Concept:The ripple factor is defined as the ratio of the RMS value of the AC ripple component to the DC (average) component in the rectified output. For an unfiltered half-wave rectifier, this ratio is \(1.21\).
Formula:$$r = \sqrt{\left(\frac{I_{\text{rms}}}{I_{\text{dc}}}\right)^2 - 1} = \sqrt{\left(\frac{I_0 / 2}{I_0 / \pi}\right)^2 - 1} = \sqrt{\left(\frac{\pi}{2}\right)^2 - 1}$$
Solution:- Evaluate the form factor \(k_f = \frac{\pi}{2} \approx 1.5708\).
- $$r = \sqrt{(1.5708)^2 - 1} = \sqrt{2.4674 - 1} = \sqrt{1.4674} \approx 1.211$$
- Thus, the ripple factor of a half-wave rectifier is \(1.21\) (meaning AC ripple exceeds the DC output).
Why other options are incorrect:- Option A: \(0.48\) is the ripple factor of an unfiltered full-wave rectifier.
- Option C: \(1.57\) is the form factor (\(I_{\text{rms}} / I_{\text{dc}}\)) of a half-wave rectifier, not the ripple factor.
- Option D: \(1.11\) is the form factor of a full-wave rectifier.
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