Concept:A single diode conducts only during the positive half-cycle when forward biased and blocks current during the negative half-cycle.
Formula:$$v_o(t) = \begin{cases} V_m \sin(\omega t) & 0 \le \omega t < \pi \\ 0 & \pi \le \omega t < 2\pi \end{cases}$$
Solution:- During the positive half-cycle, the diode is forward biased and conducts, producing a half-sine output across the load.
- During the negative half-cycle, the diode is reverse biased, cutting off current completely (\( V_{\text{load}} = 0 \)).
- The resulting waveform consists of unidirectional pulses separated by zero-voltage gaps.
Why other options are incorrect:- Option A: Describes unrectified AC input.
- Option B: Describes the output of a fully filtered and regulated DC power supply.
- Option C: Describes full-wave rectification.
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