Concept:A diode conducts when its anode is made positive relative to its cathode by an amount greater than the forward threshold voltage \( V_0 \).
Formula:$$I_{\text{load}} = \frac{V_m \sin(\omega t) - V_{\text{barrier}}}{R_f + R_L} \quad (\text{for } V_{\text{in}} > V_{\text{barrier}})$$
Solution:- During the positive half-cycle of the input \( \text{AC} \), the transformer secondary makes the anode positive with respect to the cathode.
- The diode is forward-biased, reducing its dynamic resistance and allowing current to flow through load resistor \( R_L \).
Why other options are incorrect:- Option A: During the negative half-cycle, the diode is reverse-biased and blocks current.
- Option C: At zero input voltage, no external electric field is applied, so the net current is zero.
- Option D: Reverse breakdown occurs under high reverse voltage and is avoided in normal rectifier operation.
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