Concept:In a full-wave rectifier, the ripple frequency is doubled (\( 2f = 100\text{ Hz} \)), so the peak-to-peak ripple voltage is given by \( V_r = \frac{I_{\text{dc}}}{2 f C} \).
Formula:$$V_r = \frac{I_{\text{dc}}}{2 f C} \implies C = \frac{I_{\text{dc}}}{2 f \cdot V_r}$$
Solution:- Given: \( I_{\text{dc}} = 100\text{ mA} = 0.10\text{ A} \), \( f = 50\text{ Hz} \) (so \( 2f = 100\text{ Hz} \)), and \( V_r = 2.0\text{ V} \).
- Required capacitance: \( C = \frac{0.10\text{ A}}{2 \times 50 \times 2.0\text{ V}} = \frac{0.10}{200} = 0.00050\text{ F} = 500\ \mu\text{F} \).
Why other options are incorrect:- Option A: 1000 µF is the capacitance required for a half-wave rectifier.
- Option C: 250 µF would allow a larger ripple voltage of 4 V.
- Option D: 2000 µF would reduce ripple to 0.5 V, exceeding the required 2.0 V specification.
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