The voltage over a cycle varies as
`V=V_(0)sin omega t` for `0 le t le (pi)/(omega)`
`=-V_(0)sin omega t` for `(pi)/(omega)le t le (2pi)/(omega)`
The average value of the voltage one cycle is

Options

A. `(V_(0))/(sqrt(2))`
B. `(V_(0))/(2)`
C. zero
D. `(2V_(0))/(pi)`

Explanation

To find the average value of the voltage over one complete cycle, we can follow these steps:

### Step 1: Define the voltage function
The voltage function is given as:
- \( V = V_0 \sin(\omega t) \) for \( 0 \leq t \leq \frac{\pi}{\omega} \)
- \( V = -V_0 \sin(\omega t) \) for \( \frac{\pi}{\omega} < t \leq \frac{2\pi}{\omega} \)

### Step 2: Set up the average voltage formula
The average value of voltage \( V_{\text{avg}} \) over one complete cycle (from \( t = 0 \) to \( t = \frac{2\pi}{\omega} \)) can be calculated using the formula:
\[ V_{\text{avg}} = \frac{1}{T} \int_0^T V(t) \, dt \]
where \( T = \frac{2\pi}{\omega} \).

### Step 3: Split the integral
We can split the integral into two parts:
\[ V_{\text{avg}} = \frac{1}{T} \left( \int_0^{\frac{\pi}{\omega}} V_0 \sin(\omega t) \, dt + \int_{\frac{\pi}{\omega}}^{\frac{2\pi}{\omega}} -V_0 \sin(\omega t) \, dt \right) \]

### Step 4: Calculate the first integral
For the first integral:
\[ \int_0^{\frac{\pi}{\omega}} V_0 \sin(\omega t) \, dt \]
Using the integral of sine:
\[ \int \sin(x) \, dx = -\cos(x) \]
Thus,
\[ \int_0^{\frac{\pi}{\omega}} V_0 \sin(\omega t) \, dt = V_0 \left[ -\frac{1}{\omega} \cos(\omega t) \right]_0^{\frac{\pi}{\omega}} = V_0 \left( -\frac{1}{\omega} \left( -1 - 1 \right) \right) = \frac{2V_0}{\omega} \]

### Step 5: Calculate the second integral
For the second integral:
\[ \int_{\frac{\pi}{\omega}}^{\frac{2\pi}{\omega}} -V_0 \sin(\omega t) \, dt \]
This can be calculated similarly:
\[ -V_0 \left[ -\frac{1}{\omega} \cos(\omega t) \right]_{\frac{\pi}{\omega}}^{\frac{2\pi}{\omega}} = -V_0 \left( -\frac{1}{\omega} \left( 1 - (-1) \right) \right) = \frac{2V_0}{\omega} \]

### Step 6: Combine the results
Now, we can combine both integrals:
\[ V_{\text{avg}} = \frac{1}{T} \left( \frac{2V_0}{\omega} + \frac{2V_0}{\omega} \right) = \frac{1}{\frac{2\pi}{\omega}} \left( \frac{4V_0}{\omega} \right) = \frac{2V_0}{\pi} \]

### Final Answer
Thus, the average value of the voltage over one complete cycle is:
\[ V_{\text{avg}} = \frac{2V_0}{\pi} \]