Concept:When the input AC voltage drops below the capacitor voltage, the diode turns OFF and the capacitor discharges stored charge into the load resistor \( R_L \) with an exponential decay time constant \( \tau = R_L C \).
Formula:$$v_C(t) = V_m \exp\left(-\frac{t}{R_L C}\right) \approx V_m \left(1 - \frac{t}{R_L C}\right) \quad (\text{for } t \ll R_L C)$$
Solution:- The capacitor charges to peak voltage \( V_m \) during diode conduction.
- When the diode turns OFF, the capacitor discharges through \( R_L \) according to \( v(t) = V_m e^{-t / (R_L C)} \).
- A larger time constant \( R_L C \) slows this discharge, reducing output ripple.
Why other options are incorrect:- Option B: \( V_m(1 - e^{-t/RC}) \) describes capacitor charging from a DC source, not discharging.
- Option C: The discharge is governed by an exponential decay, not an AC cosine wave.
- Option D: Linear decay is only an approximation for the initial portion of the exponential curve.
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