Statement I: To get a steady DC output from the pulsating voltage received from a full wave rectifier we can connect a capacitor across the output parallel to the load $$R_L$$.
Statement II: To get a steady DC output from the pulsating voltage received from a full wave rectifier we can connect an inductor in series with $$R_L$$.
In the light of the above statements, choose the most appropriate answer from the options given below:

We have a full-wave rectifier whose output is not a pure, steady direct voltage; instead the rectifier delivers a pulsating waveform whose envelope contains both a dc component and an ac (ripple) component. The task is to reduce or almost remove this unwanted ripple so that the voltage across the load resistor $$R_L$$ is as constant as possible.

First, recall a basic electrical fact: a capacitor opposes changes in voltage, while an inductor opposes changes in current. These two facts are the foundation of the two standard filter circuits used after a rectifier.

Let us examine Statement I. If a capacitor of capacitance $$C$$ is connected directly across the load, it charges up to the peak value of each half-cycle of the rectifier output. During the portion of each cycle in which the rectifier voltage falls below this peak, the diode(s) become reverse biased and the capacitor starts to discharge through $$R_L$$. Because the discharge takes place through the relatively large time constant $$\tau = R_L C$$, the voltage falls only a little before the next peak re-charges the capacitor. Thus the instantaneous voltage across the load remains nearly constant and the ripple amplitude

$$\Delta V \;=\; \frac{I_{\text{dc}}}{f\,C}$$

(formula for a simple capacitor filter, where $$I_{\text{dc}}$$ is the dc load current and $$f$$ is twice the mains frequency for a full-wave rectifier) can be made very small by choosing a sufficiently large $$C$$. Hence connecting a capacitor in parallel with $$R_L$$ indeed converts the pulsating output into an almost steady dc. Therefore Statement I is true.

Now consider Statement II. Instead of a capacitor shunt filter, we may insert an inductor $$L$$ in series with the load. When the rectangular-shaped rectified current attempts to rise rapidly at the start of each half-cycle, the induced emf $$-L\,\dfrac{di}{dt}$$ in the inductor opposes this change. Likewise, when the current tends to fall rapidly, the inductor releases its stored magnetic energy and tries to keep the current flowing. Mathematically the ripple factor for a simple series-inductor filter is

$$r \;=\; \frac{R_L}{3\sqrt{2}\,\omega\,L}, \qquad \text{where } \omega = 2\pi f.$$

We see that the ripple factor becomes small if $$\omega L \gg R_L$$; that is, a sufficiently large inductor dramatically smooths the current and therefore the voltage across the load. So a series inductor does give a steadier dc output. Hence Statement II is also true.

Because both Statement I and Statement II are correct, the option that declares both of them true is the right choice.

Hence, the correct answer is Option B.