Physics Electronics PMDC Conceptual Practice
PMDC Verified Question 235 of 494
Why is a \( \pi \)-section (CLC) filter more effective at reducing ripple than a simple shunt capacitor filter alone?
A
It eliminates the need for any rectifying diodes
B
It completely eliminates all electrical resistance from the load
C
It converts input AC into high-voltage radio frequencies
D
It combines two shunt capacitors with a series inductor, providing two stages of capacitive smoothing and inductive ripple attenuation
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option D: It combines two shunt capacitors with a series inductor, providing two stages of capacitive smoothing and inductive ripple attenuation
Concept:

A \( \pi \)-section (CLC) filter uses an input shunt capacitor \( C_1 \), a series inductor \( L \), and an output shunt capacitor \( C_2 \) in a Greek letter \( \pi \) shape, providing multi-stage ripple attenuation.

Formula:

$$\text{Ripple Factor (CLC Filter): } r \approx \frac{\sqrt{2}}{8\pi^3 f^3 L C_1 C_2 R_L} \propto \frac{1}{L C_1 C_2}$$

Solution:

  • The first capacitor (\( C_1 \)) provides initial peak-charge smoothing across the rectifier output.


  • The series choke (\( L \)) blocks remaining AC ripple current with high reactance.


  • The second capacitor (\( C_2 \)) bypasses any residual ripple to ground, delivering a very smooth DC output to the load.


Why other options are incorrect:

  • Option A: Rectifying diodes are still required to convert AC to DC before the filter.
  • Option B: The filter does not alter the load resistance.
  • Option C: The filter removes AC fluctuations; it does not generate radio frequencies.

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