Radio Amateur Quiz Canadian License

EN FR

Question A-003-002-007

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

Correct Answer

625 watts

Explanation

Peak Envelope Voltage (PEV) = Vpp / 2 = 500V / 2 = 250V. PEP = (PEV^2) / (2 × R) for a sinusoidal envelope peak. PEP = (250^2) / (2 × 50) = 62500 / 100 = 625 watts.

Warning: Generated by AI