An AC voltage is given by E = `underset(o)(E) ` sin 2` pi `t / T
Then , the mean value of volatage calculated over any time interval of T / 2

Options

A. (a)is always zero
B. (b)is never zero
C. (c)is always (2`underset(o)(E) ` / `pi `)
D. (d)may be zero

Answer

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 a time interval of \( \frac{T}{2} \), we can follow these steps:

### Step 1: Understand the Mean Value Formula
The mean value \( \overline{E} \) of a function \( E(t) \) over a time interval \( T \) is given by the formula:
\[ \overline{E} = \frac{1}{T} \int_0^T E(t) \, dt \]
For our case, we need to find the mean value over the interval \( \frac{T}{2} \):
\[ \overline{E} = \frac{1}{\frac{T}{2}} \int_0^{\frac{T}{2}} E(t) \, dt = \frac{2}{T} \int_0^{\frac{T}{2}} E(t) \, dt \]

### Step 2: Substitute the Expression for E(t)
Substituting \( E(t) = E_0 \sin\left(\frac{2\pi t}{T}\right) \):
\[ \overline{E} = \frac{2}{T} \int_0^{\frac{T}{2}} E_0 \sin\left(\frac{2\pi t}{T}\right) \, dt \]

### Step 3: Factor Out Constants
We can factor out \( E_0 \) from the integral:
\[ \overline{E} = \frac{2E_0}{T} \int_0^{\frac{T}{2}} \sin\left(\frac{2\pi t}{T}\right) \, dt \]

### Step 4: Evaluate the Integral
To evaluate the integral, we can use the substitution \( u = \frac{2\pi t}{T} \), which implies \( dt = \frac{T}{2\pi} du \). The limits change as follows:
- When \( t = 0 \), \( u = 0 \)
- 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: Evaluate the Integral of sin(u)
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 expression:
\[ \overline{E} = \frac{2E_0}{T} \cdot \frac{T}{2\pi} \cdot 2 = \frac{2E_0}{T} \cdot \frac{T}{\pi} = \frac{4E_0}{\pi} \]

### Final Answer
The mean value of the voltage over the time interval \( \frac{T}{2} \) is:
\[ \overline{E} = \frac{4E_0}{\pi} \]