Updated: Dec 9, 2025 | Source: 2023-2027 Question Pool | Topic: G9A
G9A09G9A

What standing wave ratio results from connecting a 50-ohm feed line to a 200-ohm resistive load?

Deep Dive: G9A09

The correct answer is A: 4:1. The standing wave ratio that results from connecting a 50-ohm feed line to a 200-ohm resistive load is 4:1. SWR = Zload/Zline (or Zline/Zload, whichever is greater) = 200/50 = 4:1. For amateur radio operators, this is a basic SWR calculation. Understanding this helps when calculating SWR.

Why Other Answers Are Wrong

Option B (1:4): Incorrect. 1:4 is the inverse - SWR is always ≥ 1:1, so it's 4:1, not 1:4. SWR is the larger ratio. Option C (2:1): Incorrect. 2:1 is too low - 200/50 = 4:1, not 2:1. Calculation error. Option D (1:2): Incorrect. 1:2 is the inverse and too low - SWR is 4:1, not 1:2. SWR is always ≥ 1:1.

Exam Tip

50Ω feed line to 200Ω load = 4:1 SWR. Think 'S'WR = 'S'imple ratio = '2'00/'5'0 = '4':1. SWR = Zload/Zline (or larger ratio) = 200/50 = 4:1. Not 1:4, not 2:1, not 1:2 - just 4:1.

Memory Aid

50Ω feed line to 200Ω load = 4:1 SWR. Think 'S'WR = '2'00/'5'0 = '4':1. SWR = Zload/Zline (or larger ratio). Basic SWR calculation.

Real-World Example

A 50-ohm feed line connected to a 200-ohm resistive load. SWR = 200/50 = 4:1. The standing wave ratio is the ratio of the load impedance to the feed line impedance (or the larger ratio). This is a basic SWR calculation - impedance mismatch creates SWR.

Source & Coverage

Question Pool: 2023-2027 Question Pool

Subelement: G9A

Reference: 2023-2027 Question Pool · G9 - Antennas and Feed Lines

Key Concepts

Standing wave ratio 50-ohm feed line 200-ohm load SWR calculation

Verified Content

Question from the official FCC General Class pool. Explanation reviewed by licensed amateur radio operators and mapped to the G9A topic.