General exam question G5B14
Subelement G5, ELECTRICAL PRINCIPLES· Group G5B
G5B14
What is the output PEP of 500 volts peak-to-peak across a 50-ohm load?
- A.8.75 watts
- B.625 watts (correct answer)
- C.2500 watts
- D.5000 watts
Why500 volts peak-to-peak across a 50-ohm load produces an output PEP of 625 watts. PEP into a resistive load from a peak-to-peak voltage is found with PEP = (V_pp)² ÷ (8 × R), so 500 volts peak-to-peak squared (250,000) divided by 400 (8 times 50 ohms) comes out to 625 watts, answer B. The other values don't follow from applying that formula to a 500 V p-p signal across 50 ohms.
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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