If the rms value of sinusoidal input to a full wave rectifier is $\dfrac{{{v_0}}}{{\sqrt 2 }}$ , then the rms value of the rectifier's output is?
A. $\dfrac{{{v_0}}}{{\sqrt 2 }}$
B. $\dfrac{{{v_0}^2}}{{\sqrt 2 }}$
C. \[\dfrac{{{v_0}^2}}{2}\]
D. $\sqrt 2 {v_0}^2$
E. $2{v_0}^2$

Hint: In this problem, the input rms is given. In a full wave rectifier, the complete cycle is rectified. The output voltage is the same as the input voltage. A full wave rectifier rectifies the negative component of the input signal while the positive component of the signal remains the same.

Complete answer:
Let us first understand what a full wave rectifier is and how it works:
Full-wave rectifiers, rectify only the negative component of the input voltage into a positive voltage, then convert it into DC, utilizing a diode bridge configuration. In contrast, half-wave rectification removes only the negative voltage component before converting to DC.

The full wave rectifier circuit consists of two power diodes which are connected to a single load resistance with each diode supplying the current to the load resistor in turns. The average output voltage is higher than for that of a half wave rectifier, the output of the full wave rectifier has much less ripple than that of the half wave rectifier. This produces a smoother output waveform. Two diodes are now used in full wave rectifiers, one for each half of the cycle.

Since, the output voltage is the same as the input voltage, therefore option A is the correct answer.

Note: In case of full wave rectifiers, the rms value of sinusoidal input to a full wave rectifier is the rms value of the rectifier's output is the same. Also, only the negative input voltage is rectified in case of full wave rectifier. In the given problem, ${v_0}$ is the peak value of the voltage. Hence any option greater than this peak voltage must be eliminated.