An ac source is of `(200)/(sqrt(2))` V, 50 Hz. The value of voltage after `(1)/(600)s` from the start is

Options

A. 200 V
B. `(200)/(sqrt(2))V`
C. 100 V
D. 50 V

Explanation

To solve the problem step by step, we will calculate the instantaneous voltage of the AC source after a given time. Here’s the detailed solution:

### Step 1: Identify the given values
- RMS Voltage, \( V_{\text{rms}} = \frac{200}{\sqrt{2}} \, \text{V} \)
- Frequency, \( f = 50 \, \text{Hz} \)
- Time, \( t = \frac{1}{600} \, \text{s} \)

### Step 2: Calculate the peak voltage \( V_0 \)
The relationship between RMS voltage and peak voltage is given by:
\[ V_{\text{rms}} = \frac{V_0}{\sqrt{2}} \]
From this, we can express \( V_0 \) as:
\[ V_0 = V_{\text{rms}} \times \sqrt{2} \]
Substituting the value of \( V_{\text{rms}} \):
\[ V_0 = \left(\frac{200}{\sqrt{2}}\right) \times \sqrt{2} = 200 \, \text{V} \]

### Step 3: Calculate angular frequency \( \omega \)
The angular frequency \( \omega \) is given by:
\[ \omega = 2\pi f \]
Substituting the value of \( f \):
\[ \omega = 2\pi \times 50 = 100\pi \, \text{rad/s} \]

### Step 4: Substitute values into the instantaneous voltage formula
The instantaneous voltage \( V(t) \) is given by:
\[ V(t) = V_0 \sin(\omega t) \]
Substituting \( V_0 \), \( \omega \), and \( t \):
\[ V(t) = 200 \sin(100\pi \times \frac{1}{600}) \]

### Step 5: Simplify the argument of the sine function
Calculating the argument:
\[ 100\pi \times \frac{1}{600} = \frac{100\pi}{600} = \frac{\pi}{6} \]

### Step 6: Calculate the sine value
Now we can find:
\[ V(t) = 200 \sin\left(\frac{\pi}{6}\right) \]
We know that:
\[ \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \]
Thus:
\[ V(t) = 200 \times \frac{1}{2} = 100 \, \text{V} \]

### Conclusion
The instantaneous voltage after \( \frac{1}{600} \) seconds from the start is:
\[ \boxed{100 \, \text{V}} \]