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

Options

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

Answer

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

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

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

### Step 3: Substitute \( \omega \) and \( t \) into the emf equation
The instantaneous value of emf can be calculated using the formula:
\[ E = E_0 \cos(\omega t) \]
Substituting the values of \( E_0 \), \( \omega \), and \( t \):
\[ E = 10 \cos(100\pi \times \frac{1}{600}) \]

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

### Step 5: Calculate \( \cos\left(\frac{\pi}{6}\right) \)
Using the known value:
\[ \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \]
Substituting this value back into the equation:
\[ E = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \, V \]

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

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