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/pi` V
B. 10 V
C. `20/pi` V
D. `10/sqrt2` V

Explanation

To find the DC component of the output voltage from a half-wave diode rectifier with a peak voltage of 10 V, we can follow these steps:

### Step 1: Understand the Half-Wave Rectifier Output
A half-wave rectifier allows only one half of the AC waveform to pass through, effectively blocking the negative half. The output voltage waveform will be a series of positive half-cycles.

### Step 2: Identify the Peak Voltage
The peak voltage (V_peak) of the output is given as 10 V. This is the maximum voltage that the rectifier will output during each half-cycle.

### Step 3: Calculate the Average (DC) Voltage
The average (DC) voltage (V_dc) for a half-wave rectifier can be calculated using the formula:
\[ V_{dc} = \frac{V_{peak}}{\pi} \]
This formula arises because the average value of the sine function over one complete cycle (0 to 2π) is 0, and we only consider the positive half-cycle (0 to π).

### Step 4: Substitute the Peak Voltage
Substituting the given peak voltage into the formula:
\[ V_{dc} = \frac{10 V}{\pi} \]

### Step 5: Simplify the Expression
Calculating the numerical value:
\[ V_{dc} \approx \frac{10}{3.14} \approx 3.18 V \]

### Final Answer
Thus, the DC component of the output voltage is approximately:
\[ V_{dc} \approx 3.18 V \]

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