Deep Dive: G5B12
The correct answer is B: 245 volts. The RMS voltage across a 50-ohm dummy load dissipating 1200 watts is 245 volts. P = V²/R, so V = √(P×R) = √(1200×50) = √60,000 = 245V. For amateur radio operators, this is a basic power/voltage calculation. Understanding this helps when calculating voltages from power.
Why Other Answers Are Wrong
Option A (173 volts): Incorrect. 173 volts is too low - that would give P = 173²/50 = 600W, not 1200W. Calculation error. Option C (346 volts): Incorrect. 346 volts is too high - that would give P = 346²/50 = 2400W, not 1200W. Calculation error. Option D (692 volts): Incorrect. 692 volts is way too high - that would give P = 692²/50 = 9600W, not 1200W. Calculation error.
Exam Tip
RMS voltage from power: V = √(P×R) = √(1200×50) = √60,000 = 245V. Think 'V'oltage = √('P'ower × 'R'esistance). Use V = √(P×R) formula. Not 173V (600W), not 346V (2400W), not 692V (9600W) - just 245V.
Memory Aid
RMS voltage from power: V = √(P×R) = √(1200×50) = 245V. Think 'V'oltage = √('P'ower × 'R'esistance). Use V = √(P×R) formula. Standard voltage calculation from power.
Real-World Example
A 50-ohm dummy load dissipates 1200 watts. RMS voltage = √(P×R) = √(1200×50) = √60,000 = 245 volts. This is the RMS voltage required to produce 1200 watts in a 50-ohm load.
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.