Concept:

The total AC power input delivered by the secondary winding to a full-wave rectifier is given by \( P_{\text{ac}} = I_{\text{rms}}^2 R_{\text{total}} \), where \( R_{\text{total}} = r_f + R_L \) and the root-mean-square current for full-wave rectification is \( I_{\text{rms}} = \frac{I_m}{\sqrt{2}} \).

Formula:

$$I_m = \frac{V_m}{r_f + R_L}$$

$$I_{\text{rms}} = \frac{I_m}{\sqrt{2}}$$

$$P_{\text{ac}} = I_{\text{rms}}^2 (r_f + R_L) = \frac{I_m^2}{2} (r_f + R_L) = \frac{V_m^2}{2 (r_f + R_L)}$$

Solution:

• Total circuit loop resistance during each conduction half-cycle: \( R_{\text{total}} = r_f + R_L = 10\ \Omega + 1500\ \Omega = 1510\ \Omega \).



• Peak instantaneous current: \( I_m = \frac{30.2\text{ V}}{1510\ \Omega} = 0.020\text{ A} = 20\text{ mA} \).



• RMS current across the full AC cycle: \( I_{\text{rms}} = \frac{20\text{ mA}}{\sqrt{2}} \).



• Total AC power input: \( P_{\text{ac}} = I_{\text{rms}}^2 \times R_{\text{total}} = \left(\frac{0.020}{\sqrt{2}}\right)^2 \times 1510 = \frac{0.00040}{2} \times 1510 = 0.00020 \times 1510 = 0.302\text{ W} = \mathbf{302\text{ mW}} \).

Why other options are incorrect:

• Option A: 151 mW erroneously divides the total AC power input by 2.
• Option C: 75 mW is one-fourth of the total AC input power.
• Option D: 604 mW results from using peak current (\( I_m^2 R \)) rather than RMS current (\( I_{\text{rms}}^2 R \)).