Concept:In a bridge rectifier, current flows through two conducting diodes in series during each half-cycle, producing a total forward drop of \( 2V_B \).
Formula:$$V_{\text{load, peak}} = V_m - 2V_B$$
Solution:- Two forward-biased diodes conduct in series with the load during each half-cycle.
- Total forward barrier drop: \( 2 V_B = 2 \times 0.7\text{ V} = 1.4\text{ V} \).
- Calculate peak load voltage:
- $$V_{\text{load, peak}} = 21.4\text{ V} - 1.4\text{ V} = 20.0\text{ V}$$
Why other options are incorrect:- Option A: \( 21.4\text{ V} \) neglects the diode voltage drops entirely (ideal model).
- Option C: \( 20.7\text{ V} \) subtracts only a single diode drop (\( 0.7\text{ V} \)), which applies to center-tapped rectifiers rather than bridge rectifiers.
- Option D: \( 19.3\text{ V} \) subtracts three diode drops (\( 2.1\text{ V} \)).
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