Concept:A center-tapped transformer produces two out-of-phase secondary voltages with respect to the center tap, alternately biasing the two diodes.
Formula:$$V_{\text{sec1}} = +V_m \sin(\omega t), \quad V_{\text{sec2}} = -V_m \sin(\omega t)$$
Solution:- During the positive half-cycle, the top terminal is positive and the bottom terminal is negative relative to the center tap.
- Diode \( D_1 \) (connected to the top terminal) is forward-biased and conducts current through the load.
- Diode \( D_2 \) (connected to the bottom terminal) is reverse-biased and blocks current.
Why other options are incorrect:- Option B: Diode \( D_2 \) conducts during the negative half-cycle, not the positive half-cycle.
- Option C: The two diodes cannot conduct simultaneously in a center-tapped transformer because its outer terminals have opposite polarities.
- Option D: If both diodes were reverse-biased, no output voltage would appear across the load.
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