Radio Amateur Quiz Canadian License

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Question A-003-002-011

An oscilloscope measures 500 volts peak-to-peak across a 50 ohm dummy load connected to the transmitter output during unmodulated carrier conditions. What would an average-reading power meter indicate under the same transmitter conditions?

Correct Answer

625 watts

Explanation

For an unmodulated carrier, average power = PEP. Peak Voltage (Vp) = Vpp / 2 = 500V / 2 = 250V. Average Power = (V_rms)^2 / R = (Vp/√2)^2 / R = Vp^2 / (2R) = (250^2) / (2 × 50) = 625 watts.

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