Concept:When two identical ideal diodes are connected in inverse-parallel (antiparallel), one diode conducts during the positive half-cycle while the other conducts during the negative half-cycle.
Formula:$$\text{Positive Half-Cycle: } D_1 \text{ ON}, D_2 \text{ OFF} \implies v_R(t) = +V_m \sin(\omega t)$$
$$\text{Negative Half-Cycle: } D_2 \text{ ON}, D_1 \text{ OFF} \implies v_R(t) = -V_m |\sin(\omega t)|$$
Solution:- During the positive half-cycle, \( D_1 \) is forward-biased and conducts current through \( R \) from left to right.
- During the negative half-cycle, \( D_2 \) is forward-biased and conducts current through \( R \) from right to left.
- Because both half-cycles pass through \( R \) with opposite polarities, no rectification occurs and the output voltage matches the original AC input wave.
Why other options are incorrect:- Option A: Half-wave rectification blocks one of the half-cycles, which does not happen here because the inverse-parallel diode conducts it.
- Option B: Full-wave rectification requires current to pass through the load in the same direction on both half-cycles.
- Option C: A constant DC level requires rectification followed by large smoothing filters.
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