Concept:With a filter capacitor, the diode conducts only when the incoming AC voltage exceeds the capacitor voltage (\( v_{\text{in}}(t) > v_C(t) \)). This short conduction angle forces all recharging current into brief, high-amplitude pulses.
Formula:$$I_{\text{diode,peak}} = I_{\text{dc}} \left( 1 + \pi \sqrt{\frac{2 V_m}{V_r}} \right) \gg I_{\text{dc}}$$
Solution:- The capacitor maintains a high voltage across the load between peaks.
- The diode turns ON only near the crest of the AC cycle when \( v_{\text{in}} \) exceeds \( v_C \).
- Because all charge delivered to the load over the full cycle must be replenished during this brief window, the diode conducts in short, high-current pulses.
Why other options are incorrect:- Option A: Full \( 180^\circ \) conduction occurs only with purely resistive loads without a filter capacitor.
- Option C: Conduction occurs near the positive peak, not during the negative half-cycle.
- Option D: Peak diode current is significantly higher than the average DC load current.
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