Question

A radiograph made using 300 mA, 0.1 second, and 75 kVp exhibits motion unsharpness, but otherwise satisfactory technical quality. The radiograph will be repeated using a shorter exposure time. Using 86 kV and 500 mA, what should be the new exposure time?

0.03 second

Answer 1

To calculate the new exposure time, you can use the equation: I1 * t1 = I2 * t2, where I1 is the initial intensity, t1 is the initial exposure time, I2 is the final intensity, and t2 is the final exposure time. Using the given values, the new exposure time should be approximately 0.05 seconds.

Step-by-step explanation:

To calculate the new exposure time, we can use the equation:

I1 * t1 = I2 * t2

where I1 is the initial intensity, t1 is the initial exposure time, I2 is the final intensity, and t2 is the final exposure time.

First, we need to calculate the initial intensity:

Initial intensity = mA * kVp = 300 mA * 75 kVp = 22500 mAs (milliamperes per second)

Next, we can plug in the values and solve for t2:

22500 mAs * 0.1 s = I2 * t2

t2 = (22500 mAs * 0.1 s) / I2

Using the new values of kVp and mA, we have:

I2 = 500 mA * 86 kVp = 43000 mAs

t2 = (22500 mAs * 0.1 s) / 43000 mAs

t2 = 0.0523 s, or approximately 0.05 s.