Question

a radiographer takes an initial radiograph using the technical factors of 50 ma at 0.46 sec. there is involuntary motion on the radiograph, but the radiographer wants to maintain ir exposure. if the exposure time is reduced to 0.115 sec, what is the new ma?

Answer 1

To maintain the same IR exposure when reducing the exposure time from 0.46 s to 0.115 s, the milliampere (mA) must be increased. The new mA setting is calculated to be 200 mA.

Step-by-step explanation:

The student's question involves calculating a new milliampere (mA) setting for a radiograph when the exposure time is reduced while maintaining the same image receptor (IR) exposure. The original settings were 50 mA at 0.46 seconds, and the new exposure time is 0.115 seconds.

To maintain the same IR exposure, the product of mA and time must remain constant (mAs constancy). Therefore, if the time is reduced, the mA must increase proportionally. The formula to find the new mA is: New mA = Old mA × (Old time / New time).

Using the values provided: New mA = 50 mA × (0.46 s / 0.115 s) = 200 mA.

The new mA setting that maintains the same IR exposure while reducing involuntary motion artifacts by shortening the exposure time is 200 mA.