The average value of alternating current for half cycle in terms of `I_(0)` is

Options

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

Answer

To find the average value of alternating current for half a cycle in terms of \( I_0 \), we can follow these steps:

### Step-by-Step Solution:

1. **Define the Current Function**:
The alternating current can be expressed as:
\[ i(t) = I_0 \sin(\omega t) \]
where \( I_0 \) is the peak current and \( \omega \) is the angular frequency.

2. **Determine the Time Interval for Half Cycle**:
For a sine wave, one complete cycle corresponds to a time period \( T = \frac{2\pi}{\omega} \). Therefore, the time for half a cycle is:
\[ t = \frac{T}{2} = \frac{\pi}{\omega} \]

3. **Set Up the Average Value Formula**:
The average value of a function over an interval \([a, b]\) is given by:
\[ \text{Average} = \frac{1}{b - a} \int_a^b i(t) \, dt \]
For half a cycle, we will integrate from \( 0 \) to \( \frac{\pi}{\omega} \).

4. **Change the Limits of Integration**:
Since we are interested in the average over the half cycle, we will integrate from \( 0 \) to \( \pi \) in terms of \( \theta \) where \( \theta = \omega t \). Thus, when \( t = 0 \), \( \theta = 0 \) and when \( t = \frac{\pi}{\omega} \), \( \theta = \pi \).

5. **Rewrite the Integral**:
The average value of current over half a cycle becomes:
\[ \text{Average} = \frac{1}{\pi} \int_0^{\pi} I_0 \sin(\theta) \, d\theta \]

6. **Integrate the Function**:
The integral of \( \sin(\theta) \) is:
\[ \int \sin(\theta) \, d\theta = -\cos(\theta) \]
Evaluating from \( 0 \) to \( \pi \):
\[ \int_0^{\pi} \sin(\theta) \, d\theta = [-\cos(\theta)]_0^{\pi} = -\cos(\pi) - (-\cos(0)) = -(-1) - (-1) = 2 \]

7. **Calculate the Average Value**:
Plugging the result of the integral back into the average formula:
\[ \text{Average} = \frac{1}{\pi} \cdot I_0 \cdot 2 = \frac{2 I_0}{\pi} \]

### Final Answer:
The average value of alternating current for half a cycle in terms of \( I_0 \) is:
\[ \text{Average} = \frac{2 I_0}{\pi} \]