What is the new exposure time when switching to 86 kV and 500 mA, given a previous setting of 300 mA at 0.1 seconds?

What is the new exposure time when switching to 86 kV and 500 mA, given a previous setting of 300 mA at 0.1 seconds?

• A. 0.12 second
• B. 0.06 second
• C. 0.03 second
• D. 0.01 second

Answer: (C)

To find the new exposure time when changing from one set of parameters to another in radiology, you can utilize the principle of maintaining the same exposure by considering the relationship between milliamperage (mA), exposure time (seconds), and kilovoltage (kV). First, calculate the previous exposure in terms of milliampere-seconds (mAs). In this case, the previous settings were 300 mA and 0.1 seconds, which gives: \[ \text{Previous mAs} = 300 \text{ mA} \times 0.1 \text{ s} = 30 \text{ mAs} \] Now, if you switch to 86 kV and 500 mA, we need to keep the exposure in terms of mAs consistent. The formula for the new exposure time can be derived from the need to preserve the same mAs: \[ \text{New mAs} = 500 \text{ mA} \times \text{New Exposure Time (s)} \] Setting the new mAs equal to the previous mAs gives: \[ 500 \text{ mA} \times \text{New Exposure Time (s)} =

To find the new exposure time when changing from one set of parameters to another in radiology, you can utilize the principle of maintaining the same exposure by considering the relationship between milliamperage (mA), exposure time (seconds), and kilovoltage (kV).

First, calculate the previous exposure in terms of milliampere-seconds (mAs). In this case, the previous settings were 300 mA and 0.1 seconds, which gives:

[

\text{Previous mAs} = 300 \text{ mA} \times 0.1 \text{ s} = 30 \text{ mAs}

]

Now, if you switch to 86 kV and 500 mA, we need to keep the exposure in terms of mAs consistent. The formula for the new exposure time can be derived from the need to preserve the same mAs:

[

\text{New mAs} = 500 \text{ mA} \times \text{New Exposure Time (s)}

]

Setting the new mAs equal to the previous mAs gives:

[

500 \text{ mA} \times \text{New Exposure Time (s)} =