The X-ray tube is moved from an FFD of 40 inches to an FFD of 60 inches. At 40 inches, a technique of 10 mAs and 70 kVp was suitable. Which milliamperage value would most likely re... The X-ray tube is moved from an FFD of 40 inches to an FFD of 60 inches. At 40 inches, a technique of 10 mAs and 70 kVp was suitable. Which milliamperage value would most likely result in a comparable exposure at the new distance of 60 inches?
Understand the Problem
The question is asking to calculate the new mAs value required to maintain comparable exposure when the FFD (Film-Focal Distance) is changed from 40 inches to 60 inches, given an initial technique of 10 mAs at 70 kVp at 40 inches.
Answer
$mAs_2 = 22.5$
Answer for screen readers
$mAs_2 = 22.5$
Steps to Solve
- Identify the relevant formula
The formula that relates mAs and distance (FFD) is the direct square law: $$ \frac{mAs_1}{mAs_2} = \frac{D_1^2}{D_2^2} $$ Where: $mAs_1$ = initial mAs $mAs_2$ = new mAs $D_1$ = initial distance $D_2$ = new distance
- Plug in the given values
We are given: $mAs_1 = 10$ $D_1 = 40$ $D_2 = 60$
Plugging these values into the formula: $$ \frac{10}{mAs_2} = \frac{40^2}{60^2} $$
- Solve for $mAs_2$
$$ \frac{10}{mAs_2} = \frac{1600}{3600} $$
$$ mAs_2 = 10 \cdot \frac{3600}{1600} $$
$$ mAs_2 = 10 \cdot \frac{36}{16} $$
$$ mAs_2 = 10 \cdot \frac{9}{4} $$
$$ mAs_2 = \frac{90}{4} $$
$$ mAs_2 = 22.5 $$
$mAs_2 = 22.5$
More Information
The new mAs required to maintain a comparable exposure at 60 inches is 22.5 mAs. Note that the kVp value is not used in this calculation because the direct square law relates only mAs and distance.
Tips
A common mistake is to incorrectly set up the direct square law formula. Ensure that the distances are squared and are in the correct positions relative to their corresponding mAs values. Another mistake is not squaring the distances.
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