Concept:Applying Kirchhoff's Voltage Law around the loop containing the transformer secondary, conducting diodes, and non-conducting diodes yields the practical Peak Inverse Voltage.
Formula:$$\text{PIV} = V_m - V_D$$
Solution:- During conduction, one diode pair connects the load across the transformer secondary.
- The reverse voltage across a non-conducting diode equals the peak supply voltage \( V_m \) minus the small forward voltage drop \( V_D \) of the conducting diode in series.
- Therefore, \( \text{PIV} = V_m - V_D \approx V_m \).
Why other options are incorrect:- Option B: \( 2V_m - V_D \) applies to center-tapped full-wave rectifiers, where each diode must block both secondary halves.
- Options C & D: Forward diode drops reduce the net reverse voltage seen by non-conducting diodes rather than increasing it.
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