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# The peak voltage in the output of a half-wave diode rectifier fed with a sinusoidal signal without filter is 10V. The DC component of the output voltage is:(A) $10/\sqrt 2 V$(B) $10/\pi V$(C) 10V(D) $20/\pi V$

Last updated date: 19th Sep 2024
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Hint We should know that the half wave rectifier passes only one of the halves of the input sine wave, which can be either positive or negative, and rejects the other half of the input sine wave. The output of the half wave rectifier is known as the pulsating DC. The ripple that is obtained by us in the output waveform can be reduced if we use a filter.

Here, the peak voltage is the output of a half- wave diode rectifier fed with a sinusoidal signal without the filter is 10V.
Hence, we can say that the dc component of the output voltage is given by:
${V_{av}} = \dfrac{{\int\limits_0^\pi {10\sin \left( {\omega t} \right)} }}{{2\pi }}$
We have to put the values to get the following expression:
$=\dfrac{10}{2\pi }(-\cos \theta )_{0}^{\pi }=\dfrac{2\times 10}{2\pi }$
$=10/\pi V$
So, we can say that the peak voltage in the output of a half-wave diode rectifier fed with a sinusoidal signal without filter is 10V will have the dc component of the output voltage as $10/\pi V$.

Hence the correct option is B.

Note We should know that the half wave rectifier always uses the principle of the PN junction diode and therefore it will convert the AC to the DC. In case of a half wave rectifier circuit, the load resistance is always connected to the PN junction diode in the series form. The alternating current is the input of the half wave rectifier.