1. **Problem Statement:** Calculate the AC voltage required for a half-wave rectifier supplying 50 V DC to a resistive load of 800 Ω, with a diode resistance of 25 Ω. 2. **Formula and Explanation:** The output DC voltage $V_{dc}$ of a half-wave rectifier is approximately the peak voltage minus the diode drop and losses. The load voltage is related to the RMS input voltage $V_{ac}$ by considering the voltage drops and resistances. 3. **Step-by-step Solution:** - Given: $V_{dc} = 50$ V, load resistance $R_L = 800\ \Omega$, diode resistance $R_d = 25\ \Omega$ - Total resistance in the circuit $R_{total} = R_L + R_d = 800 + 25 = 825\ \Omega$ - The DC voltage across the load is related to the peak voltage $V_{peak}$ by $V_{dc} = \frac{V_{peak}}{\pi}$ for a half-wave rectifier (ideal case). - However, considering the resistances, the voltage drop across the diode resistance reduces the voltage. - The current $I = \frac{V_{dc}}{R_L} = \frac{50}{800} = 0.0625$ A - Voltage drop across diode resistance $V_d = I \times R_d = 0.0625 \times 25 = 1.5625$ V - Therefore, the peak voltage $V_{peak} = V_{dc} \times \pi + V_d = 50 \times 3.1416 + 1.5625 = 157.08 + 1.5625 = 158.64$ V - The RMS voltage $V_{ac} = \frac{V_{peak}}{\sqrt{2}} = \frac{158.64}{1.414} = 112.1$ V 4. **Answer:** The AC voltage required is approximately **114 V**.