An alternating emf given by `V = V_(0) sin omegat` has peak value 10 volt and freguency `50` Hz The instantaneous emf at .

Options

A. `10V`
B. `5sqrt3 V`
C. `5V`
D. `1V`

Explanation

To find the instantaneous emf given by the equation \( V = V_0 \sin(\omega t) \), we will follow these steps:

### Step 1: Identify the given values
- Peak value \( V_0 = 10 \) volts
- Frequency \( f = 50 \) Hz

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

### Step 3: Write the equation for instantaneous emf
The equation for instantaneous emf is:
\[ V(t) = V_0 \sin(\omega t) \]
Substituting the values of \( V_0 \) and \( \omega \):
\[ V(t) = 10 \sin(100\pi t) \]

### Step 4: Substitute \( t = \frac{1}{600} \) seconds into the equation
Now, we will substitute \( t = \frac{1}{600} \) seconds into the equation:
\[ V\left(\frac{1}{600}\right) = 10 \sin\left(100\pi \times \frac{1}{600}\right) \]

### Step 5: Simplify the argument of the sine function
Calculating the argument:
\[ 100\pi \times \frac{1}{600} = \frac{100\pi}{600} = \frac{\pi}{6} \]
Thus, we have:
\[ V\left(\frac{1}{600}\right) = 10 \sin\left(\frac{\pi}{6}\right) \]

### Step 6: Calculate the sine value
The sine of \( \frac{\pi}{6} \) is:
\[ \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \]

### Step 7: Calculate the instantaneous emf
Substituting the sine value back into the equation:
\[ V\left(\frac{1}{600}\right) = 10 \times \frac{1}{2} = 5 \, \text{volts} \]

### Final Answer
The instantaneous emf at \( t = \frac{1}{600} \) seconds is \( 5 \) volts.

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