To reduce the ripples in a rectifier circuit with capacitor filter

\[\left( a \right)\] ${R_L}$ should be increased
$\left( b \right)$ Input frequency should be decreased
$\left( c \right)$ Input frequency should be increased
$\left( d \right)$ Capacitors with high capacitance should be used

Hint: The ripple factor is defined as the ratio of alternating current component to the direct current component in the rectifier output is said to be as the ripple factor. And mathematically it can be represented as $\dfrac{{{I_{a.c}}}}{{{I_{d.c}}}}$ . And by using this we will be able to answer this question.

Complete step by step answer:
We know the ripple factor of a full wave rectifier using capacitor filter is expressed as:
$ \Rightarrow r = \dfrac{1}{{4\sqrt 3 {R_L}C\upsilon }}$
Here, ${R_L}$ is the load resistance, $C$ is the capacitance, and $\upsilon $ is the input frequency.
From the above relation, we can write the relationship of ripple factor with capacitance, input frequency and load resistance, as:
\[ \Rightarrow r \propto \dfrac{1}{{{R_L}C\upsilon }}\]
Therefore, the higher the values of capacitance, input frequency and load resistance, the lower the value of ripple factor.
Hence, to reduce the ripples in a rectifier circuit with capacitor filter ${R_L}$ should be increased, Input frequency should be increased, and Capacitor with high capacitance should be used.

Therefore, the option \[\left( a \right)\] is correct.

Additional Information: Ripple factor of a rectifier $ = \dfrac{{{\text{value}}\;{\text{of}}\;{\text{a}}{\text{.c}}{\text{.}}\;{\text{component}}}}{{{\text{value}}\;{\text{of}}\;{\text{d}}{\text{.c}}{\text{.}}\;{\text{component}}}} = \dfrac{{{I_{a.c.}}}}{{{I_{d.c.}}}} = \dfrac{{{E_{a.c}}}}{{{E_{d.c}}}}$ .

Note: A single capacitor C of high value of capacitance is connected across the output of the rectifier could serve the purpose of the filter circuit.
The capacitance offers low impedance to the alternating current component $\left( {{X_C} = \dfrac{1}{{2\pi \upsilon C}}} \right)$ and offers infinite impedance to direct current component. Therefore, the alternative current component is bypassed or filtered out. The direct current component is obstructed by the capacitor. It produces a voltage drop across load resistance ${R_L}$ as a filtered direct current output, which is almost equal to direct current voltage. This type of filter is mostly used in supplying power.