Concept:For a half-wave rectifier, the output waveform repeats once every full cycle of the input AC waveform. Therefore, the ripple frequency equals the input supply frequency.
Formula:$$v(t) = V_m \sin(\omega t) = V_m \sin(2\pi f t)$$
$$f_{\text{ripple, HW}} = f_{\text{in}}$$
Solution:- Comparing \(v(t) = 250 \sin(100\pi t)\) with \(v(t) = V_m \sin(2\pi f t)\):
- $$2\pi f = 100\pi \implies f = \frac{100\pi}{2\pi} = 50\text{ Hz}$$
- For a half-wave rectifier:
- $$f_{\text{ripple}} = f_{\text{in}} = 50\text{ Hz}$$
Why other options are incorrect:- Option A: \(25\text{ Hz}\) results from dividing the supply frequency by 2.
- Option C: \(100\text{ Hz}\) would be the output ripple frequency if a full-wave rectifier were used.
- Option D: \(200\text{ Hz}\) is an incorrect multiple.
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