`epsi_(0)=epsi_(0)sinomegat`, represents an ac equation. The time in which emf becomes half of its maximum value starting from zero is:

Options

A. `(T)/(2)`
B. `(T)/(3)`
C. `(T)/(6)`
D. `(T)/(12)`

Explanation

To solve the problem of finding the time at which the EMF becomes half of its maximum value in the AC equation \( E = E_0 \sin(\omega t) \), we can follow these steps:

### Step 1: Identify the maximum value of EMF
The maximum value of EMF, denoted as \( E_{\text{max}} \), is given by \( E_0 \). Therefore, half of the maximum value is:
\[ E_{\text{half}} = \frac{E_0}{2} \]

### Step 2: Set up the equation
We need to find the time \( t \) when the EMF equals half of its maximum value:
\[ E_0 \sin(\omega t) = \frac{E_0}{2} \]

### Step 3: Simplify the equation
Dividing both sides by \( E_0 \) (assuming \( E_0 \neq 0 \)):
\[ \sin(\omega t) = \frac{1}{2} \]

### Step 4: Solve for \( \omega t \)
The sine function equals \( \frac{1}{2} \) at specific angles. The principal angle is:
\[ \omega t = \frac{\pi}{6} \]
However, sine is periodic, so it can also equal \( \frac{1}{2} \) at:
\[ \omega t = \frac{\pi}{6} + n\pi \quad (n \in \mathbb{Z}) \]

### Step 5: Substitute \( \omega \)
We know that \( \omega = \frac{2\pi}{T} \), where \( T \) is the time period. Substituting this into our equation gives:
\[ \frac{2\pi}{T} t = \frac{\pi}{6} \]

### Step 6: Solve for \( t \)
Rearranging the equation to solve for \( t \):
\[ t = \frac{T}{12} \]

### Conclusion
Thus, the time in which the EMF becomes half of its maximum value starting from zero is:
\[ t = \frac{T}{12} \]