Deep Dive: E5A11
The correct answer is C: 47.3 kHz. What is the half-power bandwidth of a resonant circuit that has a resonant frequency of 7.1 MHz and a Q of 150 is 47.3 kHz. Bandwidth = f/Q = 7.1×10^6 / 150 ≈ 47,333 Hz = 47.3 kHz. For amateur radio operators, this is important for circuit calculations. Understanding this helps when calculating bandwidth.
Why Other Answers Are Wrong
Option A (157.8 Hz): Incorrect. 157.8 Hz is too low - bandwidth = 7.1 MHz / 150 ≈ 47.3 kHz. 157.8 Hz is wrong. Option B (315.6 Hz): Incorrect. 315.6 Hz is too low - bandwidth = 7.1 MHz / 150 ≈ 47.3 kHz. 315.6 Hz is wrong. Option D (23.67 kHz): Incorrect. 23.67 kHz is too low - bandwidth = 7.1 MHz / 150 ≈ 47.3 kHz. 23.67 kHz is wrong.
Exam Tip
Half-power bandwidth = resonant frequency / Q. Think 'B'andwidth = 'F'/'Q'. Bandwidth = 7.1×10^6 / 150 ≈ 47,333 Hz = 47.3 kHz. Not 157.8 Hz, not 315.6 Hz, not 23.67 kHz - just 47.3 kHz.
Memory Aid
Half-power bandwidth = resonant frequency / Q. Think 'B'andwidth = 'F'/'Q'. Bandwidth = 7.1 MHz / 150 ≈ 47.3 kHz. Important for circuit calculations.
Real-World Example
A resonant circuit with resonant frequency 7.1 MHz and Q of 150: The half-power bandwidth is f/Q = 7.1×10^6 / 150 ≈ 47,333 Hz = 47.3 kHz. This is the bandwidth - 47.3 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.