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: Understand the Voltage Function
The voltage \( V(t) \) is defined piecewise over one complete cycle:
- For \( 0 \leq t \leq \frac{\pi}{\omega} \):
\[ V(t) = V_0 \sin(\omega t) \]
- For \( \frac{\pi}{\omega} < t \leq \frac{2\pi}{\omega} \):
\[ V(t) = -V_0 \sin(\omega t) \]

### Step 2: Set Up the Average Voltage Formula
The average value of the voltage over one cycle is given by:
\[ V_{avg} = \frac{1}{T} \int_0^T V(t) \, dt \]
where \( T = \frac{2\pi}{\omega} \).

### Step 3: Calculate the Integral
We will split the integral into two parts:
1. From \( 0 \) to \( \frac{\pi}{\omega} \)
2. From \( \frac{\pi}{\omega} \) to \( \frac{2\pi}{\omega} \)

Thus, we have:
\[ V_{avg} = \frac{1}{T} \left( \int_0^{\frac{\pi}{\omega}} V_0 \sin(\omega t) \, dt + \int_{\frac{\pi}{\omega}}^{\frac{2\pi}{\omega}} -V_0 \sin(\omega t) \, dt \right) \]

### Step 4: Evaluate the First Integral
For the first integral:
\[ \int_0^{\frac{\pi}{\omega}} V_0 \sin(\omega t) \, dt = V_0 \left[-\frac{1}{\omega} \cos(\omega t)\right]_0^{\frac{\pi}{\omega}} = V_0 \left[-\frac{1}{\omega} \cos\left(\pi\right) + \frac{1}{\omega} \cos(0)\right] \]
\[ = V_0 \left[-\frac{1}{\omega} (-1) + \frac{1}{\omega} (1)\right] = V_0 \left[\frac{2}{\omega}\right] \]

### Step 5: Evaluate the Second Integral
For the second integral:
\[ \int_{\frac{\pi}{\omega}}^{\frac{2\pi}{\omega}} -V_0 \sin(\omega t) \, dt = -V_0 \left[-\frac{1}{\omega} \cos(\omega t)\right]_{\frac{\pi}{\omega}}^{\frac{2\pi}{\omega}} = -V_0 \left[-\frac{1}{\omega} \cos\left(2\pi\right) + \frac{1}{\omega} \cos\left(\pi\right)\right] \]
\[ = -V_0 \left[-\frac{1}{\omega} (1) + \frac{1}{\omega} (-1)\right] = -V_0 \left[-\frac{2}{\omega}\right] = \frac{2V_0}{\omega} \]

### Step 6: Combine the Integrals
Now, we can combine both integrals:
\[ \int_0^{\frac{2\pi}{\omega}} V(t) \, dt = \frac{2V_0}{\omega} + \frac{2V_0}{\omega} = \frac{4V_0}{\omega} \]

### Step 7: Calculate the Average Voltage
Now substitute back into the average voltage formula:
\[ V_{avg} = \frac{1}{\frac{2\pi}{\omega}} \cdot \frac{4V_0}{\omega} = \frac{4V_0}{2\pi} = \frac{2V_0}{\pi} \]

### Final Result
The average value of the voltage over one cycle is:
\[ \boxed{\frac{2V_0}{\pi}} \]