If the kVp is increased from 70 to 80, what new mAs is required to maintain the same image receptor exposure?

To determine the new milliampere-seconds (mAs) required to maintain the same image receptor exposure when the kilovolt peak (kVp) is increased from 70 to 80, it's important to apply the 15% rule in radiography. The 15% rule states that an increase in kVp by 15% results in doubling the exposure, while decreasing the kVp by 15% halves the exposure.

In this scenario, increasing the kVp from 70 to 80 represents more than a 15% increase (specifically, about a 14.3% increase). To compensate for this increased energy reaching the receptor while maintaining the same exposure, you would need to reduce the mAs.

A general guideline to apply is that when kVp is increased, to maintain exposure, the mAs should be decreased proportionally. The precise adjustment is often calculated, but generally, a 10% increase in kVp requires a reduction of mAs by roughly 20-30% to maintain correct exposure. Since 10 mAs at 70 kVp is an initial exposure level, reducing it accordingly to maintain exposure with the higher kVp would result in needing around 7.5 mAs.

Thus,