Concept:An uncharged capacitor has zero initial voltage (\( v_C(0) = 0 \)), behaving momentarily as a short circuit (\( i_C = C \frac{dv}{dt} \)) and drawing a large initial inrush surge current until it charges.
Formula:$$I_{\text{surge}} = \frac{V_m - 2V_D}{R_{\text{transformer}} + r_f} \gg I_{\text{steady-state}}$$
Solution:- At the moment of turn-on (\( t = 0 \)), the filter capacitor is completely uncharged (\( V_C = 0 \)).
- The entire peak secondary voltage appears across the series resistance of the transformer and diodes.
- This produces a large inrush surge current, requiring diodes to be rated for peak non-repetitive surge currents (\( I_{\text{FSM}} \)) or protected by series thermistors.
Why other options are incorrect:- Option B: The forward-biased diodes conduct inrush current; they do not experience reverse breakdown.
- Option C: Inrush current begins immediately at turn-on.
- Option D: Transformer turns ratios are physically fixed by winding construction.
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