ V = 60 \sin(200 \pi t) \]
We can see that the peak voltage \( V_0 \) is the coefficient of the sine function. Therefore:
\[ V_0 = 60 \, \text{V} \]

### Step 3: Determine the Angular Frequency
In the equation \( V = V_0 \sin(\omega t) \), we have:
\[ \omega = 200 \pi \]

### Step 4: Calculate the Frequency
The relationship between angular frequency \( \omega \) and frequency \( f \) is given by:
\[ \omega = 2 \pi f \]
Thus, we can solve for \( f \):
\[ f = \frac{\omega}{2 \pi} = \frac{200 \pi}{2 \pi} = 100 \, \text{Hz} \]

### Final Answer
The peak voltage and frequency are:
- Peak Voltage: \( 60 \, \text{V} \)
- Frequency: \( 100 \, \text{Hz} \)

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