General exam question G5B06
Subelement G5, ELECTRICAL PRINCIPLES· Group G5B
G5B06
What is the PEP produced by 200 volts peak-to-peak across a 50-ohm dummy load?
- A.1.4 watts
- B.100 watts (correct answer)
- C.353.5 watts
- D.400 watts
Why200 volts peak-to-peak across a 50-ohm dummy load produces a PEP of 100 watts. PEP into a resistive load from a peak-to-peak voltage is found with PEP = (V_pp)² ÷ (8 × R), so 200 volts peak-to-peak squared (40,000) divided by 400 (8 times 50 ohms) comes out to 100 watts, answer B. The other values don't follow from applying that formula to a 200 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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