Question

If the rms value of the sinusoidal input to a full wave rectifier is $\dfrac{{{V_o}}}{{\sqrt 2 }}$, then the rms value of the rectifier’s output is
(A) $\dfrac{{{V_o}}}{{\sqrt 2 }}$
(B) \[\dfrac{{V_o^2}}{{\sqrt 2 }}\]
(C) \[\dfrac{{V_o^2}}{2}\]
(D) \[\sqrt 2 V_o^2\]

Answer 1

Hint:
The rms value is given by the value of the peak voltage divided by $\sqrt 2 $. So in case of a full wave rectifier, the peak value of the sine wave still remains the same. So the rms value is the same as the rms value of the sinusoidal input.
Formula used: In this solution we will be using the following formula,
$\Rightarrow {V_{rms}} = \dfrac{{{V_o}}}{{\sqrt 2 }}$
Where ${V_{rms}}$ is the rms voltage and ${V_o}$ is the peak to peak voltage of the sinusoidal input.

Complete step by step answer:
The circuit of a full wave rectifier consists of 2 diodes. One of each of the diodes is used for each of the half cycles. In case of the half wave rectifier, there are spaces present between the half waves, but in the cases of the full wave rectifier, these spaces are filled.
The peak voltage of the output waveform of the full wave rectifier is the same as that of the half wave rectifier. This peak voltage is the same as that of the input sinusoidal wave.
So the rms value of the full wave rectifier will be equal to the rms value of the input sinusoidal wave.
For the input sinusoidal wave, it is given in the question that the rms voltage is given as,
$\Rightarrow {V_{rms}} = \dfrac{{{V_o}}}{{\sqrt 2 }}$
So for the full wave rectifier also, the rms voltage is given as $\dfrac{{{V_o}}}{{\sqrt 2 }}$.
So the correct answer is option (A).

Note:
The half wave rectifier is a kind of rectifier that allows only one half of the cycle of an ac waveform to pass and blocks the other half cycle. These half wave rectifiers are used to convert ac voltage or current to dc and use only one diode.