Concept:The peak load voltage in a practical half-wave rectifier accounts for the diode forward barrier voltage drop (\( V_B \)).
Formula:$$I_{\text{dc}} = \frac{V_{m} - V_B}{\pi R_L}$$
Solution:- Calculate peak load voltage: \( V_{\text{peak, load}} = V_m - V_B = 32.1\text{ V} - 0.7\text{ V} = 31.4\text{ V} \).
- Calculate peak current: \( I_m = \frac{31.4\text{ V}}{10\text{ }\Omega} = 3.14\text{ A} \).
- Calculate average DC current:
- $$I_{\text{dc}} = \frac{I_m}{\pi} = \frac{3.14\text{ A}}{3.1416} \approx 1.00\text{ A}$$
Why other options are incorrect:- Option A: \( 3.14\text{ A} \) is the peak load current (\( I_m \)), not the average DC current.
- Option B: \( 0.50\text{ A} \) results from an extra division by 2.
- Option C: \( 2.00\text{ A} \) is the DC current for a full-wave rectifier under the same conditions.
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