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(5))`
C. 100 V
D. 50 V

Answer

To solve the problem, we need to find the instantaneous voltage of an AC source after a specific time. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the given parameters
- The RMS voltage \( V_{RMS} = \frac{200}{\sqrt{2}} \) V
- Frequency \( f = 50 \) Hz
- Time \( t = \frac{1}{600} \) s

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

### Step 3: Calculate the 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:
\[ 100 \pi \times \frac{1}{600} = \frac{100 \pi}{600} = \frac{\pi}{6} \]
So now we have:
\[ V(t) = 200 \sin\left(\frac{\pi}{6}\right) \]

### Step 6: Calculate \( \sin\left(\frac{\pi}{6}\right) \)
We know that:
\[ \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \]
Substituting this back into the equation gives:
\[ V(t) = 200 \times \frac{1}{2} = 100 \text{ V} \]

### Final Answer
The instantaneous voltage after \( \frac{1}{600} \) seconds from the start is \( 100 \) V.

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