Concept:For continuous inductor current in an L-section filter, the DC load current must exceed the peak AC ripple current through the inductor.
Formula:$$L_c \approx \frac{R_L}{3\omega} = \frac{R_L}{6\pi f} \implies I_{\text{critical}} = \frac{V_{\text{dc}}}{R_{L,\text{max}}} = \frac{2V_m}{3\pi \omega L}$$
Solution:- When load current exceeds \( I_{\text{critical}} \), inductor current flows continuously, maintaining an average DC output voltage of \( V_{\text{dc}} = \frac{2V_m}{\pi} \).
- If load current falls below \( I_{\text{critical}} \) (light load or open circuit), current through the inductor becomes discontinuous.
- The capacitor then charges toward the peak voltage \( V_m \), causing the DC output voltage to rise.
Why other options are incorrect:- Option A: Magnetic saturation occurs at high currents, not low currents.
- Option B: The output voltage rises toward \( V_m \) rather than dropping to zero.
- Option C: The diodes continue conducting pulse currents to maintain the capacitor charge.
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