If 85 kVp, 400 mA, and ⅛ s are used for an exposure with single-phase equipment, what new mA value is needed for a similar receptor exposure using three-phase, 12-pulse equipment?

To determine the new mA value needed for a similar receptor exposure when switching from single-phase equipment to three-phase, 12-pulse equipment, it’s essential to consider the difference in efficiency and output of these two types of imaging systems.

Single-phase equipment produces a lower average tube output compared to three-phase systems. Specifically, three-phase, 12-pulse equipment has about a 1.4 times higher output than single-phase equipment for the same applied kVp settings. This means that for the same receptor exposure, the mA (milliamperes) can be reduced when moving to a more efficient system.

Given that the original exposure settings are 400 mA and ⅛ second (which equals 0.125 seconds), we need to calculate the total mAs (milliampere-seconds) used in the single-phase exposure, which would be:

mAs = mA × s = 400 mA × 0.125 s = 50 mAs.

When using three-phase, 12-pulse equipment, since this system is more efficient, the necessary mA can be calculated by reducing the mA by the conversion factor (approximately 1.4). Thus, the new mA value would