Concept:In any full-wave rectifier circuit (center-tapped or bridge), two output pulses are generated for every single cycle of the input AC supply. Therefore, the output ripple frequency is exactly twice the input AC line frequency.
Formula:$$v(t) = V_m \sin(2\pi f_{\text{in}} t) \implies 2\pi f_{\text{in}} = 120\pi \implies f_{\text{in}} = 60\text{ Hz}$$
$$f_{\text{ripple}} = 2 \times f_{\text{in}}$$
Solution:- Calculate input frequency:
- $$f_{\text{in}} = \frac{120\pi}{2\pi} = 60\text{ Hz}$$
- Calculate output ripple frequency for full-wave rectification:
- $$f_{\text{ripple}} = 2 \times 60\text{ Hz} = 120\text{ Hz}$$
Why other options are incorrect:- Option A: \(60\text{ Hz}\) is the input supply frequency, which would be the output ripple frequency only for a half-wave rectifier.
- Option C: \(240\text{ Hz}\) represents four times the input frequency.
- Option D: \(30\text{ Hz}\) is half the input frequency.
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