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))/(sqrt2)`
B. `((2)/(pi))V_(0)`
C. `((2)/(pi))V_(0)`
D. ZERO

Answer

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

### Step 1: Define the average voltage formula
The average value of a function \( V(t) \) over a time period \( T \) is given by the formula:
\[ V_{\text{avg}} = \frac{1}{T} \int_0^T V(t) \, dt \]
In our case, the time period \( T \) for one complete cycle is \( T = \frac{2\pi}{\omega} \).

### Step 2: Break the integral into two parts
The voltage function \( V(t) \) is defined in two parts:
1. For \( 0 \leq t \leq \frac{\pi}{\omega} \):
\[ V(t) = V_0 \sin(\omega t) \]
2. For \( \frac{\pi}{\omega} < t \leq \frac{2\pi}{\omega} \):
\[ V(t) = -V_0 \sin(\omega t) \]

Thus, we can express the average voltage as:
\[ V_{\text{avg}} = \frac{1}{T} \left( \int_0^{\frac{\pi}{\omega}} V_0 \sin(\omega t) \, dt + \int_{\frac{\pi}{\omega}}^{\frac{2\pi}{\omega}} -V_0 \sin(\omega t) \, dt \right) \]

### Step 3: Calculate the first integral
Calculate the integral from \( 0 \) to \( \frac{\pi}{\omega} \):
\[ \int_0^{\frac{\pi}{\omega}} V_0 \sin(\omega t) \, dt \]
Using the integral of sine:
\[ \int \sin(x) \, dx = -\cos(x) \]
We have:
\[ \int_0^{\frac{\pi}{\omega}} V_0 \sin(\omega t) \, dt = V_0 \left[ -\frac{1}{\omega} \cos(\omega t) \right]_0^{\frac{\pi}{\omega}} = V_0 \left( -\frac{1}{\omega} \left( -1 - 1 \right) \right) = \frac{2V_0}{\omega} \]

### Step 4: Calculate the second integral
Now calculate the integral from \( \frac{\pi}{\omega} \) to \( \frac{2\pi}{\omega} \):
\[ \int_{\frac{\pi}{\omega}}^{\frac{2\pi}{\omega}} -V_0 \sin(\omega t) \, dt \]
Using the same integration method:
\[ \int_{\frac{\pi}{\omega}}^{\frac{2\pi}{\omega}} -V_0 \sin(\omega t) \, dt = -V_0 \left[ -\frac{1}{\omega} \cos(\omega t) \right]_{\frac{\pi}{\omega}}^{\frac{2\pi}{\omega}} = -V_0 \left( -\frac{1}{\omega} (1 - (-1)) \right) = -V_0 \left( -\frac{2}{\omega} \right) = \frac{2V_0}{\omega} \]

### Step 5: Combine the results
Now substitute the results of the two integrals back into the average voltage formula:
\[ V_{\text{avg}} = \frac{1}{\frac{2\pi}{\omega}} \left( \frac{2V_0}{\omega} + \frac{2V_0}{\omega} \right) = \frac{1}{\frac{2\pi}{\omega}} \left( \frac{4V_0}{\omega} \right) = \frac{4V_0}{2\pi} = \frac{2V_0}{\pi} \]

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