Ham Radio General Class Test 2026 – 400 Free Practice Questions to Pass the Exam

Question: 1 / 545

What is the output PEP from a transmitter if an oscilloscope measures 200 volts peak-to-peak across a 50-ohm dummy load connected to the transmitter output?

1.4 watts

100 watts

To determine the output PEP (Peak Envelope Power) of the transmitter, one must analyze the measured voltage across the 50-ohm dummy load. The voltage measured is peak-to-peak, so to calculate the PEP, the peak voltage is needed.

First, the peak voltage (Vpk) can be derived from the peak-to-peak voltage. The relationship is that the peak voltage is half of the peak-to-peak voltage. By taking the 200 volts peak-to-peak and dividing it by 2, the peak voltage is found to be 100 volts.

Next, to calculate the PEP, the formula used is:

\[ PEP = \frac{(V_{pk})^2}{R} \]

where:

- Vpk is the peak voltage (100 volts)

- R is the resistance (50 ohms)

Inserting the calculated peak voltage into this formula:

\[ PEP = \frac{(100)^2}{50} = \frac{10000}{50} = 200 \text{ watts} \]

However, when considering the PEP in linear metrics, it is important to note that the value since power is given in terms of RMS voltage and peak power can be discussed

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353.5 watts

400 watts

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