Question E9F07:
How does parallel conductor transmission line compare to coaxial cable with a plastic dielectric?  

Choices:  
A: Lower loss  
B: Higher SWR  
C: Smaller reflection coefficient  
D: Lower velocity factor    

Correct Answer is  A: Lower loss  

Correct Answer: A — Lower loss

Why A is correct:
- Parallel-conductor line (ladder/open-wire) places most of the E-field in air, so dielectric loss is very low. Coax with a plastic dielectric (e.g., polyethylene) has higher dielectric loss.
- Total attenuation is the sum of conductor and dielectric losses:  
  $$\alpha \approx \alpha_c + \alpha_d$$  
  For ladder line, $\alpha_d$ is minimal; for plastic-dielectric coax, $\alpha_d$ is significant.
- Useful rule of thumb at HF: 450 Ω ladder line can be around 0.1–0.2 dB per 100 ft, while common 50 Ω coax (RG-58/213) is typically several times higher, and the gap widens at higher frequencies or with high SWR.

Why the other choices are incorrect:
- B: Higher SWR — SWR depends on the load and the line’s characteristic impedance, not the line type per se.  
  $$|\Gamma| = \left|\frac{Z_L - Z_0}{Z_L + Z_0}\right|,\quad S = \frac{1+|\Gamma|}{1-|\Gamma|}$$  
  For a given $Z_L$ and $Z_0$, a parallel line does not “inherently” have higher SWR than coax.
- C: Smaller reflection coefficient — Same reason as above; $|\Gamma|$ depends on $Z_L$ and $Z_0$, not whether the line is parallel-conductor or coax.
- D: Lower velocity factor — The opposite is true. Velocity factor is approximately  
  $$VF \approx \frac{1}{\sqrt{\epsilon_r}}$$  
  Ladder/open-wire line has an effective $\epsilon_r$ near 1 (mostly air), so $VF \approx 0.95$–1.00. Plastic-dielectric coax often has $VF \approx 0.66$ (solid PE) to 0.80–0.85 (foam), which is lower.

Extra tip:
- Loss increases with SWR, but coax suffers more because its dielectric loss rises with the increased standing current/voltage; ladder line remains comparatively low-loss even at high SWR.