The peak voltage in the output of a half-wave diode rectifier fed with a sinusiodal signal without filter is `10 V`. The `dc` component of the output voltage is

Options

A. `10//sqrt(2) V`
B. `10//piV`
C. `10 V`
D. `20//piV`

Explanation

To find the DC component of the output voltage from a half-wave diode rectifier fed with a sinusoidal signal, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Peak Voltage**:
The peak voltage (V_peak) given in the problem is 10 V.

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

3. **Formula for DC Component**:
The DC component (V_dc) of the output voltage for a half-wave rectifier can be calculated using the formula:
\[ V_{dc} = \frac{V_{peak}}{\pi} \]

4. **Substitute the Peak Voltage**:
Now, substitute the given peak voltage into the formula:
\[ V_{dc} = \frac{10 \, V}{\pi} \]

5. **Calculate the Value**:
Using the approximate value of \(\pi \approx 3.14\):
\[ V_{dc} \approx \frac{10}{3.14} \approx 3.18 \, V \]

6. **Final Answer**:
Therefore, the DC component of the output voltage is approximately 3.18 V.

### Summary of the Solution:
The DC component of the output voltage from a half-wave diode rectifier with a peak voltage of 10 V is approximately 3.18 V.