The average value of an alternating voltage `V=V_(0) sin omega t ` over a full cycle is

Options

A. `V_(0)`
B. `(V_(0))/(Sqrt(2))`
C. `(2V_(0))/(pi)`
D. Zero

Answer

To find the average value of the alternating voltage given by the equation \( V = V_0 \sin(\omega t) \) over a full cycle, we can follow these steps:

### Step 1: Understand the Function
The function \( V = V_0 \sin(\omega t) \) represents a sinusoidal waveform, where \( V_0 \) is the peak voltage, and \( \omega \) is the angular frequency. The waveform oscillates between \( V_0 \) and \(-V_0\).

### Step 2: Define the Average Value Over a Cycle
The average value of a function over a period \( T \) is given by the formula:
\[ \text{Average Value} = \frac{1}{T} \int_0^T V(t) \, dt \]
For our case, \( T \) is the time period of the sine wave.

### Step 3: Calculate the Time Period
The time period \( T \) of the sine wave is given by:
\[ T = \frac{2\pi}{\omega} \]

### Step 4: Set Up the Integral
We need to evaluate the integral of \( V(t) = V_0 \sin(\omega t) \) over one full cycle:
\[ \text{Average Value} = \frac{1}{T} \int_0^T V_0 \sin(\omega t) \, dt \]

### Step 5: Substitute the Time Period
Substituting \( T \) into the equation:
\[ \text{Average Value} = \frac{\omega}{2\pi} \int_0^{2\pi/\omega} V_0 \sin(\omega t) \, dt \]

### Step 6: Evaluate the Integral
Now we evaluate the integral:
\[ \int_0^{2\pi/\omega} V_0 \sin(\omega t) \, dt = V_0 \left[-\frac{1}{\omega} \cos(\omega t)\right]_0^{2\pi/\omega} \]
Calculating the limits:
\[ = V_0 \left[-\frac{1}{\omega} (\cos(2\pi) - \cos(0))\right] = V_0 \left[-\frac{1}{\omega} (1 - 1)\right] = 0 \]

### Step 7: Conclusion
Thus, the average value over a full cycle is:
\[ \text{Average Value} = \frac{1}{T} \cdot 0 = 0 \]

Therefore, the average value of the alternating voltage \( V = V_0 \sin(\omega t) \) over a full cycle is **0**.