Concept:In a center-tapped full-wave rectifier, only one diode conducts during each half-cycle, introducing only a single forward voltage drop (\( V_D = 0.70\text{ V} \)) between the secondary winding and the load.
Formula:$$V_{\text{load,peak}} = V_m - V_D = V_m - 0.70\text{ V}$$
Solution:- Applying Kirchhoff's Voltage Law to the active half-winding loop: \( V_m - V_D - V_{\text{load}} = 0 \).
- Because only one diode conducts per half-cycle, the peak load voltage is \( V_{\text{load,peak}} = V_m - V_D = V_m - 0.70\text{ V} \).
Why other options are incorrect:- Option A: \( V_m \) assumes an ideal diode with zero threshold voltage.
- Option B: \( V_m - 2V_D \) is the peak load voltage for a bridge rectifier, where two diodes conduct in series.
- Option D: \( 2V_m - 2V_D \) incorrectly uses the full end-to-end secondary voltage.
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