Deep Dive: E5A10
The correct answer is A: 7.12 MHz. What is the resonant frequency of an RLC circuit if R is 33 ohms, L is 50 microhenries, and C is 10 picofarads is 7.12 MHz. f = 1/(2π√(LC)) = 1/(2π√(50×10^-6 × 10×10^-12)) ≈ 7.12 MHz. For amateur radio operators, this is important for circuit calculations. Understanding this helps when designing resonant circuits.
Why Other Answers Are Wrong
Option B (23.5 kHz): Incorrect. 23.5 kHz is too low - the calculation gives 7.12 MHz. 23.5 kHz is wrong. Option C (7.12 kHz): Incorrect. 7.12 kHz is too low - the calculation gives 7.12 MHz. 7.12 kHz is wrong. Option D (23.5 MHz): Incorrect. 23.5 MHz is too high - the calculation gives 7.12 MHz. 23.5 MHz is wrong.
Exam Tip
RLC resonant frequency = 1/(2π√(LC)). Think 'R'esonant = '1'/(2π√('L'C)). f = 1/(2π√(50×10^-6 × 10×10^-12)) ≈ 7.12 MHz. Not 23.5 kHz, not 7.12 kHz, not 23.5 MHz - just 7.12 MHz.
Memory Aid
RLC resonant frequency = 1/(2π√(LC)). Think 'R'esonant = '7'.12 MHz. f = 1/(2π√(50×10^-6 × 10×10^-12)) ≈ 7.12 MHz. Important for circuit calculations.
Real-World Example
An RLC circuit with R=33 ohms, L=50 microhenries, C=10 picofarads: The resonant frequency is f = 1/(2π√(LC)) = 1/(2π√(50×10^-6 × 10×10^-12)) ≈ 7.12 MHz. Note that R doesn't affect resonant frequency. This is the frequency - 7.12 MHz.
Source & Coverage
Question Pool: 2024-2028 Question Pool
Subelement: E5A
Reference: 2024-2028 Question Pool · E5 - Electrical Principles
Key Concepts
Verified Content
Question from the official FCC Extra Class pool. Explanation reviewed by licensed amateur radio operators and mapped to the E5A topic.