An AC voltage is given by `E=E_(0) sin(2pit)/(T)`. Then the mean value of voltage calculated over time interval of T/2 seconds

Options

A. is always zero
B. is never zero
C. is `(2E_(0)//pi)` always
D. may be zero

Explanation

To find the mean value of the AC voltage given by the equation \( E = E_0 \sin\left(\frac{2\pi t}{T}\right) \) over the time interval of \( \frac{T}{2} \), we can follow these steps:

### Step 1: Understand the Mean Value Formula
The mean value of a function \( E(t) \) over a time interval \( [a, b] \) is given by:
\[ \text{Mean Value} = \frac{1}{b - a} \int_a^b E(t) \, dt \]
In this case, we need to calculate the mean value over the interval \( [0, \frac{T}{2}] \).

### Step 2: Set Up the Integral
Here, \( a = 0 \) and \( b = \frac{T}{2} \). Thus, the mean value of voltage \( E \) over the interval is:
\[ \text{Mean Value} = \frac{1}{\frac{T}{2} - 0} \int_0^{\frac{T}{2}} E_0 \sin\left(\frac{2\pi t}{T}\right) \, dt \]
This simplifies to:
\[ \text{Mean Value} = \frac{2}{T} \int_0^{\frac{T}{2}} E_0 \sin\left(\frac{2\pi t}{T}\right) \, dt \]

### Step 3: Factor Out Constants
Since \( E_0 \) is a constant, we can factor it out of the integral:
\[ \text{Mean Value} = \frac{2E_0}{T} \int_0^{\frac{T}{2}} \sin\left(\frac{2\pi t}{T}\right) \, dt \]

### Step 4: Evaluate the Integral
To evaluate the integral, we can use the substitution:
Let \( u = \frac{2\pi t}{T} \), then \( du = \frac{2\pi}{T} dt \) or \( dt = \frac{T}{2\pi} du \).
When \( t = 0 \), \( u = 0 \) and when \( t = \frac{T}{2} \), \( u = \pi \).

Thus, the integral becomes:
\[ \int_0^{\frac{T}{2}} \sin\left(\frac{2\pi t}{T}\right) \, dt = \int_0^{\pi} \sin(u) \frac{T}{2\pi} \, du \]
This simplifies to:
\[ \frac{T}{2\pi} \int_0^{\pi} \sin(u) \, du \]

### Step 5: Calculate the Integral of Sine
The integral of \( \sin(u) \) from \( 0 \) to \( \pi \) is:
\[ \int_0^{\pi} \sin(u) \, du = [-\cos(u)]_0^{\pi} = -\cos(\pi) - (-\cos(0)) = 1 + 1 = 2 \]

### Step 6: Substitute Back
Now substituting back into our mean value equation:
\[ \text{Mean Value} = \frac{2E_0}{T} \cdot \frac{T}{2\pi} \cdot 2 = \frac{2E_0}{T} \cdot \frac{T}{\pi} = \frac{4E_0}{\pi} \]

### Step 7: Conclusion
Thus, the mean value of the AC voltage over the time interval \( \frac{T}{2} \) is:
\[ \text{Mean Value} = \frac{4E_0}{\pi} \]