General exam question G7B08
Subelement G7, PRACTICAL CIRCUITS· Group G7B
How is the efficiency of an RF power amplifier determined?
- A.Divide the DC input power by the DC output power
- B.Divide the RF output power by the DC input power (correct answer)
- C.Multiply the RF input power by the reciprocal of the RF output power
- D.Add the RF input power to the DC output power
WhyAn RF power amplifier's efficiency is determined by dividing the RF output power by the DC input power. An RF power amplifier's efficiency is found by dividing the useful RF output power it delivers by the DC input power it consumes to produce that output, answer B. Dividing DC input by DC output doesn't measure anything meaningful about RF conversion, multiplying RF input power by the reciprocal of RF output power is a nonsensical combination, and adding RF input to DC output mixes unrelated quantities, so A, C, and D are wrong.
Published 2026-08-28 · Last updated 2026-08-28 · Question text is verbatim from the 2023-2027 NCVEC Element 3 pool (public domain); the explanation above is original writing, see our methodology.
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