The peak value of an alternating emf E given by
`E=(E_0) cos omega t`
is 10V and freqency is 50 Hz. At time `t=(1//600)s` the instantaneous value of emf is

Options

A. 10V
B. `5sqrt(3)V`
C. `5 V`
D. 1 V

Explanation

To find the instantaneous value of the alternating emf given by the equation \( E = E_0 \cos(\omega t) \), we will follow these steps:

### Step 1: Identify the given values
- Peak value of emf, \( E_0 = 10 \, \text{V} \)
- Frequency, \( f = 50 \, \text{Hz} \)
- Time, \( t = \frac{1}{600} \, \text{s} \)

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

### Step 3: Substitute \( \omega \) and \( t \) into the emf equation
Now we substitute \( \omega \) and \( t \) into the emf equation:
\[ E = E_0 \cos(\omega t) \]
\[ E = 10 \cos(100 \pi \times \frac{1}{600}) \]

### Step 4: Simplify the argument of the cosine function
Calculating the argument of the cosine:
\[ 100 \pi \times \frac{1}{600} = \frac{100 \pi}{600} = \frac{\pi}{6} \]
Thus, we have:
\[ E = 10 \cos\left(\frac{\pi}{6}\right) \]

### Step 5: Calculate \( \cos\left(\frac{\pi}{6}\right) \)
The value of \( \cos\left(\frac{\pi}{6}\right) \) is:
\[ \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \]

### Step 6: Final calculation of \( E \)
Now substituting this value back into the equation:
\[ E = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \, \text{V} \]

### Conclusion
The instantaneous value of emf at \( t = \frac{1}{600} \, \text{s} \) is:
\[ E = 5\sqrt{3} \, \text{V} \]

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