1. **Problem Statement:** Calculate the AC voltage required to supply 50V DC to a resistive load of 800 Ω using a half-wave rectifier with a diode resistance of 25 Ω. 2. **Understanding the Circuit:** In a half-wave rectifier, the output DC voltage ($V_{dc}$) is related to the peak voltage ($V_{peak}$) of the AC input by the formula: $$V_{dc} = \frac{V_{peak}}{\pi} - I \times R_d$$ where $I$ is the current through the load and $R_d$ is the diode resistance. 3. **Calculate the Current ($I$):** Using Ohm's law for the DC side: $$I = \frac{V_{dc}}{R} = \frac{50}{800} = 0.0625\,A$$ 4. **Express $V_{dc}$ in terms of $V_{peak}$:** Rearranging the formula: $$V_{dc} = \frac{V_{peak}}{\pi} - I R_d$$ Substitute $V_{dc} = 50$, $I = 0.0625$, and $R_d = 25$: $$50 = \frac{V_{peak}}{\pi} - 0.0625 \times 25$$ Calculate the voltage drop across the diode resistance: $$0.0625 \times 25 = 1.5625$$ So: $$50 = \frac{V_{peak}}{\pi} - 1.5625$$ 5. **Solve for $V_{peak}$:** $$\frac{V_{peak}}{\pi} = 50 + 1.5625 = 51.5625$$ Multiply both sides by $\pi$: $$V_{peak} = 51.5625 \times \pi \approx 51.5625 \times 3.1416 = 161.95\,V$$ 6. **Calculate RMS Voltage ($V_{rms}$):** For a sinusoidal AC voltage: $$V_{rms} = \frac{V_{peak}}{\sqrt{2}} = \frac{161.95}{1.414} \approx 114.5\,V$$ **Final Answer:** The required AC RMS voltage is approximately **114.5 V**.