Concept:A shorted diode creates a zero-resistance path across one branch. When the opposite diode pair turns ON during the alternate half-cycle, it creates a direct short circuit across the transformer secondary winding.
Formula:$$I_{\text{short}} = \frac{V_s}{R_{\text{winding}}} \to \text{Very Large Fault Current} \implies \text{Fuse Blows / Component Failure}$$
Solution:- On the half-cycle where the shorted diode's branch is active, current flows through the short.
- On the alternate half-cycle, when another diode turns ON, the shorted diode creates a direct short circuit across the AC secondary winding.
- This draws excessive fault current, blowing the protective fuse or destroying the remaining diodes and transformer.
Why other options are incorrect:- Option B: Short circuits cause overcurrent faults, not stable DC generation.
- Option C: Efficiency drops to zero under a short-circuit fault condition.
- Option D: Short-circuiting the source does not eliminate ripple.
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