Concept:An uncharged capacitor has zero initial voltage across its terminals, causing it to draw a large instantaneous inrush current limited only by winding and diode resistances.
Formula:$$I_{\text{surge}} = \frac{V_m - V_B}{R_{\text{sec}} + r_f + R_{\text{surge}}} \gg I_{\text{steady-state}}$$
Solution:- Before power is applied, voltage across the filter capacitor is zero (\( V_C(0^-) = 0\text{ V} \)).
- Switching on at the AC voltage peak (\( V_m \)) causes the capacitor to behave momentarily like a short circuit (\( \frac{dv}{dt} \to \infty \)).
- This produces a large initial surge current through the conducting diodes until the capacitor charges toward \( V_m \).
Why other options are incorrect:- Option A: Biasing and load currents do not change the intrinsic breakdown voltage of the diode.
- Option C: Diodes conduct in forward bias during charging rather than undergoing reverse breakdown.
- Option D: Conduction begins immediately upon applying forward voltage.
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