Which technique produces a 2 milliroentgens (mR) exposure to an image receptor plate if 70 kilovolts
peak (kVp) at 25 milliampere-seconds (mAs) produces 1.6 mR?
A. 70 kVp at 50 mAs.
B. 70 kVp at 31.25 mAs.
C. 70 kVp at 27.5 mAs.
D. 70 kVp at 35 mAs.
My answer: C
Correct Answer: B
Feedback: 2 mR is 25% greater than 1.6 mR; so, a 25% increase in mAs achieves this result.
25 mAs x 1.25 = 31.25 mAs. All other responses are incorrect due to math errors or
misunderstanding the relationship of mAs to exposure.
The question states that when we use the exposure factors of 70 kVp and 25 mAs it will
produce an image receptor exposure of measuring 1.6 mR.
Next, the question is asking you what the exposure factors need to be to produce an
image receptor exposure of 2.0 mR.
1. We determine that to change exposure rate we must adjust mAs not kVp. (mAs and
exposure have a directly proportionate relationship; if mAs is doubled, exposure is
doubled)
2. Calculate the percentage of increased exposure needed. The first technical
factors produce an exposure of 1.6 mR. The new desired exposure is 2.0 mR. Divide 2.0
by 1.6 = 1.25 (that is a 25% increase from the original 1.6 mR exposure)
3. Multiple 1.25 by the original mAs : 1.25 x 25 = 31.25- This is the new mAs need to
increase the original exposure of 1.6 mR to a 2.0 mR exposure.