Concept:In a bridge rectifier, current flows through two conducting diodes in series with the load during each half-cycle, resulting in two diode forward drops (\( 2 V_D = 1.40\text{ V} \)).
Formula:$$V_{\text{load,peak}} = V_m - 2 V_D = V_m - 2(0.70\text{ V}) = V_m - 1.40\text{ V}$$
Solution:- Applying Kirchhoff's Voltage Law around the active conduction loop: \( V_m - V_{D1} - V_{\text{load}} - V_{D2} = 0 \).
- Because two diodes conduct simultaneously in series, the peak load voltage is \( V_{\text{load,peak}} = V_m - 2V_D = V_m - 1.40\text{ V} \).
Why other options are incorrect:- Option A: \( V_m \) applies only to ideal diodes with zero forward voltage drop.
- Option C: \( V_m - V_D \) is the peak load voltage for a center-tapped full-wave rectifier (which uses one diode per half-cycle).
- Option D: \( 2V_m - V_D \) is not applicable to standard bridge rectification.
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