If an exposure of 82 kVp, 300 mA, and 0.05 sec were made with 3-phase equipment, what would be the required mAs using single-phase equipment to achieve a similar exposure?

To understand the correct answer, it is essential to grasp the difference in efficiency between three-phase and single-phase equipment in radiography. Three-phase equipment delivers a more constant voltage and effectively utilizes the available power, resulting in a higher exposure for the same settings compared to single-phase equipment.

In this scenario, the original exposure settings are given as 82 kVp, 300 mA, and 0.05 seconds, which results in a certain milliamperes per second (mAs) calculation. The mAs is determined by multiplying the tube current (mA) by the exposure time (seconds). Here, it would be:

300 mA × 0.05 seconds = 15 mAs.

When transitioning from three-phase to single-phase equipment, it is generally accepted that single-phase requires approximately 1.5 to 2 times the mAs to achieve a similar exposure because of its inherent inefficiency. In the case of 15 mAs from three-phase, if we apply a factor of 2 for conversion to single-phase:

15 mAs × 2 = 30 mAs.

This calculation leads us to the conclusion that to achieve a similar exposure using single-phase equipment, a requirement of 30 mAs would be