Ham Radio General Class Practice Test

What is the output PEP from a transmitter if an oscilloscope measures 500 volts peak-to-peak across a 50-ohm resistor connected to the transmitter output?

1. 8.75 watts

2. 625 watts

3. 2500 watts

4. 5000 watts

The correct answer is: 625 watts

To determine the output PEP (Peak Envelope Power) of a transmitter based on the given peak-to-peak voltage measurement, we will utilize the formula related to power in a resistive load. First, we need to convert the peak-to-peak voltage (Vpp) to the RMS (root mean square) voltage, which can then be used to calculate power. For a sine wave, the relationship between peak-to-peak voltage and RMS voltage is given by: \[ V_{RMS} = \frac{V_{pp}}{2\sqrt{2}} \] In this case, given that the voltage measured is 500 volts peak-to-peak, the RMS voltage calculation will be: \[ V_{RMS} = \frac{500 V}{2\sqrt{2}} \approx 176.78 V \] Next, we calculate the power using the formula for power in a resistive load: \[ P = \frac{V_{RMS}^2}{R} \] Where \( R \) is the resistance, which in this scenario is 50 ohms. Plugging in the values we have: \[ P = \frac{(176.78 V)^2}{50 \Omega