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?

Why

To find the power, first halve the 200 volts peak-to-peak to get 100 volts peak, then divide by the square root of two to find the root-mean-square voltage, which represents the effective heating value. The full explanation — and one for every question — is in Pro.

Practise this section

Measurements is section A-003 of the Advanced exam: Measurements: meters, oscilloscopes, counters, signal generators, and what a reading actually tells you.

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