Physics Electronics PMDC Conceptual Practice
PMDC Verified Question 234 of 494
In an inductor-input (choke-input) smoothing filter connected to a rectifier, how does the series inductor help reduce ripple in the load current?
A
It converts DC into high-frequency AC
B
It provides zero resistance to AC ripple
C
It opposes sudden changes in current (\( V_L = -L \frac{di}{dt} \)), smoothing out current fluctuations
D
It steps up the peak voltage to infinity
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option C: It opposes sudden changes in current (\( V_L = -L \frac{di}{dt} \)), smoothing out current fluctuations
Concept:

An inductor opposes changes in current by generating a self-induced back EMF (\( V_L = -L \frac{di}{dt} \)), smoothing out fluctuations in the rectified current delivered to the load.

Formula:

$$X_L = 2\pi f L \quad (\text{High impedance to AC ripple, zero resistance to steady DC})$$

Solution:

  • The series inductor presents a high inductive reactance (\( X_L = 2\pi f L \)) to AC ripple frequencies, attenuating current fluctuations.


  • It presents near-zero resistance to steady DC (\( f = 0 \)), allowing direct current to pass freely to the load.


Why other options are incorrect:

  • Option A: The filter smooths DC; it does not convert DC back into AC.
  • Option B: The inductor presents high impedance to AC ripple, not zero resistance.
  • Option D: Passive filters cannot step up DC voltages.

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