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

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Answer

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

### Step 1: Understand the Function
The voltage function given is \( V(t) = V_0 \sin(\omega t) \), where:
- \( V_0 \) is the peak voltage,
- \( \omega \) is the angular frequency,
- \( t \) is time.

### Step 2: Define the Average Value Over a Full Cycle
The average value \( V_{avg} \) of a function over one complete cycle is defined as:
\[ V_{avg} = \frac{1}{T} \int_0^T V(t) \, dt \]
where \( T \) is the period of the function. For a sine function, the period \( T \) is given by:
\[ T = \frac{2\pi}{\omega} \]

### Step 3: Set Up the Integral
Substituting \( V(t) \) into the average value formula, we have:
\[ V_{avg} = \frac{1}{T} \int_0^T V_0 \sin(\omega t) \, dt \]
Substituting \( T \):
\[ V_{avg} = \frac{\omega}{2\pi} \int_0^{2\pi/\omega} V_0 \sin(\omega t) \, dt \]

### Step 4: Evaluate the Integral
Now, we need to evaluate the integral:
\[ \int_0^{2\pi/\omega} \sin(\omega t) \, dt \]
Using the substitution \( u = \omega t \), \( du = \omega dt \), we change the limits accordingly:
- When \( t = 0 \), \( u = 0 \)
- When \( t = \frac{2\pi}{\omega} \), \( u = 2\pi \)

Thus, the integral becomes:
\[ \int_0^{2\pi} \sin(u) \frac{du}{\omega} \]

### Step 5: Solve the Integral
The integral of \( \sin(u) \) over one complete cycle (from \( 0 \) to \( 2\pi \)) is:
\[ \int_0^{2\pi} \sin(u) \, du = 0 \]
This is because the area under the curve for \( \sin(u) \) from \( 0 \) to \( \pi \) is positive and from \( \pi \) to \( 2\pi \) is negative, canceling each other out.

### Step 6: Calculate the Average Value
Substituting back, we find:
\[ V_{avg} = \frac{V_0}{2\pi} \cdot 0 = 0 \]

### Conclusion
The average value of the alternating voltage \( V = V_0 \sin(\omega t) \) over a full cycle is:
\[ \boxed{0} \]