For an analog x-ray image using 300 mA, 0.07 second, and 95 kVp, what exposure time is needed to achieve similar receptor exposure with 500 mA and 80 kVp?

To achieve similar receptor exposure using different mA (milliamperes) and kVp (kilovolt peak) settings, you can apply the concept of the exposure maintenance formula, which states that for two different sets of exposure parameters to achieve the same receptor exposure, the product of mA, exposure time, and kVp should be considered.

In the first case, you have an exposure with 300 mA and 0.07 seconds. The total exposure can be calculated as:

[ \text{Exposure}_1 = 300 , \text{mA} \times 0.07 , \text{s} = 21 , \text{mAs} ]

For the second case with 500 mA and an unknown exposure time ( t ), you will also want to consider how the kVp affects the exposure. Generally, through the 15% rule, a 15% increase in kVp doubles the exposure (given tissue characteristics remain constant). Thus, reducing the kVp from 95 to 80 is a significant drop (more than 15%).

Calculating the necessary mAs to compensate for the decrease in kVp is essential. While the exact adjustment for kVp to mAs