Question

The peak voltage in the output of a half-wave diode rectifier fed with a sinusoidal signal without a filter is 10V. The d. c. component of the output voltage is

• A. \(\frac{10}{\pi}\) V
• B. 10 V
• C. \(\frac{20}{\pi}\) V
• D. \(\frac{10}{\sqrt 2}\) V

Answer 1

The Correct Option is A

To find the direct current (d.c.) component of the output voltage of a half-wave rectifier, we need to understand the working of the rectifier and the nature of the voltage in its output.

A half-wave rectifier allows only one half-cycle (either positive or negative) of the input AC signal to pass through while blocking the other half. This results in a waveform with a non-zero average value, which corresponds to the d.c. component. The d.c. component of the output voltage of a half-wave rectifier is given by the formula:

V_{dc} = \frac{V_{peak}}{\pi}

where:

• \(V_{dc}\) is the d.c. component of the output voltage.
• \(V_{peak}\) is the peak value of the input AC voltage.

In the given problem, the peak voltage \(V_{peak}\) is 10V. Substituting this value into the formula, we get:

V_{dc} = \frac{10}{\pi}\,V

Therefore, the d.c. component of the output voltage is \(\frac{10}{\pi}\) V, which matches the correct option.

Let's briefly consider why the other options are incorrect:

• 10 V: This would be the peak voltage itself, not the average (d.c.) component.
• \(\frac{20}{\pi}\) V: This is twice the correct answer and does not fit the d.c. component formula for a half-wave rectifier.
• \(\frac{10}{\sqrt{2}}\) V: This is the RMS (Root Mean Square) value of the voltage, not the average (d.c.) component.