Which kVp and mAs settings would be appropriate for a technique chart that indicates a setting of 50 kVp and 12.5 mAs?

Which kVp and mAs settings would be appropriate for a technique chart that indicates a setting of 50 kVp and 12.5 mAs?

• A. 125 mA, 1/30 sec
• B. 250 mA, 1/20 sec
• C. 250 mA, 1/5 sec
• D. 125 mA, 1/20 sec

Answer: (B)

The chosen answer is appropriate because it accurately represents the required kilovolt peak (kVp) and milliampere-seconds (mAs) settings based on the technique chart specifications of 50 kVp and 12.5 mAs. For the settings of 50 kVp and 12.5 mAs, the chosen option of 250 mA and 1/20 sec results in the correct mAs value. This is nearly equivalent to calculating the mAs product as follows: 1. First, convert the time from seconds to seconds for easier calculations. The fraction 1/20 seconds equals 0.05 seconds. 2. Multiply the milliampere setting (250 mA) by the time in seconds (0.05 sec): 250 mA x 0.05 sec = 12.5 mAs. This calculation directly meets the requirement of the technique chart, demonstrating that the settings can produce the desired exposure. In contrast, the other choices either do not produce the required mAs of 12.5 or do not correspond accurately to the kVp. For instance, the settings of 250 mA and 1/5 sec lead to a much higher mAs (50 m

The chosen answer is appropriate because it accurately represents the required kilovolt peak (kVp) and milliampere-seconds (mAs) settings based on the technique chart specifications of 50 kVp and 12.5 mAs.

For the settings of 50 kVp and 12.5 mAs, the chosen option of 250 mA and 1/20 sec results in the correct mAs value. This is nearly equivalent to calculating the mAs product as follows:

1. First, convert the time from seconds to seconds for easier calculations. The fraction 1/20 seconds equals 0.05 seconds.

2. Multiply the milliampere setting (250 mA) by the time in seconds (0.05 sec):

250 mA x 0.05 sec = 12.5 mAs.

This calculation directly meets the requirement of the technique chart, demonstrating that the settings can produce the desired exposure.

In contrast, the other choices either do not produce the required mAs of 12.5 or do not correspond accurately to the kVp. For instance, the settings of 250 mA and 1/5 sec lead to a much higher mAs (50 m