Deep Dive: G5C12
The correct answer is B: 14.3 microfarads. The capacitance of a 20-microfarad capacitor connected in series with a 50-microfarad capacitor is 14.3 microfarads. 1/Ctotal = 1/20 + 1/50 = 0.05 + 0.02 = 0.07. Ctotal = 1/0.07 = 14.29 µF ≈ 14.3 µF. For amateur radio operators, this is a basic series capacitance calculation. Understanding this helps when analyzing series capacitors.
Why Other Answers Are Wrong
Option A (0.07 microfarads): Incorrect. 0.07 µF is the reciprocal (1/Ctotal), not the total capacitance. You need to take 1/0.07 = 14.3 µF. Option C (70 microfarads): Incorrect. 70 µF is the sum (parallel), not series. Series capacitance is less than individual capacitors. Option D (1,000 microfarads): Incorrect. 1,000 µF is way too high - that would be 20 × 50, not the series combination. Calculation error.
Exam Tip
Series capacitance: 1/C = 1/20 + 1/50 = 0.07, C = 1/0.07 = 14.3 µF. Think 'S'eries 'C'apacitors = 'S'um of 'R'eciprocals, then 'R'eciprocal. Use 1/Ctotal = 1/C1 + 1/C2. Not 0.07 µF (reciprocal), not 70 µF (sum), not 1000 µF - just 14.3 µF.
Memory Aid
Series capacitance: 1/C = 0.07, C = 14.3 µF. Think 'S'eries 'C'apacitors = 'S'um of 'R'eciprocals. Use 1/Ctotal = 1/C1 + 1/C2, then take reciprocal. Standard series calculation.
Real-World Example
Two capacitors in series: 20 µF and 50 µF. 1/Ctotal = 1/20 + 1/50 = 0.05 + 0.02 = 0.07. Ctotal = 1/0.07 = 14.29 microfarads ≈ 14.3 microfarads. Series capacitance is less than either individual capacitor.
Source & Coverage
Question Pool: 2023-2027 Question Pool
Subelement: G5C
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 G5C topic.