Deep Dive: G5C09
The correct answer is C: 33.3 microfarads. The capacitance of three 100-microfarad capacitors connected in series is 33.3 microfarads. 1/Ctotal = 1/100 + 1/100 + 1/100 = 3/100 = 0.03. Ctotal = 1/0.03 = 33.3 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 (0.33 microfarads): Incorrect. 0.33 µF is too low - that would be if capacitance was divided by 300, not 3. Calculation error. Option B (3.0 microfarads): Incorrect. 3.0 µF is too low - series capacitance is less than individual, but 100/3 = 33.3, not 3.0. Calculation error. Option D (300 microfarads): Incorrect. 300 µF is the sum (parallel), not series. Series capacitance is less than individual capacitors.
Exam Tip
Series capacitance: 1/C = 1/100 + 1/100 + 1/100 = 3/100 = 0.03, C = 1/0.03 = 33.3 µF. Think 'S'eries 'C'apacitors = 'S'um of 'R'eciprocals, then 'R'eciprocal. Use 1/Ctotal = 1/C1 + 1/C2 + 1/C3. Not 0.33 µF, not 3.0 µF, not 300 µF (sum) - just 33.3 µF.
Memory Aid
Series capacitance: 1/C = 3/100 = 0.03, C = 33.3 µF. Think 'S'eries 'C'apacitors = 'S'um of 'R'eciprocals. Use 1/Ctotal = 1/C1 + 1/C2 + 1/C3, then take reciprocal. Standard series calculation.
Real-World Example
Three capacitors in series: 100 µF each. 1/Ctotal = 1/100 + 1/100 + 1/100 = 3/100 = 0.03. Ctotal = 1/0.03 = 33.3 microfarads. Series capacitance is less than individual capacitors (33.3 < 100).
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.