Concept:An inductor opposes changes in current. Its inductive reactance (\( X_L = 2\pi f L \)) increases with frequency, opposing AC ripple while allowing DC to pass with minimal ohmic loss.
Formula:$$X_L(\text{ac}) = 2\pi f L \gg R_{\text{choke}}, \quad X_L(\text{dc}) = 0\text{ }\Omega$$
Solution:- For DC (\( f = 0 \)), inductive reactance is zero (\( X_L = 0 \)), allowing DC current to pass with minimal resistance.
- For AC ripple frequencies (\( f = 100\text{ Hz} \) or \( 120\text{ Hz} \)), \( X_L \) is large, dropping the ripple voltage across the choke.
- This reduces the AC ripple reaching the series load resistor.
Why other options are incorrect:- Option A: Inductors block high-frequency AC and pass DC, which is the opposite of the statement.
- Option C: A passive filter inductor does not alter the semiconductor physics of diode drops.
- Option D: Inductor filters smooth current waveforms; they do not double output voltage.
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