Deep Dive: G5B05
The correct answer is A: Approximately 61 milliwatts. The watts consumed when a current of 7.0 milliamperes flows through a 1,250-ohm resistance is approximately 61 milliwatts. Power = I²R = (0.007)² × 1,250 = 0.000049 × 1,250 = 0.06125 watts = 61.25 milliwatts. For amateur radio operators, this is a basic power calculation. Understanding this helps when calculating power consumption.
Why Other Answers Are Wrong
Option B (61 watts): Incorrect. 61 watts is 1000 times too high - the answer is in milliwatts, not watts. 61 mW = 0.061W, not 61W. Option C (11 milliwatts): Incorrect. 11 milliwatts is too low - calculation gives 61 mW, not 11 mW. Calculation error. Option D (11 watts): Incorrect. 11 watts is way too high - the answer is in milliwatts. 11W is 1000 times too high.
Exam Tip
Power = I²R. 7.0 mA, 1,250Ω: P = (0.007)² × 1,250 = 0.06125W = 61.25 mW. Think 'P'ower = 'I'²'R'. Use P = I²R formula. Answer is in milliwatts (61 mW), not watts. Not 61W, not 11 mW, not 11W - just 61 mW.
Memory Aid
Power = I²R. 7.0 mA, 1,250Ω: P = (0.007)² × 1,250 = 61.25 mW. Think 'P'ower = 'I'²'R'. Use P = I²R formula. Answer is approximately 61 milliwatts.
Real-World Example
A current of 7.0 milliamperes (0.007 amperes) flows through a 1,250-ohm resistor. Power consumed = I²R = (0.007)² × 1,250 = 0.000049 × 1,250 = 0.06125 watts = 61.25 milliwatts. The resistor consumes approximately 61 milliwatts.
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.