X-Ray Certification Exposure Factors 3

Question 1

A radiographer uses 200 mA at 0.05 s. What is the resulting mAs?

Accepted Answer

10 mAs

Answer

mAs = mA × time = 200 × 0.05 = 10 mAs.

Question 2

Which combination represents equivalent mAs to 300 mA at 1/10 s?

Accepted Answer

600 mA at 1/20 s

Answer

300 × 0.1 = 30 mAs; 600 × 0.05 = 30 mAs, confirming reciprocity.

Question 3

Reducing focal spot size from large to small primarily improves which image quality factor?

Accepted Answer

Recorded detail (spatial resolution)

Answer

A smaller focal spot reduces geometric penumbra, improving spatial resolution and recorded detail.

Question 4

A chest radiograph taken at 120 kVp / 4 mAs is to be repeated at 80 kVp. Using the 15% rule, approximately what mAs is needed?

Accepted Answer

16 mAs

Answer

Each 15% kVp reduction halves receptor exposure, requiring mAs to double per step; two steps down from 120 to 80 kVp requires 4× mAs = 16 mAs.

Question 5

At a fixed mAs, what is the primary trade-off when selecting a very high mA with a very short time?

Accepted Answer

Increased focal spot blooming at high mA

Answer

Very high mA causes space charge and thermal effects that cause focal spot blooming, enlarging the effective focal spot and reducing detail.

X-Ray Certification Practice Test