Concept:A bridge rectifier requires two series diodes to conduct on each half-cycle. If one diode opens, its conduction branch is disabled for that half-cycle, while the other branch continues conducting on the opposite half-cycle, converting the circuit into a half-wave rectifier.
Formula:$$\text{One Diode Open: } f_{\text{ripple}} = f_{\text{in}} \quad (\text{Half-Wave Operation}) \quad \text{and} \quad V_{\text{dc}} = \frac{V_m}{\pi}$$
Solution:- The half-cycle that depends on the open diode cannot complete its circuit path, so load current is zero during that half-cycle.
- The opposite half-cycle uses the remaining intact diode pair, which continues conducting normally.
- As a result, the bridge rectifier operates as a half-wave rectifier with double the ripple and half the DC output power.
Why other options are incorrect:- Option A: Full-wave rectification requires all four diodes to be functional.
- Option B: The intact diode pair still conducts on the opposite half-cycle, so the output does not drop to zero entirely.
- Option D: Ripple frequency drops from \( 2f_{\text{in}} \) to \( f_{\text{in}} \); it does not quadruple.
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