Concept:A shunt capacitor acts as a low-pass filter. It charges up to the peak voltage when the rectifier output rises, and discharges slowly through the load resistor when the rectifier output drops, effectively reducing voltage ripple and providing a smoother DC level.
Formula:$$V_{\text{ripple (rms)}} \approx \frac{V_{\text{dc}}}{2\sqrt{3} f C R_L}$$
Solution:- During the peak of the rectified pulse, the capacitor rapidly charges to peak voltage \(V_m\).
- When the rectified waveform falls towards zero, the diode turns off and the capacitor supplies energy to the load \(R_L\).
- This fills in the valleys between pulses, greatly reducing ripple and smoothing the output voltage.
Why other options are incorrect:- Option A: A passive capacitor cannot amplify power or act as an active voltage amplifier.
- Option B: Current limiting is performed by series resistors, not parallel filter capacitors.
- Option D: A filter does not invert polarity; it maintains a unidirectional smoothed voltage.
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