Ham Amateur Radio Technician Practice Exam 2026 - Free Technician Practice Questions and Study Guide

To determine the output power from a 500 volts peak-to-peak signal across a 50-ohm load, you first need to calculate the RMS (Root Mean Square) voltage since power is typically calculated using RMS values for AC signals.

The relationship between peak-to-peak voltage and RMS voltage is given by:

\[ V_{RMS} = \frac{V_{peak-to-peak}}{2\sqrt{2}} \]

Substituting 500 volts for the peak-to-peak voltage, we find:

\[ V_{RMS} = \frac{500}{2\sqrt{2}} = \frac{500}{2 \times 1.414} = \frac{500}{2.828} \approx 176.78 \text{ volts} \]

Next, we can calculate the power in watts using the formula:

\[ P = \frac{V_{RMS}^2}{R} \]

Where \( R \) is the load resistance (50 ohms in this case). Plugging in the values:

\[ P = \frac{(176.78)^2}{50} \]

Calculating \( (176.78)^2 \) gives approximately 31345.448