Deep Dive: G5B14
The correct answer is B: 625 watts. The output PEP of 500 volts peak-to-peak across a 50-ohm load is 625 watts. Vpeak = Vpp/2 = 500/2 = 250V. Vrms = Vpeak/√2 = 250/1.414 = 176.8V. PEP = Vrms²/R = 176.8²/50 = 31,250/50 = 625W. For amateur radio operators, this is a basic PEP calculation. Understanding this helps when calculating PEP from peak-to-peak voltage.
Why Other Answers Are Wrong
Option A (8.75 watts): Incorrect. 8.75 watts is way too low - calculation error. 500V peak-to-peak gives much more power. Option C (2500 watts): Incorrect. 2500 watts is too high - that might be using peak voltage incorrectly. Correct calculation gives 625W. Option D (5000 watts): Incorrect. 5000 watts is way too high - that might be using peak-to-peak voltage directly. Correct calculation gives 625W.
Exam Tip
PEP from 500V peak-to-peak, 50Ω: Vpeak = 250V, Vrms = 176.8V, P = 176.8²/50 = 625W. Think 'P'EP = 'P'eak 'E'nvelope 'P'ower. Convert peak-to-peak to RMS, then use P = Vrms²/R. Not 8.75W, not 2500W, not 5000W - just 625W.
Memory Aid
PEP from 500V peak-to-peak, 50Ω = 625W. Think 'P'EP = 'P'eak 'E'nvelope 'P'ower. Vpeak = 250V, Vrms = 176.8V, P = 625W. Standard PEP calculation.
Real-World Example
500 volts peak-to-peak across a 50-ohm load. Peak voltage = 500/2 = 250V. RMS voltage = 250/√2 = 176.8V. PEP = 176.8²/50 = 31,250/50 = 625 watts. This is the PEP produced.
Source & Coverage
Question Pool: 2023-2027 Question Pool
Subelement: G5B
Reference: 2023-2027 Question Pool · G5 - Electrical Principles
Key Concepts
Verified Content
Question from the official FCC General Class pool. Explanation reviewed by licensed amateur radio operators and mapped to the G5B topic.