CRAC Class B (2025) — Questions 601–610

Inside a metal vehicle, the body acts like a Faraday shield, attenuating the rubber-duck's signal (A). Reflected waves raising the SWR (B), poor heat dissipation (C), and road noise on the audio (D) are not the characteristic problem of operating a rubber-duck inside a car.

Safety is paramount: any antenna part that might fall must stay outside the power equipment's safety clearance (at least 1 m at 35 kV and below) to avoid electrocution or short-circuit (A). The other options give incorrect or irrelevant distance criteria (10× wavelength, 1 mm/kV, 1/4 wavelength are not the correct safety rule).

A GP needs a real ground/radial plane to form the image of the element; without enough conductive area the resonance and impedance shift from theory, so connecting a large metal ground plane is the correct fix (A). Lengthening to 1/2 λ changes the antenna type (B), loading coils don't fix a missing ground (C), and a balun is not the issue for an unbalanced vertical (D).

Loading an antenna means adding reactance (typically a series inductor) to the element so it behaves electrically longer than its physical length (A). The other options describe mechanical accessories (spring, extender, guy tensioner), not electrical loading.

For a shortened vertical, top loading (scheme C) places the inductive reactance where the current is highest, giving the largest effective current moment and thus the best efficiency; middle loading (B) is next, and base loading (A) is worst. So the efficiency order is C > B > A, which is exactly option A (C-top, B-middle, A-bottom).

Gain references must be converted before comparing: 1 dBd = 2.15 dBi, so antenna B = 1 + 2.15 = 3.15 dBi. Antenna A = 6.15 dBi. The difference is 6.15 − 3.15 = 3.0 dB, which is a power ratio of 10^(3/10) ≈ 2.0. So A's signal is twice B's. Options C and D are wrong because you cannot subtract dBi and dBd directly.

0 dBd means equal to a half-wave dipole, so antenna A ≈ dipole. Convert B: 2 dBi − 2.15 = −0.15 dBd, i.e., slightly LESS than a dipole. Thus B (2 dBi) is slightly weaker than A (0 dBd). Option A is correct; C (claiming B is 2 dB stronger) and D (equivalent) are wrong.

Convert to a common reference: antenna A = 4.5 dBd = 4.5 + 2.15 = 6.65 dBi. Antenna B = 5.85 dBi. The difference is 6.65 − 5.85 = 0.8 dB, so A is 0.8 dB stronger. Correct answer is A.

Convert: antenna A = 2.9 dBd = 2.9 + 2.15 = 5.05 dBi. Antenna B = 5.85 dBi. Difference = 5.85 − 5.05 = 0.8 dB, so B is 0.8 dB stronger. Correct answer is A.

SWR measures how well the load impedance matches the line's characteristic impedance (A); perfect match means no standing wave and SWR = 1:1 (B); a mismatch reflects part of the forward energy back toward the source (C). D is wrong: you cannot cancel standing waves by adjusting only the source end — the mismatch at the load (or a matching network) must be addressed.