Deep Dive: E9F10
The correct answer is C: An inductive reactance. A 1/8-wavelength transmission line shorted at the far end presents an inductive reactance to an RF generator. The line appears as an inductor. Transmission lines shorter than 1/4 wavelength transform impedances progressively. A shorted line shorter than 1/4 wavelength appears inductive. At 1/8 wavelength, the line presents inductive reactance - it looks like an inductor. As the line gets longer (approaching 1/4 wavelength), the reactance increases. At 1/4 wavelength, it becomes very high impedance (open). The 1/8-wavelength shorted line is useful for providing inductive reactance in matching networks or filters.
Why Other Answers Are Wrong
Option A: Incorrect. A shorted line shorter than 1/4 wavelength appears inductive, not capacitive. An open line shorter than 1/4 wavelength appears capacitive. Option B: Incorrect. The impedance isn't the same as the characteristic impedance. It's a reactive impedance (inductive). Option D: Incorrect. The impedance isn't zero. It's an inductive reactance with a specific value depending on the line length and characteristic impedance.
Exam Tip
1/8λ shorted line = Inductive reactance. Remember: A 1/8-wavelength transmission line shorted at the far end presents an inductive reactance to the generator. Shorted lines shorter than 1/4λ appear inductive.
Memory Aid
**1**/**8**λ **S**horted = **I**nductive **R**eactance (think '1/8λS = IR')
Real-World Example
You have a 1/8-wavelength piece of transmission line shorted at the far end. At the input, it appears as an inductor with inductive reactance. This can be useful in matching networks where you need inductive reactance. As you make the line longer (toward 1/4 wavelength), the inductive reactance increases.
Source & Coverage
Question Pool: 2024-2028 Question Pool
Subelement: E9F
Reference: FCC Part 97.3
Key Concepts
Verified Content
Question from the official FCC Extra Class pool. Explanation reviewed by licensed amateur radio operators and mapped to the E9F topic.