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?

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?

• A. 200 mA
• B. 600 mA
• C. 0.125 s
• D. 0.25 s

Answer: (A)

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

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