Deep Dive: G7B08
The correct answer is B: Divide the RF output power by the DC input power. How the efficiency of an RF power amplifier is determined is to divide the RF output power by the DC input power. Efficiency = (RF output / DC input) × 100%. For amateur radio operators, this is a fundamental amplifier calculation. Understanding this helps when evaluating amplifier performance.
Why Other Answers Are Wrong
Option A: Incorrect. Efficiency isn't DC input divided by DC output - that would be greater than 1, which doesn't make sense. Efficiency is output divided by input. Option C: Incorrect. Efficiency isn't RF input multiplied by reciprocal of RF output - that's not the efficiency formula. Efficiency is output divided by input. Option D: Incorrect. Efficiency isn't RF input added to DC output - that's not how efficiency is calculated. Efficiency is output divided by input.
Exam Tip
RF amplifier efficiency = RF output / DC input. Think 'E'fficiency = 'E'ffective output / 'E'nergy input. Divide RF output power by DC input power. Not DC/DC, not RF input/reciprocal, not RF+DC - just RF output / DC input.
Memory Aid
RF amplifier efficiency = RF output / DC input. Think 'E'fficiency = 'E'ffective output / 'E'nergy input. Divide RF output power by DC input power. Standard efficiency calculation.
Real-World Example
An RF power amplifier: DC input = 200 watts, RF output = 100 watts. Efficiency = 100 / 200 = 0.5 = 50%. The amplifier converts 50% of DC input power to RF output power. This is how amplifier efficiency is calculated.
Source & Coverage
Question Pool: 2023-2027 Question Pool
Subelement: G7B
Reference: 2023-2027 Question Pool · G7 - Practical Circuits
Key Concepts
Verified Content
Question from the official FCC General Class pool. Explanation reviewed by licensed amateur radio operators and mapped to the G7B topic.