If a duration of 0.05 second is selected to achieve 30 mAs, what milliamperage is required?

To find the required milliamperage when given a specific duration in seconds to achieve a target milliampere-seconds (mAs), you can use the formula:

[ \text{mAs} = \text{mA} \times \text{time (seconds)} ]

In this case, you know the target mAs is 30 and the duration is 0.05 seconds. Rearranging the formula to solve for milliamperes (mA) gives:

[ \text{mA} = \frac{\text{mAs}}{\text{time}} ]

Substituting the known values into the equation:

[ \text{mA} = \frac{30 \text{ mAs}}{0.05 \text{ seconds}} ]

Calculating this yields:

[ \text{mA} = \frac{30}{0.05} = 600 ]

Thus, a milliamperage of 600 is required to achieve 30 mAs in 0.05 seconds. This reasoning confirms that the second choice is correct, as it aligns with the calculated requirement based on the principles of radiation exposure in imaging.