Concept:In a center-tapped full-wave rectifier, each diode handles one half-cycle independently. If one diode opens, the remaining diode continues to rectify its half-cycle, converting the circuit into a half-wave rectifier.
Formula:$$\text{Normal: } f_{\text{ripple}} = 2 f_{\text{in}}, \ V_{\text{dc}} = \frac{2V_m}{\pi} \longrightarrow \text{One Diode Open: } f_{\text{ripple}} = f_{\text{in}}, \ V_{\text{dc}} = \frac{V_m}{\pi}$$
Solution:- The half-winding associated with the open diode cannot deliver current to the load.
- The other half-winding and intact diode continue to supply current during their half-cycle.
- Thus, the circuit operates as a half-wave rectifier.
Why other options are incorrect:- Option A: The single diode still rectifies its active half-cycle, producing pulsating DC rather than AC.
- Option B: Output power and average DC voltage decrease by half.
- Option C: The intact diode continues conducting on its half-cycle.
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