Avarage value of current for half wave rectifier having peak value of current `I_0` is

Options

A. A. `I_0/2`
B. B. `I_0/sqrt(2)`
C. C. `I_0/pi`
D. D. `2I_0/pi`

Answer

To find the average value of current for a half-wave rectifier with a peak value of current \( I_0 \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Half-Wave Rectification**:
- In a half-wave rectifier, only one half (positive or negative) of the AC waveform is allowed to pass through, while the other half is blocked. This results in a pulsating DC output.

2. **Average Value Calculation**:
- The average value of the output current over one complete cycle (which is \( 2\pi \) radians for a sine wave) can be calculated using the formula:
\[ I_{avg} = \frac{1}{T} \int_0^T I(t) dt \]
- For a half-wave rectifier, the current \( I(t) \) is \( I_0 \sin(\omega t) \) for \( 0 \leq t \leq \frac{T}{2} \) and \( 0 \) for \( \frac{T}{2} < t < T \).

3. **Setting Up the Integral**:
- The average current over one cycle (T) can be expressed as:
\[ I_{avg} = \frac{1}{T} \int_0^{T/2} I_0 \sin(\omega t) dt \]
- Since \( T = \frac{2\pi}{\omega} \), we can substitute \( T \) into the equation.

4. **Evaluating the Integral**:
- The integral can be evaluated as follows:
\[ I_{avg} = \frac{1}{T} \cdot I_0 \int_0^{T/2} \sin(\omega t) dt \]
- The integral of \( \sin(\omega t) \) is:
\[ -\frac{1}{\omega} \cos(\omega t) \bigg|_0^{T/2} = -\frac{1}{\omega} \left( \cos(\frac{\pi}{2}) - \cos(0) \right) = -\frac{1}{\omega} (0 - 1) = \frac{1}{\omega} \]

5. **Substituting Back**:
- Now substituting back into the average current formula gives:
\[ I_{avg} = \frac{1}{T} \cdot I_0 \cdot \frac{1}{\omega} = \frac{I_0}{T \cdot \omega} \]
- Since \( T = \frac{2\pi}{\omega} \), we have:
\[ I_{avg} = \frac{I_0 \cdot \omega}{2\pi \cdot \omega} = \frac{I_0}{2\pi} \]

6. **Final Result**:
- The average value of the current for a half-wave rectifier is:
\[ I_{avg} = \frac{I_0}{\pi} \]

### Conclusion:
Thus, the average value of current for a half-wave rectifier with a peak value of current \( I_0 \) is \( \frac{I_0}{\pi} \).