Deep Dive: G5B03
The correct answer is B: 200 watts. The watts of electrical power consumed if 400 VDC is supplied to an 800-ohm load is 200 watts. Power = V²/R = 400²/800 = 160,000/800 = 200 watts. For amateur radio operators, this is a basic power calculation. Understanding this helps when calculating power consumption.
Why Other Answers Are Wrong
Option A (0.5 watts): Incorrect. 0.5 watts is way too low - 400V across 800Ω gives 200W, not 0.5W. Calculation error. Option C (400 watts): Incorrect. 400 watts is too high - that would be if power = V, but power = V²/R. Calculation error. Option D (3200 watts): Incorrect. 3200 watts is way too high - that would be if power = V×R, but power = V²/R. Calculation error.
Exam Tip
Power = V²/R. 400V, 800Ω: P = 400²/800 = 160,000/800 = 200W. Think 'P'ower = 'V'²/'R'. Use P = V²/R formula. Not V/R (0.5W), not V (400W), not V×R (3200W) - just V²/R (200W).
Memory Aid
Power = V²/R. 400V, 800Ω: P = 400²/800 = 200W. Think 'P'ower = 'V'²/'R'. Use P = V²/R formula for power calculation. Standard power formula.
Real-World Example
You supply 400 volts DC to an 800-ohm resistor. Power consumed = V²/R = 400²/800 = 160,000/800 = 200 watts. The resistor consumes 200 watts of power. This is a basic power calculation using P = V²/R.
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.