The average value of a saw-tooth voltage `V=V_(0) ((2t)/(T)-1)` Over 0 to `(T)/(2)`, (T- time period) is given by

Options

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

Answer

To find the average value of the saw-tooth voltage \( V = V_0 \left( \frac{2t}{T} - 1 \right) \) over the interval from \( 0 \) to \( \frac{T}{2} \), we can follow these steps:

### Step-by-Step Solution:

1. **Define the Average Value Formula**:
The average value \( V_{avg} \) of a function \( V(t) \) over an interval \( [a, b] \) is given by:
\[ V_{avg} = \frac{1}{b-a} \int_a^b V(t) \, dt \]
Here, \( a = 0 \) and \( b = \frac{T}{2} \).

2. **Set Up the Integral**:
Substitute \( V(t) \) into the average value formula:
\[ V_{avg} = \frac{1}{\frac{T}{2} - 0} \int_0^{\frac{T}{2}} V_0 \left( \frac{2t}{T} - 1 \right) dt \]
This simplifies to:
\[ V_{avg} = \frac{2}{T} \int_0^{\frac{T}{2}} V_0 \left( \frac{2t}{T} - 1 \right) dt \]

3. **Integrate the Function**:
Now, we need to calculate the integral:
\[ \int_0^{\frac{T}{2}} \left( \frac{2V_0 t}{T} - V_0 \right) dt \]
This can be split into two separate integrals:
\[ = \int_0^{\frac{T}{2}} \frac{2V_0 t}{T} dt - \int_0^{\frac{T}{2}} V_0 dt \]

4. **Calculate Each Integral**:
- For the first integral:
\[ \int_0^{\frac{T}{2}} \frac{2V_0 t}{T} dt = \frac{2V_0}{T} \cdot \frac{t^2}{2} \Bigg|_0^{\frac{T}{2}} = \frac{2V_0}{T} \cdot \frac{(\frac{T}{2})^2}{2} = \frac{2V_0}{T} \cdot \frac{T^2}{8} = \frac{V_0 T}{4} \]
- For the second integral:
\[ \int_0^{\frac{T}{2}} V_0 dt = V_0 \cdot t \Bigg|_0^{\frac{T}{2}} = V_0 \cdot \frac{T}{2} \]

5. **Combine the Results**:
Now substituting back into the average value formula:
\[ V_{avg} = \frac{2}{T} \left( \frac{V_0 T}{4} - V_0 \cdot \frac{T}{2} \right) \]
Simplifying this gives:
\[ V_{avg} = \frac{2}{T} \left( \frac{V_0 T}{4} - \frac{V_0 T}{2} \right) = \frac{2}{T} \left( \frac{V_0 T}{4} - \frac{2V_0 T}{4} \right) = \frac{2}{T} \left( -\frac{V_0 T}{4} \right) = -\frac{V_0}{2} \]

6. **Final Result**:
Thus, the average value of the saw-tooth voltage over the interval \( [0, \frac{T}{2}] \) is:
\[ V_{avg} = \frac{V_0}{2} \]