Deep Dive: E5A12
The correct answer is C: 31.4 kHz. What is the half-power bandwidth of a resonant circuit that has a resonant frequency of 3.7 MHz and a Q of 118 is 31.4 kHz. Bandwidth = f/Q = 3.7×10^6 / 118 ≈ 31,356 Hz ≈ 31.4 kHz. For amateur radio operators, this is important for circuit calculations. Understanding this helps when calculating bandwidth.
Why Other Answers Are Wrong
Option A (436.6 kHz): Incorrect. 436.6 kHz is too high - bandwidth = 3.7 MHz / 118 ≈ 31.4 kHz. 436.6 kHz is wrong. Option B (218.3 kHz): Incorrect. 218.3 kHz is too high - bandwidth = 3.7 MHz / 118 ≈ 31.4 kHz. 218.3 kHz is wrong. Option D (15.7 kHz): Incorrect. 15.7 kHz is too low - bandwidth = 3.7 MHz / 118 ≈ 31.4 kHz. 15.7 kHz is wrong.
Exam Tip
Half-power bandwidth = resonant frequency / Q. Think 'B'andwidth = 'F'/'Q'. Bandwidth = 3.7×10^6 / 118 ≈ 31,356 Hz ≈ 31.4 kHz. Not 436.6 kHz, not 218.3 kHz, not 15.7 kHz - just 31.4 kHz.
Memory Aid
Half-power bandwidth = resonant frequency / Q. Think 'B'andwidth = 'F'/'Q'. Bandwidth = 3.7 MHz / 118 ≈ 31.4 kHz. Important for circuit calculations.
Real-World Example
A resonant circuit with resonant frequency 3.7 MHz and Q of 118: The half-power bandwidth is f/Q = 3.7×10^6 / 118 ≈ 31,356 Hz ≈ 31.4 kHz. This is the bandwidth - 31.4 kHz.
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.