AIPMT PRELIMS 2004 Physics Rectifiers MCQ Question

To determine the direct current (d.c.) component of the output voltage from a half-wave diode rectifier fed with a sinusoidal signal, we start by analyzing the behavior of the rectifier.

Explanation of the Correct Answer

In a half-wave rectifier, the output voltage is derived from the positive half of the input sinusoidal waveform. When we have a peak voltage (V_peak) of 10V, the output voltage waveform will look like a series of pulses, each corresponding to the positive half of the input sinusoidal wave.

The d.c. component (or average value) of the output voltage for a half-wave rectifier can be calculated using the formula:

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

This formula arises because, for a half-wave rectifier, the average value of the output voltage over one complete cycle (which includes both positive and negative halves, but only the positive half is considered) is given by integrating the positive half of the sine wave and normalizing it over the period.

Given that the peak voltage $V_{peak}$ is 10V, we can substitute this into the formula:

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

Thus, the d.c. component of the output voltage is $\frac{10}{\pi}$ volts, which corresponds to Option A.

Clarification of Other Options

Let's briefly evaluate the other options:

• Option B: 10 V
  This option suggests that the d.c. component is equal to the peak voltage. However, for a half-wave rectifier, the average value is always less than the peak voltage because it only considers the positive half of the waveform. Thus, this option is incorrect.

• Option C: 20/π V
  This value does not correspond to any standard calculation for a half-wave rectifier. The average output voltage is derived specifically from the peak voltage divided by $\pi$, not doubled. Hence, this option is also incorrect.

• Option D: 10/√2 V
  This value represents the root mean square (r.m.s.) value of the peak voltage. The r.m.s. value is calculated as $V_{rms} = \frac{V_{peak}}{\sqrt{2}}$. However, the r.m.s. value is not the same as the d.c. component. Therefore, this option is also incorrect.

Summary

The correct answer is Option A: $\frac{10}{\pi}$ V because it correctly applies the formula for the d.c. component of the output voltage from a half-wave rectifier, which is derived from the peak voltage. Other options either misinterpret the relationship between peak voltage and average voltage or refer to different voltage types altogether.