Two alternating voltage generators produce emfs of the same amplitude`(E_0)` but with a phase difference of `(pi)//3`. The resultant emf is

Options

A. (A) `E_0 sin [ omega t+(pi)//3)]`
B. (B) `E_0 sin [ omega t+(pi)//6)]`
C. (C) `sqrt(3)E_0 sin [omega t+(pi)//6)]`
D. (D) `sqrt(3)E_0 sin [ omega t+(pi)//2)]`

Explanation

To find the resultant emf from two alternating voltage generators producing emfs of the same amplitude \(E_0\) but with a phase difference of \(\frac{\pi}{3}\), we can follow these steps:

### Step 1: Define the emfs
Let the first emf be represented as:
\[ E_1 = E_0 \sin(\omega t) \]
And the second emf, considering the phase difference of \(\frac{\pi}{3}\), is:
\[ E_2 = E_0 \sin\left(\omega t + \frac{\pi}{3}\right) \]

### Step 2: Write the expression for the resultant emf
The resultant emf \(E\) is the sum of the two emfs:
\[ E = E_1 + E_2 = E_0 \sin(\omega t) + E_0 \sin\left(\omega t + \frac{\pi}{3}\right) \]

### Step 3: Factor out the common amplitude
We can factor out \(E_0\):
\[ E = E_0 \left( \sin(\omega t) + \sin\left(\omega t + \frac{\pi}{3}\right) \right) \]

### Step 4: Use the sine addition formula
We can use the sine addition formula:
\[ \sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]
Let \(A = \omega t\) and \(B = \omega t + \frac{\pi}{3}\). Then:
\[ E = E_0 \left( 2 \sin\left(\frac{\omega t + \left(\omega t + \frac{\pi}{3}\right)}{2}\right) \cos\left(\frac{\omega t - \left(\omega t + \frac{\pi}{3}\right)}{2}\right) \right) \]

### Step 5: Simplify the expression
Calculating the averages:
\[ \frac{A+B}{2} = \frac{2\omega t + \frac{\pi}{3}}{2} = \omega t + \frac{\pi}{6} \]
\[ \frac{A-B}{2} = \frac{-\frac{\pi}{3}}{2} = -\frac{\pi}{6} \]
Thus, we have:
\[ E = E_0 \left( 2 \sin\left(\omega t + \frac{\pi}{6}\right) \cos\left(-\frac{\pi}{6}\right) \right) \]

### Step 6: Substitute the value of \(\cos\left(-\frac{\pi}{6}\right)\)
Using the fact that \(\cos\left(-\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\):
\[ E = E_0 \left( 2 \sin\left(\omega t + \frac{\pi}{6}\right) \cdot \frac{\sqrt{3}}{2} \right) \]
This simplifies to:
\[ E = \sqrt{3} E_0 \sin\left(\omega t + \frac{\pi}{6}\right) \]

### Final Result
The resultant emf is:
\[ E = \sqrt{3} E_0 \sin\left(\omega t + \frac{\pi}{6}\right) \]