Deep Dive: G5C08
The correct answer is D: 10.750 nanofarads. The equivalent capacitance of two 5.0-nanofarad capacitors and one 750-picofarad capacitor connected in parallel is 10.750 nanofarads. 750 pF = 0.75 nF. Parallel: Ctotal = 5.0 + 5.0 + 0.75 = 10.75 nF. For amateur radio operators, this is a basic parallel capacitance calculation. Understanding this helps when analyzing parallel capacitors.
Why Other Answers Are Wrong
Option A (576.9 nanofarads): Incorrect. 576.9 nF is way too high - that might be a calculation error or wrong formula. Parallel capacitors add, not multiply. Option B (1,733 picofarads): Incorrect. 1,733 pF = 1.733 nF, which is too low. Total should be 10.75 nF = 10,750 pF. Option C (3,583 picofarads): Incorrect. 3,583 pF = 3.583 nF, which is too low. Total should be 10.75 nF = 10,750 pF.
Exam Tip
Parallel capacitance: 5.0 nF + 5.0 nF + 0.75 nF (750 pF) = 10.75 nF. Think 'P'arallel 'C'apacitors = 'P'lus all 'C'apacitances. Parallel capacitors add directly. Convert units first (750 pF = 0.75 nF). Not 576.9 nF, not 1,733 pF, not 3,583 pF - just 10.75 nF.
Memory Aid
Parallel capacitance: 5.0 + 5.0 + 0.75 = 10.75 nF. Think 'P'arallel 'C'apacitors = 'P'lus all. Parallel capacitors add directly. Convert 750 pF = 0.75 nF first. Standard parallel calculation.
Real-World Example
Three capacitors in parallel: 5.0 nF, 5.0 nF, and 750 pF (0.75 nF). Total capacitance = 5.0 + 5.0 + 0.75 = 10.75 nanofarads. Parallel capacitors add directly - just sum the capacitances. Remember to convert units (pF to nF).
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.