Concept:In a center-tapped full-wave rectifier, the non-conducting diode is subjected to the sum of the peak secondary half-winding voltage and the peak voltage across the load.
Formula:$$\text{PIV}_{\text{center-tapped}} = 2V_m$$
Solution:- When diode \( D_1 \) conducts, its cathode is at approximately \( +V_m \).
- The anode of the non-conducting diode \( D_2 \) is at \( -V_m \) due to the secondary center-tapped winding.
- The net reverse voltage across \( D_2 \) is \( V_{\text{cathode}} - V_{\text{anode}} = V_m - (-V_m) = 2V_m \).
Why other options are incorrect:- Option A: \( V_m / 2 \) is far below the actual reverse voltage and would result in diode destruction.
- Option B: \( V_m \) is the PIV for a bridge rectifier, not a center-tapped rectifier.
- Option C: \( V_m / \sqrt{2} \) corresponds to the RMS voltage.
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