Concept:For a half-wave rectifier with a shunt capacitor filter, the peak-to-peak ripple voltage is related to the DC load current, line frequency, and capacitance by \( V_r = \frac{I_{\text{dc}}}{f C} \).
Formula:$$V_r = \frac{I_{\text{dc}}}{f C} \implies C = \frac{I_{\text{dc}}}{f \cdot V_r}$$
Solution:- Given: \( I_{\text{dc}} = 100\text{ mA} = 0.10\text{ A} \), \( f = 50\text{ Hz} \), and \( V_r = 2.0\text{ V} \).
- Required capacitance: \( C = \frac{0.10\text{ A}}{50\text{ Hz} \times 2.0\text{ V}} = \frac{0.10}{100} = 0.0010\text{ F} = 1000\ \mu\text{F} \).
Why other options are incorrect:- Option B: 500 µF is the capacitance required for a full-wave rectifier (where ripple frequency is \( 2f \)).
- Option C: 100 µF would result in a large ripple voltage of 20 V.
- Option D: 250 µF would produce a ripple voltage of 8 V.
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