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**Hint**Ripple is defined as the unwanted AC component that remains in a circuit while converting the AC voltage into a DC Voltage. Write down the ripple formula for a half wave rectifier and substitute the values for a standard AC circuit and find the value.

**Complete Step By Step Solution**

A Half wave rectifier is a type of rectifier which is tuned to allow only one half cycle of alternating transformation wave to pass and block the other half , which converts one half AC voltage into DC voltage. This rectifier requires only one single diode to achieve this operation.

In general, the ripple factor is defined as the ratio of the Root mean squared value of the AC component and the DC component RMS value that’s obtained as the output of the rectifier. The main reason behind the occurrence of ripple factor is due to the presence of unwanted AC components that are still fluctuating in a rectifier circuit during the conversion process.

Mathematically ,ripple factor is given

\[\gamma = \sqrt {{{(\dfrac{{{V_{rms}}}}{{{V_{DC}}}})}^2} - 1} \]

Where \[{V_{rms}}\] stands for AC voltage and \[{V_{DC}}\]stands for output DC voltage.

The above equation can also be rearranged to give ,Ripple factor (r)

\[r = \sqrt {(\dfrac{{{I^2}_{rms}}}{{{I^2}_{DC}}}) - 1} \]

For a half wave rectifier, we know it operates only in half cycle. Therefore the current and voltage values will be halved from the original to provide desired output.

Thus, we get, \[{I_{rms}} = {I_m}/2\] and \[{I_{DC}} = {I_m}/\pi \]. Substituting this on the above formula we get,

\[ \Rightarrow r = \sqrt {(\dfrac{{{I^2}_{rms}}}{{{I^2}_{DC}}}) - 1} \]

\[ \Rightarrow r = \sqrt {\dfrac{{(\dfrac{{{I^2}_m}}{4})}}{{(\dfrac{{{I^2}_m}}{{{\pi ^2}}})}} - 1} \]

\[ \Rightarrow r = \sqrt {(\dfrac{{{\pi ^2} - 4}}{4}) - 1} = 1.21\]( For a standard half wave-rectifier)

**Thus, option(a) is the right answer for the given question.**

**Note**In order to construct an efficient rectifier, the ripple factor must be kept very low. Usage of external components such as capacitors and inductors as primary filters will help to reduce the ripples occurring in the circuit.

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